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International GCSE Physics · Pearson Edexcel 4PH1 · Topic 8

Astrophysics

185 free, high-quality past-paper-style questions on Astrophysics for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (b) Motion in the Universe; (c) Stellar Evolution; (d) Cosmology. For when you have finished the past papers, or want to practise one topic at a time.Notes 8(b) →Notes 8(c) →Notes 8(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.877 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1IGCSE-PHY-8-0018(b) Motion in the Universe Papers 1 & 2Grade 4–51 mark
Which statement about the universe is correct?
Show mark scheme
  • B (the universe contains billions of galaxies and a galaxy contains billions of stars);
  • A is incorrect because the Milky Way is a galaxy, not a star, and the solar system is inside it
  • C is incorrect because the order of scale is reversed: galaxies are far larger than the solar system
  • D is incorrect because the solar system is inside the Milky Way, not the other way round
Theory: How are stars, galaxies and the universe related? Single · Double · Physics
2IGCSE-PHY-8-0028(b) Motion in the Universe Papers 1 & 2Grade 4–51 mark
An astronaut has a mass of 80 kg on the Earth.
The gravitational field strength on the Earth is 10 N/kg and on the Moon is 1.6 N/kg.
Which row gives the mass and the weight of the astronaut on the Moon?
Show mark scheme
  • D (mass = 80 kg, weight = 128 N);
  • A is incorrect because weight depends on g, which is smaller on the Moon (800 N is the weight on Earth)
  • B is incorrect because it confuses mass with weight; mass is measured in kg and does not change
  • C is incorrect because mass does not depend on location (80 × 1.6 ÷ 10 = 12.8 ≈ 13 kg wrongly scales the mass)
Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
3IGCSE-PHY-8-0038(b) Motion in the Universe Papers 1 & 2Grade 4–56 marks
A rover is sent to explore the surface of Mars.
The gravitational field strength on the surface of Mars is 3.7 N/kg.
(a)
State the unit of gravitational field strength.
(1)
(b)
State the relationship between weight, mass and gravitational field strength.
(1)
(c)
The weight of the rover on the surface of Mars is 3790 N.
Calculate the mass of the rover.
(2)
(d)
Explain why the weight of the rover on the Earth is greater than its weight on Mars, even though its mass is the same.
(2)
Show mark scheme
(a)
  • N/kg;
Allow
  • newtons per kilogram
  • m/s²
Do not accept
  • N
  • kg/N
(b)
  • weight = mass × gravitational field strength;
Allow
  • W = m × g
  • W = mg
  • any correct rearrangement
Do not accept
  • W = m/g
  • mass = weight × g
(c)
  • substitution / rearrangement;
  • evaluation;
  • e.g. m = 3790 ÷ 3.7
  • (m =) 1020 (kg)
Allow
  • 1024, 1024.3, 1000 (kg)
Do not accept
  • 379 (kg) from g = 10 N/kg (scores 0)
  • 14 000 (kg) from W × g (scores 0)
  • answer in N
(d)
  • (gravitational field strength / g) on the Earth is greater (than on Mars);
  • because the Earth has a greater mass (than Mars);
Allow
  • Earth pulls objects with a greater force per kg for MP1
  • reverse argument for Mars
Do not accept
  • Mars has less air / no atmosphere
  • the rover loses mass on Mars
  • Mars is further from the Sun
Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
4IGCSE-PHY-8-0048(b) Motion in the Universe Papers 1 & 2Grade 4–55 marks
The diagram shows the orbit of a comet around the Sun.
P and Q are two points on the orbit.
SunPQorbit of cometnot to scale
(a)
Name the force that causes the comet to orbit the Sun.
(1)
(b)
Describe two ways in which the orbit of this comet is different from the orbit of the Earth around the Sun.
(2)
(c(i))
State whether the comet is moving faster at P or at Q.
(1)
(c(ii))
Give a reason for your answer to (c)(i).
(1)
Show mark scheme
(a)
  • gravity / gravitational (force / attraction);
Allow
  • weight
Do not accept
  • magnetic force
  • centrifugal force
  • friction
(b)
  • any two from:
  • comet's orbit is (highly) elliptical / Earth's orbit is (almost) circular;
  • distance of comet from Sun varies greatly / Earth stays at about the same distance;
  • comet has a much longer time period / takes longer to orbit;
  • comet's speed changes greatly / Earth has (almost) constant speed;
  • comet's orbit may be in a different plane (to the planets);
Allow
  • comet goes far beyond the planets
  • oval for elliptical
Do not accept
  • the comet orbits a different star
  • the comet does not orbit the Sun
(c(i))
  • P;
Do not accept
  • Q
  • same speed at P and Q
(c(ii))
  • closer to the Sun so gravitational force is stronger / greater;
Allow
  • it has gained speed / kinetic energy as it moved towards the Sun
Do not accept
  • it is further from the Sun
  • there is less air resistance at P
Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
5IGCSE-PHY-8-0058(c) Stellar Evolution Papers 1 & 2Grade 4–55 marks
The diagram shows the stages in the life cycle of the Sun.
Two of the stages are labelled A and B.
nebulaAred giantB
(a)
Name the stages A and B.
(2)
(b)
Describe how a star forms from a nebula.
(2)
(c)
Some stars end their lives in a supernova.
Give a reason why the Sun will not become a supernova.
(1)
Show mark scheme
(a)
  • A main sequence (star);
  • B white dwarf;
Allow
  • stable star for A
Do not accept
  • red dwarf / black dwarf / neutron star for B
  • protostar for A
(b)
  • (gravitational) attraction pulls gas and dust together / nebula collapses;
  • (core) becomes hot enough for (nuclear) fusion (to start);
Allow
  • temperature and pressure increase until fusion starts
Do not accept
  • fission
  • it burns / chemical reaction
(c)
  • (mass of) the Sun is too small / not massive enough;
Allow
  • only stars more massive than the Sun become supernovae
Do not accept
  • the Sun is too hot
  • the Sun is too young
Theory: How does a star like the Sun evolve? Single · Double · Physics
6IGCSE-PHY-8-0088(c) Stellar Evolution Papers 1 & 2Grade 4–56 marks
The table gives the surface temperature of five stars.
Star / name: Rigel, Deneb, Sun, Arcturus, Antares
Surface temperature / K: 12100, 8500, 5800, 4300, 3500
(a)
Give the name of the star in the table that appears red.
(1)
(b)
Describe how the colour of a star is related to its surface temperature.
(2)
(c)
The star Vega has a similar colour to Deneb.
Suggest a value for the surface temperature of Vega.
(1)
(d)
Near the end of its life, the Sun will become a red giant.
Its outer layers will expand and cool.
Explain how the colour of the Sun will change.
(2)
Show mark scheme
(a)
  • Antares;
Do not accept
  • Arcturus
  • more than one star given
(b)
  • hotter stars are blue / white;
  • cooler stars are red / orange;
Allow
  • the higher the temperature the bluer the star (for both marks if both ends implied)
  • correct colour sequence red, orange, yellow, white, blue with increasing temperature for 2 marks
Do not accept
  • red stars are hottest
  • colour depends on distance from Earth
(c)
  • any value from 7500 to 11 000 (K);
Do not accept
  • value outside the range
  • value in °C
(d)
  • (surface) temperature decreases / it gets cooler;
  • (so) colour changes from yellow (or white) to red / orange;
Allow
  • becomes redder for MP2
Do not accept
  • becomes bluer
  • becomes red because it is bigger (for MP1)
Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
7IGCSE-PHY-8-0218(b) Motion in the Universe Papers 1 & 2Grade 4–51 mark
Gravitational force causes many objects in space to move in orbits.
Which row shows the object that each body orbits?
Show mark scheme
  • B (moon → a planet; comet → the Sun; artificial satellite → the Earth);
  • A is incorrect because comets orbit the Sun, not the Earth, and artificial satellites are put into orbit around the Earth
  • C is incorrect because a moon orbits a planet, not the Sun directly (and most artificial satellites orbit the Earth)
  • D is incorrect because comets orbit the Sun, not a planet, and artificial satellites orbit the Earth
Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
8IGCSE-PHY-8-0228(c) Stellar Evolution Papers 1 & 2Grade 4–51 mark
Stars can be classified according to their colour.
Which list gives star colours in order of increasing surface temperature (coolest first)?
Show mark scheme
  • C (red, orange, yellow, white, blue);
  • A is incorrect because it is in order of decreasing temperature (hottest first)
  • B is incorrect because orange stars are cooler than yellow stars
  • D is incorrect because blue stars are the hottest, not the coolest; the everyday link between blue and cold does not apply to stars
Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
9IGCSE-PHY-8-0238(b) Motion in the Universe Papers 1 & 2Grade 4–55 marks
A student writes five astronomical terms on cards.
galaxy planet solar system star universe
(a)
Write the five terms in order of size, starting with the smallest.
(2)
(b)
Give the name of the galaxy that our solar system is in.
(1)
(c)
A student says:
'The Milky Way and our solar system are the same thing.'
Explain why the student is wrong.
(2)
Show mark scheme
(a)
  • planet, star, solar system in this order (as the three smallest);
  • galaxy then universe (as the two largest);
Allow
  • one mark for any four in the correct relative order
  • all five in correct order but starting with the largest scores 1
Do not accept
  • solar system placed before star
(b)
  • Milky Way;
Allow
  • our galaxy is the Milky Way
Do not accept
  • Andromeda
  • the Sun
(c)
  • the Milky Way is a galaxy / contains billions of stars;
  • the solar system is (one star,) the Sun and its planets / is (a tiny) part of the Milky Way;
Allow
  • the solar system is inside the Milky Way for MP2
  • the Sun is only one of the stars in the Milky Way for MP2
Do not accept
  • the Milky Way is inside the solar system
  • the Milky Way is a star
Theory: How are stars, galaxies and the universe related? Single · Double · Physics
10IGCSE-PHY-8-0248(c) Stellar Evolution Papers 1 & 2Grade 4–55 marks
A student writes four statements about the life cycle of a star with a mass similar to the Sun.
Each statement contains one mistake.
1 The star forms from a nebula, which is pulled together by magnetic forces.
2 During the main sequence stage, nuclear fission of hydrogen releases energy in the core.
3 When the hydrogen in the core runs out, the star shrinks and becomes a red giant.
4 At the end of its life, the star explodes in a supernova, leaving a white dwarf.
(a)
For each statement, identify the mistake and write the correct word or words.
(4)
(b)
Describe what a white dwarf is.
(1)
Show mark scheme
(a)
  • statement 1: gravitational force / gravity (not magnetic);
  • statement 2: fusion (not fission);
  • statement 3: expands / gets bigger (not shrinks);
  • statement 4: no supernova / outer layers are released (drift away) / star sheds its outer layers;
Allow
  • statement 4: 'only massive stars explode as a supernova' or 'the star does not explode'
Do not accept
  • statement 4: 'it becomes a neutron star' / 'black hole'
  • statement 2: 'chemical burning'
(b)
  • (small) hot dense core (remnant) of the star;
Allow
  • small hot star that no longer has fusion / cools down over time
Do not accept
  • a small cold star
  • a star that has exploded
Theory: How does a star like the Sun evolve? Single · Double · Physics
11IGCSE-PHY-8-0268(c) Stellar Evolution Papers 1 & 2Grade 4–56 marks
Albireo looks like a single star to the naked eye.
Through a telescope it is seen to be two stars close together in the sky.
Albireo A is orange and Albireo B is blue.
Astronomers use letters to classify stars according to their colour.
Class / letter: O, B, A, F, G, K, M
Colour / name: blue, blue-white, white, yellow-white, yellow, orange, red
(a)
The Sun is a yellow star.
State the class of the Sun.
(1)
(b)
State which of the two stars of Albireo has the higher surface temperature.
Give a reason for your answer.
(2)
(c)
Describe how the surface temperature of a star changes from class O to class M.
(1)
(d)
A student says:
'Albireo A and Albireo B have different colours, so they must be at different distances from the Earth.'
Explain why the colours of the two stars do not give any information about their distances from the Earth.
(2)
Show mark scheme
(a)
  • G;
Do not accept
  • yellow (colour not class)
(b)
  • Albireo B;
  • blue stars are hotter (than orange stars) / orange stars are cooler;
Allow
  • MP2 can be scored by a correct comparison of colour order (e.g. blue is at the hot end)
  • B with no reason scores 1
Do not accept
  • Albireo A
  • blue means colder
(c)
  • temperature decreases (from O to M);
Allow
  • O hottest and M coolest
Do not accept
  • increases
(d)
  • colour (of a star) depends on (is related to) its surface temperature;
  • colour does not depend on distance / distance affects brightness (how bright it appears) not colour;
Allow
  • a star further away appears dimmer but the same colour, for MP2
Do not accept
  • colour depends on the size of the star
  • stars further away look redder (as the only point)
Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
12IGPE-057-00-018(c) Stellar Evolution Paper 1PEasy1 mark

An astronomer compares four stars by their colour. Which star has the highest surface temperature?

Show mark scheme
Marking pointMarkNotes
Star colour depends on surface temperature. The order from coolest to hottest is red, orange, yellow, white, then blue. A blue-white star is therefore the hottest. 'Red' is a tempting wrong answer because red is linked with heat in everyday life. 'Yellow' may be chosen because the Sun is yellow, but the Sun is only a medium-temperature star.✓ 1Answer D

Answer: D

Common error: Red often means 'hot' on taps and warning signs. For stars it is the opposite: red stars are the coolest and blue stars are the hottest.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
13IGPE2-061-01-018(c) Stellar Evolution Paper 1PEasy1 mark

A blacksmith heats a steel bar in a forge. As its temperature rises, the bar glows dull red, then orange, then yellow-white. Stars show the same link between colour and temperature.
Four stars are observed through a telescope. Which star has the highest surface temperature?

Show mark scheme
Marking pointMarkNotes
A star's colour depends on its surface temperature. Cool stars look red, and hotter stars look orange, then yellow, then white, then blue. A blue-white star is therefore the hottest of the four. Red is the coolest star colour, even though red is often linked with 'hot' in everyday life, for example on hot taps. A yellow star like the Sun is only medium in temperature.✓ 1Answer B

Answer: B

Common error: Do not use everyday colour labels such as red for hot and blue for cold. For stars the order runs the other way: red is coolest and blue is hottest.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
14IGPE2-061-01-048(c) Stellar Evolution Paper 1PEasy4 marks

A planetarium is preparing a short show about the life of a star with a similar mass to the Sun.

(a)

The four stages below are in the wrong order:
red giant, nebula, white dwarf, main sequence star
Write the stages in the correct order, starting with the earliest.

(2)
(b)

State what runs out in the core of the star at the end of its main-sequence stage.

(1)
(c)

When the star becomes a red giant, its surface cools. State how the colour of the star shows this change.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
nebula first, then main sequence star✓ 1
red giant third, then white dwarf last✓ 1
Part (b)
hydrogen (the fuel for nuclear fusion)✓ 1
Part (c)
the star becomes redder / changes from yellow towards red✓ 1

Exam tip: Learn the life cycle of a Sun-like star as a fixed chain: nebula, main sequence star, red giant, white dwarf. Check that each stage follows on from the one before it, for example that a red giant forms only after the core hydrogen runs out.

Theory: How does a star like the Sun evolve? Single · Double · Physics
15IGPE2-061-01-128(c) Stellar Evolution Paper 1PEasy6 marks

Infrared images from a space telescope show young stars forming inside a large cloud in space called a nebula.

(a)

State what a nebula is made of.

(1)
(b)

Describe how a protostar forms from a nebula.

(2)
(c)

State the process that must begin in the core for the protostar to become a main sequence star.

(1)
(d)

Explain why a main sequence star stays the same size for a long time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
gas and dust (mainly hydrogen)✓ 1
Part (b)
gravity pulls the gas and dust together / the cloud collapses or contracts✓ 1
the material becomes denser and its temperature increases (accept: gravitational potential energy is transferred to thermal energy)✓ 1
Part (c)
nuclear fusion of hydrogen (into helium)✓ 1
Part (d)
the inward force of gravity✓ 1
is balanced by the outward pressure (accept thermal/gas/radiation pressure) caused by the energy released by fusion / the forces are in equilibrium✓ 1

Exam tip: For main-sequence stability, name both forces and their directions: gravity acts inwards and pressure from fusion acts outwards. Then state that the two are balanced.

Theory: How does a star like the Sun evolve? Single · Double · Physics
16IGPE2-061-02-018(d) Cosmology Paper 1PEasy1 mark

Which observation is evidence that the universe is expanding?

Show mark scheme
Marking pointMarkNotes
Light from distant galaxies arrives with longer wavelengths than when it was emitted (red-shift). This shows the galaxies are moving away from us, which is consistent with an expanding universe. Blue-shift would mean the galaxies were approaching. The CMB is cooling, not warming. The largest red-shifts come from the most distant galaxies, not the closest ones.✓ 1Answer B

Answer: B

Common error: Red-shift means the wavelength increases and the source is moving away. Blue-shift means the wavelength decreases and the source is approaching. Do not swap them.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
17IGPE2-061-02-048(d) Cosmology Paper 1PEasy4 marks

An engineer is testing a spectrometer that will be fitted to a space telescope. The spectrometer will measure the spectra of light from distant galaxies.

(a)

State what is meant by red-shift.

(2)
(b)

A galaxy's light is red-shifted. State the direction in which the galaxy is moving relative to Earth.

(1)
(c)

Name the theory of the origin of the universe that is supported by red-shift measurements.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The observed wavelength of light from a galaxy is longer / its frequency is lower / its spectral lines are shifted towards the red end of the spectrum compared with the same light from a stationary source, e.g. a spectrum measured in a laboratory✓ 2
Part (b)
Away from Earth / receding✓ 1
Part (c)
The Big Bang theory✓ 1

Exam tip: A definition of red-shift needs a comparison: longer wavelength than the light from a source that is not moving.

Theory: What is red-shift in the light from galaxies? Physics only · 2P
18IGPE2-061-02-088(d) Cosmology Paper 1PEasy4 marks

A student stands at the roadside as an ambulance drives past with its siren sounding. The student uses this to help understand light from galaxies.

(a)

Compare the pitch of the siren heard by the student while the ambulance is moving away with the pitch heard while it was approaching.

(1)
(b)

Explain this difference in terms of the sound waves reaching the student while the ambulance is moving away.

(2)
(c)

State the name of the equivalent effect seen in light from a galaxy that is moving away from Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The pitch is lower while it is moving away (than while it was approaching)✓ 1
Part (b)
The source moves away between emitting successive waves, so the wavefronts are spread further apart / stretched✓ 1
giving a longer wavelength; the speed of sound is unchanged, so the frequency is lower✓ 1
Part (c)
Red-shift✓ 1

Exam tip: Pitch is linked to frequency. A longer wavelength at the same wave speed means a lower frequency.

Theory: What happens to a wave when its source is moving? Physics only · 2P
19IGPE2-061-02-118(d) Cosmology Paper 1PEasy4 marks

A satellite carries sensitive detectors that map radiation reaching it from every direction in space. The satellite detects cosmic microwave background (CMB) radiation.

(a)

State the region of the electromagnetic spectrum in which the CMB radiation is detected.

(1)
(b)

State the theory of the universe for which the CMB is evidence.

(1)
(c)

Explain why the CMB radiation is detected from all directions.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Microwaves✓ 1
Part (b)
The Big Bang theory✓ 1
Part (c)
It is radiation left over from the early, hot universe✓ 1
which filled all of space, so it now reaches us from every direction✓ 1

Common error: The CMB does not come from a single source such as the Sun or one galaxy. It fills the whole universe.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
20IGPE2-061-00-018(b) Motion in the Universe Paper 1PEasy1 mark

A comet moves around the Sun in a very elongated elliptical orbit. At which point in its orbit is the comet moving fastest?

Show mark scheme
Marking pointMarkNotes
When the comet moves towards the Sun, the Sun's gravitational pull acts partly in its direction of motion, so the comet speeds up. Its speed is greatest at the point closest to the Sun. It slows down again as it moves away. A constant speed applies only to a circular orbit, not to an elongated ellipse.✓ 1Answer A

Answer: A

Common error: Do not assume all orbits have a constant speed. Only circular orbits do. A comet speeds up as it falls towards the Sun and slows down as it moves away.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
21IGPE2-061-00-038(b) Motion in the Universe Paper 1PEasy1 mark

An astronaut has a mass of 80 kg. During a simulated Mars mission, the astronaut's weight on Mars is calculated to be 300 N.
What is the gravitational field strength on Mars?

Show mark scheme
Marking pointMarkNotes
Weight = mass × gravitational field strength, so g = W / m = 300 / 80 = 3.75 ≈ 3.8 N/kg. Option A wrongly assumes g is the same on every planet. Option B divides mass by weight. Option C multiplies weight by mass.✓ 1Answer D

Answer: D

Common error: g is not 10 N/kg everywhere. It depends on the mass and radius of the planet or moon you are standing on.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
22IGPE2-061-00-048(b) Motion in the Universe Paper 1PEasy5 marks

Satellite navigation systems used in mobile phones rely on satellites in circular orbits around the Earth. One satellite has an orbital radius of 2.66 ×107 m. It takes 12 hours to complete one orbit.

satelliter = 2.66 × 10⁷ mv = ?Earth
Not to scale
(a)

State the name of the force that keeps the satellite in its orbit.

(1)
(b)

State the relationship between orbital speed, orbital radius and time period.

(1)
(c)

Calculate the orbital speed of the satellite.
Give your answer in m/s.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Gravitational force / gravity (of the Earth)✓ 1
Part (b)
orbital speed = (2 × π × orbital radius) ÷ time period, or v = 2πr / T✓ 1
Part (c)
T = 12 × 3600 = 43 200 s✓ 1
v = (2 × π × 2.66 ×107) ÷ 43 200✓ 1
v = 3870 m/s (accept 3900 m/s)✓ 1

Common error: The time period must be in seconds when the radius is in metres. Using 12 hours directly gives an answer 3600 times too large.

Theory: How do you calculate orbital speed? Double · Physics
23IGPE-057-00-048(b) Motion in the Universe Paper 1PEasy5 marks

A space agency is designing a robotic lander to land on Titan, the largest moon of Saturn. The lander has a mass of 320 kg. The gravitational field strength on the surface of Titan is 1.4 N/kg.

(a)

State what is meant by gravitational field strength.

(1)
(b)

Calculate the weight of the lander on the surface of Titan.

(2)
(c)

Explain why the lander has a different weight on Earth but the same mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The gravitational force (weight) acting per kilogram of mass✓ 1
Part (b)
W = m × g = 320 × 1.4✓ 1
W = 448 N✓ 1
Part (c)
Mass is the amount of matter in the lander, which does not change with location✓ 1
Weight depends on the gravitational field strength, which is greater on Earth (10 N/kg) than on Titan✓ 1

Exam tip: Weight is measured in newtons and mass in kilograms. Giving the weight in kg loses the final mark.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
24IGPE-057-00-058(b) Motion in the Universe Paper 1PEasy4 marks

Astronomers study objects of very different sizes, from small moons to the whole Universe.

(a)

Describe what is meant by a galaxy.

(2)
(b)

Put these objects in order of size, starting with the smallest:
the Milky Way, Triton (a moon of Neptune), the Universe, the Solar System

(1)
(c)

State how many galaxies the Universe is thought to contain.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A large collection of billions of stars✓ 1
held together by gravitational force✓ 1
Part (b)
Triton, the Solar System, the Milky Way, the Universe✓ 1
Part (c)
Billions (of galaxies)✓ 1

Common error: Do not mix up the Solar System and the galaxy. The Solar System is one star and the objects orbiting it, and it is only a tiny part of the Milky Way.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
25IGPE3-8-14r14-018(d) Cosmology Paper 1PEasy1 mark

The hydrogen lines in the spectrum of a distant galaxy are compared with the same lines from a hydrogen lamp in a laboratory. All the lines from the galaxy are at longer wavelengths. Which conclusion is correct?

Show mark scheme
Marking pointMarkNotes
C is correct: longer wavelengths (red-shift) mean the source is moving away. A is wrong because motion towards us gives shorter wavelengths (blue-shift). B is wrong because temperature does not shift every line to a longer wavelength. D is wrong because the same lines appear, so the same element is present.✓ 1Answer C

Answer: C

Common error: Red-shift means the wavelengths are longer, not that the object is red or hot.

Theory: What is red-shift in the light from galaxies? Physics only · 2P
26IGPE3-8-14r14-028(c) Stellar Evolution Paper 1PEasy1 mark

Which row lists stars in order of increasing surface temperature?

Show mark scheme
Marking pointMarkNotes
B is correct: cooler stars are red and the hottest are blue. A is wrong because it reverses the order (assuming red means hot). C is wrong because red stars are cooler than yellow stars. D is wrong because white stars are hotter than yellow stars.✓ 1Answer B

Answer: B

Exam tip: Remember the order from cool to hot: red, yellow, white, blue.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
27IGPE3-8-14r14-058(c) Stellar Evolution Paper 1PEasy4 marks

In a space game, stars are given colours according to their surface temperature. The game includes a red dwarf, a yellow star like the Sun and a blue-white star.

(a)

Put the three stars in order of increasing surface temperature.

(1)
(b)

The yellow star has a similar mass to the Sun. Describe what happens to this star after it leaves the main sequence.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 red dwarf, yellow star, blue-white star✓ 1
Part (b)
B1 it expands and becomes a red giant✓ 1
B1 the outer layers are ejected / drift away into space✓ 1
B1 the core remains as a white dwarf✓ 1

Exam tip: Name each stage in the correct order and link the final stage to the mass of the star.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
28IGPE3-8-14r14-068(b) Motion in the Universe Paper 1PEasy5 marks

A rover on Mars has a weight of 2220 N on the surface of Mars, where the gravitational field strength is 3.7 N/kg.

(a)

Calculate the mass of the rover.

(2)
(b)

Calculate the weight of the rover on Earth, where the gravitational field strength is 10 N/kg.

(2)
(c)

State what is meant by gravitational field strength.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 mass = weight ÷ gravitational field strength = 2220 ÷ 3.7✓ 1
A1 600 kg✓ 1
Part (b)
M1 600 × 10 (ECF from (a))✓ 1
A1 6000 N✓ 1
Part (c)
B1 the force (acting) per unit mass (in a gravitational field)✓ 1

Exam tip: Write the equation and substitute before calculating, and keep the unit with each answer.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
29IGPE3-8-14r17-018(c) Stellar Evolution Paper 1PEasy1 mark

Through a telescope, the two stars of a double-star system appear orange and blue. Which statement about their surface temperatures is correct?

Show mark scheme
Marking pointMarkNotes
B is correct: hotter stars emit most strongly at shorter wavelengths, so blue stars are hotter than orange stars. A reverses the link between wavelength and temperature. C is wrong because stars in one system can have very different temperatures. D confuses apparent brightness with temperature.✓ 1Answer B

Answer: B

Common error: Brightness in the sky depends on distance and size as well as temperature, so colour is the better indicator of surface temperature.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
30IGPE3-8-14r17-038(b) Motion in the Universe Paper 1PEasy1 mark

A rock sample weighs 36 N on Earth, where the gravitational field strength is 10 N/kg. The gravitational field strength on the Moon is 1.6 N/kg. What are the mass and the weight of the rock on the Moon?

Show mark scheme
Marking pointMarkNotes
A is correct: mass = 36 ÷ 10 = 3.6 kg and does not change, and weight on the Moon = 3.6 × 1.6 = 5.76 N. B wrongly treats the mass as changing. C wrongly keeps the weight the same. D divides 36 by 1.6 instead of multiplying the mass by 1.6.✓ 1Answer A

Answer: A

Common error: Mass is the same everywhere; only weight changes with gravitational field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
31IGPE3-8-14r17-058(c) Stellar Evolution Paper 1PEasy5 marks

A library display describes the life cycle of the Sun.

(a)

State the name of the stage in which the Sun spends most of its life.

(1)
(b)

Explain why the Sun stays the same size during this stage.

(2)
(c)

After the red giant stage, the Sun will become a white dwarf. State what a white dwarf is.

(1)
(d)

State the name of the nuclear reaction that releases energy in a main sequence star.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 main sequence✓ 1
Part (b)
B1 gravity pulls matter inwards✓ 1
B1 balanced by outward pressure from the energy released by hydrogen fusion in the core✓ 1
Part (c)
B1 a small, hot, dense core left over when the outer layers have been lost, which cools over time✓ 1
Part (d)
B1 nuclear fusion✓ 1

Common error: Fusion joins small nuclei to make larger nuclei; fission splits large nuclei. Stars are powered by fusion.

Theory: How does a star like the Sun evolve? Single · Double · Physics
32IGPE3-8-14r17-088(b) Motion in the Universe Paper 1PEasy5 marks

A lunar lander has a weight of 336 N on the Moon, where the gravitational field strength is 1.6 N/kg. The gravitational field strength on Earth is 10 N/kg.

(a)

Calculate the mass of the lander.

(2)
(b)

Calculate the weight of the lander on Earth.

(1)
(c)

Explain why the lander's mass is the same on the Moon as on Earth but its weight is different.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 mass = weight ÷ gravitational field strength = 336 ÷ 1.6✓ 1
A1 210 kg✓ 1
Part (b)
A1 210 × 10 = 2100 N (ECF from (a))✓ 1
Part (c)
B1 mass is the amount of matter and does not depend on location✓ 1
B1 weight is the force of gravity, which depends on the gravitational field strength at that location✓ 1

Exam tip: Use W = m × g and rearrange for the unknown quantity; check the units are N and kg.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
33IGPE3-8-15r19-018(b) Motion in the Universe Paper 1PEasy1 mark

A Mars rover has a mass of 180 kg. The gravitational field strength at the surface of Mars is 3.7 N/kg. What is the weight of the rover on Mars?

Show mark scheme
Marking pointMarkNotes
Weight = mass × gravitational field strength = 180 × 3.7 = 666 N, so C is correct. A uses the field strength of the Earth (10 N/kg). B divides mass by field strength (180 ÷ 3.7). D confuses the numerical value of the mass in kg with a force.✓ 1Answer C

Answer: C

Common error: Weight is a force found by multiplying mass by field strength; dividing gives a meaningless quantity.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
34IGPE3-8-15r19-048(c) Stellar Evolution Paper 1PEasy1 mark

Which row shows the stages in the life of a star with a mass similar to that of the Sun, in the correct order?

Show mark scheme
Marking pointMarkNotes
A star like the Sun forms from a nebula, becomes a protostar, spends most of its life on the main sequence, expands to a red giant and then ends as a white dwarf; B is correct. A and D describe the life of a much more massive star (supernova). C has protostar and main sequence in the wrong order.✓ 1Answer B

Answer: B

Exam tip: Decide first whether the star is similar to the Sun or much more massive, since the two have different endings.

Theory: How does a star like the Sun evolve? Single · Double · Physics
35IGPE3-8-15r19-058(b) Motion in the Universe Paper 1PEasy7 marks

The International Space Station (ISS) travels in a circular orbit at an average speed of 7.7 × 103 m/s. The radius of the orbit, measured from the centre of the Earth, is 6.78 × 106 m.

ISSr = 6.78 × 10⁶ m7.7 × 10³ m/sEarth
Not to scale
(a)

Calculate the distance travelled by the ISS in one complete orbit. Give your answer to 3 significant figures.

(2)
(b)

Calculate the time taken for one orbit. Give your answer in minutes to 3 significant figures. Use the unrounded distance from (a).

(3)
(c)

The ISS moves at a constant speed. Explain why the ISS is accelerating.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
use of circumference = 2πr✓ 1
4.26 × 107 m✓ 1
Part (b)
use of time = distance ÷ speed✓ 1
5.53 × 103 s (accept 5532 s)✓ 1
92.2 minutes (ECF from (a))✓ 1
Part (c)
velocity is a vector, so it depends on both speed and direction✓ 1
the direction of motion is changing all the time, so the velocity changes and the ISS accelerates✓ 1

Exam tip: Convert the unit last: find the time in seconds from metres and m/s, then divide by 60.

Theory: How do you calculate orbital speed? Double · Physics
36IGPE3-8-17r27-018(d) Cosmology Paper 1PEasy1 mark

Which statement about the cosmic microwave background radiation (CMBR) is correct?

Show mark scheme
Marking pointMarkNotes
B is correct: the radiation was released in the hot, dense early universe and the expansion of space has stretched it to microwave wavelengths. A is wrong because the CMBR reaches us from every direction almost equally. C is wrong because the CMBR is not made by galaxy collisions. D is wrong because the radiation is red-shifted (wavelength increased), not blue-shifted, and it is not from supernovae.✓ 1Answer B

Answer: B

Common error: The CMBR is a red-shifted relic of the early universe, not radiation from stars or galaxies that exist now.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
37IGPE3-8-17r27-038(c) Stellar Evolution Paper 1PEasy1 mark

Which row shows the stages the Sun will pass through after it leaves the main sequence?

Show mark scheme
Marking pointMarkNotes
The Sun has too little mass to explode. It will swell into a red giant, shed its outer layers as a planetary nebula and leave a white dwarf core. B and C describe the path of a star much more massive than the Sun, which ends in a supernova. D has the stages in the wrong order: a white dwarf is the end point, not a stage before the red giant.✓ 1Answer A

Answer: A

Common error: A supernova needs a star far more massive than the Sun; the Sun never becomes a neutron star or black hole.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
38IGPE3-8-17r27-088(d) Cosmology Paper 1PEasy5 marks

Light from a hydrogen atom has a wavelength of 656 nm in the laboratory. The same spectral line from a distant galaxy is observed at a wavelength of 689 nm.
Use: red-shift z = Δλ ÷ λ and recession speed v = z × c, where c = 3.00×108 m/s.

(a)

Calculate the change in wavelength.

(1)
(b)

Calculate the red-shift z. Use your unrounded value in part (c).

(2)
(c)

Calculate the recession speed of the galaxy. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δλ = 689 − 656 = 33 nm✓ 1
Part (b)
z = Δλ ÷ λ = 33 ÷ 656 (ECF from (a))✓ 1
= 0.0503✓ 1
Part (c)
v = z × c = 0.0503… × 3.00×108✓ 1
= 1.51×107 m/s✓ 1

Common error: The red-shift z has no unit; do not include nm in z.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
39IGPE3-8-17r27-108(c) Stellar Evolution Paper 1PEasy6 marks

The table gives data for four stars. A star with a lower absolute magnitude is more luminous.

StarSurface temperature / KAbsolute magnitude
P3500−6.0
Q12 000+11.5
R5800+4.8
S20 000−3.5

(a)

State which star is a white dwarf. Give a reason for your answer.

(2)
(b)

Star R is a main sequence star like the Sun. State what is happening in its core.

(2)
(c)

Star P is cool but very luminous. Explain how this is possible.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
star Q✓ 1
hot but very dim (very high absolute magnitude / low luminosity)✓ 1
Part (b)
hydrogen nuclei are fusing to form helium✓ 1
energy is released (accept: nuclear fusion)✓ 1
Part (c)
star P must be very large (a red giant or supergiant), with a huge surface area✓ 1
a large area emits a large amount of energy in total, despite the low temperature of each part of its surface✓ 1

Exam tip: Read the table for both temperature and luminosity before naming a type of star.

Theory: How do you draw the Hertzsprung–Russell diagram? Physics only · 2P
40IGPE3-8-15r41-018(c) Stellar Evolution Paper 1PEasy1 mark

Astronomers classify stars by colour. Which list shows stars in order of decreasing surface temperature?

Show mark scheme
Marking pointMarkNotes
Blue-white stars have the hottest surfaces, yellow stars (like the Sun) are cooler and red stars are the coolest, so the list blue-white, yellow, red is in decreasing order. The list starting with red is the reverse order, a common belief because red is linked with heat in everyday life. The list starting with yellow wrongly puts a yellow star above a blue-white one. The list with red second wrongly puts red above yellow.✓ 1Answer B

Answer: B

Common error: Red is the colour of the coolest stars, not the hottest; the order of colour from coolest to hottest is red, orange, yellow, white, blue-white.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
41IGPE3-8-15r41-028(c) Stellar Evolution Paper 1PEasy1 mark

Which sequence correctly shows the stages in the life cycle of the Sun?

Show mark scheme
Marking pointMarkNotes
The Sun forms from a nebula, contracts to a protostar, spends most of its life on the main sequence, expands to a red giant when its core hydrogen is used up, then sheds its outer layers and leaves a white dwarf. The first sequence is for a star of much greater mass than the Sun, which would end in a supernova. The second puts the red giant before the main sequence. The third swaps the last two stages.✓ 1Answer D

Answer: D

Exam tip: Learn two separate pathways after the main sequence: low-mass stars like the Sun and stars of much greater mass.

Theory: How does a star like the Sun evolve? Single · Double · Physics
42IGPE3-8-15r41-058(b) Motion in the Universe Paper 1PEasy5 marks

A rover is sent to Mars. Its mass is 850 kg. The gravitational field strength is 10 N/kg on Earth and 3.7 N/kg on the surface of Mars.

(a)

Calculate the weight of the rover on the surface of Mars.

(2)
(b)

Calculate how much greater the weight of the rover is on Earth than on Mars.

(2)
(c)

State what happens to the mass of the rover during the journey from Earth to Mars.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × gravitational field strength✓ 1
A1 850 × 3.7 = 3145 N✓ 1
Part (b)
M1 weight on Earth = 850 × 10 = 8500 N✓ 1
A1 8500 − 3145 = 5355 N (ECF from (a))✓ 1
Part (c)
it stays the same / remains 850 kg✓ 1

Common error: Do not change the mass when the location changes; only the weight changes.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
43IGPE3-8-15r41-068(b) Motion in the Universe Paper 1PEasy5 marks

In a space-exploration video game, a character of mass 70 kg has a weight of 112 N on the planet Zorb. The gravitational field strength on Earth is 10 N/kg.

(a)

Calculate the gravitational field strength on Zorb.

(2)
(b)

Suggest why the character can jump higher on Zorb than on Earth when the leg muscles push on the ground with the same force.

(2)
(c)

Calculate the weight of the character on Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 g = weight ÷ mass✓ 1
A1 112 ÷ 70 = 1.6 N/kg✓ 1
Part (b)
the weight (downward force) is smaller on Zorb✓ 1
so the resultant upward force and the upward acceleration are greater✓ 1
Part (c)
70 × 10 = 700 N✓ 1

Exam tip: Rearrange the equation first (g = W ÷ m) and show the substitution; it earns the method mark even if the arithmetic slips.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
44IGPE3-8-15r42-018(c) Stellar Evolution Paper 1PEasy1 mark

Astronomers observe four stars. One is red, one is yellow, one is white and one is blue. Which statement about the surface temperatures of the stars is correct?

Show mark scheme
Marking pointMarkNotes
The colour of the light from a star depends on its surface temperature: cooler stars look red, hotter stars look yellow, then white, and the hottest look blue. So the blue star is the hottest. The red star is the coolest, not the hottest, so the first option is wrong. The yellow star is cooler than the blue star, so the second option is wrong. Distance affects how bright a star looks, not its colour, so the last option is wrong.✓ 1Answer C

Answer: C

Common error: Do not link red with 'hot' from everyday life: for stars, red means the coolest surface and blue means the hottest.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
45IGPE3-8-15r42-048(b) Motion in the Universe Paper 1PEasy1 mark

A rover on the surface of Mars has a weight of 5550 N. The gravitational field strength at the surface of Mars is 3.7 N/kg. What is the mass of the rover?

Show mark scheme
Marking pointMarkNotes
Weight = mass × gravitational field strength, so mass = 5550 ÷ 3.7 = 1500 kg. Dividing by 10 N/kg (the value for Earth) gives 555 kg, which is wrong because the rover is on Mars. Multiplying 5550 by 3.7 gives about 20 500, which is not a mass. Dividing 3.7 by 5550 inverts the equation.✓ 1Answer B

Answer: B

Exam tip: Rearrange W = mg carefully and use the g value for the planet named in the question, not 10 N/kg.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
46IGPE3-8-15r42-058(c) Stellar Evolution Paper 1PEasy4 marks

Three stars, P, Q and R, are observed. Star P is red, star Q is the same colour as the Sun and star R is blue-white. The surface temperature of star P is 2900 K and the surface temperature of the Sun is 5800 K.

(a)

State which star has the highest surface temperature.

(1)
(b)

State the colour of a star that has a lower surface temperature than the Sun.

(1)
(c)

Calculate how many times greater the surface temperature of the Sun is than the surface temperature of star P.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
star R✓ 1
Part (b)
red (accept orange)✓ 1
Part (c)
M1 5800 ÷ 2900✓ 1
A1 2.0 (accept 2)✓ 1

Exam tip: 'How many times greater' means divide the larger value by the smaller one.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
47IGPE3-8-15r42-068(b) Motion in the Universe Paper 1PEasy6 marks

A Mars rover has a mass of 1040 kg. The gravitational field strength is 10 N/kg on Earth and 3.7 N/kg on the surface of Mars.

(a)

Calculate the weight of the rover on Earth.

(2)
(b)

Calculate the weight of the rover on the surface of Mars.

(2)
(c)

Explain why the weight of the rover is different on Mars but its mass is not.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × gravitational field strength / 1040 × 10✓ 1
A1 10 400 N✓ 1
Part (b)
M1 1040 × 3.7✓ 1
A1 3848 N✓ 1
Part (c)
mass is the amount of matter in the rover, which does not change with location✓ 1
weight is the force of gravity on the rover, which depends on the gravitational field strength, and this is smaller on Mars✓ 1

Common error: Weight is a force measured in newtons; mass is measured in kilograms. Do not give a weight in kg.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
48IGPE3-8-17r46-018(b) Motion in the Universe Paper 1PEasy1 mark

A mountaineer's ice axe has a mass of 2.0 kg. The axe is taken to the Moon, where the gravitational field strength is 1.6 N/kg. Which row gives the mass and the weight of the axe on the Moon?

Show mark scheme
Marking pointMarkNotes
Mass does not depend on location, so it stays 2.0 kg. Weight = mass × gravitational field strength = 2.0 × 1.6 = 3.2 N. A mass of 2.0 kg with weight 20 N wrongly keeps the Earth weight. A mass of 0.32 kg wrongly treats mass as changing along with weight. A mass of 12.5 kg comes from dividing the Earth weight of 20 N by 1.6.✓ 1Answer C

Answer: C

Common error: Mass is the amount of matter in an object and stays the same everywhere; only the weight changes with the gravitational field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
49IGPE3-8-17r46-038(c) Stellar Evolution Paper 1PEasy1 mark

Which sequence shows the stages the Sun will pass through after its main sequence stage?

Show mark scheme
Marking pointMarkNotes
The Sun does not have enough mass to become a red supergiant or to explode as a supernova. After the main sequence it expands to a red giant, throws off its outer layers and leaves a white dwarf. The two sequences with a supernova describe stars of much greater mass. The sequence beginning with a protostar is wrong because a protostar forms before the main sequence stage.✓ 1Answer D

Answer: D

Common error: A supernova, neutron star or black hole needs a star of much greater mass than the Sun. Do not use these for the Sun's future.

Theory: How does a star like the Sun evolve? Single · Double · Physics
50IGPE3-8-17r46-058(b) Motion in the Universe Paper 1PEasy5 marks

A relay unit for a lunar telecoms link is placed on the surface of the Moon. The gravitational field strength on the Moon is 1.6 N/kg. A spring balance on the Moon shows that the weight of the unit is 136 N.

(a)

Calculate the mass of the unit.

(2)
(b)

Calculate the weight the unit would have on Earth, where the gravitational field strength is 10 N/kg. Use your answer to (a) (ECF from (a)).

(2)
(c)

State why the mass of the unit is the same on the Moon as on Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 mass = weight ÷ gravitational field strength✓ 1
A1 136 ÷ 1.6 = 85 kg✓ 1
Part (b)
M1 weight = 85 × 10✓ 1
A1 850 N✓ 1
Part (c)
Mass is the amount of matter in the unit, which does not depend on the gravitational field strength✓ 1

Exam tip: Write the equation W = mg first, then rearrange it, and use the gravitational field strength given for the place in question.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
51IGPE3-8-17r46-088(c) Stellar Evolution Paper 1PEasy5 marks

Albireo is a double star. Star A looks orange and star B looks blue-white. The surface temperature of star A is about 4400 K and that of star B is about 13000 K. The Sun looks yellow-white and has a surface temperature of about 5800 K.

(a)

State which of the two stars has the higher surface temperature.

(1)
(b)

Explain how the colour of a star shows its surface temperature.

(2)
(c)

State whether the surface temperature of the Sun is higher or lower than that of (i) star A and (ii) star B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
star B (the blue-white star)✓ 1
Part (b)
hotter stars emit most strongly at shorter wavelengths, so they look bluer✓ 1
cooler stars emit most strongly at longer wavelengths, so they look orange or red✓ 1
Part (c)
(i) higher than star A✓ 1
(ii) lower than star B✓ 1

Exam tip: Compare stars using the temperatures in the question, and write the comparison out: 'hotter' or 'cooler' with the star named.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
52IGPE3-8-16r48-018(b) Motion in the Universe Paper 1PEasy1 mark

Which row correctly shows what each object orbits?

Show mark scheme
Marking pointMarkNotes
The Moon orbits Earth, a comet orbits the Sun in an elongated orbit and a navigation satellite orbits Earth. The Moon does not orbit the Sun directly in the sense used here, comets are not satellites of Earth, and navigation satellites do not orbit the Sun.✓ 1Answer B

Answer: B

Common error: Moons orbit planets, comets orbit the Sun and artificial satellites orbit the body they were launched to circle.

Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
53IGPE3-8-16r48-028(c) Stellar Evolution Paper 1PEasy1 mark

Which sequence shows the stages in the life cycle of the Sun?

Show mark scheme
Marking pointMarkNotes
The Sun is a star of modest mass, so it leaves the main sequence to become a red giant and ends as a white dwarf. A supernova and a neutron star or black hole need a much more massive star. The red giant stage cannot come before the main sequence.✓ 1Answer A

Answer: A

Common error: Do not send the Sun through a supernova; that path is only for stars much more massive than the Sun.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
54IGPE3-8-16r48-078(c) Stellar Evolution Paper 1PEasy5 marks

A theatre company stages a play about the life of the Sun.

(a)

The Sun is currently a stable main-sequence star. Explain why it is stable.

(2)
(b)

State the two stages that follow the main sequence in the life of the Sun, in the correct order.

(2)
(c)

Explain why the Sun will not end its life as a supernova.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The inward force of gravity tries to collapse the star✓ 1
This is balanced by the outward pressure from the energy released by nuclear fusion✓ 1
Part (b)
Red giant✓ 1
White dwarf✓ 1
Part (c)
The Sun does not have enough mass (is not massive enough)✓ 1

Common error: Say that fusion pressure and gravity are balanced; 'fusion stops gravity' is not enough.

Theory: How does a star like the Sun evolve? Single · Double · Physics
55IGPE3-8-16r48-108(b) Motion in the Universe Paper 1PEasy5 marks

A volcanologist studies ice volcanoes on Enceladus, a moon of Saturn, and gives a talk comparing moons and comets.

(a)

State the object that Enceladus orbits.

(1)
(b)

State what a comet orbits and describe the shape of its orbit.

(2)
(c)

State where in its orbit a comet moves fastest and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Saturn✓ 1
Part (b)
It orbits the Sun✓ 1
The orbit is a very elongated ellipse✓ 1
Part (c)
When it is closest to the Sun✓ 1
The gravitational force/pull from the Sun is strongest there, so it speeds up as it falls towards the Sun✓ 1

Common error: A comet orbits the Sun, not a planet; its speed changes around the orbit because the gravitational pull changes.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
56IGCSE-PHY-8-0068(b) Motion in the Universe Papers 1 & 2Grade 6–71 mark
A communications satellite orbits the Earth once every 24.0 hours.
The orbital radius of the satellite is 42 200 km.
What is the orbital speed of the satellite?
Show mark scheme
  • A (3070 m/s);
  • B is incorrect because it omits 2π (distance taken as the radius, 4.22 × 10⁷ ÷ 86 400)
  • C is incorrect because the radius was left in km while the time was in seconds
  • D is incorrect because the time was left in hours instead of being converted to seconds
Theory: How do you calculate orbital speed? Double · Physics
57IGCSE-PHY-8-0078(b) Motion in the Universe Papers 1 & 2Grade 6–76 marks
The International Space Station (ISS) is an artificial satellite in a circular orbit around the Earth.
The orbital radius of the ISS, measured from the centre of the Earth, is 6780 km.
The orbital speed of the ISS is 7.66 km/s.
EarthISSrvelocitynot to scale
(a)
State the name of the force that keeps the ISS in orbit and the direction in which this force acts on the ISS.
(2)
(b)
Calculate the time period of the orbit of the ISS.
Give your answer in minutes.
(4)
Show mark scheme
(a)
  • gravitational (force / pull of Earth);
  • towards the centre of the Earth;
Allow
  • gravity / weight for MP1
  • downwards / towards the Earth for MP2
Do not accept
  • centrifugal force
  • away from the Earth
  • in the direction of motion
(b)
  • rearrangement;
  • substitution;
  • evaluation in seconds;
  • conversion to minutes;
  • e.g. T = 2 × π × r ÷ v
  • T = 2 × π × 6780 ÷ 7.66 (= 5561 s)
  • (T =) 92.7 (minutes)
Allow
  • 93, 92.69… (minutes)
  • units consistent in km and km/s or m and m/s
  • ECF from incorrect evaluation for conversion mark
  • 5561 (s) scores 3
Do not accept
  • T = v × 2πr
  • T = 2πv ÷ r
Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
58IGCSE-PHY-8-0098(b) Motion in the Universe Papers 1 & 2Grade 6–76 marks
The table gives data for some bodies in the solar system.
Mass and radius are given as multiples of the values for the Earth.
Body / name: Mercury, Venus, Earth, Moon, Mars, Jupiter
Mass / Earth masses: 0.055, 0.815, 1.00, 0.012, 0.107, 318
Radius / Earth radii: 0.383, 0.949, 1.00, 0.273, 0.532, 11.2
Surface gravitational field strength / N/kg: 3.7, 8.9, 9.8, 1.6, 3.7, 24.8
(a)
Give the names of the two bodies in the table that have the same surface gravitational field strength.
(1)
(b)
Jupiter has 318 times the mass of the Earth, but its surface gravitational field strength is only about 2.5 times that of the Earth.
Use the data to suggest why.
(2)
(c)
A student says:
'The greater the mass of a planet, the greater the gravitational field strength at its surface.'
Evaluate this statement using the data in the table.
(3)
Show mark scheme
(a)
  • Mercury and Mars;
Do not accept
  • only one named
(b)
  • Jupiter has a (much) larger radius / is much bigger;
  • (so) surface is further from the centre (of the planet), reducing g;
Allow
  • radius 11.2 times Earth's quoted for MP1
  • g also depends on radius / size for MP2
Do not accept
  • Jupiter is a gas so has less gravity
  • Jupiter is further from the Sun
(c)
  • any three from:
  • trend supported for most bodies, e.g. Moon has smallest mass and smallest g / Jupiter largest mass and largest g;
  • Mars has (about twice) more mass than Mercury but the same g;
  • (because) Mars has a larger radius (than Mercury);
  • conclusion: g depends on radius as well as mass, so statement is not always true;
Allow
  • comparison of Venus and Earth (mass ratio 0.815, g ratio 0.91) as a valid trend example
Do not accept
  • unsupported 'the student is right/wrong' with no data
Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
59IGCSE-PHY-8-0108(b) Motion in the Universe Papers 1 & 2Grade 6–75 marks
Io is one of the moons of Jupiter.
Io moves in an almost circular orbit with a speed of 17.3 km/s.
The time period of Io's orbit is 1.77 days.
JupiterIoorbit of Ionot to scale
(a)
State the force that causes Io to orbit Jupiter.
(1)
(b)
Calculate the orbital radius of Io.
Give your answer in km.
(3)
(c)
Describe one difference between the orbit of Io and the orbit of Jupiter.
(1)
Show mark scheme
(a)
  • gravity / gravitational (force / attraction of Jupiter);
Do not accept
  • magnetic force
  • centripetal force on its own
(b)
  • conversion of time to seconds;
  • rearrangement and substitution;
  • evaluation;
  • e.g. T = 1.77 × 24 × 60 × 60 = 152 928 (s)
  • r = 17.3 × 152 928 ÷ (2 × π)
  • (r =) 421 000 (km)
Allow
  • 420 000, 421 069 (km)
  • 4.21 × 10⁵ (km)
  • ECF from incorrect/no time conversion: 4.87 (km) using 1.77 days scores 1
  • answer in m (4.21 × 10⁸ m) scores 2
Do not accept
  • r = 2π × v × T
(c)
  • Io orbits a planet / Jupiter but Jupiter orbits the Sun;
Allow
  • Io's orbit is much smaller / has a much shorter period
Do not accept
  • Io's orbit is elliptical but Jupiter's is circular
Theory: How do you calculate orbital speed? Double · Physics
60IGCSE-PHY-8-0118(b) Motion in the Universe Papers 1 & 2Grade 6–79 marks
Astronauts on the Moon test a light gate.
A small steel ball is released from rest and falls a measured distance s through the light gate.
The light gate measures the speed v of the ball.
The astronauts repeat this for different distances s.
The graph shows v² plotted against s, with a line of best fit.
There is no atmosphere on the Moon.
0.00.20.40.60.81.01.21.40.00.51.01.52.02.53.03.54.04.5distance fallen s / mv² / m²/s²
(a)
State the equation linking final speed, initial speed, acceleration and distance moved.
(1)
(b)
Explain why the graph of v² against s is a straight line through the origin.
(2)
(c)
Use the graph to determine the acceleration of the ball on the Moon.
(4)
(d)
A scientific instrument has a mass of 4.5 kg.
Use your answer to (c) to calculate how much less the instrument weighs on the Moon than on the Earth.
(g on Earth = 10 N/kg)
(2)
Show mark scheme
(a)
  • (final speed)² = (initial speed)² + (2 × acceleration × distance moved);
Allow
  • v² = u² + 2as
Do not accept
  • v = u + at
  • v² = u + 2as
(b)
  • (initial speed / u is zero so) v² = 2as;
  • acceleration is constant (so v² proportional to s);
Allow
  • no air resistance so acceleration constant for MP2
  • gradient = 2a / 2g
Do not accept
  • because speed is constant
(c)
  • large triangle / two points far apart read from line;
  • gradient value in range 3.1 to 3.4;
  • idea that gradient = 2 × acceleration;
  • evaluation in range 1.55 to 1.70 (m/s²);
  • e.g. gradient = (3.89 − 0) ÷ (1.20 − 0) = 3.24
  • a = 3.24 ÷ 2
  • (a =) 1.6 (m/s²)
Allow
  • use of a single data point read from the line e.g. 3.88/(2 × 1.20) = 1.62 scores MP3 and MP4 only
  • N/kg as unit
Do not accept
  • gradient taken as acceleration (3.2 m/s²) for MP4
  • use of a=v/t
(d)
  • calculation of either weight;
  • evaluation of difference;
  • e.g. Moon: 4.5 × 1.62 = 7.3 (N); Earth: 4.5 × 10 = 45 (N)
  • (difference =) 38 (N)
Allow
  • 37.7, 37.71 (N)
  • ECF from (c): 45 − 4.5 × a
Do not accept
  • difference in mass
Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
61IGCSE-PHY-8-0258(b) Motion in the Universe Papers 1 & 2Grade 6–71 mark
Satellites X and Y are in circular orbits around the Earth.
The orbital radius of Y is 4 times the orbital radius of X.
The time period of Y is 8 times the time period of X.
How does the orbital speed of Y compare with the orbital speed of X?
Show mark scheme
  • D (half the speed of X);
  • A is incorrect: this comes from dividing the period ratio by the radius ratio (8 ÷ 4) instead of radius ratio ÷ period ratio
  • B is incorrect: this uses only the radius ratio and ignores the longer time period
  • C is incorrect: this uses only the period ratio (1 ÷ 8) and ignores the larger radius
Theory: How do you calculate orbital speed? Double · Physics
62IGCSE-PHY-8-0278(b) Motion in the Universe Papers 1 & 2Grade 6–76 marks
The graph shows how the gravitational field strength, g, varies with distance from the centre of the Earth.
The dashed line shows the Earth's surface.
024681012141618200246810Earth's surfacedistance from centre of the Earth / 1000 kmgravitational field strength / N/kg
(a)
Describe how g changes as the distance from the centre of the Earth increases.
(2)
(b(i))
A weather satellite orbits at a height of 3600 km above the Earth's surface.
Use the graph to determine the gravitational field strength at the satellite.
(1)
(b(ii))
The mass of the satellite is 1500 kg.
Calculate the gravitational force (weight) acting on the satellite in orbit.
(2)
(c)
State the effect of this gravitational force on the motion of the satellite.
(1)
Show mark scheme
(a)
  • (g) decreases (as distance increases);
  • rate of decrease gets less / curve not straight / not (inversely) proportional to distance / doubling distance gives about a quarter of g;
Allow
  • falls quickly near the Earth and more slowly further away, for MP2
Do not accept
  • g decreases linearly
  • g becomes zero
(b(i))
  • (read at 10 000 km) 4.0 (N/kg);
Allow
  • 3.9–4.1 (N/kg)
Do not accept
  • a value read at 3.6 (thousand km) on the axis (height used instead of distance from centre)
  • 9.8 (surface value)
(b(ii))
  • substitution;
  • evaluation;
  • e.g. W = 1500 × 4.0
  • (W =) 6000 (N)
Allow
  • ECF from b(i)
  • 5850–6150 (N) from a reading in the range 3.9–4.1
  • 14 700 (N) (using 9.8 N/kg) scores 1 for substitution of a value of g
Do not accept
  • 6.0 (N) (POT error with mass in tonnes) for evaluation
(c)
  • keeps it in orbit (around the Earth) / makes it move in a circle (rather than a straight line);
Allow
  • pulls it towards (the centre of) the Earth
Do not accept
  • makes it fall to the ground
  • balances a centrifugal force
Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
63IGCSE-PHY-8-0288(b) Motion in the Universe Papers 1 & 2Grade 6–78 marks
A student investigates the relationship between the weight and the mass of objects in the laboratory.
The student hangs slotted masses on a newtonmeter, as shown in the diagram, and records the reading for each mass.
Mass / kg: 0.10, 0.20, 0.30, 0.40, 0.50, 0.60
Weight / N: 1.0, 2.0, 2.9, 3.9, 4.9, 5.9
newtonmeterslotted masseson a hangerclamp standnot to scale
(a)
Give two things the student should do to make sure each newtonmeter reading is accurate.
(2)
(b)
The student plots a graph of weight (y-axis) against mass (x-axis).
The graph is a straight line through the origin.
State what the gradient of this graph represents and give its unit.
(2)
(c)
Use the data in the table to determine a value for the gravitational field strength in the laboratory.
(2)
(d)
The unit of gravitational field strength can also be written as m/s².
Use the relationship force = mass × acceleration to show that 1 N/kg is the same as 1 m/s².
(2)
Show mark scheme
(a)
  • any two from:
  • check newtonmeter reads zero before adding masses / zero the newtonmeter;
  • read scale at eye level / perpendicular to the scale;
  • wait for masses to stop moving (oscillating) before reading;
  • newtonmeter vertical / not touching the stand;
Allow
  • use a newtonmeter with a smaller scale division (higher resolution)
Do not accept
  • repeat readings (alone, improves reliability not accuracy of a single reading)
  • use a balance
(b)
  • (gravitational) field strength / g;
  • N/kg;
Allow
  • m/s² for MP2
  • weight per unit mass for MP1
Do not accept
  • gravity for MP1
  • N for MP2
  • acceleration (alone) for MP1
(c)
  • substitution into g = W ÷ m (any pair of values) or gradient calculation;
  • evaluation;
  • e.g. g = 5.9 ÷ 0.60
  • (g =) 9.8 (N/kg)
Allow
  • 9.67–10 (N/kg) from any pair of data or a gradient
  • 10 (N/kg) from 1.0 ÷ 0.10 or 2.0 ÷ 0.20
Do not accept
  • 10 N/kg quoted from memory with no working (scores 0)
(d)
  • 1 N = 1 kg × 1 m/s² (from F = m × a);
  • so N/kg = (kg × m/s²) ÷ kg = m/s²;
Allow
  • argument using units only: N = kg m/s², N/kg = kg m/s² ÷ kg
Do not accept
  • 'they both measure g' with no unit working
Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
64IGCSE-PHY-8-0298(b) Motion in the Universe Papers 1 & 2Grade 6–76 marks
A spacecraft called an orbiter moves in a circular orbit around the Moon.
The orbiter is 100 km above the Moon's surface.
The radius of the Moon is 1740 km.
The time period of the orbit is 118 minutes.
Moonorbiter1740 km100 kmnot to scale
(a)
Calculate the orbital speed of the orbiter.
Give your answer in m/s.
(4)
(b)
A satellite in a low orbit around the Earth has an orbital speed of about 7800 m/s.
Suggest why the orbital speed of the orbiter around the Moon is much smaller.
(2)
Show mark scheme
(a)
  • orbital radius = 1840 km / 1.84 × 10⁶ m;
  • T = 7080 s;
  • substitution into v = 2πr/T;
  • evaluation;
  • e.g. v = 2 × π × 1.84 × 10⁶ ÷ 7080
  • (v =) 1630 (m/s)
Allow
  • 1633 (m/s), 1600 (m/s)
  • 1540 (m/s) (using r = 1740 km) scores 3
  • 1.63 (m/s) or 1630 km/s POT error scores 3
  • 98 000 (using T in minutes and r in m) scores 3
Do not accept
  • 89 (m/s) (using 100 km as radius) for MP1
(b)
  • Moon has a (much) smaller mass (than the Earth);
  • (so) gravitational force on the orbiter / gravitational field strength (at the orbit) is weaker;
Allow
  • Moon's gravitational field is weaker for MP2
  • (so) a smaller speed is enough for gravity to keep it in orbit, as an alternative to MP2 if linked to weaker gravity
Do not accept
  • there is no gravity on the Moon / in space
  • the orbiter has a smaller mass
  • the Moon has no atmosphere (no air resistance)
Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
65IGCSE-PHY-8-0308(c) Stellar Evolution Papers 1 & 2Grade 6–76 marks
In the year 1054, astronomers in China recorded a 'guest star' that suddenly appeared in the sky.
It was bright enough to be seen in daylight for several weeks and then slowly faded.
Today, telescopes pointed at the same position show the Crab Nebula: an expanding cloud of gas and dust with a neutron star at its centre.
(a)
Name the event that the astronomers saw in 1054.
(1)
(b)
Astronomers know that the star that produced the Crab Nebula had a much greater mass than the Sun.
Explain how the information above shows this.
(2)
(c)
Before this event, the core of the star collapsed.
Explain why the core collapsed.
(2)
(d)
A neutron star is one type of object that can be left after this event.
Name the other type of object that can be left.
(1)
Show mark scheme
(a)
  • supernova;
Do not accept
  • nova
  • red giant / red supergiant
  • white dwarf
(b)
  • it exploded as a supernova / it left a neutron star (at the centre);
  • stars with a mass similar to the Sun do not explode / end as a white dwarf (after losing their outer layers gently);
Allow
  • only stars much more massive than the Sun become supernovae / neutron stars, for MP2
Do not accept
  • it was very bright so it was very big (with no reference to supernova or remnant)
  • the Sun will become a neutron star
(c)
  • (nuclear) fusion (in the core) stopped / fuel for fusion ran out / no more outward pressure (from energy released);
  • inward gravitational force no longer balanced / gravity pulled the core inwards;
Allow
  • the outward force no longer balanced the inward force of gravity scores 2
Do not accept
  • fission for fusion
  • the star ran out of oxygen / stopped burning (chemical)
(d)
  • black hole;
Do not accept
  • white dwarf
  • red dwarf
  • nebula
Theory: How does a star like the Sun evolve? Single · Double · Physics
66IGPE-057-00-038(d) Cosmology Paper 1PMedium1 mark

A spectral line has a wavelength of 656 nm when measured in a laboratory. In the light from a distant galaxy, the same line is measured at 669 nm.
Speed of light = 3.0 × 108 m/s
What is the speed at which the galaxy is moving away from Earth?

Show mark scheme
Marking pointMarkNotes
Use Δλ/λ0 = v/c.
Δλ = 669 − 656 = 13 nm.
v = (13/656) × 3.0 × 108 = 5.9 × 106 m/s.
Why the other options are wrong:
• A is just the ratio Δλ/λ0, without multiplying by c.
• B divides by the observed wavelength (669 nm) instead of the laboratory wavelength.
• D uses 669/656 and forgets to subtract to find the change in wavelength.		✓ 1Answer C

Answer: C

Common error: Always divide the change in wavelength by the laboratory (unshifted) wavelength λ₀, not by the observed wavelength.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
67IGPE2-061-01-058(c) Stellar Evolution Paper 1PMedium5 marks

A student uses a small telescope and a phone camera to photograph four stars. She records their colours:
Star A – red    Star B – white    Star C – yellow    Star D – blue

(a)

List the four stars in order of increasing surface temperature.

(2)
(b)

The surface temperature of the Sun is about 5800 K. State which of the four stars has a colour most similar to the Sun.

(1)
(c)

Star A is a red giant. It emits much more light in total than star D, even though star D is much hotter.
Explain how this is possible.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A (red) coolest and D (blue) hottest, at the correct ends of the list✓ 1
C (yellow) placed before B (white) in the middle, giving A, C, B, D✓ 1
Part (b)
C / the yellow star✓ 1
Part (c)
star A is much larger / has a much bigger surface area than star D✓ 1
so, although each square metre of A emits less light, the total light emitted from its whole surface is greater; an argument using L = 4πr²σT⁴, with the larger r² outweighing the smaller T⁴, also earns this mark✓ 1

Common error: A hotter star is not automatically more luminous. Size matters as well, so mention surface area when comparing a cool giant with a hot small star.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
68IGPE2-061-02-038(d) Cosmology Paper 1PMedium1 mark

The cosmic microwave background (CMB) radiation was emitted in the early universe as much shorter wavelength radiation. Why is it detected today as microwaves?

Show mark scheme
Marking pointMarkNotes
As space expanded, the wavelength of the radiation travelling through it was stretched. Radiation once at much shorter wavelengths now lies in the microwave region. Electromagnetic waves do not slow down in a vacuum, and a slower wave would not have a shorter wavelength anyway. Satellites above the atmosphere also detect the CMB. Dust re-emission would not produce such a uniform signal from every direction.✓ 1Answer C

Answer: C

Exam tip: Link the CMB to the expansion of the universe. The same stretching of wavelengths that causes red-shift also explains why the early radiation is now in the microwave region.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
69IGPE2-061-02-078(d) Cosmology Paper 1PMedium6 marks

A planetarium is writing a short presentation for visitors about the Big Bang theory.

(a)

Explain how observations of red-shift and of the cosmic microwave background (CMB) radiation provide evidence for the Big Bang theory.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
Light from distant galaxies is red-shifted / has longer wavelengths than when emitted✓ 1
so the galaxies are moving away from us✓ 1
More distant galaxies show a greater red-shift / are moving away faster✓ 1
so the universe is expanding and must once have been concentrated in a very small, hot, dense region✓ 1
CMB radiation is detected from all directions in space / is almost uniform across the sky✓ 1
It is left-over radiation from the hot early universe, stretched into the microwave region by the expansion✓ 1

Exam tip: In a 6-mark answer, build a logical chain: observation, then what it shows, then the link to the Big Bang. Cover both red-shift and the CMB.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
70IGPE2-061-00-058(b) Motion in the Universe Paper 1PMedium5 marks

A crewed space station orbits the Earth in a circular orbit of radius 6.79 ×106 m. Its orbital speed is 7.66 km/s. Astronauts on board see several sunrises every day.

space stationr = 6.79 × 10⁶ m7.66 km/sEarth
Not to scale
(a)

Calculate the time taken for the space station to complete one orbit.
Give your answer in minutes.

(3)
(b)

The space station moves at a constant speed.
Explain why the space station is still accelerating.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2πr / v, using v = 7660 m/s✓ 1
T = (2 × π × 6.79 ×106) ÷ 7660 = 5570 s✓ 1
T = 5570 ÷ 60 = 93 minutes✓ 1
Part (b)
Its direction of motion is constantly changing, so its velocity changes✓ 1
There is a resultant force (gravity) acting towards the centre of the Earth✓ 1

Exam tip: Convert km/s to m/s before substituting. Then check the unit asked for. Here the answer must be converted from seconds to minutes.

Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
71IGPE2-061-00-088(b) Motion in the Universe Paper 1PMedium6 marks

A weather and communications satellite is placed in a geostationary orbit, so it stays above the same point on the Earth's equator. Its orbital speed is 3.07 ×103 m/s.
Radius of the Earth = 6.37 ×106 m.

satelliter = ?3.07 × 10³ m/sh = ?Earth
Not to scale (radius of the Earth = 6.37 × 10⁶ m)
(a)

State the time period of a geostationary orbit and explain why it must have this value.

(2)
(b)

Calculate the orbital radius of the satellite.

(3)
(c)

Calculate the height of the satellite above the Earth's surface.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
24 hours / 1 day✓ 1
It must match the time for the Earth to rotate once, so the satellite stays above the same point✓ 1
Part (b)
r = vT / 2π, using T = 24 × 3600 = 86 400 s✓ 1
r = (3.07 ×103 × 86 400) ÷ (2 × π)✓ 1
r = 4.22 ×107 m✓ 1
Part (c)
4.22 ×107 − 6.37 ×106 = 3.58 ×107 m✓ 1

Common error: The orbital radius is measured from the centre of the Earth, not from the surface. Subtract the Earth's radius to find the height above the ground.

Theory: How do you calculate orbital speed? Double · Physics
72IGPE-057-00-068(b) Motion in the Universe Paper 1PMedium6 marks

A navigation satellite moves in a circular orbit around the Earth. The orbital radius is 26 600 km and the satellite takes 12 hours to complete one orbit.

satelliter = 26 600 kmv = ?Earth
Not to scale
(a)

State the equation linking orbital speed, orbital radius and time period.

(1)
(b)

Calculate the orbital speed of the satellite in m/s.

(3)
(c)

The satellite moves at a constant speed. Explain why there must be a resultant force acting on it.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
orbital speed = (2 × π × orbital radius) ÷ time period, or v = 2πr/T✓ 1
Part (b)
Converts units: r = 2.66 × 107 m and T = 12 × 3600 = 43 200 s✓ 1
v = (2 × π × 2.66 × 107) ÷ 43 200✓ 1
v = 3870 m/s (accept answers from 3860 m/s to 3900 m/s)✓ 1
Part (c)
The direction of the satellite's motion is always changing, so its velocity is changing. This means it is accelerating✓ 1
An acceleration (change in velocity) needs a resultant force, by F = ma. Also accept: the force is Earth's gravitational pull acting towards the centre of the orbit✓ 1

Exam tip: Convert kilometres to metres and hours to seconds before you substitute. That gives the speed directly in m/s.

Theory: How do you calculate orbital speed? Double · Physics
73IGPE-057-00-118(d) Cosmology Paper 1PMedium5 marks

A galaxy is moving away from the Earth at a speed of 1.5 × 107 m/s. A hydrogen spectral line has a wavelength of 486 nm when measured in a laboratory.
Speed of light = 3.0 × 108 m/s

(a)

Explain what is meant by red-shift.

(2)
(b)

Calculate the wavelength of this hydrogen line when it is observed in the light from the galaxy.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The observed wavelength of light from the source is increased (lines move towards the red end of the spectrum)✓ 1
because the source is moving away from the observer✓ 1
Part (b)
Δλ = λ0 × v/c = 486 × (1.5 × 107 ÷ 3.0 × 108)✓ 1
Δλ = 24.3 nm✓ 1
Observed wavelength = 486 + 24.3 = 510 nm (510.3 nm)✓ 1

Common error: Δλ is the change in wavelength, not the observed wavelength. Add it to the laboratory value to get the wavelength that is actually observed.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
74IGPE3-8-14r14-038(b) Motion in the Universe Paper 1PMedium1 mark

A rover has a mass of 180 kg on Earth, where the gravitational field strength is 10 N/kg. The gravitational field strength on Mars is 3.7 N/kg. Which row gives the mass and weight of the rover on Mars?

Show mark scheme
Marking pointMarkNotes
A is correct: mass does not depend on location, and weight = 180 × 3.7 = 666 N. B is wrong because it treats mass as changing with field strength. C is wrong because it takes the weight to be the same as on Earth. D is wrong because it divides 180 by 3.7 instead of multiplying.✓ 1Answer A

Answer: A

Common error: Mass is the amount of matter and stays the same everywhere; only the weight changes with gravitational field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
75IGPE3-8-14r14-048(b) Motion in the Universe Paper 1PMedium1 mark

A satellite moves at constant speed in a circular orbit around the Earth. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
D is correct: velocity includes direction, which changes continuously, so there is a resultant force (gravity) towards the Earth. A is wrong because it ignores direction. B is wrong because gravity acts at orbital heights. C is wrong because gravity changes the direction, not the speed, in a circular orbit.✓ 1Answer D

Answer: D

Common error: Constant speed does not mean constant velocity; a change in direction is a change in velocity.

Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
76IGPE3-8-14r14-088(d) Cosmology Paper 1PMedium6 marks

Light from the Andromeda galaxy shows spectral lines at shorter wavelengths than the same lines measured in a laboratory. Light from most other distant galaxies shows the opposite effect.

(a)

State the name of the effect seen in the light from the Andromeda galaxy.

(1)
(b)

State what this effect shows about the motion of Andromeda relative to the Earth.

(1)
(c)

Explain how the light from most other distant galaxies supports the idea that the Universe is expanding.

(2)
(d)

A student says: 'The blue-shift of Andromeda proves the Universe is not expanding.' Evaluate this statement.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 blue-shift✓ 1
Part (b)
B1 it is moving towards the Earth / the Milky Way✓ 1
Part (c)
B1 their spectral lines are shifted to longer wavelengths (red-shift), so they are moving away from us✓ 1
B1 the more distant the galaxy, the greater the red-shift, so the faster it is receding✓ 1
Part (d)
B1 Andromeda is a nearby galaxy whose motion is dominated by gravitational attraction to the Milky Way✓ 1
B1 the great majority of galaxies still show red-shift, so one blue-shifted galaxy does not disprove expansion✓ 1

Exam tip: In an evaluation, give a reason on each side before reaching a conclusion.

Theory: What is red-shift in the light from galaxies? Physics only · 2P
77IGPE3-8-14r14-108(b) Motion in the Universe Paper 1PMedium6 marks

A comet moves around the Sun in a highly elliptical orbit. The comet takes 12 years to complete one orbit, and the total length of its path in one orbit is 4.00 × 1012 m.

Suncometlength of one orbit = 4.00 × 10¹² m
Not to scale
(a)

State where in its orbit the comet moves fastest.

(1)
(b)

Explain why the speed of the comet changes as it moves around its orbit.

(2)
(c)

Calculate the average speed of the comet over one orbit. Use 1 year = 3.15 × 107 s. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 when it is closest to the Sun✓ 1
Part (b)
B1 the gravitational attraction of the Sun is stronger when the comet is closer, and it pulls the comet forwards as it approaches, so it speeds up✓ 1
B1 as it moves away, the Sun's pull acts against its motion, so it slows down✓ 1
Part (c)
M1 time = 12 × 3.15 × 107 = 3.78 × 108 s✓ 1
M1 speed = distance ÷ time = 4.00 × 1012 ÷ (3.78 × 108)✓ 1
A1 1.06 × 104 m/s (10600 m/s)✓ 1

Common error: Convert the period to seconds before using speed = distance ÷ time with a distance in metres.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
78IGPE3-8-14r14-128(b) Motion in the Universe Paper 1PMedium5 marks

A base on Titan stores water in a sealed tank. The tank and its water have a mass of 300 kg and a weight of 405 N on the surface of Titan.

(a)

Calculate the gravitational field strength on the surface of Titan.

(2)
(b)

The tank and its water are taken to Earth, where the gravitational field strength is 10 N/kg. State the mass of the tank and calculate its weight on Earth.

(2)
(c)

Suggest why the water in the tank is easier to carry on Titan than on Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 405 ÷ 300✓ 1
A1 1.35 N/kg✓ 1
Part (b)
B1 mass is still 300 kg✓ 1
A1 weight = 300 × 10 = 3000 N✓ 1
Part (c)
B1 the weight is much smaller on Titan, so less force is needed to lift it✓ 1

Common error: Do not change the mass when the location changes; only the weight depends on gravitational field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
79IGPE3-8-14r17-028(c) Stellar Evolution Paper 1PMedium1 mark

Four stars have these colours: P is red, Q is blue-white, R is yellow and S is orange. Which row lists the stars in order of increasing surface temperature?

Show mark scheme
Marking pointMarkNotes
A is correct: red is coolest, then orange, yellow and blue-white is hottest. B lists the order of decreasing temperature. C swaps orange and yellow. D places orange cooler than red.✓ 1Answer A

Answer: A

Exam tip: Remember the colour order from cool to hot: red, orange, yellow, white, blue.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
80IGPE3-8-14r17-098(c) Stellar Evolution Paper 1PMedium6 marks

Astronomers study stars much more massive than the Sun.

(a)

Describe what happens to a star much more massive than the Sun after it leaves the main sequence, until it becomes a stellar remnant.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 hydrogen in the core runs out✓ 1
B1 the core contracts and the outer layers expand and cool, forming a red supergiant✓ 1
B1 heavier elements are formed by fusion in the core, up to iron✓ 1
B1 the star collapses and explodes as a supernova✓ 1
B1 the core left behind is very dense and may be a neutron star✓ 1
B1 if the core is massive enough, it becomes a black hole✓ 1

Exam tip: For a six-mark "describe" question, give a sequence of at least six distinct stages in the correct order. Accept equivalent wording.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
81IGPE3-8-14r17-108(d) Cosmology Paper 1PMedium5 marks

Scientists have two main pieces of evidence for the Big Bang theory.

(a)

State the part of the electromagnetic spectrum in which the cosmic microwave background radiation (CMBR) is detected.

(1)
(b)

Explain how the CMBR supports the Big Bang theory.

(2)
(c)

Astronomers observe that light from more distant galaxies shows greater red-shift. State what this tells us about the galaxies and explain how this supports the theory.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 microwaves✓ 1
Part (b)
B1 the radiation was emitted when the early universe was very hot and dense✓ 1
B1 as the universe expanded its wavelength stretched into the microwave region✓ 1
Part (c)
B1 more distant galaxies are moving away faster✓ 1
B1 consistent with the universe expanding from a single small point✓ 1

Common error: Red-shift is a stretching of wavelength towards the red end; it is not caused by the galaxy being red.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
82IGPE3-8-14r17-118(b) Motion in the Universe Paper 1PMedium4 marks

A comet moves around the Sun in an elliptical orbit.

Suncomet
Not to scale
(a)

State the part of its orbit where the comet moves fastest.

(1)
(b)

Explain why the speed of the comet changes during its orbit.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 when it is closest to the Sun✓ 1
Part (b)
B1 the Sun's gravitational pull is stronger when the comet is closer✓ 1
B1 as the comet approaches the Sun, the force has a component along its motion so it speeds up✓ 1
B1 as it moves away the force acts against its motion so it slows down✓ 1

Exam tip: Link every change in speed to the direction of the gravitational force relative to the motion.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
83IGPE3-8-15r19-028(b) Motion in the Universe Paper 1PMedium1 mark

A comet moves around the Sun in an elliptical orbit. At which point is the speed of the comet greatest, and why?

Suncomet
Not to scale
Show mark scheme
Marking pointMarkNotes
As the comet falls towards the Sun it loses gravitational potential energy and gains kinetic energy, so it is fastest at the nearest point; C is correct. A is wrong because the speed changes along an elliptical path. B is wrong because a weaker force means less speed gained, not more. D is wrong because the gravitational force is never zero on the orbit.✓ 1Answer C

Answer: C

Exam tip: Link a change in speed to energy transfers: gravitational potential energy to kinetic energy on the way in, and the reverse on the way out.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
84IGPE3-8-15r19-038(d) Cosmology Paper 1PMedium1 mark

A hydrogen spectral line has a wavelength of 656 nm in a laboratory. In light from galaxy P the line is observed at 660 nm. In light from galaxy Q the same line is observed at 666 nm. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Both lines are shifted to longer wavelength (red-shift), so both galaxies are moving away. Q has the larger shift (10 nm compared with 4 nm), so it is receding faster; D is correct. A confuses red-shift with approach. B reverses the comparison. C ignores that the size of the shift depends on speed.✓ 1Answer D

Answer: D

Common error: A bigger shift in wavelength means a faster recession, not a slower one.

Theory: What is red-shift in the light from galaxies? Physics only · 2P
85IGPE3-8-15r19-088(c) Stellar Evolution Paper 1PMedium7 marks

Stars change over millions of years. Most of the life of a star is spent in a stable stage in which hydrogen nuclei fuse to form helium.

(a)

State the name of this stage and explain why the star is stable.

(3)
(b)

Describe what happens to a star of the same mass as the Sun when the hydrogen in its core is used up.

(3)
(c)

State one possible final stage of a star with a mass much greater than the Sun.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
main sequence✓ 1
the inward force of gravity acts on the star✓ 1
this is balanced by the outward pressure produced by the energy released in fusion✓ 1
Part (b)
the star expands and cools✓ 1
it becomes a red giant✓ 1
the outer layers are lost and the core shrinks to become a white dwarf✓ 1
Part (c)
neutron star or black hole✓ 1

Common error: A red giant is larger and cooler than the main sequence star it came from, not smaller and hotter.

Theory: How does a star like the Sun evolve? Single · Double · Physics
86IGPE3-8-15r19-118(b) Motion in the Universe Paper 1PMedium7 marks

Mars orbits the Sun in an almost circular path of radius 2.28 × 1011 m. Mars takes 687 days to complete one orbit.

Marsr = 2.28 × 10¹¹ mSun
Not to scale
(a)

Show that 687 days is about 5.94 × 107 s.

(2)
(b)

Calculate the orbital speed of Mars. Give your answer to 3 significant figures. Use the unrounded time.

(3)
(c)

The orbital speed of the Earth is 2.99 × 104 m/s. Suggest why Mars moves more slowly than the Earth.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
use of 687 × 24 × 3600✓ 1
59 356 800 s, which is 5.94 × 107 s to 3 significant figures✓ 1
Part (b)
distance in one orbit = 2πr✓ 1
use of speed = distance ÷ time✓ 1
2.41 × 104 m/s✓ 1
Part (c)
the gravitational force from the Sun is weaker at the greater distance of Mars✓ 1
so a lower speed is needed to stay in orbit (accept: planets further from the Sun move more slowly)✓ 1

Common error: Convert days to seconds with 24 hours per day and 3600 s per hour before using speed = distance ÷ time.

Theory: How do you calculate orbital speed? Double · Physics
87IGPE3-8-15r19-128(b) Motion in the Universe Paper 1PMedium8 marks

A Formula 1 car completes a lap of a circuit of length 5.0 km at an average speed of 225 km/h. The International Space Station (ISS) travels at 7.7 × 103 m/s in its orbit.

(a)

Show that 225 km/h is 62.5 m/s.

(2)
(b)

Calculate the time taken by the car to complete one lap.

(2)
(c)

Calculate how many times greater the speed of the ISS is than the average speed of the car. Give your answer to 3 significant figures.

(2)
(d)

The ISS is pulled towards the Earth by gravity but does not fall to the ground. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
225 × 1000 ÷ 3600 (or 225 ÷ 3.6)✓ 1
62.5 m/s✓ 1
Part (b)
use of time = distance ÷ speed with 5000 m✓ 1
80 s✓ 1
Part (c)
use of 7.7 × 103 ÷ 62.5✓ 1
123✓ 1
Part (d)
the ISS is falling towards the Earth all the time✓ 1
it moves sideways so quickly that the surface of the Earth curves away beneath it (the ISS stays at the same height)✓ 1

Exam tip: Keep units consistent: change km/h to m/s before comparing with a speed given in m/s.

Theory: How do you calculate orbital speed? Double · Physics
88IGPE3-8-17r27-028(d) Cosmology Paper 1PMedium1 mark

A hydrogen line has a wavelength of 600 nm in the laboratory. In the light from galaxy P the line is observed at 636 nm. In the light from galaxy Q the same line is observed at 672 nm. Which conclusion is best supported by these data?

Show mark scheme
Marking pointMarkNotes
The shifts are 36 nm for P and 72 nm for Q, so Q's red-shift is twice as large. A larger red-shift means a greater recession speed, and for galaxies the speed is proportional to distance, so Q is about twice as far away. A is wrong because both wavelengths have increased, so both galaxies are receding. B reverses the relationship between red-shift and distance. C is wrong because the data say nothing about age, and a different red-shift indicates a different distance.✓ 1Answer D

Answer: D

Exam tip: Compare the change in wavelength, not the observed wavelength, when ranking red-shifts.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
89IGPE3-8-17r27-098(d) Cosmology Paper 1PMedium5 marks

The blue line of hydrogen has a wavelength of 486 nm in the laboratory. A galaxy is receding from the Earth at a speed of 9.0×106 m/s. The speed of light is 3.00×108 m/s.

(a)

Calculate the red-shift z of the light from this galaxy, using z = v ÷ c.

(2)
(b)

Calculate the wavelength at which the blue line is observed. Give your answer to 3 significant figures.

(2)
(c)

State the direction in which the line has moved along the visible spectrum.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
z = 9.0×106 ÷ 3.00×108✓ 1
= 0.030✓ 1
Part (b)
Δλ = z × λ = 0.030 × 486 = 14.58 nm (ECF from (a))✓ 1
observed wavelength = 486 + 14.58 = 501 nm✓ 1
Part (c)
towards the red end (longer wavelength)✓ 1

Exam tip: Add the change in wavelength to the laboratory wavelength; the change alone is not the observed wavelength.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
90IGPE3-8-17r27-118(c) Stellar Evolution Paper 1PMedium7 marks

A student wrote these notes about the life cycle of the Sun.

The Sun began as a cloud of dust and gas that was pulled together by magnetic force. In its core, hydrogen nuclei join together by nuclear fission to release energy. In about five billion years the Sun will swell into a red supergiant. It will then explode as a supernova and leave behind a neutron star.

(a)

The notes contain five errors. Identify each error and write the correct statement.

(5)
(b)

Explain why the Sun stays the same size for billions of years during the main sequence.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
magnetic force should be gravitational force (gravity)✓ 1
fission should be fusion✓ 1
red supergiant should be red giant✓ 1
the Sun will not explode as a supernova; it sheds its outer layers as a planetary nebula (accept: it does not have enough mass for a supernova)✓ 1
neutron star should be white dwarf✓ 1
Part (b)
the inward force of gravity is balanced by the outward pressure from the hot gas and radiation produced by fusion✓ 1
so the Sun is stable / does not shrink or expand✓ 1

Common error: A star changes size only when the balance between gravity and the outward pressure from fusion is upset.

Theory: How does a star like the Sun evolve? Single · Double · Physics
91IGPE3-8-17r27-148(c) Stellar Evolution Paper 1PMedium6 marks

A star has a mass about twenty times the mass of the Sun.

(a)

Describe what happens to this star after it leaves the main sequence.

(3)
(b)

Explain why supernovae are important for the formation of planets such as the Earth.

(2)
(c)

State why the Sun will not end its life as a neutron star or a black hole.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
it swells into a red supergiant✓ 1
it explodes as a supernova✓ 1
the core that remains becomes a neutron star or a black hole✓ 1
Part (b)
elements heavier than iron are made in a supernova (and the explosion scatters elements into space)✓ 1
the scattered material can later be pulled together by gravity to form new stars and planets✓ 1
Part (c)
the Sun does not have enough mass✓ 1

Exam tip: Use the star's mass to decide which life-cycle path it follows.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
92IGPE3-8-15r41-038(b) Motion in the Universe Paper 1PMedium1 mark

A Mars rover has a mass of 900 kg on Earth. The gravitational field strength is 10 N/kg on Earth and 3.7 N/kg on Mars. Which row gives the mass and the weight of the rover on Mars?

Show mark scheme
Marking pointMarkNotes
Mass is the amount of matter and does not change with place, so it stays 900 kg. Weight = mass × gravitational field strength = 900 × 3.7 = 3330 N. The row with 333 kg wrongly treats mass as changing with location. The row with 9000 N uses the field strength of Earth instead of Mars. The row with 243 N divides the mass by the field strength instead of multiplying.✓ 1Answer A

Answer: A

Common error: Mass is measured in kg and stays the same everywhere; weight is a force in N that depends on the local gravitational field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
93IGPE3-8-15r41-078(b) Motion in the Universe Paper 1PMedium6 marks

A satellite used to monitor reservoirs in a regional water-supply network moves in a circular orbit around the Earth. The radius of the orbit is 8.0 × 106 m. One orbit takes 7100 s.

satelliter = 8.0 × 10⁶ mv = ?Earth
Not to scale
(a)

State the force that keeps the satellite in its circular orbit.

(1)
(b)

Calculate the number of complete orbits made by the satellite in one day (24 hours).

(2)
(c)

Calculate the orbital speed of the satellite. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
gravitational force (attraction) from the Earth, directed towards the centre of the Earth✓ 1
Part (b)
time in one day = 24 × 3600 = 86 400 s✓ 1
86 400 ÷ 7100 = 12.2, so 12 complete orbits✓ 1
Part (c)
M1 orbital speed = 2πr ÷ T✓ 1
M1 (2 × π × 8.0 × 106) ÷ 7100✓ 1
A1 7080 m/s (7.08 × 103 m/s)✓ 1

Exam tip: Convert hours to seconds before dividing, and read 'complete orbits' as a whole number rounded down.

Theory: How do you calculate orbital speed? Double · Physics
94IGPE3-8-15r41-098(c) Stellar Evolution Paper 1PMedium7 marks

The Sun is a star in the main-sequence stage of its life cycle.

(a)

State the name of the cloud of gas and dust from which stars such as the Sun form.

(1)
(b)

Describe how a protostar becomes a main-sequence star.

(3)
(c)

Describe what will happen to the Sun when the hydrogen in its core is used up.

(2)
(d)

Explain why the Sun will not end its life as a supernova.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
nebula✓ 1
Part (b)
gravity pulls the gas and dust together so the protostar contracts✓ 1
the core becomes hotter and its pressure increases✓ 1
when the core is hot enough, hydrogen nuclei fuse and the star becomes stable as the outward pressure balances gravity✓ 1
Part (c)
it expands and cools to become a red giant✓ 1
the outer layers drift away and the core is left as a white dwarf✓ 1
Part (d)
the Sun does not have enough mass (only stars of much greater mass explode as supernovae)✓ 1

Exam tip: Name each stage in the correct order and say what changes between stages, such as 'expands' or 'contracts'.

Theory: How does a star like the Sun evolve? Single · Double · Physics
95IGPE3-8-15r41-108(c) Stellar Evolution Paper 1PMedium6 marks

A space-strategy game sorts stars into classes by colour. Star W is blue-white, star X is red, star Y is yellow and star Z is orange.

(a)

List the four stars in order of surface temperature, starting with the hottest.

(2)
(b)

State what the colour of a star shows about the star.

(1)
(c)

Star X is a red giant that gives out much more energy each second than the Sun. Suggest why a star with a cooler surface can give out more energy each second.

(2)
(d)

State how the colour of the Sun will change as it becomes a red giant.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
W is hottest and X is coolest✓ 1
Y is hotter than Z, giving the order W, Y, Z, X✓ 1
Part (b)
its surface temperature✓ 1
Part (c)
a red giant has a very large surface area✓ 1
each square metre emits less energy but the area is so much greater that the total energy emitted each second is greater✓ 1
Part (d)
it will change from yellow to red as its surface cools✓ 1

Exam tip: Colour tells you temperature, not size or brightness; use surface area to explain differences in the total energy emitted.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
96IGPE3-8-15r41-118(d) Cosmology Paper 1PMedium6 marks

Most scientists accept the Big Bang theory as the best explanation for the origin of the Universe.

(a)

State what the Big Bang theory says about the beginning of the Universe.

(2)
(b)

Explain how the red-shift of light from distant galaxies supports the Big Bang theory.

(2)
(c)

Cosmic microwave background radiation (CMB) is detected from all directions in space. Explain why the CMB is evidence for the Big Bang theory.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the Universe began from a very small region that was extremely hot and dense✓ 1
space has been expanding ever since✓ 1
Part (b)
the wavelength of light from distant galaxies is longer than expected (shifted towards the red end) so the galaxies are moving away from us✓ 1
this shows that the Universe is expanding, so it must have been much smaller in the past✓ 1
Part (c)
the CMB is radiation emitted when the Universe was very hot, shortly after the Big Bang✓ 1
as the Universe expanded the wavelength of this radiation stretched into the microwave region✓ 1

Exam tip: Link each piece of evidence to a statement about the early Universe, not just to a description of what is observed.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
97IGPE3-8-15r42-028(d) Cosmology Paper 1PMedium1 mark

A particular spectral line has a wavelength of 656.3 nm in a laboratory. In light from the Andromeda galaxy the same line is measured at a wavelength of 655.9 nm. Which conclusion is correct?

Show mark scheme
Marking pointMarkNotes
The observed wavelength is shorter than the laboratory value, which is a blue-shift. A blue-shift means the source is moving towards the observer, so Andromeda is approaching the Earth. Moving away would give a longer wavelength (red-shift), so the second option is wrong. Stretching of space would increase the wavelength, not decrease it, so the third option is wrong. Temperature changes the colour of the whole spectrum, not the position of one identified spectral line, so the last option is wrong.✓ 1Answer A

Answer: A

Exam tip: Shorter wavelength means blue-shift and motion towards you; longer wavelength means red-shift and motion away.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
98IGPE3-8-15r42-038(c) Stellar Evolution Paper 1PMedium1 mark

A star with a mass about twenty times the mass of the Sun ends its life as a neutron star. Which row shows the stages of its life in the correct order?

Show mark scheme
Marking pointMarkNotes
A cloud of gas and dust (nebula) collapses to form a protostar, which begins fusing hydrogen and becomes a main sequence star. A star of this mass then swells into a red supergiant, explodes as a supernova and leaves a neutron star. The first row is the life cycle of a star like the Sun (red giant, planetary nebula, white dwarf). The second row puts the main sequence before the protostar. The third row puts the protostar before the nebula it forms from.✓ 1Answer D

Answer: D

Common error: Learn the two routes separately: Sun-sized stars end as white dwarfs, while much more massive stars go through a supernova.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
99IGPE3-8-15r42-078(d) Cosmology Paper 1PMedium6 marks

Light from the Andromeda galaxy is blue-shifted, but light from most other galaxies is red-shifted.

(a)

State what is meant by blue-shift and what it shows about the motion of Andromeda relative to the Earth.

(2)
(b)

State what the red-shift of light from most other galaxies shows about the way these galaxies are moving and about the Universe.

(2)
(c)

Suggest why the light from Andromeda is blue-shifted when most galaxies show red-shift.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the wavelengths of the light observed are shorter than the wavelengths emitted✓ 1
Andromeda is moving towards the Earth✓ 1
Part (b)
most galaxies are moving away from the Earth✓ 1
the Universe is expanding✓ 1
Part (c)
Andromeda is relatively close to the Milky Way (it is in the same group of galaxies)✓ 1
the gravitational attraction between the two galaxies is large enough to pull them together, outweighing the effect of the expansion✓ 1

Exam tip: Link each shift to a direction: red-shift means moving away, blue-shift means moving towards.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
100IGPE3-8-15r42-088(c) Stellar Evolution Paper 1PMedium5 marks

A star has a mass about fifteen times the mass of the Sun. It is in the main sequence stage of its life.

(a)

Explain why the star stays the same size for millions of years during the main sequence.

(2)
(b)

State the name of the stage that follows the main sequence for a star of this mass.

(1)
(c)

Describe what happens to the star in a supernova and name one object that may be left behind.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
gravity pulls the material of the star inwards✓ 1
this is balanced by the outward pressure produced by the energy released from fusion in the core, so the forces are balanced✓ 1
Part (b)
red supergiant✓ 1
Part (c)
the star explodes, throwing its outer layers into space✓ 1
a neutron star or a black hole is left behind✓ 1

Common error: A main sequence star is stable because the forces balance, not because fusion has stopped.

Theory: How does a star like the Sun evolve? Single · Double · Physics
101IGPE3-8-15r42-108(c) Stellar Evolution Paper 1PMedium6 marks

A railway company lays out a scale model of the Solar System along a straight track. In the model, 1 m of track represents 1.0 × 1010 m in space. The Sun is at one end of the track.
Distances from the Sun: Earth 1.5 × 1011 m; Neptune 4.5 × 1012 m; Proxima Centauri (the nearest star other than the Sun) 4.0 × 1016 m.

(a)

Calculate the distance along the track at which Earth should be placed.

(2)
(b)

Calculate the distance along the track at which Neptune should be placed.

(2)
(c)

Calculate the distance along the track at which Proxima Centauri should be placed, and give your answer in metres.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 1.5 × 1011 ÷ 1.0 × 1010✓ 1
A1 15 m✓ 1
Part (b)
M1 4.5 × 1012 ÷ 1.0 × 1010✓ 1
A1 450 m✓ 1
Part (c)
M1 4.0 × 1016 ÷ 1.0 × 1010✓ 1
A1 4.0 × 106 m (accept 4 000 000 m or 4000 km)✓ 1

Exam tip: Write powers of ten carefully: dividing 10¹⁶ by 10¹⁰ gives 10⁶.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
102IGPE3-8-15r42-118(c) Stellar Evolution Paper 1PMedium5 marks

The Sun is a star in the main sequence stage of its life. It fuses hydrogen into helium in its core.

(a)

The list shows four stages in the life of the Sun, in the wrong order: white dwarf, main sequence, red giant, protostar.
Write the stages in the correct order, starting with the earliest.

(2)
(b)

State the process that releases energy in the core of the Sun.

(1)
(c)

Explain why the Sun will not end its life as a supernova.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
protostar then main sequence✓ 1
red giant then white dwarf✓ 1
Part (b)
nuclear fusion✓ 1
Part (c)
the Sun does not have enough mass✓ 1
so the gravitational force is not great enough for the core to collapse violently / its core does not reach the temperatures needed to fuse heavy elements✓ 1

Common error: Do not say that the Sun 'burns' fuel: energy comes from nuclear fusion, not from chemical burning.

Theory: How does a star like the Sun evolve? Single · Double · Physics
103IGPE3-8-17r46-028(c) Stellar Evolution Paper 1PMedium1 mark

Albireo is a double star. Through a small telescope one star looks orange and the other looks blue-white. Which statement about the two stars is correct?

Show mark scheme
Marking pointMarkNotes
A hotter surface emits its radiation with a peak at shorter wavelengths, so the blue-white star is the hotter one. The statement that orange light has more energy is false, since blue light has the higher frequency and energy. Being close together in the sky says nothing about temperature. Colour is set by surface temperature, not by travel time of the light.✓ 1Answer A

Answer: A

Common error: Do not link colour to distance. A star's colour tells you its surface temperature, with blue the hottest and red the coolest.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
104IGPE3-8-17r46-068(b) Motion in the Universe Paper 1PMedium7 marks

On the Moon an astronaut lifts an equipment case of mass 18 kg through a vertical height of 2.5 m onto a platform. The gravitational field strength on the Moon is 1.6 N/kg. On Earth the gravitational field strength is 10 N/kg.

(a)

Calculate the weight of the case on the Moon.

(2)
(b)

Calculate the gain in gravitational potential energy of the case when it is lifted onto the platform.

(2)
(c)

The same amount of energy is used to lift the case on Earth. Calculate the height through which the case would be raised.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = mg = 18 × 1.6✓ 1
A1 28.8 N✓ 1
Part (b)
M1 energy = weight × height = 28.8 × 2.5✓ 1
A1 72 J (ECF from (a))✓ 1
Part (c)
M1 weight on Earth = 18 × 10 = 180 N✓ 1
M1 height = 72 ÷ 180✓ 1
A1 0.40 m (ECF from (b))✓ 1

Common error: Keep the mass at 18 kg for both places. Only the gravitational field strength changes, so the weight and the energy change.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
105IGPE3-8-17r46-098(c) Stellar Evolution Paper 1PMedium5 marks

Astronomers use surface temperature to place a star in a colour class.
Red: below 3500 K
Orange: 3500 K to 5000 K
Yellow: 5000 K to 6500 K
White: 6500 K to 10000 K
Blue-white: above 10000 K
Star X has a surface temperature of 4300 K. Star Y has a surface temperature of 12900 K.

(a)

Give the colour class of (i) star X and (ii) star Y.

(2)
(b)

Calculate how many times greater the temperature of star Y is than the temperature of star X.

(2)
(c)

Explain why the temperatures must be in kelvin when they are used to find the ratio in (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
(i) orange✓ 1
(ii) blue-white✓ 1
Part (b)
M1 12900 ÷ 4300✓ 1
A1 3.0✓ 1
Part (c)
the kelvin scale starts at absolute zero, so a ratio of kelvin temperatures is meaningful, whereas the zero of the Celsius scale is arbitrary✓ 1

Common error: Do not divide temperatures in °C. A ratio only makes sense when both temperatures are on the kelvin scale.

Theory: How are stars classified by colour, and what does the colour tell you? Single · Double · Physics
106IGPE3-8-17r46-118(c) Stellar Evolution Paper 1PMedium6 marks

The life cycle of a star has these stages, listed here in a random order: white dwarf, red giant, protostar, main sequence star, nebula.

(a)

Write the stages in the order in which the Sun passes through them.

(2)
(b)

State the process that releases energy in a main sequence star.

(2)
(c)

Explain why the Sun will not end its life as a neutron star or a black hole.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
nebula, then protostar, then main sequence star✓ 1
red giant, then white dwarf✓ 1
Part (b)
nuclear fusion✓ 1
hydrogen nuclei join to form helium✓ 1
Part (c)
the Sun does not have enough mass✓ 1
so its core will not collapse any further than a white dwarf (and there will be no supernova)✓ 1

Common error: The order of the stages depends on the mass of the star. State which path the Sun follows and why.

Theory: How does a star like the Sun evolve? Single · Double · Physics
107IGPE3-8-17r46-128(c) Stellar Evolution Paper 1PMedium9 marks

The Sun will spend about 10 billion years (1.0 × 1010 years) as a main sequence star and has so far spent 4.6 billion years in this stage. A star with 20 times the mass of the Sun spends about 10 million years (1.0 × 107 years) as a main sequence star.

(a)

Calculate how many more billion years the Sun will remain a main sequence star.

(2)
(b)

Calculate the percentage of its main sequence lifetime that the Sun has already used.

(2)
(c)

Explain why the more massive star has a much shorter main sequence lifetime even though it contains more hydrogen.

(3)
(d)

Calculate how many times longer the main sequence lifetime of the Sun is than that of the star with 20 times the Sun's mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 10 − 4.6✓ 1
A1 5.4 billion years✓ 1
Part (b)
M1 4.6 ÷ 10 × 100✓ 1
A1 46%✓ 1
Part (c)
greater mass gives greater gravitational compression of the core, so a higher core temperature and pressure✓ 1
the rate of fusion is much higher✓ 1
so the hydrogen is used up faster than in the Sun✓ 1
Part (d)
M1 (1.0 × 1010) ÷ (1.0 × 107)✓ 1
A1 1000 times✓ 1

Exam tip: In 'explain' answers about stellar lifetimes, link mass to core temperature, then to rate of fusion, then to the time the fuel lasts.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
108IGPE3-8-16r48-038(d) Cosmology Paper 1PMedium1 mark

A spectral line has a wavelength of 600 nm in a laboratory. In light from a distant galaxy the same line has a wavelength of 630 nm. The speed of light is 3.0 × 108 m/s. At what speed is the galaxy moving away from Earth?

Show mark scheme
Marking pointMarkNotes
The red-shift is z = (630 − 600) ÷ 600 = 0.050, so the speed is 0.050 × 3.0 × 108 = 1.5 × 107 m/s. Dividing by the observed wavelength gives 1.4 × 107 m/s. A power of ten error gives 1.5 × 106 m/s. Dividing the two wavelengths and multiplying by c gives 3.2 × 108 m/s, which uses 630 ÷ 600 instead of the fractional change.✓ 1Answer C

Answer: C

Common error: The red-shift uses the change in wavelength divided by the original (laboratory) wavelength, not the ratio of the two wavelengths.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
109IGPE3-8-16r48-058(b) Motion in the Universe Paper 1PMedium5 marks

Io is a moon of Jupiter with many active volcanoes. A volcanologist plans to monitor Io's eruptions.

(a)

State the object that Io orbits.

(1)
(b)

Io takes 1.77 days to complete one orbit. Calculate this time in seconds.

(2)
(c)

The volcanologist observes Io for 53.1 days. Calculate the number of orbits Io completes in this time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Jupiter✓ 1
Part (b)
M1 1.77 × 24 × 3600✓ 1
A1 1.53 × 105 s (152 928 s)✓ 1
Part (c)
M1 53.1 ÷ 1.77✓ 1
A1 30 (orbits)✓ 1

Exam tip: Keep the same unit for both times when dividing; days divided by days gives a pure number of orbits.

Theory: How do you calculate orbital speed? Double · Physics
110IGPE3-8-16r48-088(b) Motion in the Universe Paper 1PMedium5 marks

A navigation satellite used by ships orbits at a radius of 2.66 × 107 m from the centre of the object it orbits. It completes one orbit in 4.32 × 104 s.

(a)

State the object that the satellite orbits.

(1)
(b)

Calculate the orbital speed of the satellite. Give your answer to 3 significant figures.

(3)
(c)

A geostationary satellite used for ship television has an orbital period of 24 hours. State why it stays above the same point on the Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Earth✓ 1
Part (b)
M1 v = 2πr/T✓ 1
M1 2π × 2.66 × 107 ÷ 4.32 × 104✓ 1
A1 3870 m/s (3.87 × 103 m/s)✓ 1
Part (c)
Its period is the same as the Earth's rotation period (and it orbits in the same direction)✓ 1

Exam tip: In v = 2πr/T the distance is the radius of the orbit measured from the centre, not the height above the surface.

Theory: How do you calculate orbital speed? Double · Physics
111IGPE3-8-16r48-128(d) Cosmology Paper 1PMedium6 marks

A stage show about the early universe tells the audience about the evidence for the Big Bang.

(a)

Explain how the red-shift of light from distant galaxies is evidence that the universe is expanding.

(2)
(b)

Describe cosmic microwave background radiation and explain how it supports the Big Bang theory.

(2)
(c)

Describe the Big Bang theory.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Light from almost all distant galaxies is shifted to longer wavelengths (red-shifted)✓ 1
This shows the galaxies are moving away from us, and the more distant ones are moving away faster✓ 1
Part (b)
Microwave radiation detected from all directions in space✓ 1
It is radiation from the early hot universe whose wavelength has been stretched by the expansion of the universe✓ 1
Part (c)
The universe began from a very small, hot, dense region✓ 1
and has been expanding ever since✓ 1

Common error: Do not say the galaxies are moving through space away from a centre; the evidence is that almost all distant galaxies show red-shift.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
112IGCSE-PHY-8-0128(b) Motion in the Universe Papers 1 & 2Grade 8–98 marks
The table gives data about the orbits of four planets around the Sun.
Planet / name: Mercury, Earth, Mars, Jupiter
Orbital radius / 10⁹ m: 57.9, 150, 228, 778
Time period / days: 88.0, 365, 687, 4330
A student makes this claim:
'Planets further from the Sun must travel faster, because they have a longer distance to travel in each orbit.'
(a)
Calculate the orbital speed of Mercury.
Give your answer in km/s.
(3)
(b)
Evaluate the student's claim.
Use calculations based on the data in the table to support your answer.
(3)
(c)
Suggest why the orbital speed of a planet depends on its distance from the Sun.
(2)
Show mark scheme
(a)
  • conversion of time to seconds;
  • substitution;
  • evaluation in km/s;
  • e.g. T = 88.0 × 24 × 3600 = 7 603 200 (s)
  • v = 2 × π × 57.9 × 10⁹ ÷ 7 603 200
  • (v =) 47.8 (km/s)
Allow
  • 47.85, 48 (km/s)
  • 47 800 m/s scores 2
  • ECF from incorrect time conversion
Do not accept
  • v = 2πrT
(b)
  • any three from:
  • orbital speed of at least one other planet correctly calculated (Earth 29.9 / Mars 24.1 / Jupiter 13.1 km/s);
  • statement that further planets have lower orbital speeds;
  • idea that time period increases by a greater factor than orbital radius / circumference (e.g. Jupiter radius ×13.4 but period ×49);
  • conclusion that the claim is incorrect;
Allow
  • ECF from (a)
  • speeds in m/s
Do not accept
  • unsupported conclusion with no calculation or comparison
(c)
  • gravitational force (of Sun) is weaker further from the Sun;
  • (so) a lower speed is needed to stay in orbit / planet is accelerated less;
Allow
  • gravity pulls less strongly
Do not accept
  • there is less air resistance further out
  • outer planets are bigger so slower
Theory: How do you calculate orbital speed? Double · Physics
113IGCSE-PHY-8-0138(c) Stellar Evolution Papers 1 & 2Grade 8–96 marks
A star forms with a mass about 20 times the mass of the Sun.
(a)
Describe and explain the life cycle of this star, from its formation to its final stage.
You should refer to the roles of gravity and nuclear fusion in your answer.
(6)
Show mark scheme
(a)
  • Indicative content:
  • MP1. star forms from a nebula / cloud of gas and dust;
  • MP2. gravity pulls the gas and dust together / nebula collapses and heats up;
  • MP3. (hydrogen) fusion starts in the core → main-sequence star;
  • MP4. star is stable because (outward) force/pressure from fusion energy balances (inward) gravity;
  • MP5. massive star uses up its hydrogen faster / spends less time on main sequence (than the Sun);
  • MP6. when hydrogen in the core runs out, the star expands to become a red supergiant;
  • MP7. fusion of heavier elements in the core;
  • MP8. fusion stops and gravity causes the core to collapse;
  • MP9. supernova explosion;
  • MP10. leaves a neutron star;
  • MP11. or a black hole if the remaining core is massive enough (gravity so strong light cannot escape);
  • MP12. material thrown out forms new nebulae;
Allow
  • award 1 mark per MP to a maximum of 6, but use levels to judge overall quality
  • red giant for red supergiant
  • max 4 marks if no reference to gravity or fusion
Do not accept
  • white dwarf as final stage of this star
  • fission for fusion
  • star 'burns' fuel (chemical combustion)
Level 0 (0)No rewardable content.
Level 1 (1–2)Names some stages, possibly out of order, with little or no explanation; may include incorrect final stage.
Level 2 (3–4)Stages mostly in correct order including supernova; some explanation of the role of gravity or fusion at one or two stages.
Level 3 (5–6)Complete correct sequence nebula → main sequence → red supergiant → supernova → neutron star / black hole, with linked explanations of the role of gravity and fusion at several stages.
Theory: How does a star like the Sun evolve? Single · Double · Physics
114IGCSE-PHY-8-0318(b) Motion in the Universe Papers 1 & 2Grade 8–96 marks
Satellites in the Global Positioning System (GPS) orbit the Earth in circular orbits.
A GPS satellite orbits at a height of 20 200 km above the Earth's surface.
The radius of the Earth is 6370 km.
The orbital speed of the satellite is 3870 m/s.
EarthGPS satellite6370 km20 200 kmnot to scale
(a)
Determine the number of times the satellite orbits the Earth in one day.
(4)
(b)
A student says:
'GPS satellites always stay above the same point on the Earth's surface, because the Earth spins at the same rate as they orbit.'
Use your answer to (a) to evaluate the student's statement.
(2)
Show mark scheme
(a)
  • orbital radius = 26 570 km / 2.657 × 10⁷ m;
  • rearrangement and substitution into T = 2πr ÷ v;
  • T = 43 100 s / 12.0 hours;
  • (number of orbits =) 2.0;
  • e.g. T = 2 × π × 2.657 × 10⁷ ÷ 3870 = 43 138 s
  • 86 400 ÷ 43 138 = 2.00
Allow
  • 43 000–43 200 (s), 2 / 2.00 / 'twice'
  • ECF for MP3 and MP4 from an incorrect radius: r = 20 200 km gives T = 32 800 s and 2.6 orbits (scores 3)
  • alternative route: distance travelled in a day = 3870 × 86 400 = 3.34 × 10⁸ m, divided by 2πr (1.67 × 10⁸ m) = 2.0, for MP2–MP4
Do not accept
  • answer given as a time with no number of orbits for MP4
  • T = 2πv ÷ r
(b)
  • Earth rotates once (per day) while satellite completes two orbits / satellite has a shorter period than the Earth's rotation (24 h);
  • so it does not stay above the same point / the student is wrong;
Allow
  • ECF from (a) if number of orbits ≠ 1
Do not accept
  • student is right
  • conclusion with no comparison of periods for MP1
Theory: How do you calculate orbital speed? Double · Physics
115IGCSE-PHY-8-0328(b) Motion in the Universe Papers 1 & 2Grade 8–98 marks
Astronomers measure the brightness of a distant star over 20 days.
A planet orbits this star in a circular orbit.
Each time the planet passes between the star and the Earth, the brightness of the star decreases slightly.
The graph shows the results.
0246810121416182098.599.099.5100.0100.5time / daysbrightness / % of normal
(a)
State the force that causes the planet to orbit the star.
(1)
(b)
Use the graph to determine the time period of the orbit of the planet.
Explain how you used the graph to make your value as accurate as possible.
(3)
(c)
The orbital speed of the planet is 95 km/s.
The orbital radius of the planet Mercury around the Sun is 5.8 × 10¹⁰ m.
A student says that the planet is much closer to its star than Mercury is to the Sun.
Evaluate the student's statement.
Use a calculation in your answer.
(4)
Show mark scheme
(a)
  • gravity / gravitational (force of the star);
Do not accept
  • magnetic force
  • centrifugal force
(b)
  • time between consecutive dips (centres) read from the graph;
  • (T =) 5.5 days;
  • measured over several dips (first to last) and divided by number of orbits (3);
Allow
  • 5.4–5.6 (days)
  • MP1 for e.g. 7.5 − 2.0 = 5.5
Do not accept
  • 11 days (time for two dips counted as one orbit)
  • width of one dip
(c)
  • conversion of T to seconds (475 200 s);
  • substitution into r = vT ÷ 2π (with v in m/s);
  • evaluation of r;
  • comparison with Mercury and conclusion (student correct);
  • e.g. r = 95 000 × 475 200 ÷ (2 × π)
  • (r =) 7.2 × 10⁹ (m)
  • 7.2 × 10⁹ m is (about 8 times) less than 5.8 × 10¹⁰ m so student is correct
Allow
  • ECF from (b)
  • 7.1–7.3 × 10⁹ (m) from T = 5.4–5.6 days
  • r in km: 7.2 × 10⁶ km compared with 5.8 × 10⁷ km
  • MP4: consistent conclusion from an incorrect r
Do not accept
  • 7.2 × 10⁶ m (km/s not converted) for MP3
Theory: How do you calculate orbital speed? Double · Physics
116IGCSE-PHY-8-0338(b) Motion in the Universe Papers 1 & 2Grade 8–96 marks
The table gives information about four objects that move in orbits.
Object / name: Titan, Neptune, Comet Hale–Bopp, Hubble Space Telescope
Type of object / type: moon, planet, comet, artificial satellite
Shape of orbit / description: almost circular, almost circular, very elliptical, almost circular
Time period / approximate: 16 days, 165 years, 2500 years, 95 minutes
(a)
Use the information in the table and your own knowledge to compare the orbits of these four objects.
In your answer you should explain the role of gravity.
(6)
Show mark scheme
(a)
  • Indicative content:
  • MP1. gravitational force keeps each object in orbit / provides the force towards the body it orbits;
  • MP2. moon (Titan) orbits a planet / Saturn;
  • MP3. planet (Neptune) and comet both orbit the Sun;
  • MP4. artificial satellite (Hubble) orbits the Earth;
  • MP5. moon, planet and satellite have (almost) circular orbits so distance (and speed) stays (nearly) constant;
  • MP6. comet's orbit is very elliptical so its distance from the Sun changes a lot;
  • MP7. comet moves fastest when closest to the Sun because the gravitational force is strongest there;
  • MP8. comet moves slowly far from the Sun / spends most of its time far from the Sun;
  • MP9. for objects orbiting the Sun, a larger orbit means a longer time period (Neptune 165 years, the comet, which travels much further from the Sun, 2500 years);
  • MP10. comet's orbit may cross the orbits of planets / comet is not in the same plane (tilted orbit);
Allow
  • award 1 mark per MP to a maximum of 6, but use levels to judge overall quality
  • Saturn need not be named for MP2
  • max 4 marks if there is no reference to gravity
Do not accept
  • centrifugal force balances gravity
  • comet orbits the Earth
  • speed of comet constant
Level 0 (0)No rewardable content.
Level 1 (1–2)Isolated statements about one or two objects, e.g. what they orbit or orbit shape, with little or no comparison or reference to gravity.
Level 2 (3–4)Some comparison of what the objects orbit and the shapes of their orbits; some reference to gravity or to changes in speed or period, but links are incomplete.
Level 3 (5–6)Coherent comparison covering what each object orbits, orbit shapes and time periods, with linked explanation of the role of gravity including the changing speed of the comet.
Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
117IGPE2-061-01-118(b) Motion in the Universe Paper 1PHard6 marks

A space telescope has found a planet orbiting a white dwarf.
The planet's orbital radius is 3.0 ×109 m and its orbital period is 1.4 days.

planetr = 3.0 × 10⁹ mv = ?white dwarf
Not to scale
(a)

State the equation linking orbital speed, orbital radius and time period.

(1)
(b)

Calculate the orbital speed of the planet in m/s.

(3)
(c)

Before it became a white dwarf, the star was a red giant with a radius of about 1.5 ×1011 m.
Suggest why astronomers think the planet was not at its present orbital radius during the red giant stage.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
orbital speed = (2 × π × orbital radius) ÷ time period✓ 1
Part (b)
T = 1.4 × 24 × 3600 = 120 960 s✓ 1
v = (2 × π × 3.0 ×109) ÷ 120 960✓ 1
= 1.6 ×105 m/s (accept 1.56 ×105 m/s)✓ 1
Part (c)
the present orbital radius (3.0 ×109 m) is much smaller than the red giant's radius, so the planet would have been inside the star / engulfed (and likely destroyed)✓ 1
so the planet must have orbited further out and moved inwards later, OR it formed after the red giant stage (e.g. from leftover material), OR it was captured later✓ 1

Common error: The period must be in seconds to give a speed in m/s. Using 1.4 days directly gives an answer that is wrong by a factor of 86 400.

Theory: How do you calculate orbital speed? Double · Physics
118IGPE2-061-00-138(b) Motion in the Universe Paper 1PHard7 marks

A space probe sends back data on two moons of Jupiter. Both moons have almost circular orbits.
Moon 1: orbital radius 4.22 ×108 m, time period 1.77 days.
Moon 2: orbital radius 1.88 ×109 m, time period 16.7 days.

moon 1, 4.22 × 10⁸ mmoon 2, 1.88 × 10⁹ mJupiter
Not to scale
(a)

Calculate the orbital speed of moon 1.

(3)
(b)

Calculate the orbital speed of moon 2.

(2)
(c)

Explain why moon 1 has a higher orbital speed than moon 2.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 1.77 × 24 × 3600 = 152 900 s✓ 1
v = (2 × π × 4.22 ×108) ÷ 152 900✓ 1
v = 1.73 ×104 m/s✓ 1
Part (b)
T = 16.7 × 86 400 = 1.443 ×106 s✓ 1
v = (2 × π × 1.88 ×109) ÷ 1.443 ×106 = 8.19 ×103 m/s✓ 1
Part (c)
Moon 1 is closer to Jupiter, so the gravitational force per kg on it is stronger✓ 1
It must travel faster to stay in orbit rather than fall towards Jupiter✓ 1

Exam tip: Convert days to seconds in one step by multiplying by 86 400. Keep extra figures until the final answer, then round to 3 significant figures.

Theory: How do you calculate orbital speed? Double · Physics
119IGPE-057-00-098(c) Stellar Evolution Paper 1PHard6 marks

Astronomers are observing a main sequence star with about 25 times the mass of the Sun.

(a)

Describe the stages in the life of this star after the hydrogen in its core runs out.

(4)
(b)

State the two possible final states of this star and explain what decides which one forms.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
• The core contracts and the outer layers expand and cool, so the star becomes a red supergiant✓ 1
• Heavier elements, up to iron, are made by fusion in the core✓ 1
• When fusion in the core stops, the core collapses and the star explodes as a supernova✓ 1
• The outer layers are blown into space as a nebula, leaving behind a very dense core (remnant)✓ 1
Part (b)
• Neutron star or black hole (both needed)✓ 1
• Which one forms depends on the mass of the remaining core: a more massive core (more than about 3 solar masses) collapses further to form a black hole, and a less massive core forms a neutron star✓ 1

Exam tip: Give the stages in the correct order and link each change to its cause, such as fusion stopping and gravity making the core collapse. A list of stage names alone will not earn full marks.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
120IGPE-057-00-138(b) Motion in the Universe Paper 1PHard7 marks

Miranda and Oberon are two moons of the planet Uranus. Both have approximately circular orbits.
Miranda: orbital radius 129 900 km, time period 1.413 days
Oberon: orbital radius 583 500 km, time period 13.46 days

Miranda, 129 900 kmOberon, 583 500 kmUranus
Not to scale
(a)

Calculate the orbital speed of Oberon in m/s.

(3)
(b)

A student says: 'Oberon travels further in each orbit, so it must move faster than Miranda.'
Evaluate this statement. Your answer should include a calculation of the orbital speed of Miranda.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
Converts units: T = 13.46 × 24 × 3600 = 1.163 × 106 s and r = 5.835 × 108 m✓ 1
v = (2 × π × 5.835 × 108) ÷ 1.163 × 106✓ 1
v = 3150 m/s✓ 1
Part (b)
Converts units: T = 1.413 × 24 × 3600 = 1.221 × 105 s✓ 1
v = (2 × π × 1.299 × 108) ÷ 1.221 × 105 = 6690 m/s✓ 1
Miranda is faster (about twice as fast as Oberon), so the statement is wrong✓ 1
The gravitational force from Uranus is weaker further away, so a moon further out moves more slowly (and its longer orbit takes much more time)✓ 1

Exam tip: In 'evaluate' questions, finish with a clear judgement and back it with both your calculated values and a physics reason.

Theory: How do you calculate orbital speed? Double · Physics
121IGPE3-8-14r14-118(c) Stellar Evolution Paper 1PHard6 marks

Describe the life cycle of a star with a mass about 20 times that of the Sun, from the cloud of gas and dust to its final stage. Explain how the end of its life differs from that of the Sun.

(a)

Answer in full sentences, in order.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 gravity pulls a cloud of gas and dust (nebula) together to form a protostar✓ 1
B1 as the protostar contracts it heats up until nuclear fusion of hydrogen begins✓ 1
B1 a stable main-sequence star: the outward pressure from fusion balances the inward gravitational force✓ 1
B1 when the hydrogen is used up it expands to become a red supergiant✓ 1
B1 it then collapses and explodes as a supernova, leaving a neutron star or a black hole✓ 1
B1 the Sun becomes a red giant, loses its outer layers and ends as a white dwarf, without a supernova✓ 1

Exam tip: In a 6-mark answer, keep the stages in order and link each change to a cause, such as gravity or fusion running out.

Theory: How does a star like the Sun evolve? Single · Double · Physics
122IGPE3-8-14r14-148(b) Motion in the Universe Paper 1PHard8 marks

The star Alpha is 4.2 light-years from the Earth. The speed of light is 3.00 × 108 m/s and 1 year = 3.15 × 107 s.

(a)

State what is meant by a light-year.

(1)
(b)

Calculate the distance to Alpha in metres. Give your answer to 3 significant figures.

(3)
(c)

Suggest why astronomers use light-years rather than kilometres for distances to stars.

(1)
(d)

A probe travels to Alpha at a constant speed of 2.0 × 104 m/s. Calculate the time taken in years. Give your answer to 3 significant figures. Use the unrounded distance.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the distance travelled by light in one year✓ 1
Part (b)
M1 distance light travels in 1 year = 3.00 × 108 × 3.15 × 107 = 9.45 × 1015 m✓ 1
M1 4.2 × 9.45 × 1015✓ 1
A1 3.97 × 1016 m✓ 1
Part (c)
B1 distances in km would be extremely large numbers, so light-years are more convenient✓ 1
Part (d)
M1 time = distance ÷ speed = 3.969 × 1016 ÷ (2.0 × 104)✓ 1
A1 1.98 × 1012 s✓ 1
A1 6.30 × 104 years (63000 years)✓ 1

Exam tip: Check that the units match before dividing: metres with m/s gives seconds, then convert to years.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
123IGPE3-8-14r17-138(b) Motion in the Universe Paper 1PHard9 marks

A diver of mass 55 kg stands on a 3.0 m high platform above a pool on the Moon, where the gravitational field strength is 1.6 N/kg. She steps off with no initial speed. Ignore air resistance.

diver, mass 55 kgpool3.0 m
Diagram not to scale
(a)

Calculate the weight of the diver on the Moon.

(2)
(b)

Calculate the speed at which she reaches the pool surface. Use v2 = u2 + 2as and give your answer to 3 significant figures.

(3)
(c)

Calculate the kinetic energy of the diver as she reaches the pool surface. Use the unrounded value of v2 from (b).

(2)
(d)

A diver of the same mass steps off the same platform on Earth. Explain why her speed reaching the water is greater than on the Moon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = m × g = 55 × 1.6✓ 1
A1 88 N✓ 1
Part (b)
M1 u = 0, a = 1.6 m/s2, s = 3.0 m substituted✓ 1
M1 v2 = 2 × 1.6 × 3.0 = 9.6✓ 1
A1 3.10 m/s✓ 1
Part (c)
M1 KE = ½ × 55 × 9.6✓ 1
A1 264 J (accept 264 to 265 J)✓ 1
Part (d)
B1 the gravitational field strength on Earth is greater so her acceleration is greater✓ 1
B1 more gravitational potential energy is transferred to kinetic energy over the same height✓ 1

Common error: Use the unrounded value of v² in later calculations to avoid carrying a rounding error into your answer.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
124IGPE3-8-14r17-148(c) Stellar Evolution Paper 1PHard7 marks

Star X has a mass of 0.8 times the mass of the Sun. Star Y has a mass of 25 times the mass of the Sun.

(a)

State the type of stellar remnant that Star X will become and one possible remnant for Star Y.

(2)
(b)

Star Y has much more hydrogen than Star X but spends less time on the main sequence. Explain why.

(3)
(c)

The early universe contained almost only hydrogen and helium. Explain why Earth contains elements such as iron.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 X becomes a white dwarf✓ 1
B1 Y becomes a neutron star or a black hole✓ 1
Part (b)
B1 greater mass means greater gravitational force squeezing the core, giving higher core temperature and pressure✓ 1
B1 hydrogen fusion therefore happens much faster✓ 1
B1 so the hydrogen is used up in a shorter time✓ 1
Part (c)
B1 heavier elements were made by fusion in earlier stars and in supernova explosions✓ 1
B1 these were thrown out into space and later formed part of the cloud from which the Sun and planets formed✓ 1

Exam tip: In "explain why" questions, link cause to effect: more mass leads to more pressure, which leads to faster fusion, which leads to a shorter life.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
125IGPE3-8-15r19-078(b) Motion in the Universe Paper 1PHard8 marks

A comet takes 200 years to complete one orbit of the Sun. The length of its orbital path is 2.52 × 1013 m. Use 1 year = 3.15 × 107 s.

Suncometlength of one orbit = 2.52 × 10¹³ m
Not to scale
(a)

State the shape of the path of a comet around the Sun.

(1)
(b)

Calculate the average speed of the comet in its orbit.

(3)
(c)

The speed of the comet is 4.4 × 104 m/s at its closest approach to the Sun and much lower when it is far away. Explain, in terms of energy, why its speed changes in this way.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
elliptical (accept a very elongated oval)✓ 1
Part (b)
convert the period to seconds: 200 × 3.15 × 107 = 6.3 × 109 s✓ 1
use of speed = distance ÷ time✓ 1
4.0 × 103 m/s✓ 1
Part (c)
as the comet moves towards the Sun it loses gravitational potential energy✓ 1
this energy is transferred to kinetic energy, so the speed increases✓ 1
the speed is greatest at the closest approach where the potential energy is least✓ 1
as it moves away it gains potential energy and loses kinetic energy, so it is slowest when furthest from the Sun✓ 1

Exam tip: In an energy explanation name both stores and say which one increases and which decreases.

Theory: How are the orbits of comets, moons and planets different? Single · Double · Physics
126IGPE3-8-15r19-138(b) Motion in the Universe Paper 1PHard6 marks

The International Space Station (ISS) orbits the Earth in a circular path with an average speed of 7.66 × 103 m/s. One orbit takes 92.0 minutes. The radius of the Earth is 6.37 × 106 m.

ISSr = ?7.66 × 10³ m/sh = ?Earth
Not to scale (radius of the Earth = 6.37 × 10⁶ m)
(a)

Show that the time for one orbit is 5520 s.

(1)
(b)

Calculate the radius of the orbit of the ISS, measured from the centre of the Earth. Give your answer to 3 significant figures.

(3)
(c)

Calculate the height of the ISS above the surface of the Earth. Use the unrounded radius from (b) and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
92.0 × 60 = 5520 s✓ 1
Part (b)
distance in one orbit = speed × time = 7660 × 5520 = 4.23 × 107 m (accept 4.228 × 107 m)✓ 1
use of 2πr = distance in one orbit✓ 1
6.73 × 106 m✓ 1
Part (c)
height = orbit radius − radius of the Earth✓ 1
3.60 × 105 m (ECF from (b))✓ 1

Common error: The radius of an orbit is measured from the centre of the Earth, so subtract the Earth's radius to find the height above the surface.

Theory: How do you calculate orbital speed? Double · Physics
127IGPE3-8-17r27-048(c) Stellar Evolution Paper 1PHard1 mark

The table shows data for two stars.

Star X: surface temperature 5800 K, absolute magnitude +4.8
Star Y: surface temperature 16 000 K, absolute magnitude +11.8

A smaller absolute magnitude means a more luminous star. Star Y is much hotter than star X but much less luminous. Which statement best explains this?

Show mark scheme
Marking pointMarkNotes
Luminosity depends on both surface temperature and surface area. Star Y is hotter, so for the same area it would emit more, so it must have a far smaller area: it is a white dwarf. A is wrong because absolute magnitude is defined for a standard distance, so distance has no effect on it. B is wrong because a larger area would increase the total emitted energy. D is wrong because red-shift concerns the motion of the star, not its absolute magnitude.✓ 1Answer C

Answer: C

Exam tip: Absolute magnitude compares stars as if they were at the same distance, so differences in it come from the stars themselves.

Theory: What is absolute magnitude? Physics only · 2P
128IGPE3-8-17r27-068(b) Motion in the Universe Paper 1PHard9 marks

Astronomers measured the recession speed v and the distance d of four galaxies.

Galaxyd / Mpcv / km/s
A201400
B453150
C805600
D1208400

(a)

Show that the data support the idea that v is directly proportional to d.

(2)
(b)

State the value of the Hubble constant H0 given by the data, in km/s per Mpc.

(1)
(c)

Predict the recession speed of a galaxy at a distance of 150 Mpc.

(2)
(d)

1 Mpc = 3.09×1019 km. Convert H0 to units of s−1. Give your answer to 3 significant figures.

(2)
(e)

Use your unrounded value of H0 from (d) to estimate the age of the universe in years, using age = 1/H0. 1 year = 3.15×107 s. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
calculates v ÷ d for at least two galaxies✓ 1
the ratio is constant, 70 km/s per Mpc, for all four galaxies so v is proportional to d✓ 1
Part (b)
H0 = 70 km/s per Mpc✓ 1
Part (c)
v = H0 × d = 70 × 150 (ECF from (b))✓ 1
= 10 500 km/s (1.05×104 km/s)✓ 1
Part (d)
H0 = 70 ÷ 3.09×1019✓ 1
= 2.27×10−18 s−1✓ 1
Part (e)
age = 1/H0 = 4.41×1017 s (ECF from (d))✓ 1
= 1.40×1010 years✓ 1

Common error: H₀ in km/s per Mpc must have the distance unit cancelled before the value is used as 1/H₀ to give a time.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
129IGPE3-8-15r41-148(b) Motion in the Universe Paper 1PHard7 marks

A lander is searching Mars for ice, a possible source of water for future visitors. The lander has a mass of 1200 kg. The gravitational field strength on Mars is 3.7 N/kg. During its descent the thrusters produce an upward force of 6000 N.

thrust 6000 Nweight = ?lander
Forces on the lander during the descent (not to scale)
(a)

Calculate the weight of the lander on Mars.

(2)
(b)

Calculate the resultant force on the lander during the descent.

(2)
(c)

State whether the lander speeds up or slows down. Give a reason.

(1)
(d)

Calculate the acceleration of the lander. Use the resultant force from (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × gravitational field strength✓ 1
A1 1200 × 3.7 = 4440 N✓ 1
Part (b)
M1 resultant force = 6000 − 4440✓ 1
A1 1560 N upwards (ECF from (a))✓ 1
Part (c)
slows down because the resultant force acts upwards, opposite to its downward motion✓ 1
Part (d)
M1 acceleration = resultant force ÷ mass✓ 1
A1 1560 ÷ 1200 = 1.3 m/s2✓ 1

Exam tip: Subtract the forces acting in opposite directions to find the resultant force before using force = mass × acceleration.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
130IGPE3-8-15r42-138(d) Cosmology Paper 1PHard7 marks

The Big Bang theory describes the origin of the Universe.

(a)

Explain how observations of red-shift and of the cosmic microwave background radiation (CMBR) support the Big Bang theory.

(6)
(b)

State the region of the electromagnetic spectrum in which the CMBR is detected.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
light from distant galaxies is red-shifted (wavelength is longer than expected)✓ 1
this shows the galaxies are moving away from us✓ 1
the more distant the galaxy, the greater the red-shift, so the faster it moves away✓ 1
so the Universe is expanding, and working backwards everything was once in a very small region✓ 1
CMBR is radiation that is detected from all directions in space✓ 1
it is the remains of the radiation from the hot early Universe, stretched by the expansion to microwave wavelengths✓ 1
(accept other valid, correctly explained points, up to 6 marks)—
Part (b)
microwave✓ 1

Common error: Do not say the Big Bang was an explosion into empty space: it was the start of the expansion of space itself.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
131IGPE3-8-17r46-078(b) Motion in the Universe Paper 1PHard6 marks

A food scientist uses a spring balance that was calibrated on Earth. Its scale is marked in grams, so that a force of 1.0 N is shown as 100 g. On the Moon the scientist hangs a sample of flour from the balance and the scale reads 240 g. The gravitational field strength on Earth is 10 N/kg and on the Moon is 1.6 N/kg.

(a)

Calculate the force pulling the sample down on the spring.

(2)
(b)

Calculate the true mass of the flour sample.

(2)
(c)

The scientist now uses a beam balance with standard masses on the Moon. Explain why this gives the true mass of the sample.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 240 g converted to 0.24 kg and multiplied by 10 N/kg, or 240 ÷ 100✓ 1
A1 2.4 N✓ 1
Part (b)
M1 mass = 2.4 ÷ 1.6✓ 1
A1 1.5 kg (ECF from (a))✓ 1
Part (c)
The beam balance compares the sample with standard masses✓ 1
the gravitational pull on both sides is reduced by the same factor, so they balance when the masses are equal✓ 1

Exam tip: A spring balance measures force, whatever its scale is marked in. Convert the reading to a force before using the Moon's field strength.

Theory: Why is gravitational field strength different on the Moon and other planets? Single · Double · Physics
132IGPE3-8-16r48-118(d) Cosmology Paper 1PHard6 marks

In a planetarium show a spectral line from a distant galaxy has a laboratory wavelength of 650 nm. The galaxy is moving away from Earth at 1.8 × 107 m/s. The speed of light is 3.0 × 108 m/s.

(a)

Calculate the red-shift z of the galaxy.

(2)
(b)

Calculate the wavelength at which the line is observed.

(3)
(c)

State the direction in which the wavelength of the line has shifted, and what this shows about the motion of the galaxy.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 z = v/c✓ 1
A1 1.8 × 107 ÷ 3.0 × 108 = 0.060✓ 1
Part (b)
M1 change in wavelength = zλ = 0.060 × 650✓ 1
M1 observed wavelength = 650 + 39✓ 1
A1 689 nm (ECF from (a))✓ 1
Part (c)
Towards longer wavelengths (the red end), showing the galaxy is moving away from us✓ 1

Common error: The observed wavelength is the laboratory wavelength plus the change; do not stop at the change in wavelength.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
133IGCSE-PHY-8-0148(d) Cosmology Paper 2 onlyGrade 4–51 mark
A galaxy is moving away from the Earth.
Light from the galaxy is compared with light from the same element in a laboratory on Earth.
Which row describes the light observed from the galaxy?
Show mark scheme
  • C (wavelength longer, frequency lower);
  • A is incorrect because this happens when the source moves towards the observer (blue-shift)
  • B is incorrect because wavelength and frequency cannot both increase when the speed of light is unchanged (v = f × λ)
  • D is incorrect because a change in frequency must be accompanied by a change in wavelength
Theory: What happens to a wave when its source is moving? Physics only · 2P
134IGCSE-PHY-8-0348(c) Stellar Evolution Paper 2 onlyGrade 4–51 mark
Which row correctly describes the axes of a Hertzsprung–Russell (HR) diagram?
Show mark scheme
  • A (temperature increasing to the left; brightest at the top);
  • B is incorrect because on an HR diagram surface temperature increases from right to left
  • C is incorrect because a larger positive absolute magnitude means a dimmer star, so the largest positive values are at the bottom
  • D is incorrect because the x-axis shows surface temperature (or colour), not distance from the Earth
Theory: How do you draw the Hertzsprung–Russell diagram? Physics only · 2P
135IGCSE-PHY-8-0358(d) Cosmology Paper 2 onlyGrade 4–56 marks
A train travels at a constant speed along a straight track and passes through a station without stopping.
The driver sounds the train's horn, which emits sound with a frequency of 400 Hz.
A passenger stands still on the platform.
(a)
Describe how the frequency of the sound heard by the passenger compares with 400 Hz
• as the train approaches the station
• as the train moves away from the station.
(2)
(b)
The speed of sound in air is 340 m/s.
Calculate the wavelength of the sound emitted by the horn.
(2)
(c)
Some stars orbit another star.
As one of these stars moves around its orbit, it moves towards the Earth for part of the time and away from the Earth for the rest of the time.
Describe how the observed wavelength of a line in the spectrum of this star changes as it moves around its orbit.
(2)
Show mark scheme
(a)
  • higher (than 400 Hz) as the train approaches;
  • lower (than 400 Hz) as the train moves away;
Allow
  • higher / lower pitch
  • 'higher then lower' with no reference to 400 Hz scores 1
Do not accept
  • louder / quieter (loudness is not frequency)
  • stays at 400 Hz
  • lower as it approaches and higher as it moves away (scores 0)
(b)
  • substitution;
  • evaluation;
  • e.g. λ = 340 ÷ 400
  • (λ =) 0.85 (m)
Allow
  • 85 cm
  • rearrangement of v = f × λ seen for MP1
Do not accept
  • 136 000 (m) (340 × 400) for evaluation
  • 1.18 (m) (400 ÷ 340) for evaluation
(c)
  • (observed) wavelength is shorter (than the reference wavelength) when the star moves towards the Earth;
  • (observed) wavelength is longer when the star moves away from the Earth;
Allow
  • blue-shift for MP1 and red-shift for MP2
  • wavelength alternately increases and decreases / changes periodically, with no link to direction of motion, scores 1
Do not accept
  • longer when moving towards the Earth
  • the star changes colour because its temperature changes
Theory: What happens to a wave when its source is moving? Physics only · 2P
136IGPE2-061-00-108(d) Cosmology Paper 2PEasy5 marks

A student stands at the side of a road. An ambulance with its siren sounding drives towards her, passes her and then moves away from her.

(a)

Describe how the pitch of the siren heard by the student changes as the ambulance passes her and moves away.

(1)
(b)

Explain, in terms of waves, why the pitch heard by the student is lower when the ambulance is moving away.

(2)
(c)

A similar effect occurs with light from distant galaxies.
State what the red-shift of almost all distant galaxies tells us about the universe.

(1)
(d)

State one other piece of evidence that supports the Big Bang theory.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The pitch becomes lower (than when the ambulance was approaching)✓ 1
Part (b)
The source moves away between emitting successive wavefronts, so the wavefronts reaching the student are further apart / the wavelength is longer✓ 1
The speed of sound is unchanged, so the frequency heard is lower (f = v/λ)✓ 1
Part (c)
The universe is expanding / (almost all) galaxies are moving away from us. Also accept: more distant galaxies are moving away faster✓ 1
Part (d)
Cosmic microwave background (CMB) radiation. Also accept: the observed abundance of hydrogen and helium in the universe✓ 1

Common error: A source moving away gives a longer observed wavelength and a lower observed frequency. Do not say that the source itself emits a different frequency: only the frequency received changes.

Theory: What happens to a wave when its source is moving? Physics only · 2P
137IGPE3-8-15r19-098(d) Cosmology Paper 2PEasy6 marks

A hospital physicist visits an observatory and is shown the spectrum of a distant galaxy. A hydrogen spectral line has a wavelength of 600 nm when measured in a laboratory. The galaxy is moving away from the Earth at a speed of 4.5 × 106 m/s. The speed of light is 3.0 × 108 m/s.

(a)

State what is meant by red-shift.

(1)
(b)

Calculate the wavelength at which the line is observed.

(3)
(c)

Explain how the red-shift of light from distant galaxies supports the Big Bang theory.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the wavelength of light from a source that is moving away appears longer (shifted towards the red end of the spectrum)✓ 1
Part (b)
use of change in wavelength ÷ original wavelength = speed ÷ speed of light✓ 1
change in wavelength = 600 × 4.5 × 106 ÷ 3.0 × 108 = 9.0 nm✓ 1
609 nm✓ 1
Part (c)
light from almost all galaxies is red-shifted, so the galaxies are moving away from us (the universe is expanding)✓ 1
so the matter in the universe must once have been in a much smaller region, which is consistent with a Big Bang✓ 1

Exam tip: The observed wavelength is the original wavelength plus the change, so add the shift to 600 nm.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
138IGCSE-PHY-8-0168(d) Cosmology Paper 2 onlyGrade 6–76 marks
The Andromeda galaxy is moving towards the Milky Way.
The velocity of Andromeda towards the Earth is 3.0 × 10⁵ m/s.
A line in the spectrum of sodium has a reference wavelength of 589.0 nm when measured in a laboratory.
The speed of light is 3.0 × 10⁸ m/s.
(a)
Explain whether the wavelength of this sodium line observed in light from Andromeda is longer or shorter than 589.0 nm.
(2)
(b)
Calculate the observed wavelength of this sodium line in light from Andromeda.
Give your answer to 4 significant figures.
(3)
(c)
State how the frequency of this line observed in light from Andromeda compares with the frequency measured in the laboratory.
(1)
Show mark scheme
(a)
  • (observed wavelength is) shorter;
  • (because) galaxy / source is moving towards the observer / Earth;
Allow
  • blue-shift for MP1
  • waves are squashed / compressed for MP2 if linked to motion towards
Do not accept
  • longer / red-shift (scores 0)
  • shorter because the galaxy is closer (for MP2)
(b)
  • rearrangement / substitution;
  • evaluation of change in wavelength;
  • subtraction from reference wavelength;
  • e.g. Δλ = 589.0 × 3.0 × 10⁵ ÷ 3.0 × 10⁸
  • Δλ = 0.589 (nm)
  • (λ =) 588.4 (nm)
Allow
  • 588.41 (nm)
  • 589.6 (nm) from adding Δλ scores 2
Do not accept
  • 589.0 (nm) / no change (scores max 1 for the rearrangement)
(c)
  • higher / increased;
Do not accept
  • lower
  • the same
Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
139IGCSE-PHY-8-0178(d) Cosmology Paper 2 onlyGrade 6–77 marks
The Big Bang theory describes how the universe began.
One piece of evidence for the Big Bang theory is cosmic microwave background (CMB) radiation.
(a)
Describe how the universe has evolved according to the Big Bang theory.
(2)
(b)
Describe what CMB radiation is.
(2)
(c)
Explain how the existence of CMB radiation supports the Big Bang theory.
(3)
Show mark scheme
(a)
  • started from a (very) small, hot, dense point / region;
  • has expanded (and cooled) ever since;
Allow
  • about 13.8 / 14 billion years ago
  • singularity
Do not accept
  • an explosion of matter into existing space (for MP1)
  • the universe is contracting
(b)
  • microwave radiation (detected) from all directions (in space);
  • left over from the early universe / Big Bang;
Allow
  • low-temperature / about 2.7 K radiation filling the universe for MP1
Do not accept
  • radiation from microwave ovens / the Sun
  • radiation from nearby stars
(c)
  • any three from:
  • early universe was very hot and emitted (high-energy / short-wavelength) radiation;
  • expansion of the universe has stretched the wavelength (of this radiation) / universe has cooled;
  • so the radiation is now in the microwave region (as predicted by the theory);
  • radiation is (almost) uniform / same in all directions, as expected if it came from the whole early universe;
Allow
  • CMB was predicted by Big Bang theory before it was discovered
Do not accept
  • CMB is caused by galaxies moving away (red-shift of starlight)
Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
140IGCSE-PHY-8-0198(c) Stellar Evolution Paper 2 onlyGrade 6–76 marks
The HR diagram shows the main sequence and the present position of the Sun, S.
30 00010 00060003000−10−50+5+10+15surface temperature / Kabsolute magnitudemain sequenceStemperature scale not linear
(a)
The Sun will become a red giant.
Its surface temperature will decrease and its absolute magnitude will decrease (it will be about 1000 times brighter at the standard distance).
Explain how the position of the Sun on the HR diagram will change.
(3)
(b)
The Sun will then become a white dwarf.
The white dwarf will have a higher surface temperature than the Sun has now but will be about 1000 times less bright at the standard distance.
Describe and explain the position of the white dwarf on the HR diagram compared with S and with the main sequence.
(3)
Show mark scheme
(a)
  • (lower temperature so) moves to the right;
  • (brighter so) moves up / absolute magnitude decreases (becomes more negative);
  • ends above the main sequence / in the upper right (red giant region);
Allow
  • moves up and to the right for MP1 and MP2
Do not accept
  • moves down because it is cooler
  • moves left because it is bigger
(b)
  • (higher temperature so) to the left of S;
  • (less bright so) below S / absolute magnitude larger (more positive);
  • (so) below / to the left of the main sequence / in the lower left (white dwarf region);
Allow
  • lower left of S for MP1 and MP2
Do not accept
  • white dwarf is on the main sequence
  • above S because it is hotter (for MP2)
Theory: How do you draw the Hertzsprung–Russell diagram? Physics only · 2P
141IGCSE-PHY-8-0368(d) Cosmology Paper 2 onlyGrade 6–76 marks
A space probe is travelling directly away from the Earth at a speed of 14.0 km/s.
It sends radio signals to the Earth with a frequency of 8.40 GHz.
The speed of radio waves is 3.0 × 10⁸ m/s.
The equation used for the red-shift of light from galaxies also applies to the radio waves from the probe.
(a)
Calculate the wavelength of the radio waves emitted by the probe.
(2)
(b)
Calculate the change in wavelength of the radio waves received on the Earth.
(3)
(c)
State how the frequency of the radio waves received on the Earth compares with 8.40 GHz.
(1)
Show mark scheme
(a)
  • substitution;
  • evaluation;
  • e.g. λ = 3.0 × 10⁸ ÷ 8.40 × 10⁹
  • (λ =) 0.0357 (m)
Allow
  • 0.036 (m), 3.57 × 10⁻² (m), 3.6 cm
Do not accept
  • 35.7 (m) (GHz taken as MHz) for evaluation
(b)
  • rearrangement / Δλ = λ0 × v ÷ c;
  • substitution with v in m/s;
  • evaluation;
  • e.g. Δλ = 0.0357 × 14 000 ÷ 3.0 × 10⁸
  • (Δλ =) 1.67 × 10⁻⁶ (m)
Allow
  • ECF from (a)
  • 1.7 × 10⁻⁶ (m)
  • 1.67 × 10⁻⁹ (m) (v left in km/s) scores 2
Do not accept
  • 0.0357 + 1.67 × 10⁻⁶ given as the change in wavelength
(c)
  • lower / decreased;
Allow
  • slightly less
Do not accept
  • higher
  • the same
Theory: What happens to a wave when its source is moving? Physics only · 2P
142IGCSE-PHY-8-0378(c) Stellar Evolution Paper 2 onlyGrade 6–77 marks
The table gives the surface temperature and the absolute magnitude of four stars, P, Q, R and S.
Star / letter: P, Q, R, S
Surface temperature / K: 4000, 20000, 5800, 9500
Absolute magnitude / no unit: −2, −4, +4.8, +12
(a)
Seen from the Earth, star Q appears much less bright than star R.
Suggest how this is possible even though star Q has the more negative absolute magnitude.
(2)
(b)
One of the stars is a white dwarf.
Identify this star and explain how the data show that it is a white dwarf.
(3)
(c)
State the group of stars on the HR diagram to which star P belongs and give the colour of star P.
(2)
Show mark scheme
(a)
  • Q is (much) further away from the Earth (than R);
  • absolute magnitude is the brightness at a standard distance / how bright a star appears (from Earth) also depends on its distance;
Allow
  • Q gives out more light but it spreads out over a greater distance, for MP2
Do not accept
  • Q is cooler (it is hotter)
  • absolute magnitude of Q is larger so it is dimmer
(b)
  • S;
  • high (surface) temperature / hot / white;
  • large positive absolute magnitude / very dim (at standard distance) / below the main sequence;
Allow
  • hot but dim so it must be small, for MP3
Do not accept
  • S because it is closest to the Earth
(c)
  • red giants (or supergiants);
  • red / orange(-red);
Allow
  • giants
Do not accept
  • red dwarf
  • white
Theory: What is absolute magnitude? Physics only · 2P
143IGCSE-PHY-8-0388(d) Cosmology Paper 2 onlyGrade 6–710 marks
A student uses a balloon to model the expansion of the universe.
The student draws five ink dots, A to E, on a partly inflated balloon.
The dots represent galaxies.
The student measures the distance of each dot from dot A along the surface of the balloon.
The student then inflates the balloon further and measures the distances again.
Dot / name: B, C, D, E
Distance from A before / cm: 2.0, 3.5, 5.0, 6.5
Distance from A after / cm: 3.0, 5.3, 7.4, 9.8
ABCDEballoonink dots drawn on surfacenot to scale
(a)
Describe how the student could measure the distance between two dots along the curved surface of the balloon.
(2)
(b)
Use the data to show that the increase in distance of each dot from A is approximately proportional to its original distance from A.
(3)
(c)
Explain how the results of this model are linked to the red-shift of light from galaxies at different distances from the Earth.
(2)
(d)
Another student says:
'All the dots moved away from dot A, so dot A must be at the centre of the expansion. This means the Earth is at the centre of the universe.'
Evaluate this statement.
(2)
(e)
State what this model suggests about the universe in the past.
(1)
Show mark scheme
(a)
  • use string / flexible tape measure along the surface (between the dots);
  • then measure the string (marked length) with a ruler;
Allow
  • use a tape measure that bends to fit the surface scores 2
Do not accept
  • ruler held straight across between dots (ignores curve)
(b)
  • calculates increases: 1.0, 1.8, 2.4, 3.3 (cm);
  • calculates increase ÷ original distance (or after ÷ before) for at least three dots;
  • ratio is (approximately) constant ≈ 0.5 (or 1.5) so proportional;
Allow
  • ratios 0.48–0.51 or 1.48–1.51
  • comparison using doubling e.g. D is 2.5 × B's distance and its increase is 2.4 × B's increase, for MP2 and MP3
Do not accept
  • 'the increases get bigger' alone for MP3
(c)
  • more distant galaxies move away faster (greater recession velocity);
  • so (light from) more distant galaxies has a greater red-shift;
Allow
  • more distant dots moved further in the same time, linked to galaxies, for MP1
Do not accept
  • galaxies further away have a smaller red-shift
  • red-shift due to galaxies being hotter
(d)
  • (an observer on) any dot would see all other dots moving away (with the same pattern);
  • so there is no centre (on the surface) / A is not special, so the statement is wrong;
Allow
  • the expansion is of space itself, everywhere at once, for MP1
Do not accept
  • agreeing with the statement
(e)
  • galaxies were closer together / universe was smaller (and denser, hotter);
Allow
  • universe began from a single point (Big Bang)
Do not accept
  • universe was the same size
  • universe was bigger
Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
144IGPE-057-00-028(c) Stellar Evolution Paper 2PMedium1 mark

Star P has an absolute magnitude of −2. Star Q has an absolute magnitude of +5.
Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Absolute magnitude is the brightness a star would have if it were placed at a standard distance of 10 parsecs. A more negative value means a brighter star. P (−2) would therefore look brighter than Q (+5) at that standard distance. Absolute magnitude gives no information about how far away the stars really are, so options A and D are wrong. Option B reverses the magnitude scale.✓ 1Answer C

Answer: C

Exam tip: On the magnitude scale, smaller (more negative) numbers mean brighter stars. Absolute magnitude removes the effect of distance, so it compares how much light the stars actually give out.

Theory: What is absolute magnitude? Physics only · 2P
145IGPE2-061-01-028(c) Stellar Evolution Paper 2PMedium1 mark

A Hertzsprung–Russell (HR) diagram is drawn with absolute magnitude on the vertical axis, with the brightest stars at the top. Surface temperature is on the horizontal axis and decreases from left to right.
In which region of the diagram are white dwarfs found?

Show mark scheme
Marking pointMarkNotes
White dwarfs are very hot, so they lie on the left, where temperature is high. They are also very small, so they emit little light in total and lie near the bottom. This places them in the bottom left. The top right is where red giants and supergiants are found. The bottom right holds cool, dim red dwarfs: choosing it assumes a dim star must be cool. The top left holds hot, very luminous main-sequence stars.✓ 1Answer C

Answer: C

Exam tip: On this diagram, temperature increases to the LEFT. Check the direction of each axis before placing any group of stars.

Theory: How do you draw the Hertzsprung–Russell diagram? Physics only · 2P
146IGPE2-061-01-078(c) Stellar Evolution Paper 2PMedium6 marks

A science museum is designing an interactive touch-screen display. It follows the life of a star with a mass about 25 times the mass of the Sun.

(a)

Describe the life cycle of this star, from its formation to what remains after it has died.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
a nebula (cloud) of gas and dust is pulled together by gravity (accept: forms a protostar)✓ 1
the core becomes hot enough for hydrogen nuclei to fuse into helium, so it becomes a main sequence star✓ 1
the star is stable because the outward pressure from the energy released by fusion balances the inward pull of gravity✓ 1
when the hydrogen in the core runs out, the star expands to become a red supergiant (accept: it fuses heavier elements in its core)✓ 1
the core collapses and the star explodes as a supernova✓ 1
the remnant is a neutron star, or a black hole if the remaining core is very massive✓ 1

Exam tip: In 6-mark description answers, give the stages in a clear time order and link each stage to its cause. For example, hydrogen running out in the core leads to expansion. A list of stage names on its own will not score full marks.

Theory: How does a star like the Sun evolve? Single · Double · Physics
147IGPE2-061-01-108(c) Stellar Evolution Paper 2PMedium7 marks

A space-telescope team records data for five stars:
Star P: surface temperature 30 000 K, absolute magnitude −6
Star Q: surface temperature 3500 K, absolute magnitude −5
Star R: surface temperature 12 000 K, absolute magnitude +12
Star S: surface temperature 5800 K, absolute magnitude +5
Star T: surface temperature 3500 K, absolute magnitude +13

(a)

State which star is most likely to be a white dwarf.

(1)
(b)

State which star is most likely to be a red supergiant.

(1)
(c)

Star S is similar to the Sun. Name the group of stars on an HR diagram that star S belongs to.

(1)
(d)

Stars Q and T have the same surface temperature but very different absolute magnitudes.
Explain why.

(2)
(e)

Describe where star R would be plotted on an HR diagram. The diagram has temperature decreasing from left to right and the brightest stars at the top.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R✓ 1
Part (b)
Q✓ 1
Part (c)
main sequence✓ 1
Part (d)
at the same temperature, each square metre of surface emits the same amount of light✓ 1
Q is much larger than T / has a much bigger surface area, so it emits much more light in total✓ 1
Part (e)
towards the left, because it has a high temperature✓ 1
near the bottom, because it has a large positive absolute magnitude / is faint✓ 1

Exam tip: Two stars with the same temperature have the same colour. If their absolute magnitudes differ, the reason must be size, so link this to surface area.

Theory: How do you draw the Hertzsprung–Russell diagram? Physics only · 2P
148IGPE2-061-01-138(c) Stellar Evolution Paper 2PMedium7 marks

A street lamp far away can look dimmer than a weak torch held close to you. Astronomers use absolute magnitude to avoid this problem when comparing stars.

(a)

State what is meant by the absolute magnitude of a star.

(2)
(b)

Explain why two stars that look equally bright from Earth can have different absolute magnitudes.

(2)
(c)

Star X has an absolute magnitude of −3. Star Y has an absolute magnitude of +7.
State which star emits more light.

(1)
(d)

A decrease of 5 in absolute magnitude means that a star is 100 times more luminous.
Calculate how many times more luminous star X is than star Y.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the brightness a star would have✓ 1
if it were seen from a standard distance (10 parsecs)✓ 1
Part (b)
how bright a star looks depends on both its light output and its distance from Earth✓ 1
a more luminous star that is further away can look the same as a less luminous star that is closer✓ 1
Part (c)
X✓ 1
Part (d)
the difference is 10, which is two steps of 5✓ 1
100 × 100 = 10 000 times✓ 1

Common error: Each step of 5 magnitudes multiplies the luminosity by 100, so two steps give 100 × 100. Adding the factors to get 200 is a frequent mistake.

Theory: What is absolute magnitude? Physics only · 2P
149IGPE2-061-02-028(d) Cosmology Paper 2PMedium1 mark

A hydrogen line has a wavelength of 656 nm when measured in a laboratory. In light from a distant galaxy the same line is observed at 669 nm.
Speed of light = 3.0 ×108 m/s
What is the speed at which the galaxy is moving away from Earth?

Show mark scheme
Marking pointMarkNotes
Δλ = 669 − 656 = 13 nm.
v = (Δλ/λ0) × c = (13/656) × 3.0 ×108 = 5.9 ×106 m/s.
The equation uses the emitted (laboratory) wavelength λ0 = 656 nm. Dividing by the observed wavelength gives 5.8 ×106 m/s, which is wrong. Using 669/656 × c gives a speed greater than light. 2.0 ×10−2 is only the ratio Δλ/λ0, not yet multiplied by c.		✓ 1Answer B

Answer: B

Exam tip: In Δλ/λ₀ = v/c, λ₀ is always the wavelength from the stationary source, which here is the laboratory value.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
150IGPE2-061-02-108(d) Cosmology Paper 2PMedium6 marks

An astronomy club measured the fractional change in wavelength, Δλ/λ0, for light from four galaxies.
Galaxy A: distance 50 million light-years, Δλ/λ0 = 0.0037
Galaxy B: distance 100 million light-years, Δλ/λ0 = 0.0075
Galaxy C: distance 200 million light-years, Δλ/λ0 = 0.0150
Galaxy D: distance 300 million light-years, Δλ/λ0 = 0.0225
Speed of light = 3.0 ×108 m/s

(a)

Describe the relationship between distance and Δλ/λ0 shown by the data.

(2)
(b)

Calculate the speed at which galaxy D is moving away from Earth.

(2)
(c)

A club member says the data prove that the Milky Way is at the centre of the universe. Evaluate this claim.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δλ/λ0 increases as distance increases✓ 1
Δλ/λ0 is (directly) proportional to distance, e.g. it doubles when the distance doubles from B to C✓ 1
Part (b)
v = 0.0225 × 3.0 ×108✓ 1
v = 6.8 ×106 m/s✓ 1
Part (c)
The claim is not justified✓ 1
because in an expanding universe an observer in any galaxy would see all other galaxies moving away, with more distant ones moving faster✓ 1

Exam tip: To show proportionality, quote a pair of values from the data: doubling the distance doubles the red-shift.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
151IGPE2-061-02-128(d) Cosmology Paper 2PMedium6 marks

An astronomer studies two galaxies.
Galaxy X: a line emitted at 600 nm is red-shifted by 6.0 nm.
Galaxy Y: a line emitted at 450 nm is red-shifted by 18 nm.
Speed of light = 3.0 ×108 m/s

(a)

Calculate the speed at which galaxy X is moving away from Earth.

(2)
(b)

Calculate the speed at which galaxy Y is moving away from Earth.

(2)
(c)

Suggest which galaxy is further from Earth. Explain your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = (6.0/600) × 3.0 ×108✓ 1
v = 3.0 ×106 m/s✓ 1
Part (b)
v = (18/450) × 3.0 ×108✓ 1
v = 1.2 ×107 m/s✓ 1
Part (c)
Galaxy Y✓ 1
because it has the greater fractional red-shift / recession speed, and more distant galaxies move away faster✓ 1

Common error: Compare the fractional change Δλ/λ₀, not Δλ alone. Lines with different emitted wavelengths cannot be compared directly by Δλ.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
152IGPE2-061-00-028(d) Cosmology Paper 2PMedium1 mark

A hydrogen spectral line has a wavelength of 486.0 nm when measured in a laboratory. In the light from a distant galaxy, the same line is measured at 495.7 nm.
The speed of light is 3.0 ×108 m/s.
Which row gives the speed of the galaxy and its direction of motion relative to Earth?

Show mark scheme
Marking pointMarkNotes
Change in wavelength = 495.7 − 486.0 = 9.7 nm. Use v = (Δλ / λ0) × c = (9.7 / 486.0) × 3.0 ×108 ≈ 6.0 ×106 m/s. The observed wavelength is longer, so the light is red-shifted and the galaxy is moving away from Earth. Option A gets the direction wrong. Option B forgets to divide by λ0. Option D uses the ratio of the two wavelengths instead of the change in wavelength.✓ 1Answer C

Answer: C

Exam tip: Always subtract first to find the change in wavelength. An increase in wavelength (red-shift) means the source is moving away.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
153IGPE2-061-00-098(d) Cosmology Paper 2PMedium6 marks

Astronomers study a calcium absorption line in the spectrum of a distant galaxy. In a laboratory, this line has a wavelength of 393.4 nm. The galaxy is moving away from Earth at 1.8 ×107 m/s.
Speed of light = 3.0 ×108 m/s.

(a)

State what is meant by red-shift.

(2)
(b)

Calculate the wavelength of this calcium line as observed from Earth.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
The observed wavelength of light is increased / shifted towards the red end of the spectrum✓ 1
because the source is moving away from the observer✓ 1
Part (b)
Δλ / λ0 = v / c✓ 1
Δλ = 393.4 × (1.8 ×107 ÷ 3.0 ×108)✓ 1
Δλ = 23.6 nm✓ 1
observed wavelength = 393.4 + 23.6 = 417 nm✓ 1

Exam tip: The equation gives the change in wavelength, not the observed wavelength. For a red-shift, add the change to the laboratory value.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
154IGPE2-061-00-128(d) Cosmology Paper 2PMedium6 marks

A space telescope records spectra from two galaxies, A and B.
Galaxy A: a spectral line with a laboratory wavelength of 500 nm is red-shifted by 12 nm.
Galaxy B: a spectral line with a laboratory wavelength of 600 nm is red-shifted by 30 nm.
Speed of light = 3.0 ×108 m/s.

(a)

Calculate the speed at which galaxy A is moving away from Earth.

(3)
(b)

A student says that galaxy B must be further away than galaxy A because its spectral line is shifted by more nanometres.
Evaluate this statement.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = (Δλ / λ0) × c✓ 1
v = (12 ÷ 500) × 3.0 ×108✓ 1
v = 7.2 ×106 m/s✓ 1
Part (b)
The reasoning is flawed: the size of the shift in nm also depends on the laboratory wavelength, so the fractional shift Δλ/λ0 must be compared✓ 1
For B, Δλ/λ0 = 30/600 = 0.05, which is greater than 0.024 for A. Accept: B's speed is 1.5 ×107 m/s compared with 7.2 ×106 m/s for A. Either way, B is moving away faster✓ 1
Recession speed is proportional to distance (Hubble's law), so B is further away. The conclusion is correct but the reason given is not✓ 1

Common error: Compare galaxies using the fractional change Δλ/λ₀, not the observed wavelength or the shift in nm alone.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
155IGPE3-8-14r14-078(b) Motion in the Universe Paper 2PMedium8 marks

A student compares the orbits of Earth and Mars around the Sun. Mars has an orbital radius of 2.28 × 1011 m and takes 5.94 × 107 s to complete one orbit. Earth has an orbital radius of 1.50 × 1011 m and takes 3.16 × 107 s. Assume both orbits are circular. The student claims: 'Mars has the larger orbit, so it travels at the greater speed.'

Earth, 1.50 × 10¹¹ mMars, 2.28 × 10¹¹ mSun
Not to scale
(a)

Calculate the orbital speed of Mars. Give your answer to 3 significant figures.

(3)
(b)

Calculate the orbital speed of Earth. Give your answer to 3 significant figures.

(2)
(c)

Evaluate the student's claim.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 distance in one orbit = 2π × 2.28 × 1011✓ 1
M1 speed = distance ÷ time✓ 1
A1 2.41 × 104 m/s (24100 m/s)✓ 1
Part (b)
M1 2π × 1.50 × 1011 ÷ (3.16 × 107)✓ 1
A1 2.98 × 104 m/s (29800 m/s)✓ 1
Part (c)
B1 Mars does travel further in one orbit because its orbit is larger✓ 1
B1 but its speed (2.41 × 104 m/s) is less than Earth's (2.98 × 104 m/s), so the claim is not supported (ECF from (a) and (b))✓ 1
B1 the Sun's gravitational pull is weaker at a greater distance, so a lower speed is needed to stay in orbit✓ 1

Common error: A larger orbit means a longer path, but the period is also much longer, so the speed can be lower.

Theory: How do you calculate orbital speed? Double · Physics
156IGPE3-8-14r14-098(d) Cosmology Paper 2PMedium6 marks

The Universe is thought to have begun with the Big Bang.

(a)

State what is meant by cosmic microwave background radiation.

(1)
(b)

Explain how the cosmic microwave background radiation supports the Big Bang theory.

(2)
(c)

A galaxy is 9.46 × 1024 m from the Earth and is moving away at a constant speed of 2.00 × 107 m/s. Estimate how long ago the galaxy and the Earth were together, assuming the speed has never changed. Give your answer in years to 3 significant figures. Use 1 year = 3.15 × 107 s and the unrounded time in seconds.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 microwave radiation that reaches the Earth from all directions in space✓ 1
Part (b)
B1 the radiation was emitted at very high energy (short wavelength) soon after the Big Bang✓ 1
B1 the expansion of the Universe has stretched its wavelength into the microwave region✓ 1
Part (c)
M1 time = distance ÷ speed = 9.46 × 1024 ÷ (2.00 × 107)✓ 1
A1 4.73 × 1017 s✓ 1
A1 1.50 × 1010 years (15000000000 years)✓ 1

Exam tip: Convert seconds to years last, and keep extra digits until the final answer.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
157IGPE3-8-14r17-048(b) Motion in the Universe Paper 2PMedium1 mark

The International Space Station orbits in a circle of radius 6.78 × 106 m at a constant speed of 7.7 × 103 m/s. Which is closest to the time for one orbit?

Show mark scheme
Marking pointMarkNotes
B is correct: T = 2πr ÷ v = 2π × 6.78 × 106 ÷ 7.7 × 103 ≈ 5530 s ≈ 92 minutes. A leaves out the factor 2π. C confuses the station with a geostationary satellite. D leaves the answer in seconds but labels it minutes.✓ 1Answer B

Answer: B

Exam tip: Orbital speed = circumference ÷ period, so T = 2πr ÷ v. Convert seconds to minutes at the end.

Theory: How do you calculate orbital speed? Double · Physics
158IGPE3-8-14r17-068(d) Cosmology Paper 2PMedium6 marks

Light from a distant galaxy contains a spectral line from hydrogen. In a laboratory the line has wavelength 656 nm. In the light from the galaxy it is observed at 659.28 nm. Speed of light = 3.0 × 108 m/s.

(a)

Calculate the change in wavelength.

(1)
(b)

Calculate the speed at which the galaxy is moving, using z = Δλ ÷ λ = v ÷ c.

(3)
(c)

State the direction in which the galaxy is moving relative to Earth and give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 659.28 − 656 = 3.28 nm✓ 1
Part (b)
M1 z = 3.28 ÷ 656 = 0.0050 (ECF from (a))✓ 1
M1 v = z × c = 0.0050 × 3.0 × 108✓ 1
A1 1.5 × 106 m/s✓ 1
Part (c)
B1 away from Earth✓ 1
B1 the wavelength has increased (red-shift)✓ 1

Exam tip: Show z as a ratio first: change in wavelength divided by the wavelength measured in the laboratory.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
159IGPE3-8-14r17-078(b) Motion in the Universe Paper 2PMedium6 marks

An Earth-observation satellite photographs a wind farm as it passes overhead. It moves in a circular orbit of radius 7.0 × 106 m and takes 98 minutes for one orbit.

satelliter = 7.0 × 10⁶ mv = ?Earth
Not to scale
(a)

Show that the orbital period is 5880 s.

(1)
(b)

Calculate the orbital speed of the satellite. Give your answer to 3 significant figures.

(3)
(c)

The satellite's speed is constant but its velocity is changing. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 98 × 60 = 5880 s✓ 1
Part (b)
M1 distance for one orbit = 2π × 7.0 × 106✓ 1
M1 speed = 2πr ÷ T = 2π × 7.0 × 106 ÷ 5880✓ 1
A1 7480 m/s✓ 1
Part (c)
B1 velocity includes direction, and the direction of motion changes continuously✓ 1
B1 gravitational force acts towards the centre of the Earth, causing acceleration without a change in speed✓ 1

Common error: A satellite in a circular orbit has constant speed but changing velocity, because velocity is a vector.

Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
160IGPE3-8-15r19-108(d) Cosmology Paper 2PMedium7 marks

Light from a hydrogen source in a laboratory has a wavelength of 480 nm. The same spectral line is observed at 486 nm in light from galaxy X and at 492 nm in light from galaxy Y. The speed of light is 3.0 × 108 m/s.

(a)

Calculate the speed at which galaxy X is moving away from the Earth. Give your answer to 3 significant figures.

(3)
(b)

Determine the speed at which galaxy Y is moving away from the Earth.

(2)
(c)

Galaxy Y is found to be further from the Earth than galaxy X. State what this suggests about the speed of recession of galaxies and their distance.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
change in wavelength = 6 nm✓ 1
use of speed = (change in wavelength ÷ original wavelength) × speed of light✓ 1
3.75 × 106 m/s✓ 1
Part (b)
change in wavelength is 12 nm, which is twice that of galaxy X (or use the same equation with 12 nm)✓ 1
7.5 × 106 m/s✓ 1
Part (c)
more distant galaxies are moving away faster✓ 1
the speed of recession increases with distance, which suggests that the universe is expanding✓ 1

Exam tip: If the original wavelength is the same, the change in wavelength is proportional to the speed, so doubling the shift doubles the speed.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
161IGPE3-8-17r27-058(b) Motion in the Universe Paper 2PMedium5 marks

Cosmologists use the Hubble constant to estimate the age of the universe. A measured value of the Hubble constant is 2.2×10−18 per second.

(a)

The age of the universe can be estimated as 1/H0, where H0 is the Hubble constant. Calculate this age in seconds. Give your answer to 3 significant figures.

(2)
(b)

Convert your answer to part (a) into years. Use your unrounded value from (a). 1 year = 3.15×107 s. Give your answer to 3 significant figures.

(2)
(c)

State one assumption made when the age is calculated as 1/H0.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
age = 1 ÷ H0 (with H0 = 2.2×10−18 s−1)✓ 1
= 4.55×1017 s (4.545… s)✓ 1
Part (b)
divide by 3.15×107 s (ECF from (a))✓ 1
= 1.44×1010 years✓ 1
Part (c)
the universe has expanded at a constant rate / the galaxies have always moved at constant speed (accept equivalent)✓ 1

Exam tip: Keep the unrounded value from one part to use in the next, and round only the final answer.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
162IGPE3-8-17r27-128(b) Motion in the Universe Paper 2PMedium6 marks

A satellite used to monitor crop health moves in a circular orbit around the Earth. The radius of the orbit is 7.0×106 m and the satellite takes 5800 s to complete one orbit.

satelliter = 7.0 × 10⁶ mv = ?Earth
Not to scale
(a)

Calculate the orbital speed of the satellite using v = 2πr ÷ T. Give your answer to 3 significant figures.

(2)
(b)

The speed of the satellite is constant but the satellite is accelerating. Explain why.

(2)
(c)

A second satellite orbits at a smaller radius. State how its orbital speed compares with that of the first satellite and give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = 2 × π × 7.0×106 ÷ 5800✓ 1
= 7.58×103 m/s✓ 1
Part (b)
the direction of motion keeps changing, so the velocity changes✓ 1
acceleration is the rate of change of velocity, so the satellite accelerates (towards the centre of the Earth)✓ 1
Part (c)
its orbital speed is greater✓ 1
the gravitational force from the Earth is stronger at a smaller radius, so a higher speed is needed to stay in orbit✓ 1

Exam tip: The unit of the orbital speed follows the unit of radius, so convert to metres before using v = 2πr ÷ T.

Theory: What keeps moons, planets, satellites and comets in orbit? Single · Double · Physics
163IGPE3-8-15r41-048(d) Cosmology Paper 2PMedium1 mark

A line in the spectrum of a laboratory lamp has a wavelength of 500 nm. In the light from a distant galaxy the same line has a wavelength of 502 nm. The speed of light is 3.0 × 108 m/s. At what speed is the galaxy moving away from the Earth?

Show mark scheme
Marking pointMarkNotes
The change in wavelength is 2 nm, so change in wavelength ÷ reference wavelength = 2 ÷ 500 = 0.004. Speed of galaxy = 0.004 × 3.0 × 108 = 1.2 × 106 m/s. The value 6.0 × 108 m/s comes from multiplying the speed of light by 2 and ignoring the reference wavelength. The value 7.5 × 1010 m/s comes from inverting the ratio (500 ÷ 2). The value 1.2 × 103 m/s comes from using the speed of light in km/s but giving the answer in m/s.✓ 1Answer C

Answer: C

Exam tip: The wavelength change is divided by the reference (laboratory) wavelength, so the nanometre units cancel.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
164IGPE3-8-15r42-098(b) Motion in the Universe Paper 2PMedium5 marks

A navigation satellite orbits the Earth in a circular path of radius 7.0 × 106 m. It takes 5830 s to complete one orbit.

satelliter = 7.0 × 10⁶ mv = ?Earth
Not to scale
(a)

The speed of the satellite is constant but its velocity is changing. Explain why.

(2)
(b)

Calculate the orbital speed of the satellite. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
velocity is speed in a given direction✓ 1
the direction of motion changes continuously as the satellite moves round the circle✓ 1
Part (b)
M1 distance for one orbit = 2πr = 2 × π × 7.0 × 106✓ 1
M1 speed = distance ÷ time (÷ 5830)✓ 1
A1 7540 m/s (unrounded 7544 m/s)✓ 1

Exam tip: Use the radius, not the diameter, in 2πr, and keep the period in seconds.

Theory: How do you calculate orbital speed? Double · Physics
165IGPE3-8-17r46-048(d) Cosmology Paper 2PMedium1 mark

Which statement about the cosmic microwave background radiation is correct?

Show mark scheme
Marking pointMarkNotes
The background radiation was released early in the history of the universe and has been stretched to microwave wavelengths by the expansion of the universe, and it arrives from all directions. It is not re-emitted starlight, since it is far too uniform and too cool. A supernova remnant would give a signal from a single region of the sky. An origin in the atmosphere is wrong because the same radiation is detected from space.✓ 1Answer B

Answer: B

Exam tip: For the cosmic microwave background, link three ideas: very early universe, expansion of space, and wavelength now in the microwave region.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
166IGPE3-8-16r48-048(c) Stellar Evolution Paper 2PMedium1 mark

The Hubble constant is 2.3 × 10−18 s−1. The age of the universe can be estimated as 1 ÷ H0. Use 1 year = 3.15 × 107 s. What is the estimated age of the universe?

Show mark scheme
Marking pointMarkNotes
1 ÷ (2.3 × 10−18) = 4.3 × 1017 s, and dividing by 3.15 × 107 s per year gives 1.4 × 1010 years. Leaving the answer as 4.3 × 1017 forgets to convert from seconds. Dividing by seconds in a day gives 5.0 × 1012. Multiplying H0 by the seconds in a year gives 7.2 × 10−11.✓ 1Answer D

Answer: D

Exam tip: The unit s⁻¹ means 1 ÷ H₀ is in seconds; convert to years before comparing.

Theory: Which units do I need for Topic 8? Single · Double · Physics
167IGPE3-8-16r48-068(d) Cosmology Paper 2PMedium7 marks

Light from a quasar contains a spectral line that has a wavelength of 500 nm when measured in a laboratory. The same line from the quasar is detected at a frequency of 5.0 × 1014 Hz. The speed of light is 3.0 × 108 m/s.

(a)

Calculate the observed wavelength of the line in metres.

(3)
(b)

Calculate the red-shift z of the quasar.

(2)
(c)

Calculate the speed at which the quasar is moving away from Earth.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 v = fλ✓ 1
M1 λ = 3.0 × 108 ÷ 5.0 × 1014✓ 1
A1 6.0 × 10−7 m✓ 1
Part (b)
M1 change in wavelength = 600 − 500 = 100 nm✓ 1
A1 z = 100 ÷ 500 = 0.20 (ECF from (a))✓ 1
Part (c)
M1 v = zc = 0.20 × 3.0 × 108✓ 1
A1 6.0 × 107 m/s (ECF from (b))✓ 1

Exam tip: Convert wavelength and frequency to SI units (m, Hz) before using v = fλ.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
168IGPE3-8-16r48-138(b) Motion in the Universe Paper 2PMedium6 marks

A shipping company tracks its vessels using two satellites. Both orbit Earth in circular orbits.

SatelliteOrbital radius / mOrbital period / s
A6.8 × 1065.5 × 103
B4.2 × 1078.6 × 104

A, 6.8 × 10⁶ mB, 4.2 × 10⁷ mEarth
Not to scale
(a)

Calculate the orbital speed of satellite A. Give your answer to 3 significant figures.

(2)
(b)

Calculate the orbital speed of satellite B. Give your answer to 3 significant figures.

(2)
(c)

Suggest why satellite B moves more slowly than satellite A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 v = 2πr/T = 2π × 6.8 × 106 ÷ 5.5 × 103✓ 1
A1 7770 m/s✓ 1
Part (b)
M1 v = 2π × 4.2 × 107 ÷ 8.6 × 104✓ 1
A1 3070 m/s✓ 1
Part (c)
B is further from Earth so the gravitational force/field strength on it is smaller✓ 1
A smaller force needs only a smaller speed for the satellite to stay in orbit✓ 1

Exam tip: Give each speed to the same number of significant figures so the two can be compared fairly.

Theory: How do you calculate orbital speed? Double · Physics
169IGCSE-PHY-8-0188(d) Cosmology Paper 2 onlyGrade 8–910 marks
Astronomers measure the wavelength of a hydrogen line in the light from three galaxies, P, Q and R.
The reference wavelength of this line is 486.1 nm.
The speed of light is 3.0 × 10⁸ m/s.
Galaxy / name: P, Q, R
Distance from Earth / 10²⁴ m: 1.5, 3.0, 4.5
Observed wavelength / nm: 491.6, 497.1, 502.6
Recession velocity / 10⁶ m/s: 3.4, 6.8, not given
(a)
Calculate the recession velocity of galaxy R.
(3)
(b)
Use the data to show that the recession velocity of a galaxy is directly proportional to its distance from the Earth.
(2)
(c)
Explain why this relationship provides evidence that the universe is expanding.
(2)
(d)
Assume galaxy R has always moved away from the Earth at its present velocity.
Estimate the time, in years, since galaxy R and the Milky Way were at the same place.
(1 year = 3.2 × 10⁷ s)
(3)
Show mark scheme
(a)
  • change in wavelength calculated;
  • substitution / rearrangement;
  • evaluation;
  • e.g. Δλ = 502.6 − 486.1 = 16.5 (nm)
  • v = 16.5 × 3.0 × 10⁸ ÷ 486.1
  • (v =) 1.0 × 10⁷ (m/s)
Allow
  • 1.02 × 10⁷, 1.018 × 10⁷, 10 183 090 (m/s)
  • 9.8 × 10⁶ m/s from using 502.6 as denominator scores 2
Do not accept
  • 3.1 × 10⁸ m/s (ratio of wavelengths × c)
(b)
  • v ÷ d calculated for at least two galaxies;
  • values the same / constant (≈ 2.3 × 10⁻¹⁸), so v ∝ d;
Allow
  • distance doubles (P to Q) and velocity doubles; distance ×3 (P to R) and velocity ×3
  • ECF from (a)
Do not accept
  • 'velocity increases as distance increases' alone (not proportional) for MP2
(c)
  • (all) galaxies are moving away (from us / red-shifted);
  • more distant galaxies move faster so galaxies are moving apart from each other / space is expanding;
Allow
  • red-shift increases with distance for MP1
Do not accept
  • galaxies are expanding
  • the Milky Way is at the centre of the universe
(d)
  • use of time = distance ÷ speed;
  • evaluation in seconds;
  • conversion to years;
  • e.g. t = 4.5 × 10²⁴ ÷ 1.02 × 10⁷
  • (t =) 4.4 × 10¹⁷ (s)
  • (t =) 1.4 × 10¹⁰ (years)
Allow
  • 1.38 × 10¹⁰, 1.41 × 10¹⁰, 14 billion (years)
  • ECF from (a)
  • use of 3.15 × 10⁷ s per year (1.40 × 10¹⁰ years)
  • use of data for P or Q (same time, 1.4 × 10¹⁰ years, because v ∝ d)
Do not accept
  • multiplying distance by velocity
Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
170IGCSE-PHY-8-0208(d) Cosmology Paper 2 onlyGrade 8–98 marks
The diagram shows two absorption lines measured in a laboratory, and the same two lines in the spectrum of light from galaxy G.
One line is from calcium and the other is from sodium.
Laboratory wavelengths: 393.4 nm (calcium) and 589.0 nm (sodium)
Wavelengths in light from galaxy G: 409.1 nm (calcium) and 612.6 nm (sodium)
laboratory393.4 nm (calcium)589.0 nm (sodium)galaxy G409.1 nm612.6 nm350400450500550600650λ / nm
(a)
Explain how the diagram shows that galaxy G is moving away from the Earth.
(2)
(b)
A student says:
'The galaxy has only one velocity, so both lines should be shifted by the same number of nanometres.'
Use the data to show that the student is wrong, and explain why the two lines are shifted by different amounts.
(4)
(c)
A line of magnesium has a laboratory wavelength of 517.3 nm.
Determine the wavelength of this line in the light from galaxy G.
(2)
Show mark scheme
(a)
  • lines (in light from G) are at longer wavelengths / shifted towards the red end;
  • (red-shift) occurs when the source moves away from the observer;
Allow
  • frequency of the lines is lower for MP1
Do not accept
  • the galaxy is red
  • lines shifted to shorter wavelengths
(b)
  • both changes in wavelength calculated (15.7 nm and 23.6 nm), showing they are different;
  • Δλ/λ0 calculated for both lines (0.0399 and 0.0401);
  • Δλ/λ0 is the same (≈ 0.040) for both lines / both give the same velocity;
  • (because) Δλ/λ0 = v/c, so Δλ is proportional to λ0 / longer reference wavelength gives a bigger shift;
Allow
  • velocities from both lines (1.20 × 10⁷ m/s and 1.197 × 10⁷ m/s, using c = 3.0 × 10⁸ m/s) for MP2 and MP3
  • λ/λ0 = 1.040 for both lines for MP2
  • values to 2 s.f. (0.040) for MP3 but MP2 needs values to at least 3 s.f.
Do not accept
  • 'the lines are different colours' with no reference to proportion
  • the galaxy has two velocities
(c)
  • change in wavelength = 0.040 × 517.3 / evaluation of Δλ = 20.7 (nm);
  • (λ =) 538.0 (nm);
  • e.g. Δλ = 0.0400 × 517.3 = 20.7 (nm)
  • λ = 517.3 + 20.7
Allow
  • 537.9 to 538.1 (nm)
  • 538 (nm)
  • ECF from ratio in (b)
  • 517.3 × 1.04 for both marks
Do not accept
  • 517.3 + 15.7 or 517.3 + 23.6 (copying one shift) (scores 0)
  • 496.6 (nm) (subtracting)
Theory: What is red-shift in the light from galaxies? Physics only · 2P
171IGCSE-PHY-8-0398(d) Cosmology Paper 2 onlyGrade 8–98 marks
A spiral galaxy is seen edge-on from the Earth.
The galaxy is moving away from the Earth and is also rotating about its centre.
Astronomers measure the wavelength of a hydrogen line in light from three points on the galaxy: X, C (the centre) and Y.
The reference wavelength of this hydrogen line is 656.3 nm.
The speed of light is 3.0 × 10⁸ m/s.
Point / name: X, C, Y
Observed wavelength / nm: 664.7, 664.0, 663.3
XCYgalaxy as seen from the Earth (edge-on)not to scale
(a)
Explain why the wavelength observed from X is longer than the wavelength observed from Y.
(3)
(b)
Determine the speed at which the edge of the galaxy is rotating about its centre.
(4)
(c)
Explain why the astronomers needed to measure the wavelength at the centre C as well as at X and Y.
(1)
Show mark scheme
(a)
  • galaxy rotating so X is moving away from the Earth and Y towards the Earth (relative to the centre);
  • X has the greater velocity away from the Earth (recession velocity);
  • greater velocity away gives greater red-shift / larger increase in wavelength;
Allow
  • Y has less red-shift because its rotation is towards us, for MP3
Do not accept
  • X is further away from the Earth (as the only reason)
  • Y is blue-shifted (since its wavelength is still longer than 656.3 nm)
(b)
  • Δλ due to rotation = 0.7 nm (from X − C or C − Y or (X − Y) ÷ 2);
  • use of Δλ ÷ λ0 = v ÷ c;
  • substitution;
  • evaluation;
  • e.g. v = 3.0 × 10⁸ × 0.7 ÷ 656.3
  • (v =) 3.2 × 10⁵ (m/s)
Allow
  • 3 × 10⁵ (m/s) (1 s.f., since 0.7 nm has 1 s.f.)
  • λ0 = 664.0 nm used: 3.16 × 10⁵ (m/s)
  • 6.4 × 10⁵ (m/s) (using X − Y = 1.4 nm) scores 3
  • 3.5 × 10⁶ or 3.8 × 10⁶ (m/s) (recession velocity of C or X) scores 2
Do not accept
  • 3.2 × 10⁻⁴ / 320 (m/s) (POT error) for evaluation
(c)
  • red-shift at C is due to whole galaxy moving away / to subtract (remove) the galaxy's recession velocity;
Allow
  • to compare X and Y with the centre to find the shift due to rotation only
Do not accept
  • to get an average
  • because C is the brightest
Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
172IGCSE-PHY-8-0408(d) Cosmology Paper 2 onlyGrade 8–99 marks
A quasar is a very distant, very bright object at the centre of a galaxy.
A spectral line in the light from a quasar is observed at a wavelength of 529.9 nm.
The quasar is moving away from the Earth at 2.7 × 10⁷ m/s.
The speed of light is 3.0 × 10⁸ m/s.
(a)
Determine the reference wavelength of this spectral line.
(4)
(b)
Light from galaxy G has a fractional change in wavelength, Δλ/λ0, of 0.030.
State and explain whether the quasar or galaxy G is further from the Earth.
(2)
(c)
Calculate the frequency of the light in this spectral line when it is received on the Earth.
State how this frequency compares with the frequency of the light emitted by the quasar.
(3)
Show mark scheme
(a)
  • v ÷ c = 0.090;
  • λ = λ0 + 0.090 λ0 / λ = 1.090 × λ0 / (λ − λ0) ÷ λ0 = 0.090;
  • λ0 = 529.9 ÷ 1.090;
  • (λ0 =) 486 (nm);
Allow
  • 486.1, 486.15 (nm)
  • any correct route, e.g. λ0 = λ ÷ (1 + v/c), or λ = λ0 + λ0 × v/c solved for λ0
  • 482 (nm) (using Δλ = 0.090 × 529.9, i.e. observed wavelength as reference) scores 2
Do not accept
  • 5890 (nm) (529.9 ÷ 0.090) for evaluation
  • 577.6 (nm) (adding the shift) for evaluation
(b)
  • quasar (is further away);
  • it has the greater red-shift so it is moving away faster, and more distant galaxies move away faster (have greater red-shift);
Allow
  • quasar's red-shift (0.090) is three times that of G, for MP2 if linked to distance
Do not accept
  • galaxy G
  • quasar because it is brighter
(c)
  • substitution into f = c ÷ λ;
  • evaluation;
  • (received frequency is) lower (than emitted);
  • e.g. f = 3.0 × 10⁸ ÷ 529.9 × 10⁻⁹
  • (f =) 5.66 × 10¹⁴ (Hz)
Allow
  • 5.7 × 10¹⁴, 5.661 × 10¹⁴ (Hz)
  • POT error from nm conversion (e.g. 5.66 × 10⁵ Hz) loses the evaluation mark only
  • comparison with the emitted frequency 3.0 × 10⁸ ÷ 486 × 10⁻⁹ = 6.17 × 10¹⁴ Hz for MP3
Do not accept
  • higher (than emitted) for MP3
  • 6.17 × 10¹⁴ Hz (reference wavelength used) for evaluation
Theory: What happens to a wave when its source is moving? Physics only · 2P
173IGPE2-061-01-038(c) Stellar Evolution Paper 2PHard1 mark

Two stars, P and Q, are photographed with a smartphone camera. Star P appears brighter in the night sky than star Q.
Star P has an absolute magnitude of +6. Star Q has an absolute magnitude of −1.
Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Absolute magnitude compares stars as if they were all at the same standard distance. A smaller (more negative) value means the star is more luminous. Q, at −1, therefore emits more light than P, at +6. P still appears brighter from Earth, so P must be much closer to us. Option A confuses how bright a star looks with how much light it emits. Option C assumes a bigger magnitude number means brighter. Option D gives the wrong conclusion about distance.✓ 1Answer B

Answer: B

Common error: On the magnitude scale, a smaller or more negative number means a brighter star. Apparent brightness also depends on distance, so it cannot be used on its own to compare how much light stars emit.

Theory: What is absolute magnitude? Physics only · 2P
174IGPE2-061-01-088(c) Stellar Evolution Paper 2PHard8 marks

A phone's battery life depends on both the energy stored and the power drawn. A star's main-sequence lifetime can be estimated in the same way.
For the Sun:
energy available from core hydrogen fusion = 1.2 ×1044 J
power output = 3.8 ×1026 W
1 year = 3.15 ×107 s

(a)

State the equation linking power, energy transferred and time taken.

(1)
(b)

Calculate the main-sequence lifetime of the Sun in years.

(3)
(c)

Star Z is more massive than the Sun. It has 15 times as much energy available from fusion as the Sun, and its power output is 10 000 times greater than the Sun's.
Calculate the main-sequence lifetime of star Z in years.

(2)
(d)

A student says: 'More massive stars live longer because they have more fuel.'
Evaluate this statement using your answers.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
power = energy transferred ÷ time taken✓ 1
Part (b)
time = 1.2 ×1044 ÷ 3.8 ×1026 = 3.2 ×1017 s✓ 1
÷ 3.15 ×107✓ 1
= 1.0 ×1010 years✓ 1
Part (c)
time = (15 × 1.2 ×1044) ÷ (1.0 ×104 × 3.8 ×1026) = 4.7 ×1014 s✓ 1
÷ 3.15 ×107 = 1.5 ×107 years✓ 1
(alternative: lifetime = 1.0 ×1010 × 15 ÷ 10 000 = 1.5 ×107 years, full marks)—
Part (d)
the statement is incorrect: star Z has more fuel but a far shorter lifetime (about 107 years compared with 1010 years)✓ 1
because its power output increases much more than its fuel supply, so it uses up its fuel faster✓ 1

Common error: Rearranging power = energy ÷ time gives time = energy ÷ power. A frequent slip is to multiply energy by power, so check that the answer is in seconds before converting to years.

Theory: How does a star much more massive than the Sun evolve? Double · Physics
175IGPE2-061-02-068(d) Cosmology Paper 2PHard6 marks

A calcium absorption line has a wavelength of 393 nm when measured in a laboratory. A galaxy is moving away from Earth at 2.4 ×107 m/s.
Speed of light = 3.0 ×108 m/s
Use the equation: change in wavelength ÷ laboratory wavelength = speed of galaxy ÷ speed of light

(a)

Calculate the change in wavelength, Δλ, of the calcium line in light from this galaxy.

(3)
(b)

Calculate the wavelength of the calcium line observed on Earth.

(1)
(c)

A second galaxy is much further away from Earth. Suggest how the change in wavelength of its calcium line compares with your calculated value of Δλ for the first galaxy. Give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δλ/λ0 = v/c✓ 1
Δλ = 393 × (2.4 ×107 / 3.0 ×108)✓ 1
Δλ = 31 nm (allow 31.4 nm)✓ 1
Part (b)
393 + 31 = 424 nm (allow 424.4 nm)✓ 1
Part (c)
The change in wavelength is greater✓ 1
because more distant galaxies are moving away faster✓ 1

Common error: Δλ is only the change in wavelength. To find the observed wavelength for a receding galaxy, add Δλ to the laboratory value.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
176IGPE2-061-02-098(d) Cosmology Paper 2PHard5 marks

Light from a distant galaxy contains a spectral line observed on Earth at a wavelength of 575 nm. The galaxy is moving away from Earth at 4.5 ×107 m/s. An astronomer wants to identify the element by finding the wavelength of the line when emitted.
Speed of light = 3.0 ×108 m/s

(a)

Show that v/c for this galaxy is 0.15.

(1)
(b)

Calculate the emitted wavelength, λ0, of the spectral line.

(3)
(c)

A student calculates Δλ = 0.15 × 575 nm. Explain why this method is incorrect.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
4.5 ×107 / 3.0 ×108 = 0.15✓ 1
Part (b)
Observed wavelength = λ0 + Δλ = λ0 + 0.15λ0 = 1.15λ0✓ 1
λ0 = 575 / 1.15✓ 1
λ0 = 500 nm✓ 1
Part (c)
Δλ is a fraction of the emitted wavelength λ0, not of the observed wavelength✓ 1

Common error: Multiplying the observed wavelength by v/c gives about 489 nm, which is wrong. The ratio v/c always applies to the emitted wavelength.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
177IGPE2-061-02-138(d) Cosmology Paper 2PHard6 marks

An engineer designs a spectrometer that can detect a change in wavelength as small as 0.020 nm. It is used to measure a spectral line with an emitted wavelength of 500 nm.
Speed of light = 3.0 ×108 m/s

(a)

Calculate the smallest speed of a moving source that this spectrometer can detect using this line.

(3)
(b)

Light from the nearby Andromeda galaxy is blue-shifted. State what this tells us about the motion of Andromeda.

(1)
(c)

Suggest why the blue-shift of Andromeda does not disprove the idea that the universe is expanding.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = (Δλ/λ0) × c✓ 1
v = (0.020/500) × 3.0 ×108✓ 1
v = 1.2 ×104 m/s✓ 1
Part (b)
Andromeda is moving towards Earth / the Milky Way✓ 1
Part (c)
Andromeda is close to the Milky Way, so gravitational attraction between the two galaxies pulls them together✓ 1
Almost all distant galaxies are red-shifted, so the overall pattern still shows expansion✓ 1

Exam tip: The expansion trend comes from many distant galaxies. One nearby exception can be explained by local gravity and does not overturn the overall evidence.

Theory: How are stars, galaxies and the universe related? Single · Double · Physics
178IGPE2-061-00-118(d) Cosmology Paper 2PHard6 marks

A science museum is preparing a display about the origin of the universe. The display shows spectra of light from galaxies at different distances from Earth, next to the spectrum of the same elements measured in a laboratory.

(a)

Explain how observations of the light from distant galaxies provide evidence for the Big Bang theory.

(6)
Show mark scheme
Marking pointMarkNotes
Part (a)
• Spectral lines from galaxies appear at longer wavelengths (lower frequencies) than the same lines measured in a laboratory✓ 1
• This red-shift shows that the galaxies are moving away from us✓ 1
• (Almost) all distant galaxies show red-shift, whichever direction we look✓ 1
• More distant galaxies show a greater red-shift, so they are receding faster✓ 1
• This means the universe is expanding✓ 1
• Tracing the expansion back in time shows the universe began from a very small, very hot, dense region: the Big Bang✓ 1

Exam tip: Build a logical chain: observation (wavelength increased), then meaning (moving away), then pattern (further away means faster), then conclusion (expanding, so it started from a tiny, hot, dense state).

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
179IGPE3-8-15r19-148(d) Cosmology Paper 2PHard9 marks

A hydrogen spectral line has a wavelength of 650 nm in a laboratory. In light from a distant galaxy the line is observed at 656.5 nm. The galaxy is 1.30 × 1024 m from the Earth. The speed of light is 3.0 × 108 m/s. Use 1 year = 3.15 × 107 s.

(a)

Calculate the speed at which the galaxy is moving away from the Earth.

(3)
(b)

Assume that the galaxy has always moved at this speed. Calculate the time it has taken to travel from the Earth's position to its present distance. Give your answer to 3 significant figures in seconds.

(2)
(c)

Convert your answer to (b) into years. Use the unrounded value from (b) and give your answer to 3 significant figures.

(2)
(d)

State the assumption made in this calculation and state what the time calculated is an estimate of.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
change in wavelength = 6.5 nm✓ 1
use of speed = (change in wavelength ÷ original wavelength) × speed of light✓ 1
3.0 × 106 m/s✓ 1
Part (b)
use of time = distance ÷ speed✓ 1
4.33 × 1017 s✓ 1
Part (c)
divide by 3.15 × 107 s✓ 1
1.38 × 1010 years✓ 1
Part (d)
the speed of the galaxy has been constant (accept: the galaxy has always moved at the same speed)✓ 1
an estimate of the age of the universe (the time since the Big Bang)✓ 1

Exam tip: Keep the unrounded answer from one part on your calculator when it is needed in the next part.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
180IGPE3-8-15r41-088(d) Cosmology Paper 2PHard9 marks

Light from a spiral galaxy is analysed. A particular spectral line has a wavelength of 600 nm when measured in the laboratory. The table gives the wavelength of this line in the light from three places in the galaxy.
Centre of the galaxy: 606.0 nm
Left-hand edge: 606.6 nm
Right-hand edge: 605.4 nm
Speed of light = 3.0 × 108 m/s

(a)

State what the wavelength measured at the centre of the galaxy shows about the motion of the galaxy relative to the Earth.

(1)
(b)

Calculate the speed at which the centre of the galaxy is moving relative to the Earth.

(3)
(c)

Explain why the line from the left-hand edge has a longer wavelength than the line from the right-hand edge.

(2)
(d)

Determine the speed at which the left-hand edge moves around the centre of the galaxy.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
the galaxy is moving away from the Earth (the wavelength has increased, red-shift)✓ 1
Part (b)
M1 change in wavelength ÷ reference wavelength = speed of galaxy ÷ speed of light✓ 1
M1 (6.0 ÷ 600) = v ÷ (3.0 × 108)✓ 1
A1 3.0 × 106 m/s✓ 1
Part (c)
the galaxy is rotating, so the left-hand edge moves away from the Earth faster than the centre does✓ 1
the right-hand edge moves around towards the Earth, which reduces its recession, so its increase in wavelength is smaller✓ 1
Part (d)
M1 extra wavelength change due to rotation = 606.6 − 606.0 = 0.6 nm✓ 1
M1 v = (0.6 ÷ 600) × 3.0 × 108✓ 1
A1 3.0 × 105 m/s (300 000 m/s)✓ 1
Accept finding the speed of the edge as 3.3 × 106 m/s and subtracting 3.0 × 106 m/s from (b).—

Common error: A rotating galaxy gives a spread of wavelengths: subtract the motion of the whole galaxy before finding how fast the edges rotate.

Theory: How do you use \dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}? Physics only · 2P
181IGPE3-8-15r41-138(d) Cosmology Paper 2PHard8 marks

Light from two distant galaxies, P and Q, contains a spectral line that has a wavelength of 500 nm in the laboratory. In the light from galaxy P the line has a wavelength of 501.5 nm. The speed of light is 3.0 × 108 m/s.

(a)

Calculate the speed at which galaxy P is moving away from the Earth.

(3)
(b)

Galaxy Q is moving away from the Earth at 2.7 × 106 m/s. Calculate the wavelength of the line in the light from galaxy Q.

(3)
(c)

Galaxy Q is moving away three times as fast as galaxy P. Explain why astronomers conclude that galaxy Q is probably further from the Earth than galaxy P.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 change in wavelength ÷ reference wavelength = speed of galaxy ÷ speed of light✓ 1
M1 (1.5 ÷ 500) × 3.0 × 108✓ 1
A1 9.0 × 105 m/s✓ 1
Part (b)
M1 change in wavelength ÷ 500 = (2.7 × 106) ÷ (3.0 × 108)✓ 1
M1 change in wavelength = 0.009 × 500 = 4.5 nm✓ 1
A1 wavelength = 500 + 4.5 = 504.5 nm✓ 1
Part (c)
more distant galaxies are observed to move away faster (have greater red-shift)✓ 1
this is because the Universe (space) is expanding, so the greater the distance the faster the recession✓ 1

Common error: The final answer is a wavelength, not a change in wavelength; add the change to the laboratory value.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
182IGPE3-8-15r42-128(b) Motion in the Universe Paper 2PHard7 marks

Mars orbits the Sun in a nearly circular path of radius 2.28 × 1011 m at an average orbital speed of 2.41 × 104 m/s. One Earth day is 86 400 s.

Marsr = 2.28 × 10¹¹ m2.41 × 10⁴ m/sSun
Not to scale
(a)

Calculate the time Mars takes to complete one orbit of the Sun. Give your answer in seconds to 3 significant figures.

(3)
(b)

Convert your unrounded answer to part (a) into Earth days. Give your answer to 3 significant figures.

(2)
(c)

An Earth year is 365 days. Calculate how many days longer a Martian year is than an Earth year. Use the unrounded value from part (b) and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 distance for one orbit = 2πr✓ 1
M1 time = distance ÷ speed✓ 1
A1 5.94 × 107 s (unrounded 5.944 × 107 s)✓ 1
Part (b)
M1 5.944 × 107 ÷ 86 400✓ 1
A1 688 days✓ 1
Part (c)
M1 687.99 − 365✓ 1
A1 323 days✓ 1

Exam tip: Keep the unrounded value on your calculator between parts so that the final answer is not affected by early rounding.

Theory: How do you calculate orbital speed? Double · Physics
183IGPE3-8-15r42-148(b) Motion in the Universe Paper 2PHard6 marks

A communications satellite used by airlines is placed in a geostationary orbit. It travels at an orbital speed of 3.07 × 103 m/s.

satelliter = ?3.07 × 10³ m/sh = ?Earth
Not to scale (radius of the Earth = 6.4 × 10⁶ m)
(a)

State why the time for one orbit of a geostationary satellite is 24 hours.

(1)
(b)

Calculate the radius of the orbit, measured from the centre of the Earth. Give your answer to 3 significant figures.

(3)
(c)

The radius of the Earth is 6.4 × 106 m. Calculate the height of the satellite above the surface of the Earth. Use the unrounded value from part (b) and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the satellite must take the same time as the Earth takes to rotate once, so it stays above the same point on the Earth✓ 1
Part (b)
M1 rearrange v = 2πr ÷ T to r = vT ÷ 2π✓ 1
M1 T = 24 × 3600 = 86 400 s✓ 1
A1 4.22 × 107 m (unrounded 4.2216 × 107 m)✓ 1
Part (c)
M1 4.2216 × 107 − 6.4 × 106✓ 1
A1 3.58 × 107 m (ECF from (b))✓ 1

Exam tip: Convert hours to seconds before using the orbital speed equation.

Theory: How do you calculate orbital speed? Double · Physics
184IGPE3-8-17r46-138(d) Cosmology Paper 2PHard8 marks

The cosmic microwave background radiation (CMB) is detected from all directions in space. When the radiation was released, its peak wavelength was about 1.0 × 10−6 m. Today its peak wavelength is about 1.0 × 10−3 m. The speed of electromagnetic waves is 3.0 × 108 m/s.

(a)

State the region of the electromagnetic spectrum in which the CMB is detected today.

(1)
(b)

Calculate the factor by which the peak wavelength has increased.

(2)
(c)

Explain how the change in wavelength is evidence that the universe has expanded.

(2)
(d)

Calculate the peak frequency of the CMB today.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
microwave (region)✓ 1
Part (b)
M1 1.0 × 10−3 ÷ 1.0 × 10−6✓ 1
A1 1000✓ 1
Part (c)
the wavelength of the radiation has been stretched✓ 1
space itself expanded while the radiation travelled, so the universe is now much larger than when it was released✓ 1
Part (d)
M1 wave speed = frequency × wavelength, or f = v ÷ λ✓ 1
M1 3.0 × 108 ÷ 1.0 × 10−3✓ 1
A1 3.0 × 1011 Hz✓ 1

Exam tip: Give the answer to (d) with a unit and in standard form, since the numbers are very large.

Theory: What is the Big Bang theory, and what is the evidence for it? Physics only · 2P
185IGPE3-8-16r48-098(d) Cosmology Paper 2PHard10 marks

Four galaxies are observed. The table shows the distance of each from Earth and the speed at which each is moving away. The data lie on a straight line through the origin.

GalaxyDistance / 1023 mSpeed / 105 m/s
P1.02.2
Q2.55.5
R4.08.8
S6.013.2
The speed of light is 3.0 × 108 m/s. 1 year = 3.15 × 107 s.

(a)

Use the data for one galaxy to determine the Hubble constant H0 = speed ÷ distance, in s−1.

(3)
(b)

Estimate the age of the universe in years, using age = 1 ÷ H0. Give your answer to 2 significant figures.

(3)
(c)

State one assumption made in the estimate in (b).

(1)
(d)

Another galaxy has a red-shift z = 0.011. Calculate its distance from Earth in metres. Give your answer to 2 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 H0 = v/d with values from one galaxy✓ 1
M1 converts powers of ten, e.g. 5.5 × 105 ÷ 2.5 × 1023✓ 1
A1 2.2 × 10−18 s−1✓ 1
Part (b)
M1 age = 1 ÷ 2.2 × 10−18 = 4.5 × 1017 s (ECF from (a))✓ 1
M1 divides by 3.15 × 107 s per year✓ 1
A1 1.4 × 1010 years✓ 1
Part (c)
The universe has expanded at a constant rate / galaxies have moved at constant speed since the start✓ 1
Part (d)
M1 v = zc = 0.011 × 3.0 × 108 = 3.3 × 106 m/s✓ 1
M1 d = v ÷ H0 = 3.3 × 106 ÷ 2.2 × 10−18✓ 1
A1 1.5 × 1024 m (ECF from (a))✓ 1

Common error: Convert the table headings (10²³ m, 10⁵ m/s) to full numbers before dividing, or the power of ten in H₀ will be wrong.

Theory: How does red-shift depend on a galaxy's distance? Physics only · 2P
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