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Cambridge IGCSE Physics 0625 · Core and Extended · Topic 6
133 free, high-quality past-paper-style questions on Space physics for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Stars, galaxies and the expanding Universe; Earth, Moon and the Solar System. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which statement correctly describes a light-year?
| Marking point | Mark | Notes |
|---|---|---|
| A light-year is a unit of distance. It is the distance light travels through space in one year. It is not a time, despite the word 'year', and it is not a speed. | ✓ 1 | Answer A |
Answer: A
Common error: The word 'year' tempts many students to call a light-year a unit of time. It measures distance.
The Sun is a star.
Which two elements make up most of the Sun?
| Marking point | Mark | Notes |
|---|---|---|
| The Sun consists mostly of hydrogen and helium. Hydrogen nuclei fuse in its core to form helium. Oxygen and nitrogen make up most of the Earth's atmosphere. Iron and nickel make up most of the Earth's core. | ✓ 1 | Answer C |
Answer: C
Exam tip: Link the Sun's composition to its energy source: hydrogen is the fuel for fusion and helium is the product.
Which statement about the Milky Way is correct?
| Marking point | Mark | Notes |
|---|---|---|
| The Milky Way is our galaxy. It contains many billions of stars, and the Sun is one of them. The Universe contains many billions of galaxies. The Solar System is only the Sun and the objects orbiting it, and the Sun is not at the centre of the galaxy. | ✓ 1 | Answer C |
Answer: C
Common error: Do not mix up the Solar System (one star and its planets) with a galaxy (billions of stars).
Light from a distant galaxy is observed on Earth to be redshifted.
What does this mean?
| Marking point | Mark | Notes |
|---|---|---|
| Redshift is an increase in the observed wavelength of electromagnetic radiation from a source that is moving away (receding). The wavelength is shifted towards the red, long-wavelength end of the spectrum. A source moving towards us would show a decrease in wavelength. | ✓ 1 | Answer B |
Answer: B
Exam tip: Redshift has two parts: the observed wavelength increases, and the source is receding. Include both in a definition.
Astronomers studying the Solar System describe the Sun as an ordinary, mid-sized star.
Name the two elements that account for most of the mass of the Sun.
Almost all of the energy emitted by the Sun is carried by electromagnetic waves from three neighbouring regions of the spectrum. Name these three regions.
State the type of nuclear reaction that releases energy in the core of the Sun, and name the element that acts as its fuel.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| hydrogen | ✓ 1 | |
| helium | ✓ 1 | |
| Part (b) | ||
| infrared | ✓ 1 | |
| visible light (accept 'light') | ✓ 1 | |
| ultraviolet | ✓ 1 | |
| Part (c) | ||
| nuclear fusion | ✓ 1 | |
| hydrogen (nuclei fuse to form helium) | ✓ 1 | |
Common error: Fission and fusion are easily confused. Stars release energy by fusion, which joins light nuclei such as hydrogen together. Fission splits heavy nuclei.
What causes day and night on the Earth?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth spins on its axis once in about 24 hours. The half of the Earth facing the Sun has day and the half facing away has night. The yearly orbit causes the seasons, not day and night. The Sun does not orbit the Earth; it only appears to move across the sky. The Moon passing in front of the Sun causes an occasional solar eclipse, not night. | ✓ 1 | Answer B |
Answer: B
Common error: Do not mix up rotation (spinning on the axis, 24 hours, day and night) with orbit (travelling around the Sun, one year).
About how long does the Moon take to complete one orbit of the Earth?
| Marking point | Mark | Notes |
|---|---|---|
| The Moon orbits the Earth in about one month (roughly 27 to 29 days), and it goes through a full cycle of phases in that time. One day is the Earth's rotation period. One year is the Earth's orbital period around the Sun. One week is roughly the time between one main phase and the next, not a full orbit. | ✓ 1 | Answer C |
Answer: C
Exam tip: Learn three periods together: Earth's rotation is about 24 hours, the Moon's orbit is about 1 month and the Earth's orbit is about 365 days.
Why does the Earth have seasons?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth's axis is tilted. As the Earth orbits the Sun, each hemisphere spends part of the year tilted towards the Sun. During that time it has longer days and receives more direct sunlight, which gives summer. The Earth's orbit is nearly circular, so the change in distance is too small to cause seasons. Also, the two hemispheres have opposite seasons at the same time, which a change in distance could not explain. | ✓ 1 | Answer C |
Answer: C
Common error: Many students say summer happens because the Earth is closer to the Sun. If that were true, both hemispheres would have summer at the same time.
The average distance from the Sun to the Earth is 1.5×1011 m. The speed of light is 3.0×108 m/s.
About how long does light take to travel from the Sun to the Earth?
| Marking point | Mark | Notes |
|---|---|---|
| time = distance ÷ speed = 1.5×1011 ÷ 3.0×108 = 500 s, which is a little over 8 minutes. 50 s and 5000 s come from mistakes with the powers of ten. 4.5×1019 s comes from multiplying distance by speed instead of dividing. | ✓ 1 | Answer B |
Answer: B
Common error: Rearrange speed = distance ÷ time to get time = distance ÷ speed. Never multiply the distance by the speed.
A rover of mass 900 kg is sent to Mars. The gravitational field strength on the surface of Mars is 3.7 N/kg. The gravitational field strength on the surface of the Earth is 9.8 N/kg.
Calculate the weight of the rover on the Earth.
Calculate the weight of the rover on Mars.
Suggest why the gravitational field strength on the surface of Mars is smaller than on the surface of the Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg = 900 × 9.8 | ✓ 1 | |
| = 8800 N (8820 N) | ✓ 1 | |
| Part (b) | ||
| 900 × 3.7 = 3300 N (3330 N) | ✓ 1 | |
| Part (c) | ||
| Mars has a smaller mass than the Earth | ✓ 1 | |
Common error: The rover's mass is still 900 kg on Mars. Only its weight changes, because weight depends on the gravitational field strength.
The Solar System is made up of the Sun and many objects that orbit it.
State the number of planets in the Solar System.
Name two types of object, other than planets, that orbit the Sun directly.
State the force that keeps the planets in orbit around the Sun.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| eight (8) | ✓ 1 | |
| Part (b) | ||
| One mark for each different type, from: dwarf planets (e.g. Pluto); asteroids; comets . Accept 'minor planets' as one type, but not together with dwarf planets or asteroids. Do not credit moons or natural satellites, because they orbit planets. | ✓ 2 | |
| Part (c) | ||
| the gravitational force (attraction) of the Sun on the planets / gravity | ✓ 1 | |
Common error: Moons are not a correct answer to 'orbit the Sun' because they orbit planets. For the force, magnetism and 'centripetal force' on its own do not earn the mark: name gravity as the force involved.
During an evening cricket match in Sydney the floodlights are switched on as the Sun sets for the spectators. What causes the Sun to appear to set?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth spins once in about 24 hours, so a place turns away from the Sun and it appears to set: A is correct. B is the cause of the year, not the day. C is the old Earth-centred idea. D describes an eclipse, which is rare and short. | ✓ 1 | Answer A |
Answer: A
Common error: Day and night come from the Earth's spin; the seasons come from its tilt and orbit.
Why is it colder in the UK in December than in June?
| Marking point | Mark | Notes |
|---|---|---|
| The axis is tilted about 23.5°, so in December the northern hemisphere leans away from the Sun: days are shorter and the Sun's rays strike at a shallow angle: C is correct. A is the most common misconception; the Earth's distance varies only slightly and is least in early January. B is false because the Sun's output is steady. D confuses the Moon with the Sun's effect. | ✓ 1 | Answer C |
Answer: C
Common error: Seasons are caused by the tilt of the Earth's axis, not by changes in its distance from the Sun.
An astronaut of mass 75 kg stands on the Moon, where the gravitational field strength is 1.6 N/kg. On Earth (9.8 N/kg) the astronaut's weight is 735 N. Which row gives the astronaut's mass and weight on the Moon?
| Marking point | Mark | Notes |
|---|---|---|
| Mass does not depend on location, so it stays 75 kg. Weight = mass × g = 75 × 1.6 = 120 N: A is correct. B wrongly says mass falls. C says weight is unchanged. D divides mass by g (75 ÷ 1.6) instead of multiplying. | ✓ 1 | Answer A |
Answer: A
Common error: Mass stays the same everywhere; weight changes with the local gravitational field strength.
The Solar System contains planets, dwarf planets, asteroids and comets.
State where the main asteroid belt is found. [1]
Describe two differences between the orbit of a comet and the orbit of a planet. [2]
State how the speed of a comet changes as it moves towards the Sun, and explain why. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| between the orbits of Mars and Jupiter | ✓ 1 | |
| Part (b) | ||
| a comet's orbit is a long, stretched ellipse whereas a planet's orbit is almost circular | ✓ 1 | |
| a comet's distance from the Sun varies hugely and takes it far beyond the planets | ✓ 1 | |
| Part (c) | ||
| its speed increases | ✓ 1 | |
| the Sun's gravitational pull does work on it, so it gains kinetic energy (the field is stronger nearer the Sun) | ✓ 1 | |
Common error: Planets move in nearly circular orbits; comets follow very elongated ellipses.
The Earth rotates about its axis and orbits the Sun.
State the time taken for the Earth to rotate once about its axis. [1]
State the time taken for the Earth to orbit the Sun once, in days. [1]
Explain why, in June, places in the northern hemisphere have longer hours of daylight than places in the southern hemisphere. [3]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 hours | ✓ 1 | |
| Part (b) | ||
| about 365 days | ✓ 1 | |
| Part (c) | ||
| the Earth's axis is tilted (about 23.5°) relative to its orbit | ✓ 1 | |
| in June the northern hemisphere is tilted towards the Sun | ✓ 1 | |
| so more than half of each day's path of a northern place is in sunlight | ✓ 1 | |
Exam tip: Mention the tilt of the axis in any seasons answer and say which hemisphere leans towards the Sun.
Stars such as the Sun are formed in space.
State what a star forms from. [1]
Describe how the cloud becomes a stable star. [3]
Name the process that releases energy in a stable star. [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a cloud of gas and dust (a nebula) | ✓ 1 | |
| Part (b) | ||
| gravity pulls the gas and dust together | ✓ 1 | |
| the particles gain kinetic energy as it contracts, so the temperature rises | ✓ 1 | |
| when the core is hot enough hydrogen nuclei fuse, and the outward pressure balances gravity so the star is stable | ✓ 1 | |
| Part (c) | ||
| nuclear fusion (of hydrogen into helium) | ✓ 1 | |
Common error: Stars shine because of nuclear fusion, not burning in the chemical sense.
A student sorts four astronomical objects by size. Which list is in order from smallest to largest?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: the Moon is smaller than the Earth, the Earth is far smaller than the Sun (a star), and the Sun is one star among billions in the Milky Way galaxy. B puts the Earth before the Moon, which reverses their sizes. C and D put the Sun above the Milky Way, wrongly treating a star as larger than the galaxy that contains it. | ✓ 1 | Answer A |
Answer: A
Common error: A galaxy is a collection of billions of stars, so it is always larger than any single star or planet in it.
Every day a city has a period of daylight followed by a period of darkness. What is the cause?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: as the Earth spins, a given place moves from the side facing the Sun to the side facing away. B is wrong because one orbit of the Sun takes about 365 days. C is wrong because the Sun does not orbit the Earth. D is wrong because the Moon takes about 27 days to orbit the Earth. | ✓ 1 | Answer A |
Answer: A
Common error: Do not mix up rotation (spinning on an axis, 24 hours) with orbit (travelling round another body, about 365 days for the Earth).
An astronaut has a mass of 70 kg on the Earth, where the gravitational field strength is 9.8 N/kg. On the Moon the gravitational field strength is 1.6 N/kg. Which row gives the astronaut's mass and weight on the Moon?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: mass does not depend on location, so it stays 70 kg, and weight = 70 × 1.6 = 112 N. B wrongly makes mass change with location. C keeps the Earth weight (70 × 9.8). D changes the mass to 429 kg and also uses the Earth weight. | ✓ 1 | Answer A |
Answer: A
Common error: Mass (kg) is the same everywhere; only weight (N) depends on gravitational field strength.
A star is 8.0 light-years from the Earth. What does this statement mean?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: a light-year is the distance light travels in one year, so light from the star needs 8.0 years to arrive. B, C and D treat the light-year as a unit of time for an orbit, a spacecraft journey or the star's age, but it is a unit of distance. | ✓ 1 | Answer A |
Answer: A
Common error: A light-year is a unit of distance, not of time.
A shop owner in Lisbon notices that the shop window is in sunlight for only part of each day.
State the time taken for the Earth to rotate once about its axis.
State the approximate time taken for the Earth to orbit the Sun once.
Explain why the shop window is in darkness for part of each day.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 hours | ✓ 1 | |
| Part (b) | ||
| 365 days (about one year) | ✓ 1 | |
| Part (c) | ||
| the Earth rotates about its axis | ✓ 1 | |
| only half of the Earth faces the Sun at any time, so the window faces away from the Sun for part of each rotation | ✓ 1 | |
Common error: Day and night are caused by the Earth's spin, not by its orbit round the Sun.
A coastal town is building a tidal power station. The tides are caused mainly by the gravitational force of the Moon on the oceans. The Moon takes 27.3 days to orbit the Earth.
State the type of force that the Moon exerts on the oceans.
State the name given to a body, such as the Moon, that orbits a planet.
Calculate the number of complete orbits the Moon makes round the Earth in 365 days.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gravitational force (attraction) | ✓ 1 | |
| Part (b) | ||
| moon / natural satellite | ✓ 1 | |
| Part (c) | ||
| 365 / 27.3 = 13.4 | ✓ 1 | |
| 13 complete orbits | ✓ 1 | |
Exam tip: When asked for complete orbits, round down to a whole number after the division.
It is winter in the northern hemisphere in December. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: Earth's axis is tilted at about 23.5°, so in December the northern hemisphere leans away from the Sun. A is wrong: the seasons are caused by tilt, not by distance. C is wrong: the Moon does not cause the seasons. D is wrong: Earth's rotation period stays at 24 hours all year. | ✓ 1 | Answer B |
Answer: B
Common error: Explaining the seasons by Earth's distance from the Sun. The cause is the tilt of the axis.
A theatre company holds a themed performance each time the Moon completes one orbit of Earth. Approximately how long is this between performances?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: the Moon takes about 27 days, roughly one month, to orbit Earth. A is the time Earth takes to rotate once. B is wrong; one week is only about a quarter of the orbit time. D is the time for Earth to orbit the Sun. | ✓ 1 | Answer C |
Answer: C
Exam tip: Learn the three basic periods: Earth's rotation (24 hours), Moon's orbit (about a month) and Earth's orbit (about a year).
Which list shows these objects in order of increasing size?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: the Earth is a planet inside the Solar System, which is a tiny part of the Milky Way galaxy, which is one of many galaxies in the Universe. B swaps the Solar System and the Milky Way. C puts Earth larger than the Solar System. D puts the Milky Way larger than the Universe. | ✓ 1 | Answer A |
Answer: A
Exam tip: Remember the order as planet, Solar System, galaxy, Universe.
A planetarium show describes how stars form and change.
Name the cloud of gas and dust from which stars form.
Describe the balance of forces that keeps the Sun stable.
State what happens to hydrogen nuclei in the core of the Sun.
Name the final stage in the life of the Sun, after it has been a red giant.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 nebula | ✓ 1 | |
| Part (b) | ||
| B1 gravity pulls the material inwards | ✓ 1 | |
| B1 outward pressure from the hot gas and radiation produced by fusion balances the inward pull | ✓ 1 | |
| Part (c) | ||
| B1 they fuse (join together) to form helium nuclei | ✓ 1 | |
| B1 energy is released | ✓ 1 | |
| Part (d) | ||
| B1 white dwarf | ✓ 1 | |
Exam tip: 'Stable star' answers need two opposing effects: gravity inwards and pressure outwards.
Proxima Centauri is a star that is 4.2 light-years from the Sun. The speed of light is 3.00 × 108 m/s and one year is 3.15 × 107 s.
State what is meant by a light-year.
Calculate the distance from the Sun to Proxima Centauri in metres. Give your answer to 3 significant figures.
A radio signal travels at the speed of light. State the time taken for a signal to travel from the Sun to Proxima Centauri.
The Milky Way is a galaxy. State what a galaxy contains.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the distance travelled by light in one year | ✓ 1 | |
| Part (b) | ||
| M1 1 light-year = 3.00 × 108 × 3.15 × 107 = 9.45 × 1015 m | ✓ 1 | |
| M1 distance = 4.2 × 9.45 × 1015 | ✓ 1 | |
| A1 3.97 × 1016 m | ✓ 1 | |
| Part (c) | ||
| B1 4.2 years | ✓ 1 | |
| Part (d) | ||
| B1 billions of stars (with gas and dust), held together by gravity | ✓ 1 | |
Common error: Treating a light-year as a unit of time. It is a unit of distance.
Which statement correctly explains why there are seasons on Earth?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: because the axis is tilted, each hemisphere is tilted towards the Sun for part of the year, giving more direct sunlight and longer days. A is wrong: Earth's distance from the Sun varies only slightly and does not cause the seasons. B is wrong: the Moon does not change Earth's distance from the Sun. D is wrong: Earth's rotation rate is constant at about one rotation per 24 hours. | ✓ 1 | Answer C |
Answer: C
Common error: Do not blame the seasons on Earth being nearer to or further from the Sun; the cause is the tilt of the axis.
Which statement about the Moon is correct?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: the Moon's orbital period about Earth is about 27 to 28 days, roughly one month. A is wrong: the Moon only reflects sunlight. C is wrong: 24 hours is the time for Earth to rotate once on its axis. D is wrong: gravitational field strength depends on mass, and the Moon's is about 1.6 N/kg. | ✓ 1 | Answer B |
Answer: B
Common error: Keep the Earth's rotation time (24 hours) separate from the Moon's orbital time (about a month).
Star X is 6 light-years from Earth. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: a light-year is the distance light travels in one year, so light from X takes 6 years to reach us. B is wrong: a light-year is a distance, not an orbital time. C is wrong: that is the distance light travels in 6 seconds, not 6 years. D is wrong: a light-year is defined by the travel of light, not of Earth. | ✓ 1 | Answer A |
Answer: A
Common error: A light-year is a unit of distance, even though the word "year" appears in it.
The Earth rotates on its axis and orbits the Sun, and the Moon orbits the Earth.
Explain why it is warmer in the northern hemisphere in June than in December.
Explain why the Moon appears to change shape during one month.
The Earth rotates through 360° in 24 hours. Calculate the angle through which the Earth rotates in 5.0 hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The Earth's axis is tilted (relative to its orbit) | ✓ 1 | |
| in June the northern hemisphere is tilted towards the Sun, so the Sun's rays are more direct / days are longer | ✓ 1 | |
| Part (b) | ||
| Half of the Moon is always lit by the Sun and the Moon only reflects sunlight | ✓ 1 | |
| as the Moon orbits the Earth we see different amounts of the lit half | ✓ 1 | |
| Part (c) | ||
| M1 angle = 360 × 5.0 / 24 | ✓ 1 | |
| A1 75° | ✓ 1 | |
Exam tip: For an explanation of the seasons, mention both the tilt of the axis and the effect on the Sun's rays or the day length.
Proxima Centauri is the nearest star to the Sun after the Sun itself.
State what is meant by a galaxy.
Name the galaxy that contains the Sun.
Proxima Centauri is 4.2 light-years from the Earth. Calculate this distance in metres. The speed of light is 3.0 × 108 m/s and 1 year = 3.15 × 107 s. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A very large collection of stars (with gas and dust) held together by gravity | ✓ 1 | |
| Part (b) | ||
| The Milky Way | ✓ 1 | |
| Part (c) | ||
| M1 1 light-year = 3.0 × 108 × 3.15 × 107 = 9.45 × 1015 m | ✓ 1 | |
| M1 distance = 4.2 × 9.45 × 1015 = 3.969 × 1016 m | ✓ 1 | |
| A1 4.0 × 1016 m | ✓ 1 | |
Common error: A light-year is the distance light travels in one year, so multiply the speed of light by a time in seconds.
A marine biologist records the tide times on a coastal research station and notes that the Moon is visible at night. Which statement explains why the Moon can be seen from Earth?
| Marking point | Mark | Notes |
|---|---|---|
| The Moon is a non-luminous body; it is seen because sunlight is reflected from its surface. It has no fusion (that happens only in stars), it does not rely on light from the oceans, and its surface is not hot enough to glow visibly. | ✓ 1 | Answer B |
Answer: B
Common error: Planets and moons are seen by reflected light; only stars are luminous.
A star is 8.6 light-years from the Earth. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A light-year is a distance, the distance light travels in one year, so light from a star 8.6 light-years away took 8.6 years to arrive. A light-year says nothing about the star's age, no spacecraft travels at the speed of light, and 8.6 seconds is far too short. | ✓ 1 | Answer A |
Answer: A
Common error: A light-year is a unit of distance, not of time.
An architect designing a planetarium dome labels a star that is 6.8 light-years from the Earth. Speed of light = 3.00 × 108 m/s. 1 year = 3.16 × 107 s.
State what is meant by a light-year.
Calculate the distance to the star in metres. Give your answer to 3 significant figures.
The Sun is 1.5 × 1011 m from the Earth. Calculate the time taken for sunlight to reach the Earth, in minutes, to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The distance travelled by light in a vacuum in one year | ✓ 1 | |
| Part (b) | ||
| M1 distance in one light-year = 3.00 × 108 × 3.16 × 107 = 9.48 × 1015 m | ✓ 1 | |
| M1 multiply by 6.8 | ✓ 1 | |
| A1 distance = 6.45 × 1016 m | ✓ 1 | |
| Part (c) | ||
| M1 time = distance / speed = 1.5 × 1011 / 3.00 × 108 = 500 s | ✓ 1 | |
| A1 time = 8.3 minutes | ✓ 1 | |
Exam tip: Convert seconds to minutes at the end by dividing by 60.
An architect designs a lunar habitat module of mass 850 kg. The gravitational field strength is 9.8 N/kg on the Earth and 1.6 N/kg on the Moon.
Calculate the weight of the module on the Earth.
State what happens to the mass of the module when it is taken to the Moon.
Suggest why the gravitational field strength on the Moon is less than on the Earth.
Calculate the weight of the module on the Moon.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = mass × gravitational field strength = 850 × 9.8 | ✓ 1 | |
| A1 weight = 8330 N | ✓ 1 | |
| Part (b) | ||
| It stays the same (850 kg) | ✓ 1 | |
| Part (c) | ||
| The Moon has a smaller mass than the Earth | ✓ 1 | |
| Part (d) | ||
| M1 weight = 850 × 1.6 | ✓ 1 | |
| A1 weight = 1360 N | ✓ 1 | |
Common error: Mass is measured in kg and does not change with location; weight is a force in N and does.
A volcano observatory in northern Iceland records almost continuous daylight in June but only a few hours of daylight in December. What is the cause of this difference?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth's axis is tilted, so in June the northern hemisphere leans towards the Sun and has long days, and in December it leans away and has short days. The Earth's distance from the Sun changes only slightly and is not the cause. The rotation rate is constant at one turn in 24 hours. The Moon does not block sunlight for months at a time. | ✓ 1 | Answer B |
Answer: B
Common error: Seasons and day length are caused by the tilt of the Earth's axis, not by changes in distance from the Sun.
Which list shows these objects in order of increasing size?
| Marking point | Mark | Notes |
|---|---|---|
| The Moon is smaller than the Earth, the Earth is far smaller than the Sun, the Sun is one star among billions in the Milky Way galaxy, and the Milky Way is one of a huge number of galaxies in the Universe. The other lists swap the Moon and Earth, the Sun and Earth, or the Sun and the galaxy, or place the Universe before the galaxy. | ✓ 1 | Answer A |
Answer: A
Exam tip: Remember the chain: moon, planet, star, galaxy, Universe, with each item far larger than the one before.
The same side of the Moon always faces the Earth. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| If the Moon rotates exactly once on its axis during each orbit of the Earth, it keeps turning at the same rate as its position changes, so one face always points towards the Earth. If it did not rotate at all, we would see all sides over each orbit. A 24-hour turn would show every side during a month. The Moon does move: it orbits the Earth about every 27 days. | ✓ 1 | Answer D |
Answer: D
Common error: The Moon does rotate, but its rotation period equals its orbital period.
A hammer-throw ball has a mass of 4.0 kg. The gravitational field strength at the surface of the Moon is 1.6 N/kg.
What is the weight of the ball on the Moon?
| Marking point | Mark | Notes |
|---|---|---|
| Weight = mass × gravitational field strength = 4.0 × 1.6 = 6.4 N. 2.5 N comes from dividing mass by field strength. 39 N uses the field strength of the Earth (9.8 N/kg) instead of that of the Moon. 4.0 N is the mass in kg, not a weight in newtons. | ✓ 1 | Answer B |
Answer: B
Common error: Mass is measured in kg and stays the same everywhere; weight is a force in newtons that depends on the local gravitational field strength.
Which list shows the objects in order of increasing size?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth is a planet within the Solar System, the Solar System lies within the Milky Way galaxy, and the Milky Way is one of very many galaxies in the Universe, so this is the correct increasing order. The other lists place the Solar System before the Earth, put a galaxy below the Solar System, or start with the galaxy. | ✓ 1 | Answer D |
Answer: D
Exam tip: Remember the nesting: planet, then planetary system, then galaxy, then Universe.
A star is 12 light-years from the Earth. Which statement about this star is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A light-year is the distance travelled by light in one year, so a star 12 light-years away is a distance that light covers in 12 years. A light-year is a unit of distance, so it says nothing about the age of the star or the time it takes to orbit the galaxy, and 12 light-years is vastly more than 12 times the Earth-Sun distance. | ✓ 1 | Answer A |
Answer: A
Common error: A light-year is a unit of distance, not a unit of time.
A student studies the Sun and nearby stars. Light travels at 3.00 × 108 m/s and 1 year = 3.16 × 107 s.
State what happens to hydrogen nuclei in the core of the Sun to release energy.
Calculate the distance, in metres, that light travels in one year (one light-year).
A nearby star is 4.2 light-years from the Earth. Calculate the distance to this star in metres. Give your answer to 3 significant figures and use 9.48 × 1015 m for one light-year.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| hydrogen nuclei join together / fuse | ✓ 1 | |
| to form helium (nuclei) and release energy | ✓ 1 | |
| Part (b) | ||
| M1 distance = speed × time = 3.00 × 108 × 3.16 × 107 | ✓ 1 | |
| A1 9.48 × 1015 m | ✓ 1 | |
| Part (c) | ||
| M1 4.2 × 9.48 × 1015 (ECF from (b) if consistent) | ✓ 1 | |
| A1 3.98 × 1016 m (3.9816 × 1016) | ✓ 1 | |
Exam tip: Write the unit of every answer; a light-year is a distance, so the answer in (c) must be in metres.
A gymnast of mass 60 kg trains on the Earth and wants to know how her weight would change on Mars. The gravitational field strength is 9.8 N/kg on the Earth and 3.7 N/kg on Mars.
Calculate the weight of the gymnast on the Earth.
State the mass of the gymnast on Mars.
Calculate the weight of the gymnast on Mars.
Explain why a jump with the same take-off speed would reach a greater height on Mars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = mg = 60 × 9.8 | ✓ 1 | |
| A1 588 N | ✓ 1 | |
| Part (b) | ||
| 60 kg (mass does not change) | ✓ 1 | |
| Part (c) | ||
| M1 60 × 3.7 | ✓ 1 | |
| A1 222 N | ✓ 1 | |
| Part (d) | ||
| her weight / the downward force on her is smaller on Mars | ✓ 1 | |
| so she slows down more gradually / decelerates less, and so rises higher | ✓ 1 | |
Exam tip: Mass stays constant; only weight changes with the local gravitational field strength.
A rail company imagines a train that could travel to the Moon. The distance from the Earth to the Moon is 3.84 × 105 km. The train travels at a constant speed of 250 km/h. Light travels at 3.00 × 108 m/s.
Calculate the time, in hours, the train would take to travel to the Moon.
Convert your answer to (a) into days.
State the time that the Moon takes to orbit the Earth once, in days.
Light from the Sun reflects off the Moon. Calculate the time light takes to travel from the Moon to the Earth, a distance of 3.84 × 108 m. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = d / v = 3.84 × 105 / 250 | ✓ 1 | |
| A1 1536 hours | ✓ 1 | |
| Part (b) | ||
| A1 1536 / 24 = 64 days (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 27 to 29 days (accept 27.3 or about one month) | ✓ 1 | |
| Part (d) | ||
| M1 t = d / v = 3.84 × 108 / 3.00 × 108 | ✓ 1 | |
| A1 1.28 s | ✓ 1 | |
Exam tip: Check that the units in distance and speed match before dividing; here km goes with km/h and m goes with m/s.
The Moon appears to change shape during each month, from new moon to full moon and back again.
What causes these phases?
| Marking point | Mark | Notes |
|---|---|---|
| The Sun always lights half of the Moon. As the Moon orbits the Earth, we see different amounts of that lit half, so its shape appears to change. The Earth's shadow only falls on the Moon during a lunar eclipse, which is rare. The Moon does not produce its own light; it reflects sunlight. | ✓ 1 | Answer C |
Answer: C
Common error: The phases are not caused by the Earth's shadow. That explanation describes a lunar eclipse, which can only happen at full moon.
The four planets nearest the Sun are small and rocky. The four outer planets are large and mostly made of gas.
Which statement about the formation of the Solar System explains this?
| Marking point | Mark | Notes |
|---|---|---|
| In the accretion model, the Solar System formed from a rotating cloud of gas and dust. Close to the young Sun the temperature was high, so only materials such as rock and metal could clump together. Light gases were pushed outwards and collected into the large outer planets. Option C has the effect the wrong way round: the gas ended up in the outer Solar System. | ✓ 1 | Answer B |
Answer: B
Exam tip: Link the planet types to temperature and distance from the Sun: hot near the Sun means rocky planets, cooler further out means gaseous planets.
The Earth spins on its axis and also orbits the Sun. The Earth's axis is tilted.
State the approximate time taken for
(i) one rotation of the Earth on its axis
(ii) one orbit of the Earth around the Sun.
Explain why the Sun appears to move across the sky from east to west during the day.
In June the northern hemisphere is tilted towards the Sun. Explain why the northern hemisphere is warmer in June than in December.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (i) 24 hours / 1 day | ✓ 1 | |
| (ii) 365 days / 365¼ days / 1 year | ✓ 1 | |
| Part (b) | ||
| The Earth rotates on its axis | ✓ 1 | |
| from west to east, so the Sun appears to move in the opposite direction | ✓ 1 | |
| Part (c) | ||
| Any two from: | — | |
| days are longer, so there are more hours of sunlight; | — | |
| the Sun is higher in the sky, so its rays strike the ground more directly and more energy is received per unit area; | — | |
| sunlight passes through less atmosphere, so less energy is absorbed or scattered before reaching the ground | ✓ 2 | |
Common error: Do not explain the seasons by saying the Earth is closer to the Sun in summer. Explain them using the tilt, the length of the day and the angle of the Sun's rays.
A space probe travels directly away from Mars after taking off from its surface.
What happens to the gravitational force of Mars on the probe?
| Marking point | Mark | Notes |
|---|---|---|
| The gravitational field strength of Mars decreases as the distance from Mars increases, so the force on the probe gets smaller as the probe moves away. The force does not suddenly drop to zero, because gravity does not need an atmosphere to act. The probe's speed has no effect on the gravitational force. The mass of Mars is constant, but the field it produces is weaker further away, so the force is not constant. | ✓ 1 | Answer C |
Answer: C
Common error: Gravity does not stop at the edge of the atmosphere. It gets weaker with distance but never becomes exactly zero.
An Earth-observation satellite orbits in a circle of radius 7.0 × 106 m with a period of 5.8 × 103 s. What is its orbital speed?
| Marking point | Mark | Notes |
|---|---|---|
| Distance in one orbit = 2πr, so v = 2πr ÷ T = 2π × 7.0 × 106 ÷ 5.8 × 103 = 7.6 × 103 m/s: C is correct. A leaves out the factor 2π (uses r ÷ T). B uses πr ÷ T, the half circumference. D uses 4πr ÷ T. | ✓ 1 | Answer C |
Answer: C
Common error: The distance for one orbit is the circumference 2πr, not the radius.
A star is 8.6 light-years from Earth. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A light-year is a distance, the distance light travels in one year, so light from 8.6 light-years away took 8.6 years to arrive: A is correct. B confuses distance with the star's age. C confuses it with an orbital period. D confuses the distance with a speed. | ✓ 1 | Answer A |
Answer: A
Common error: A light-year is a unit of distance, not of time.
The average distance from the Sun to the Earth is 1.5 × 1011 m. Light travels at 3.0 × 108 m/s.
Calculate the time in seconds taken for light from the Sun to reach the Earth. [2]
Convert this time to minutes, giving your answer to 3 significant figures. [1]
Define a light-year. [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time = distance ÷ speed | ✓ 1 | |
| 1.5 × 1011 ÷ 3.0 × 108 = 500 s | ✓ 1 | |
| Part (b) | ||
| 500 ÷ 60 = 8.33 minutes | ✓ 1 | |
| Part (c) | ||
| the distance travelled by light in a vacuum in one year | ✓ 1 | |
Exam tip: Check the units of each quantity before dividing, and convert seconds to minutes by dividing by 60.
Which process releases the energy that a star such as the Sun radiates while it is on the main sequence?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: hydrogen nuclei fuse into helium, releasing energy. A is the process used in nuclear power stations on Earth, not in stars. C is wrong because a star has no oxygen supply for burning and chemical reactions release far too little energy. D is wrong because decay is not the main source of a star's output. | ✓ 1 | Answer B |
Answer: B
Common error: Fusion joins small nuclei together; fission splits large nuclei. Stars use fusion.
A rover of mass 160 kg is lowered onto Mars, where the gravitational field strength is 3.7 N/kg.
Explain the difference between the mass and the weight of the rover.
Calculate the weight of the rover on Mars.
The rover is held by a cable that pulls upwards with a constant force of 640 N. Calculate the acceleration of the rover.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| mass is the amount of matter in the rover, and it is the same everywhere | ✓ 1 | |
| weight is the force of gravity acting on the rover, which depends on the gravitational field strength | ✓ 1 | |
| Part (b) | ||
| W = mg = 160 × 3.7 | ✓ 1 | |
| 592 N | ✓ 1 | |
| Part (c) | ||
| resultant force = 640 − 592 = 48 N (ECF from (b)) | ✓ 1 | |
| a = F / m = 48 / 160 | ✓ 1 | |
| 0.3 m/s2 upwards | ✓ 1 | |
Exam tip: Weight acts downwards. When another force acts upwards, subtract before using F = ma.
An epidemiology student gives a talk about the night sky. She says that the nearest star beyond the Sun, Proxima Centauri, is 4.2 light-years from the Earth. Use 1 light-year = 9.5×1012 km.
State what is meant by a galaxy.
State what is meant by a light-year.
State the time taken for light from Proxima Centauri to reach the Earth.
Calculate the distance from the Earth to Proxima Centauri in km. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a very large collection of stars (with gas and dust) held together by gravity | ✓ 1 | |
| Part (b) | ||
| the distance travelled by light in one year (in a vacuum) | ✓ 1 | |
| Part (c) | ||
| 4.2 years | ✓ 1 | |
| Part (d) | ||
| distance = 4.2 × 9.5×1012 | ✓ 1 | |
| 3.99×1013 km, which is 4.0×1013 km to 2 significant figures | ✓ 1 | |
Exam tip: Multiply the number of light-years by the distance in one light-year; keep the unit in km as asked.
Stars are formed from clouds of gas and dust in space.
Describe how a star forms from a cloud of gas and dust.
State what is balanced in a star that is stable on the main sequence.
State what the Sun will become when it has used up most of its hydrogen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gravity pulls the gas and dust together | ✓ 1 | |
| the material heats up as it contracts and becomes a protostar | ✓ 1 | |
| when the temperature is high enough, nuclear fusion of hydrogen begins | ✓ 1 | |
| Part (b) | ||
| the inward force of gravity is balanced by the outward pressure from the energy released by fusion | ✓ 1 | |
| Part (c) | ||
| a red giant (and later a white dwarf) | ✓ 1 | |
Common error: A star is stable when gravity pulling inwards equals the outward pressure from fusion, not when gravity is zero.
Which row describes what will happen to the Sun after it leaves the main sequence?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: the Sun has too little mass to become a supernova, so it swells into a red giant, then sheds its outer layers and shrinks into a white dwarf. A and D describe the fate of stars much more massive than the Sun. B has the stages in the wrong order: a white dwarf is the final stage here. | ✓ 1 | Answer C |
Answer: C
Exam tip: Link the fate of a star to its mass: only stars much more massive than the Sun end as supernovae.
A navigation satellite used by ships orbits Earth in a circular path of radius 7.00 × 106 m. It takes 5800 s to complete one orbit. What is its average orbital speed?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: speed = 2πr / T = 2π × 7.00 × 106 / 5800 = 7.58 × 103 m/s. A uses r / T and forgets the circumference. B uses πr / T, forgetting the factor 2. D converts the period into minutes but leaves the speed in m/s. | ✓ 1 | Answer C |
Answer: C
Common error: Using the radius, not the circumference 2πr, as the distance travelled in one orbit.
A laser pulse sent from Earth to a reflector on the Moon returns to Earth after 2.6 s. The speed of light is 3.0 × 108 m/s. What distance does this give between Earth and the Moon?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: the pulse travels there and back, so distance = 3.0 × 108 × 2.6 / 2 = 3.9 × 108 m. C forgets to halve the time. A divides the speed by the time. D multiplies the full time by 2 instead of halving it. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a reflected signal, the measured time covers the journey twice; halve it before using distance = speed × time.
A volcano observatory photographs an ash plume at the same time every day. Earth takes 365 days to orbit the Sun, at an average orbital radius of 1.50 × 1011 m.
State the time taken for Earth to rotate once on its axis.
Calculate the average orbital speed of Earth around the Sun. Give your answer in m/s to 3 significant figures.
State the force that keeps Earth in orbit and the direction in which it acts.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 hours (1 day) | ✓ 1 | |
| Part (b) | ||
| M1 distance travelled in one orbit = 2πr = 9.42 × 1011 m | ✓ 1 | |
| M1 time in seconds = 365 × 24 × 3600 = 3.15 × 107 s | ✓ 1 | |
| A1 speed = 2.99 × 104 m/s | ✓ 1 | |
| Part (c) | ||
| B1 gravitational force (attraction between Earth and Sun) | ✓ 1 | |
| B1 towards the Sun (towards the centre of the orbit) | ✓ 1 | |
Exam tip: Convert days into seconds before using speed = distance / time, because the orbital radius is given in metres.
A theatre uses a lamp and a ball on a stand to show how the appearance of the Moon changes. The Moon orbits Earth at an average radius of 3.84 × 108 m in 27.3 days.
State why the real Moon can be seen from Earth.
Calculate the average orbital speed of the Moon. Give your answer in m/s to 3 significant figures.
Explain why the shape of the lit part of the Moon, as seen from Earth, changes during the month.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 it reflects light from the Sun | ✓ 1 | |
| Part (b) | ||
| M1 distance = 2πr = 2π × 3.84 × 108 m | ✓ 1 | |
| M1 time = 27.3 × 24 × 3600 = 2.36 × 106 s | ✓ 1 | |
| A1 speed = 1.02 × 103 m/s | ✓ 1 | |
| Part (c) | ||
| B1 half of the Moon is always lit by the Sun | ✓ 1 | |
| B1 as the Moon orbits Earth, we see a different amount of the lit half from Earth | ✓ 1 | |
Common error: Saying that the shape of the Moon is caused by Earth's shadow. Phases come from the view of the lit half; Earth's shadow only causes eclipses.
Io is a moon of Jupiter with active volcanoes. A volcano-monitoring probe orbits it, and a comet is also observed passing through the Solar System. The gravitational field strength on Io is 1.8 N/kg. Rock thrown out by a volcano has a mass of 4.0 kg.
Describe the shape of the orbit of a comet around the Sun.
State how the speed of a comet changes as it moves from far away towards the Sun, and explain why.
State where the asteroid belt is found in the Solar System.
Calculate the weight of the 4.0 kg rock on Io.
Calculate the weight of the same rock on Earth, where g = 9.8 N/kg.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 a long, thin ellipse (the Sun is not at the centre) | ✓ 1 | |
| Part (b) | ||
| B1 the speed increases | ✓ 1 | |
| B1 the Sun's gravitational pull on the comet is stronger when it is closer | ✓ 1 | |
| Part (c) | ||
| B1 between the orbits of Mars and Jupiter | ✓ 1 | |
| Part (d) | ||
| M1 weight = mass × gravitational field strength = 4.0 × 1.8 | ✓ 1 | |
| A1 7.2 N | ✓ 1 | |
| Part (e) | ||
| M1 weight = 4.0 × 9.8 | ✓ 1 | |
| A1 39.2 N | ✓ 1 | |
Common error: Saying the mass of the rock changes on Io. Mass stays the same; only the weight depends on the field strength.
Which row shows the stages in the life cycle of a star with a mass much greater than the mass of the Sun?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: a star much more massive than the Sun becomes a red supergiant, explodes as a supernova and leaves a neutron star or black hole. A is the life cycle of a star like the Sun. B wrongly ends a massive star as a white dwarf without a supernova. C names a red giant (the stage for Sun-like stars) but then a supernova, which mixes the two life cycles. | ✓ 1 | Answer D |
Answer: D
Exam tip: Learn the two routes as separate chains: Sun-sized stars end as white dwarfs, much more massive stars end after a supernova.
An astronaut has a mass of 80 kg on Earth, where the gravitational field strength is 9.8 N/kg. The gravitational field strength on the Moon is 1.6 N/kg. Which row gives the astronaut's mass and weight on the Moon?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: mass is 80 kg everywhere, and the Moon weight is 80 × 1.6 = 128 N. B wrongly scales the mass by 1.6/9.8. C keeps the Earth weight (80 × 9.8 = 784 N). D divides Earth weight by 1.6 to get a mass of 490 kg and keeps the weight at 784 N. | ✓ 1 | Answer A |
Answer: A
Common error: Mass does not change with location; only weight changes with the gravitational field strength.
Vitamin D is made in the skin when ultraviolet radiation from the Sun reaches it. The Sun is a star.
State the nuclear process that releases energy in the Sun and name the two elements involved.
Describe what will happen to a star of the same mass as the Sun at the end of its life.
The Sun is 1.5 × 1011 m from the Earth. Ultraviolet radiation travels at 3.0 × 108 m/s. Calculate the time taken for the radiation to reach the Earth. Give your answer in minutes to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Nuclear fusion | ✓ 1 | |
| hydrogen nuclei fuse to form helium | ✓ 1 | |
| Part (b) | ||
| It expands to become a red giant | ✓ 1 | |
| it then shrinks / loses its outer layers and becomes a white dwarf | ✓ 1 | |
| Part (c) | ||
| M1 time = distance / speed = 1.5 × 1011 / 3.0 × 108 = 500 s | ✓ 1 | |
| A1 8.3 minutes | ✓ 1 | |
Exam tip: If the answer is requested in minutes, divide your time in seconds by 60 as the last step.
Planets and comets orbit the Sun. A planetary scientist is preparing a mission to Mars.
Describe the shape of the orbit of a comet around the Sun.
State the position in its orbit where a comet is moving fastest.
Explain why the gravitational force of the Sun keeps a planet in a nearly circular orbit.
An astronaut has a mass of 68 kg. The gravitational field strength on Mars is 3.7 N/kg. Calculate the weight of the astronaut on Mars. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The orbit is a long, narrow ellipse (elongated / not circular) | ✓ 1 | |
| Part (b) | ||
| Closest to the Sun | ✓ 1 | |
| Part (c) | ||
| The force acts towards the Sun, at right angles to the planet's velocity | ✓ 1 | |
| it changes the direction of the velocity (providing the centripetal force) without changing the speed | ✓ 1 | |
| Part (d) | ||
| M1 W = mg = 68 × 3.7 = 251.6 N | ✓ 1 | |
| A1 252 N | ✓ 1 | |
Exam tip: Weight is a force in newtons, so use W = mg with g in N/kg.
A water company uses an Earth-observation satellite to check the levels in its reservoirs. The satellite orbits the Earth once every 100 minutes.
State the time taken for the Earth to rotate once on its axis.
Calculate the number of complete orbits the satellite makes in one day.
Explain why the satellite does not pass over the same part of the Earth's surface on each orbit.
Calculate the angle through which the Earth rotates during one orbit of the satellite.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 hours | ✓ 1 | |
| Part (b) | ||
| M1 1 day = 24 × 60 = 1440 minutes; 1440 / 100 = 14.4 | ✓ 1 | |
| A1 14 complete orbits | ✓ 1 | |
| Part (c) | ||
| The Earth rotates beneath the satellite's orbit during the time of one orbit | ✓ 1 | |
| Part (d) | ||
| M1 angle = 360 × 100 / 1440 | ✓ 1 | |
| A1 25° | ✓ 1 | |
Exam tip: For "complete" orbits, give the whole number below the calculated value rather than rounding up.
In June it is summer in the northern hemisphere of the Earth. Which statement correctly explains this?
| Marking point | Mark | Notes |
|---|---|---|
| Seasons arise from the tilt of the Earth's axis (about 23°), which changes how directly and for how long a hemisphere is lit. The Earth is actually closest to the Sun in early January, the Moon does not alter the hemisphere's distance from the Sun, and the rotation rate stays essentially constant. | ✓ 1 | Answer D |
Answer: D
Common error: Seasons are not caused by the Earth's changing distance from the Sun.
A planet orbits a star in a circular path of radius 2.4 × 1011 m. One orbit takes 6.0 × 107 s. What is the orbital speed of the planet?
| Marking point | Mark | Notes |
|---|---|---|
| Orbital speed = 2πr / T = 2π × 2.4 × 1011 / 6.0 × 107 ≈ 2.5 × 104 m/s. The value 4.0 × 103 m/s comes from r / T, leaving out the 2π; 1.3 × 104 m/s uses π instead of 2π; 5.0 × 104 m/s uses 4π. | ✓ 1 | Answer C |
Answer: C
Exam tip: The distance travelled in one orbit is the circumference 2πr, not the radius.
Which sequence correctly shows the stages in the life of a star with a mass similar to that of the Sun?
| Marking point | Mark | Notes |
|---|---|---|
| A star of the Sun's mass forms from a nebula, becomes a protostar, spends most of its life as a main sequence star, expands into a red giant and then shrinks to a white dwarf. Red supergiants, supernovae and neutron stars belong to much more massive stars, and the protostar stage comes before the main sequence. | ✓ 1 | Answer D |
Answer: D
Common error: Only stars much more massive than the Sun end in a supernova.
A comet moves around the Sun in a long elliptical orbit. At which point in its orbit is the comet moving fastest?
| Marking point | Mark | Notes |
|---|---|---|
| The Sun's gravitational pull is strongest when the comet is closest, so the comet speeds up as it falls inwards and is fastest at closest approach. It is slowest when furthest away. Constant speed applies only to a circular orbit. | ✓ 1 | Answer C |
Answer: C
Exam tip: For an elliptical orbit, speed is greatest closest to the Sun and least furthest away.
A marine biologist studying tides notes that the Moon orbits the Earth once every 27.3 days. Assume the orbit is circular with radius 3.84 × 108 m. (1 day = 86 400 s)
Explain why the Moon appears to change shape during one orbit.
Calculate the orbital speed of the Moon in m/s. Give your answer to 3 significant figures.
The Moon's speed is constant but its velocity is not. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The Moon reflects sunlight and half of it is always lit | ✓ 1 | |
| As the Moon orbits, different amounts of the lit half face the Earth | ✓ 1 | |
| Part (b) | ||
| M1 time for one orbit = 27.3 × 86 400 = 2.36 × 106 s | ✓ 1 | |
| M1 use of speed = 2πr / T | ✓ 1 | |
| A1 speed = 1020 m/s (1.02 × 103 m/s) | ✓ 1 | |
| Part (c) | ||
| Velocity is a vector and its direction changes continuously as the Moon moves round the orbit | ✓ 1 | |
| The gravitational force of the Earth acts towards the Earth's centre, at right angles to the motion, changing direction but not speed | ✓ 1 | |
Common error: Speed is the size of the velocity only; a change of direction is a change of velocity.
A logistics company names its delivery vehicles after planets. The orbital periods of four planets are: Mercury 0.24 years, Earth 1.00 year, Mars 1.88 years, Jupiter 11.9 years.
State the name of the force that keeps the planets in orbit around the Sun.
Explain two reasons why a planet further from the Sun has a longer orbital period.
Calculate how many complete orbits Jupiter makes in 47.6 Earth years.
State where the asteroid belt is found in the Solar System.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Gravitational force (gravity) of the Sun | ✓ 1 | |
| Part (b) | ||
| The orbit is longer, so the planet has further to travel | ✓ 1 | |
| The Sun's gravitational pull is weaker, so the planet moves more slowly | ✓ 1 | |
| Part (c) | ||
| M1 orbits = 47.6 / 11.9 | ✓ 1 | |
| A1 4 orbits | ✓ 1 | |
| Part (d) | ||
| Between the orbits of Mars and Jupiter | ✓ 1 | |
Exam tip: Link longer period to both a larger orbit and a lower orbital speed.
A satellite used to monitor ocean temperature orbits the Earth in a circular path at a constant speed of 7.7 × 103 m/s. It takes 5.4 × 103 s to complete one orbit.
State the force that keeps the satellite in its orbit.
Calculate the radius of the satellite's orbit. Give your answer to 2 significant figures.
A different satellite is geostationary. Explain why it always appears at the same point in the sky.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The gravitational attraction of the Earth | ✓ 1 | |
| Part (b) | ||
| M1 rearrange orbital speed = 2πr / T to give r = vT / 2π | ✓ 1 | |
| M1 substitute: 7.7 × 103 × 5.4 × 103 / 2π | ✓ 1 | |
| A1 radius = 6.6 × 106 m | ✓ 1 | |
| Part (c) | ||
| Its orbital period is 24 hours, equal to the Earth's rotation period | ✓ 1 | |
| It orbits above the equator in the same direction as the Earth rotates | ✓ 1 | |
Exam tip: State the distance for one orbit as 2πr before substituting any numbers.
A theatre planetarium show states that a star is 12 light-years from Earth. What does this mean?
| Marking point | Mark | Notes |
|---|---|---|
| A light-year is a distance, the distance light travels through space in one year, so a star 12 light-years away has light that takes 12 years to arrive. The distance has nothing to do with the age of the star or any orbital period, and 12 seconds would be a far smaller distance. | ✓ 1 | Answer D |
Answer: D
Common error: A light-year is a unit of distance, not a unit of time.
An asteroid orbits the Sun in a circular path of radius \(4.0\times10^{11}\) m at a constant speed of \(1.7\times10^{4}\) m/s. Use 1 year = \(3.2\times10^{7}\) s.
Calculate the time for one orbit of the asteroid, in years. Give your answer to 2 significant figures.
State the region of the Solar System in which most asteroids orbit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| \(T=\dfrac{2\pi r}{v}=\dfrac{2\pi\times4.0\times10^{11}}{1.7\times10^{4}}=1.48\times10^{8}\) s | ✓ 1 | |
| divide by \(3.2\times10^{7}\) | ✓ 1 | |
| 4.6 years | ✓ 1 | |
| Part (b) | ||
| between the orbits of Mars and Jupiter | ✓ 1 | |
Common error: The period is found from circumference divided by speed; do not use the radius alone as the distance travelled in one orbit.
Which sequence shows the stages in the life of a star with the same mass as the Sun?
| Marking point | Mark | Notes |
|---|---|---|
| A star like the Sun forms from a nebula, becomes a protostar, spends most of its life as a main sequence star, expands to a red giant and finally shrinks to a white dwarf. The supernova and black hole route applies only to stars much more massive than the Sun. The other sequences put the nebula after the protostar, or the red giant before the main sequence. | ✓ 1 | Answer A |
Answer: A
Common error: Do not mix up the end of a Sun-like star (white dwarf) with the end of a very massive star (supernova, then neutron star or black hole).
In June, a city in the northern hemisphere has more hours of daylight than in December. What is the reason for this?
| Marking point | Mark | Notes |
|---|---|---|
| The Earth's axis is tilted relative to its orbit, so in June the northern hemisphere leans towards the Sun and days there are longer. The distance from the Sun does not cause the seasons (the Earth is actually slightly further from the Sun in June), the Earth spins at an almost constant rate all year, and the Sun's output does not change with the months. | ✓ 1 | Answer D |
Answer: D
Common error: Seasons are caused by the tilt of the Earth's axis, not by changes in the distance between the Earth and the Sun.
The Moon orbits the Earth and the Earth spins on its axis. Give the Moon's orbital period as 27 days and the Earth's rotation period as 24 hours.
State why the Moon can be seen from the Earth.
Calculate the angle through which the Moon moves along its orbit in 9 days.
Calculate the time taken for the Earth to rotate through an angle of 75°.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it reflects light from the Sun (it does not produce its own light) | ✓ 1 | |
| Part (b) | ||
| M1 360 × 9 / 27 | ✓ 1 | |
| A1 120° | ✓ 1 | |
| Part (c) | ||
| M1 (75 / 360) × 24 | ✓ 1 | |
| A1 5.0 hours | ✓ 1 | |
Common error: Compare each angle with the full 360° turn and the full period; the time is proportional to the angle.
Mercury takes 88 days to orbit the Sun once. The Earth takes 365 days to orbit the Sun once.
State the name of the force that keeps the planets in orbit around the Sun.
Calculate the number of orbits that Mercury completes in one Earth year. Give your answer to 3 significant figures.
Suggest why the surface of Mercury is much hotter than the surface of the Earth.
Describe the motion of Mercury if the gravitational force from the Sun suddenly disappeared.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gravity / gravitational force (of the Sun) | ✓ 1 | |
| Part (b) | ||
| M1 365 / 88 | ✓ 1 | |
| A1 4.15 | ✓ 1 | |
| Part (c) | ||
| Mercury is closer to the Sun so it receives more radiation per unit area (than the Earth) | ✓ 1 | |
| Part (d) | ||
| it would move in a straight line (along the tangent to its orbit) | ✓ 1 | |
| at constant speed | ✓ 1 | |
Common error: Without a force towards the Sun a planet does not fly off outwards; it carries on in a straight line in the direction it was moving.
The Sun is a star that was formed from a cloud of gas and dust.
State the name of the stage in its life that the Sun is in now.
Describe how a protostar forms from a nebula (a cloud of gas and dust).
After the main sequence, the Sun will go through two more stages. State the two stages in the correct order.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| main sequence | ✓ 1 | |
| Part (b) | ||
| gravity pulls the gas and dust together (so the cloud contracts) | ✓ 1 | |
| the temperature rises as the material is compressed, forming a protostar | ✓ 1 | |
| Part (c) | ||
| red giant | ✓ 1 | |
| white dwarf | ✓ 1 | |
Common error: Do not give a supernova or a black hole as the end of the Sun; these are for stars much more massive than the Sun.
Engineers on Earth send radio signals to a rover on Mars. The distance between the Earth and Mars varies from a minimum of 5.6×1010 m to a maximum of 4.0×1011 m. Radio waves travel at 3.0×108 m/s.
Explain why the distance between the Earth and Mars varies.
Calculate the minimum time for a signal to travel from the Earth to Mars.
A signal is sent to the rover and a reply is sent straight back. Calculate the maximum time, in minutes, between sending the signal and receiving the reply.
Suggest why the rover cannot be steered around obstacles in real time by a driver on Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Both planets orbit the Sun at different distances / with different orbital periods / speeds | ✓ 1 | |
| so sometimes they are on the same side of the Sun (closest) and sometimes on opposite sides (furthest) | ✓ 1 | |
| Part (b) | ||
| t = d ÷ v = 5.6×1010 ÷ 3.0×108 | ✓ 1 | |
| = 190 s (187 s) | ✓ 1 | |
| Part (c) | ||
| t = (2 × 4.0×1011) ÷ 3.0×108 = 2670 s | ✓ 1 | |
| 2670 ÷ 60 = 44 minutes (44.4 min) | ✓ 1 | |
| Part (d) | ||
| The signals take minutes to arrive, so the rover could hit an obstacle before a command reaches it / the rover must make some decisions itself | ✓ 1 | |
Common error: A reply involves a journey there and back, so the distance must be doubled. Using only the one-way distance loses the mark.
The Moon orbits the Earth once every 27 days. Its orbit can be treated as a circle of radius 3.9 × 108 m.
Convert 27 days into seconds. [2]
Calculate the average orbital speed of the Moon. [3]
The Moon's speed is almost constant, yet there must be a resultant force on it. Explain why, and state the origin of this force. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 27 × 24 × 3600 | ✓ 1 | |
| = 2.33 × 106 s (2332800 s) | ✓ 1 | |
| Part (b) | ||
| v = 2πr ÷ T | ✓ 1 | |
| = 2π × 3.9 × 108 ÷ 2332800 | ✓ 1 | |
| = 1050 m/s (1.05 × 103 m/s) | ✓ 1 | |
| Part (c) | ||
| its direction of motion keeps changing, so its velocity changes and it accelerates, which needs a resultant force | ✓ 1 | |
| the force is the gravitational attraction of the Earth, directed towards the Earth | ✓ 1 | |
Common error: Constant speed does not mean constant velocity: a changing direction means acceleration.
Which statement describes the cosmic microwave background radiation (CMBR)?
| Marking point | Mark | Notes |
|---|---|---|
| The CMBR is microwave radiation that reaches us almost equally from every direction. It does not come from one source such as the Sun, the Milky Way or a supernova. Its intensity peaks at a frequency of about 160 GHz. It is radiation left over from the early Universe, and its wavelength has been stretched into the microwave region as the Universe expanded. The Sun option is wrong because the CMBR is detected day and night in all directions. The Milky Way option is wrong because the CMBR is not confined to one direction and is not produced by stars. The gamma option confuses the CMBR with high-energy radiation from violent events. | ✓ 1 | Answer C |
Answer: C
Exam tip: Link the three ideas together: microwave region, a peak frequency, and the same intensity in every direction. Together they point to an origin in the early Universe, not a nearby source.
Which observation provides evidence that the Universe began in a very hot, dense state and has been expanding ever since?
| Marking point | Mark | Notes |
|---|---|---|
| The cosmic microwave background radiation is left over from the hot early Universe, its wavelength stretched by expansion, and supports the Big Bang model. The direction of planetary orbits, star colours and the Moon's rotation say nothing about the Universe's origin. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link the cosmic microwave background radiation to the early hot Universe when asked for evidence.
A star stays the same size for billions of years during its stable stage.
Why is the star stable?
| Marking point | Mark | Notes |
|---|---|---|
| In a stable star, gravity pulls the material inwards. The very high temperature in the core, maintained by the fusion of hydrogen, produces an outward force. These two forces are equal and opposite, so the star neither collapses nor expands. | ✓ 1 | Answer D |
Answer: D
Exam tip: Name both forces and their directions: inward gravitational force and outward force from the high core temperature. Then state that they are balanced.
A very massive star swells to become a red supergiant.
What happens to this star next?
| Marking point | Mark | Notes |
|---|---|---|
| A red supergiant explodes as a supernova. This forms a nebula that contains hydrogen and heavier elements, with a neutron star or a black hole at its centre. A planetary nebula with a white dwarf is the end stage of a less massive star, which becomes a red giant rather than a red supergiant. | ✓ 1 | Answer A |
Answer: A
Common error: Keep the two paths separate: red giant leads to a planetary nebula and a white dwarf; red supergiant leads to a supernova and a neutron star or black hole.
A comet moves in a very elliptical orbit around the Sun.
Where is the comet's speed greatest, and why?
| Marking point | Mark | Notes |
|---|---|---|
| As the comet falls in towards the Sun, its gravitational potential energy decreases. The total energy is conserved, so its kinetic energy increases. The comet therefore moves fastest at its closest point to the Sun. Having the most potential energy far away means the comet has the least kinetic energy there, so it is slowest. Constant total energy does not mean constant speed, because energy moves between the potential and kinetic stores. | ✓ 1 | Answer B |
Answer: B
Exam tip: For elliptical orbits, use conservation of energy: closer to the Sun means less gravitational potential energy, more kinetic energy and a greater speed.
The average distances of four planets from the Sun are listed below.
Venus: 1.1×1011 m
Earth: 1.5×1011 m
Saturn: 1.4×1012 m
Neptune: 4.5×1012 m
Which planet has the lowest average orbital speed?
| Marking point | Mark | Notes |
|---|---|---|
| The Sun's gravitational field strength decreases with distance, and planets further from the Sun have lower orbital speeds. Neptune is furthest away, so it orbits the slowest. Venus is closest, so it has the highest speed, not the lowest. Being in the middle of the group gives Earth neither extreme. A planet's own mass does not decide its orbital speed. | ✓ 1 | Answer D |
Answer: D
Exam tip: Remember the trend: the further a planet is from the Sun, the weaker the Sun's field, the lower its orbital speed and the longer its orbital period.
Astronomers measure distances to stars and galaxies in light-years.
Speed of light = 3.0 × 108 m/s.
1 year = 3.15 × 107 s.
State what is meant by a light-year.
Show that one light-year is approximately 9.5 × 1015 m.
A nearby star is 4.2 light-years from the Earth. State how long ago the light now reaching the Earth left this star.
The diameter of the Milky Way is about 100 000 light-years. Calculate this diameter in metres.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the distance travelled by light (in a vacuum) in one year | ✓ 1 | |
| Part (b) | ||
| distance = speed × time = 3.0 × 108 × 3.15 × 107 | ✓ 1 | |
| = 9.45 × 1015 m ≈ 9.5 × 1015 m | ✓ 1 | |
| Part (c) | ||
| 4.2 years ago | ✓ 1 | |
| Part (d) | ||
| 100 000 × 9.5 × 1015 (or 100 000 × 9.45 × 1015) | ✓ 1 | |
| = 9.5 × 1020 m (accept 9.45 × 1020 m) | ✓ 1 | |
Exam tip: In a 'show that' question, write out the full calculation and give the answer to more significant figures than the value you were given.
For a distant galaxy, the Hubble constant is given by H0 = v/d. Here v is the speed at which the galaxy recedes and d is its distance from the Earth.
Which is the unit of H0?
| Marking point | Mark | Notes |
|---|---|---|
| H0 = v/d, so its unit is (m/s) ÷ m. The metres cancel, leaving 1/s, or per second. Its value is about 2.2 × 10−18 /s. m/s ignores the division by distance. s is the unit of 1/H0, not of H0. m/s2 comes from dividing a speed by a time instead of by a distance. | ✓ 1 | Answer D |
Answer: D
Exam tip: Find a derived unit by putting the units into the defining equation and cancelling.
Stars form, change and eventually die. This question is about a star with a mass similar to the Sun's.
Describe how a star is formed.
Explain why the star then remains stable for a very long time.
Describe what happens to this star when most of the hydrogen in its core has been used up.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a cloud of gas and dust in space (nebula) is pulled together / collapses under gravity | ✓ 1 | |
| forming a protostar, which heats up until the temperature is high enough for nuclear fusion of hydrogen to begin | ✓ 1 | |
| Part (b) | ||
| the inward pull of gravity is balanced by an outward force / pressure, so the star stays the same size (equilibrium) | ✓ 1 | |
| the outward force / pressure comes from the very hot core (gas or radiation pressure), kept hot by energy released in hydrogen fusion | ✓ 1 | |
| Part (c) | ||
| it expands (and its surface cools) to become a red giant | ✓ 1 | |
| its outer layers are thrown off as a planetary nebula, leaving a (hot, dense) white dwarf at the centre | ✓ 1 | |
Exam tip: Check the mass of the star in the question. A star like the Sun becomes a red giant, not a red supergiant, and does not explode as a supernova.
An astronomer studies the light from a distant galaxy. She finds that the galaxy is moving away from the Earth at 2.1 × 107 m/s.
Hubble constant H0 = 2.2 × 10−18 per second.
1 light-year = 9.5 × 1015 m.
State what is meant by redshift.
Calculate the distance of the galaxy from the Earth in metres.
Convert your answer to part (b) into light-years.
Astronomers can also find the distance to a far galaxy without using redshift. State what is measured to do this.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| an increase in the observed wavelength of electromagnetic radiation (light) from a source that is moving away / receding (accept: spectral lines shifted towards the red / longer-wavelength end) | ✓ 1 | |
| Part (b) | ||
| d = v / H0 = 2.1 × 107 / 2.2 × 10−18 | ✓ 1 | |
| = 9.5 × 1024 m | ✓ 1 | |
| Part (c) | ||
| 9.5 × 1024 / 9.5 × 1015 | ✓ 1 | |
| = 1.0 × 109 light-years (allow error carried forward from (b)) | ✓ 1 | |
| Part (d) | ||
| the (apparent) brightness of a supernova in that galaxy (accept: the brightness of another standard candle of known luminosity, e.g. a Cepheid variable star) | ✓ 1 | |
Exam tip: Rearrange H₀ = v/d to d = v/H₀ before you substitute the values. This avoids errors with the powers of ten.
Venus orbits the Sun with an average orbital radius of 1.08×1011 m. Its orbital period is 225 days.
Convert the orbital period of Venus into seconds.
Calculate the average orbital speed of Venus.
The Earth's average orbital speed is 3.0×104 m/s. Explain why the orbital speed of Venus is greater than that of the Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 225 × 24 × 3600 = 1.94×107 s | ✓ 1 | |
| Part (b) | ||
| v = 2πr/T = (2π × 1.08×1011) ÷ 1.94×107 | ✓ 1 | |
| = 3.5×104 m/s | ✓ 1 | |
| Part (c) | ||
| Venus is closer to the Sun (it has a smaller orbital radius) than the Earth, so the Sun's gravitational field strength is greater at its orbit | ✓ 1 | |
| The greater gravitational pull (per kg) means Venus must travel faster to stay in its circular orbit around the Sun | ✓ 1 | |
Exam tip: In orbital-speed questions, write v = 2πr/T first, substitute the values with the period in seconds, and only then work out the answer.
According to the Big Bang theory, the Universe began as an extremely small, extremely hot region and has been expanding ever since. Two kinds of observation support this theory.
Describe what is observed in the light from distant galaxies, and state what this shows about the galaxies.
Describe the cosmic microwave background radiation (CMBR).
Explain how the CMBR supports the Big Bang theory.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the light is redshifted: observed wavelengths are longer (frequencies lower) than for the same source measured on Earth | ✓ 1 | |
| so the galaxies are moving away from us; more distant galaxies recede faster / the Universe is expanding | ✓ 1 | |
| Part (b) | ||
| it is electromagnetic radiation in the microwave region; accept 'of a particular frequency' or 'a range of frequencies with a peak' | ✓ 1 | |
| it reaches us from every direction / from all points in space, with almost the same intensity | ✓ 1 | |
| Part (c) | ||
| it is radiation left over from the early Universe, emitted when the Universe was very hot and dense, so it originally had a much shorter wavelength | ✓ 1 | |
| as the Universe expanded, this wavelength was stretched (redshifted) into the microwave region | ✓ 1 | |
Common error: Writing only 'the Universe is expanding' does not describe the evidence. State what is actually observed: longer wavelengths in light from distant galaxies, and microwave radiation arriving from all directions.
The spectrum of light from a distant galaxy shows the same pattern of lines as light from a laboratory lamp, but every line is at a longer wavelength. What does this show?
| Marking point | Mark | Notes |
|---|---|---|
| Longer wavelengths mean a redshift, which occurs when the source is moving away: A is correct. B would give a blueshift (shorter wavelengths). C is wrong because temperature changes the brightness distribution, not a uniform shift of all lines. D is wrong because the speed of light in a vacuum is constant; redshift is a change in wavelength. | ✓ 1 | Answer A |
Answer: A
Exam tip: Redshift means longer wavelength and a source moving away; a larger shift means a faster speed.
Which sequence shows the likely future stages of the Sun?
| Marking point | Mark | Notes |
|---|---|---|
| A star of the Sun's mass swells into a red giant and then sheds its outer layers, leaving a white dwarf: A is correct. B and D are routes for stars of much greater mass. C has the stages in the wrong order. | ✓ 1 | Answer A |
Answer: A
Common error: The final stages depend on the mass of the star: only very massive stars end as supernovae.
Astronomers measure the light from distant galaxies.
Explain what is meant by the redshift of light from a distant galaxy. [2]
A galaxy is 3.0 × 1024 m from the Earth. Calculate its recession speed using v = H0d, where H0 = 2.2 × 10−18 s−1. [2]
Galaxies that are further away show a greater redshift. Explain how this supports the idea that the Universe is expanding. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the wavelengths of light from the galaxy are longer than those from a stationary source | ✓ 1 | |
| this shows the galaxy is moving away from the Earth | ✓ 1 | |
| Part (b) | ||
| v = 2.2 × 10−18 × 3.0 × 1024 | ✓ 1 | |
| = 6.6 × 106 m/s (6600000 m/s) | ✓ 1 | |
| Part (c) | ||
| more distant galaxies are moving away faster | ✓ 1 | |
| this is what would happen if space itself is expanding, so all galaxies move apart | ✓ 1 | |
Exam tip: Link greater redshift to greater speed of recession, then link speed rising with distance to expansion.
Astronomers study two kinds of evidence about the origin of the Universe.
Describe the cosmic microwave background radiation (CMBR). [2]
Explain how the CMBR supports the Big Bang theory. [2]
A hydrogen line has wavelength 656 nm in the laboratory but 689 nm in light from a distant galaxy. Calculate the speed of the galaxy relative to the Earth using v = c × Δλ ÷ λ, where c = 3.0 × 108 m/s. [3]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it is radiation in the microwave region of the spectrum | ✓ 1 | |
| it reaches the Earth from all directions in space | ✓ 1 | |
| Part (b) | ||
| it was emitted very soon after the Big Bang as high-energy, short-wavelength radiation | ✓ 1 | |
| the expansion of the Universe has stretched its wavelength into the microwave region | ✓ 1 | |
| Part (c) | ||
| Δλ = 689 − 656 = 33 nm | ✓ 1 | |
| Δλ ÷ λ = 33 ÷ 656 = 0.0503 | ✓ 1 | |
| v = 0.0503 × 3.0 × 108 = 1.5 × 107 m/s (1.509 × 107 m/s unrounded) | ✓ 1 | |
Exam tip: Use the laboratory wavelength as λ and the change in wavelength as Δλ; both must be in the same unit.
Light from a distant galaxy is analysed. Every spectral line has a longer wavelength than the same line measured in a laboratory. What does this show?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: a longer received wavelength is a red-shift, which occurs when the source is receding. B would give a shorter wavelength (blue-shift). C is wrong because temperature would change the colour distribution, not shift every line by the same fraction. D is wrong because light in a vacuum always travels at the same speed. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link red-shift to recession: longer wavelength means lower frequency and a source moving away.
A satellite used to map the spread of crop disease moves in a circular orbit of radius 7.0×106 m. It completes one orbit in 6000 s. What is its orbital speed?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: v = 2πr/T = 2π × 7.0×106 / 6000 = 7.3×103 m/s. B uses r/T and leaves out 2π. C uses πr/T, which is the distance for half an orbit. D divides by 100 instead of 6000, mishandling the period. | ✓ 1 | Answer A |
Answer: A
Common error: One orbit covers the circumference 2πr, not the radius, so the distance travelled per orbit is 2πr.
Mars moves round the Sun in a circular orbit of radius 2.3×1011 m. It takes 687 days to complete one orbit. The speed of light is 3.0×108 m/s.
Explain why the velocity of Mars is changing even though its speed is constant.
Calculate the orbital speed of Mars in m/s. Give your answer to 2 significant figures.
Calculate the time taken for light to travel from the Sun to Mars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| velocity is a vector, so it has direction as well as magnitude | ✓ 1 | |
| the direction of motion of Mars changes continuously as it moves round the orbit | ✓ 1 | |
| Part (b) | ||
| T = 687 × 24 × 3600 = 5.94×107 s | ✓ 1 | |
| v = 2πr / T = 2π × 2.3×1011 / 5.94×107 | ✓ 1 | |
| 2.4×104 m/s (24 347 m/s unrounded) | ✓ 1 | |
| Part (c) | ||
| t = d / c = 2.3×1011 / 3.0×108 | ✓ 1 | |
| 767 s (770 s to 2 significant figures) | ✓ 1 | |
Exam tip: Convert the orbital period to seconds before using v = 2πr/T if the speed is wanted in m/s.
Light from a distant galaxy contains the same spectral lines as light from a laboratory sample of hydrogen, but each line has a longer wavelength. What does this show about the galaxy?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: longer observed wavelengths (redshift) show that the galaxy is moving away from us. A would give a shorter wavelength (blueshift). C is wrong: temperature changes the colour of the continuous spectrum, not a uniform shift of every line. D is wrong: the speed of light in space is constant. | ✓ 1 | Answer B |
Answer: B
Common error: Linking redshift to the temperature of a star. Redshift is about the motion of the source.
Earth orbits the Sun at an average speed of 2.99 × 104 m/s. Mars orbits the Sun at an average radius of 2.28 × 1011 m and takes 687 days to complete one orbit.
Calculate the average orbital speed of Mars. Give your answer in m/s to 3 significant figures.
Explain why the orbital speed of Mars is less than that of Earth.
A rover on Mars has a mass of 180 kg. The gravitational field strength on Mars is 3.7 N/kg. Calculate the weight of the rover on Mars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance = 2πr = 2π × 2.28 × 1011 m | ✓ 1 | |
| M1 time = 687 × 24 × 3600 = 5.94 × 107 s | ✓ 1 | |
| A1 speed = 2.41 × 104 m/s | ✓ 1 | |
| Part (b) | ||
| B1 Mars is further from the Sun, so the Sun's gravitational pull (field strength) on it is weaker | ✓ 1 | |
| B1 a smaller force is needed to keep it in orbit, so it moves more slowly | ✓ 1 | |
| Part (c) | ||
| M1 weight = mass × gravitational field strength = 180 × 3.7 | ✓ 1 | |
| A1 666 N | ✓ 1 | |
Exam tip: Link orbital speed to the strength of the Sun's gravitational field at that distance; do not just say that the orbit is longer.
Light from galaxy P shows a greater redshift than light from galaxy Q. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: redshift is a shift to longer wavelengths caused by recession, and a greater redshift means a greater recession speed, which corresponds to a greater distance. A reverses the relationship. B is wrong: a redshift means P is moving away, not towards Earth. C is wrong: the speed of light does not change; the wavelength changes. | ✓ 1 | Answer D |
Answer: D
Exam tip: Link the three ideas in order: larger redshift, faster recession, greater distance.
A probe moves in a circular orbit of radius r with orbital period T. It is moved to a new circular orbit of radius 4r, where its orbital period is 8T. How does its orbital speed in the new orbit compare with that in the original orbit?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: speed = 2πr/T, so new speed = 2π(4r)/(8T) = half the original. A uses only the factor 4 for the radius. C comes from cancelling the radius and period factors without dividing 4 by 8. D divides by 4 instead of using the ratio 4/8. | ✓ 1 | Answer B |
Answer: B
Common error: Substitute both new values into v = 2πr/T before comparing; do not just scale one quantity.
The Sun is a star of average mass. Some stars have a mass many times that of the Sun.
State the process by which energy is released in the core of a main sequence star.
Explain why a star remains stable throughout its time on the main sequence.
Describe what happens to a star much more massive than the Sun after the main sequence stage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Hydrogen nuclei fuse together to form helium (nuclear fusion) | ✓ 1 | |
| Part (b) | ||
| The inward force of gravity acts to collapse the star | ✓ 1 | |
| This is balanced by the outward pressure from the energy released by fusion | ✓ 1 | |
| Part (c) | ||
| It expands to become a red supergiant | ✓ 1 | |
| It explodes as a supernova | ✓ 1 | |
| The core remains as a neutron star or a black hole | ✓ 1 | |
Common error: A star much more massive than the Sun does not become a white dwarf; it ends as a neutron star or black hole.
A star much more massive than the Sun reaches the end of its time as a main-sequence star. Which sequence of stages follows?
| Marking point | Mark | Notes |
|---|---|---|
| A very massive star expands into a red supergiant, then collapses and explodes as a supernova, leaving a neutron star or, for the largest masses, a black hole. The red giant to white dwarf route is the end of a star of about the Sun's mass. A protostar is the stage before a star joins the main sequence, and a supernova cannot come before the supergiant stage. | ✓ 1 | Answer C |
Answer: C
Common error: The final stage depends on mass: Sun-sized stars end as white dwarfs, much more massive stars end as neutron stars or black holes.
Galaxy P is three times as far from the Earth as galaxy Q. Assume that the speed at which a galaxy moves away from us is directly proportional to its distance.
What is the value of (speed of P) : (speed of Q)?
| Marking point | Mark | Notes |
|---|---|---|
| The speed of recession is given by v = H0d, so speed is directly proportional to distance. Three times the distance means three times the speed, a ratio of 3 : 1. 1 : 3 inverts the ratio, 9 : 1 squares the distance factor, and 1 : 1 ignores the dependence on distance. | ✓ 1 | Answer C |
Answer: C
Common error: Directly proportional means the quantities change by the same factor; do not square the factor unless the relationship says so.
Astronomers measure how fast distant galaxies are moving. Use a Hubble constant of H0 = 2.5 × 10−18 s−1 and 1 year = 3.16 × 107 s.
Describe the change in the light from distant galaxies that shows that the Universe is expanding.
A galaxy is moving away from the Earth at 7.5 × 106 m/s. Use v = H0d to calculate its distance from the Earth. Give your answer to 2 significant figures.
Estimate the time since the Big Bang using 1 / H0. Give your answer in seconds to 2 significant figures.
Convert your answer to (c) into years. Give your answer to 3 significant figures and use the unrounded value.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the wavelength (of the light) is increased / red-shifted | ✓ 1 | |
| this shows the galaxies are moving away from us (so the Universe is expanding) | ✓ 1 | |
| Part (b) | ||
| M1 d = v / H0 = 7.5 × 106 / 2.5 × 10−18 | ✓ 1 | |
| A1 3.0 × 1024 m | ✓ 1 | |
| Part (c) | ||
| M1 1 / (2.5 × 10−18) | ✓ 1 | |
| A1 4.0 × 1017 s | ✓ 1 | |
| Part (d) | ||
| M1 4.0 × 1017 / 3.16 × 107 | ✓ 1 | |
| A1 1.27 × 1010 years (1.2658 × 1010) | ✓ 1 | |
Exam tip: To convert seconds to years divide by the number of seconds in a year; check that the size of your answer is sensible for the age of the Universe.
Light from two distant galaxies, X and Y, is analysed. Light from galaxy X shows a greater redshift than light from galaxy Y.
Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A greater redshift means a greater recession speed, so X is moving away faster. The speed of a galaxy is proportional to its distance (v = H0d), so a faster galaxy is also further away. Both galaxies are redshifted, so both are moving away. | ✓ 1 | Answer B |
Answer: B
Common error: Do not assume a nearer object must be moving faster. In the expanding Universe, more distant galaxies recede faster.
A galaxy is 2.0 × 108 light-years from the Earth.
1 light-year = 9.5 × 1015 m.
The Hubble constant H0 = 2.2 × 10−18 per second.
At what speed is the galaxy moving away from the Earth?
| Marking point | Mark | Notes |
|---|---|---|
| First convert the distance to metres: d = 2.0 × 108 × 9.5 × 1015 = 1.9 × 1024 m. Then v = H0d = 2.2 × 10−18 × 1.9 × 1024 = 4.2 × 106 m/s. Each wrong option comes from a specific mistake: • 4.4 × 10−10 m/s: the distance was not converted from light-years to metres. • 0.13 m/s: the speed of light was wrongly used as the light-year conversion factor. • 8.6 × 1041 m/s: the distance was divided by H0 instead of multiplied. | ✓ 1 | Answer C |
Answer: C
Common error: Convert distances to metres before using H₀ in per second. Otherwise the speed will not come out in m/s.
Mars orbits the Sun at an average radius of 2.28×1011 m. Its orbital period is 687 days.
What is the average orbital speed of Mars?
| Marking point | Mark | Notes |
|---|---|---|
| First convert the period to seconds: T = 687 × 24 × 3600 = 5.94×107 s. Then v = 2πr/T = (2π × 2.28×1011) ÷ 5.94×107 = 2.4×104 m/s. Leaving out the 2π gives 3.8×103 m/s. Using πr (half the circumference) gives 1.2×104 m/s. Leaving T in days gives 2.1×109 m/s. | ✓ 1 | Answer C |
Answer: C
Common error: Always convert the period into seconds before using v = 2πr/T if you want the speed in m/s.
A comet of mass 2.0×1014 kg orbits the Sun. At its furthest point from the Sun its speed is 900 m/s. As it travels from its furthest point to its closest point, it loses 4.5×1022 J of gravitational potential energy. Assume that no energy is lost.
Describe the shape of the comet's orbit and the position of the Sun within it.
State the energy change as the comet moves towards the Sun.
Calculate the kinetic energy of the comet at its furthest point from the Sun.
Calculate the speed of the comet at its closest point to the Sun.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The orbit is an ellipse / very elongated | ✓ 1 | |
| The Sun is not at the centre of the orbit; it is near one end (at one focus) | ✓ 1 | |
| Part (b) | ||
| Gravitational potential energy decreases | ✓ 1 | |
| and is converted into kinetic energy, which increases | ✓ 1 | |
| Part (c) | ||
| KE = ½mv2 = 0.5 × 2.0×1014 × 9002 | ✓ 1 | |
| = 8.1×1019 J | ✓ 1 | |
| Part (d) | ||
| KE at closest point = 8.1×1019 + 4.5×1022 = 4.5×1022 J (4.508×1022 J) | ✓ 1 | |
| v = √(2 × KE ÷ m) = √(2 × 4.508×1022 ÷ 2.0×1014) | ✓ 1 | |
| = 2.1×104 m/s | ✓ 1 | |
Exam tip: Add the starting kinetic energy to the loss of potential energy to find the final kinetic energy. Then rearrange KE = ½mv² to find v.
The Hubble constant is about 2.2 × 10−18 s−1. Taking 1 year as 3.15 × 107 s, which value is the best estimate of the age of the Universe, given by 1 ÷ H0?
| Marking point | Mark | Notes |
|---|---|---|
| 1 ÷ H0 = 1 ÷ 2.2 × 10−18 = 4.5 × 1017 s; dividing by 3.15 × 107 s per year gives 1.44 × 1010 years: B is correct. A multiplies by the seconds in a year instead of dividing. C converts using 60 × 24 × 365 (forgetting the 60 s in a minute). D leaves the answer in seconds but labels it years. | ✓ 1 | Answer B |
Answer: B
Common error: 1 ÷ H₀ comes out in seconds; convert to years by dividing by the number of seconds in a year.
A communications satellite is in a geostationary orbit of radius 4.2 × 107 m about the centre of the Earth.
Describe three features of a geostationary orbit. [3]
Calculate the orbital speed of the satellite. [3]
A second satellite orbits much closer to the Earth. Explain why its orbital period is shorter. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the orbit period is 24 hours, equal to the Earth's rotation period | ✓ 1 | |
| the orbit is above the equator | ✓ 1 | |
| the satellite moves in the same direction as the Earth's rotation, so it stays above a fixed point | ✓ 1 | |
| Part (b) | ||
| orbital period T = 24 × 3600 = 8.64 × 104 s | ✓ 1 | |
| v = 2πr ÷ T | ✓ 1 | |
| = 2π × 4.2 × 107 ÷ 8.64 × 104 = 3054 m/s (3.1 × 103 m/s) | ✓ 1 | |
| Part (c) | ||
| the gravitational force and field strength are greater closer to the Earth, so it must travel faster | ✓ 1 | |
| it also has a smaller circumference to cover, so each orbit takes less time | ✓ 1 | |
Common error: Convert the period into seconds before using v = 2πr ÷ T.
The age of the Universe can be estimated as 1/H0, where H0 is the Hubble constant, assuming the expansion rate has always been the same. Later measurements show that the true value of H0 is 25% larger than the value first used. By what percentage does the estimated age change?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: new age = 1/(1.25 H0) = 0.80 × old age, a decrease of 20%. A assumes the age falls by the same percentage that H0 rose. C and D have the age increasing, wrongly treating age as proportional to H0 instead of inversely proportional. | ✓ 1 | Answer B |
Answer: B
Common error: Age is inversely proportional to H₀, so a larger H₀ means a younger Universe, and the percentage change is not the same size.
Light from a distant galaxy is red-shifted. The Hubble constant is H0 = 2.3×10−18 s−1. The galaxy is 6.0×1023 m from the Earth. Use 1 year = 3.15×107 s.
Describe what is meant by red-shift and state what it indicates about the galaxy.
Calculate the speed at which the galaxy is moving away from the Earth.
The age of the Universe can be estimated as 1/H0. Calculate this age in years, assuming the expansion rate has always been the same. Give your answer to 2 significant figures.
Suggest why the estimate in (c) would be wrong if the rate of expansion has changed over time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the wavelengths of the light received are longer than those emitted (shifted towards the red end) | ✓ 1 | |
| the galaxy is moving away from the Earth | ✓ 1 | |
| Part (b) | ||
| v = H0 d = 2.3×10−18 × 6.0×1023 | ✓ 1 | |
| 1.38×106 m/s | ✓ 1 | |
| Part (c) | ||
| 1/H0 = 1 / 2.3×10−18 = 4.35×1017 s | ✓ 1 | |
| divide by 3.15×107 s per year | ✓ 1 | |
| 1.4×1010 years (1.38×1010 years unrounded) | ✓ 1 | |
| Part (d) | ||
| the estimate assumes a constant speed of expansion, so a changing rate would give a different time | ✓ 1 | |
Common error: H₀ has units of s⁻¹, so 1/H₀ is a time in seconds and must be converted to years.
A satellite carries signals for card payments in a retail chain. It is in a geostationary orbit, which is a circular orbit of radius 4.2×107 m from the centre of the Earth, with a period of 24 hours. The satellite is 3.6×107 m above a card terminal and the speed of light is 3.0×108 m/s.
State two features of a geostationary orbit that keep the satellite above the same point on the Earth.
Calculate the orbital speed of the satellite in m/s. Give your answer to 2 significant figures.
A signal is sent from the terminal, which is directly below the satellite, up to the satellite and then straight back down to the ground. Calculate the time taken.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the orbital period is 24 hours, equal to the rotation period of the Earth | ✓ 1 | |
| the satellite moves in the same direction as the Earth's rotation (above the equator) | ✓ 1 | |
| Part (b) | ||
| T = 24 × 3600 = 86 400 s | ✓ 1 | |
| v = 2πr / T = 2π × 4.2×107 / 86 400 | ✓ 1 | |
| 3.1×103 m/s (3054 m/s unrounded) | ✓ 1 | |
| Part (c) | ||
| distance = 2 × 3.6×107 = 7.2×107 m | ✓ 1 | |
| t = d / c = 7.2×107 / 3.0×108 = 0.24 s | ✓ 1 | |
Exam tip: The signal travels up and back down, so use twice the height above the ground, not the orbital radius.
The Hubble constant is 2.5 × 10−18 s−1. One year is 3.15 × 107 s. Using 1 / H0 as an estimate, what is the age of the Universe in years?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: 1 / H0 = 1 / (2.5 × 10−18) = 4.0 × 1017 s, and dividing by 3.15 × 107 s per year gives 1.27 × 1010 years. A multiplies H0 by the number of seconds in a year. C converts seconds into hours instead of years. D leaves the answer in seconds. | ✓ 1 | Answer B |
Answer: B
Common error: Forgetting that 1 / H₀ comes out in seconds; it must be divided by the seconds in a year.
The Hubble constant is H0 = 2.1 × 10−18 s−1. A galaxy is a distance of 3.0 × 1024 m from Earth. The light from this galaxy is redshifted.
State what is meant by redshift.
Calculate the speed at which the galaxy is moving away from Earth, using v = H0 d. Give your answer in m/s to 2 significant figures.
Use 1 / H0 to estimate the age of the Universe in years. One year is 3.15 × 107 s. Give your answer to 3 significant figures.
State one piece of evidence for the Big Bang theory other than redshift.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the wavelength of light received from the galaxy is longer than when it was emitted | ✓ 1 | |
| B1 caused by the galaxy moving away from Earth | ✓ 1 | |
| Part (b) | ||
| M1 v = 2.1 × 10−18 × 3.0 × 1024 | ✓ 1 | |
| A1 6.3 × 106 m/s | ✓ 1 | |
| Part (c) | ||
| M1 age = 1 / H0 = 1 / (2.1 × 10−18) | ✓ 1 | |
| A1 4.76 × 1017 s | ✓ 1 | |
| A1 1.51 × 1010 years | ✓ 1 | |
| Part (d) | ||
| B1 cosmic microwave background radiation | ✓ 1 | |
Common error: Leaving the age in seconds. Divide by the number of seconds in a year to give years.
A star has a mass 20 times the mass of the Sun. Volcanic rock on Earth contains iron and other heavy elements.
Describe the stages in the life of this star after it leaves the main sequence.
Explain why the Sun will not end its life as a neutron star.
Explain how heavy elements, such as iron, came to be in the rock on Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 it expands to become a red supergiant | ✓ 1 | |
| B1 it explodes as a supernova | ✓ 1 | |
| B1 the core remains as a neutron star or a black hole | ✓ 1 | |
| Part (b) | ||
| B1 the Sun does not have enough mass to form a supernova (to collapse its core that far) | ✓ 1 | |
| B1 it will become a red giant and then a white dwarf instead | ✓ 1 | |
| Part (c) | ||
| B1 they were made by nuclear fusion in earlier, massive stars (and in supernova explosions) | ✓ 1 | |
| B1 supernovae scattered them into space, where they became part of the nebula that formed the Solar System | ✓ 1 | |
Exam tip: Use the word 'mass' in any answer about different star fates; the mass decides the path.
Two research teams estimate the age of the Universe using age ≈ 1/H0, where H0 is the Hubble constant. Team 2 obtains a larger value of H0 than team 1. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: age ≈ 1/H₀, so a larger H₀ gives a smaller age. A confuses a faster expansion rate with a longer time. C is wrong because the estimate is calculated from H₀. D gets the right trend but gives the wrong proportionality; a directly proportional age would increase with H₀. | ✓ 1 | Answer B |
Answer: B
Exam tip: If the formula is 1/H₀, a bigger denominator means a smaller answer; check the reason as well as the trend.
Astronomers observe the spectrum of light from a galaxy. A hydrogen line that has a wavelength of 656 nm in the laboratory is detected from the galaxy. The galaxy is 4.5 × 1023 m from the Earth. The Hubble constant is H0 = 2.2 × 10−18 /s. The speed of light is 3.0 × 108 m/s.
Calculate the recession speed of the galaxy.
A second galaxy shows a hydrogen line with twice the shift in wavelength. State how far away it is compared with the first galaxy and state what this suggests about the Universe.
Calculate the increase in wavelength of the hydrogen line from the first galaxy. Use 9.9 × 105 m/s for the speed and give your answer in nm to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v = H0d = 2.2 × 10−18 × 4.5 × 1023 | ✓ 1 | |
| A1 9.9 × 105 m/s | ✓ 1 | |
| Part (b) | ||
| It is twice as far away | ✓ 1 | |
| the Universe is expanding (galaxies further away recede faster) | ✓ 1 | |
| Part (c) | ||
| M1 Δλ / λ = v / c | ✓ 1 | |
| M1 Δλ = 656 × 9.9 × 105 / 3.0 × 108 = 2.1648 nm | ✓ 1 | |
| A1 2.2 nm | ✓ 1 | |
Exam tip: The wavelength shift is a small fraction of the original wavelength, so check that your answer is much smaller than 656 nm.
Astronomers estimate the age of the Universe from the Hubble constant. A recent measurement gives H0 = 2.3 × 10−18 /s. Use age = 1 / H0 and 1 year = 3.15 × 107 s.
Calculate the age of the Universe in seconds. Give your answer to 3 significant figures.
Convert your answer to years. Use the unrounded value from (a) and give your answer to 3 significant figures.
A later, more precise experiment gives a larger value of H0. State and explain the effect on the calculated age.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 age = 1 / (2.3 × 10−18) = 4.3478 × 1017 s | ✓ 1 | |
| A1 4.35 × 1017 s | ✓ 1 | |
| Part (b) | ||
| M1 years = 4.3478 × 1017 / 3.15 × 107 | ✓ 1 | |
| A1 1.38 × 1010 years | ✓ 1 | |
| Part (c) | ||
| The calculated age decreases | ✓ 1 | |
| age = 1 / H0, so age is inversely proportional to H0 | ✓ 1 | |
Common error: The age is 1 divided by H₀; do not multiply H₀ by anything to get a time.
Galaxy X is 2.0 × 1024 m from Earth and is moving away at 4.4 × 106 m/s. Galaxy Y is 5.0 × 1024 m from Earth. Assuming the Universe expands uniformly, what is the speed at which Galaxy Y is moving away from Earth?
| Marking point | Mark | Notes |
|---|---|---|
| Recession speed is proportional to distance (v = H0d). The distance ratio is 5.0 / 2.0 = 2.5, so the speed is 2.5 × 4.4 × 106 = 1.1 × 107 m/s. Keeping 4.4 × 106 m/s assumes speed does not depend on distance; 1.8 × 106 m/s divides by 2.5; 2.2 × 107 multiplies by 5.0 instead of the ratio 2.5. | ✓ 1 | Answer B |
Answer: B
Common error: Use the ratio of distances, since speed is directly proportional to distance.
Jupiter moves around the Sun in a circular orbit of radius 7.8 × 1011 m at an orbital speed of 1.3 × 104 m/s. Use 1 year = 3.16 × 107 s.
Calculate the orbital period of Jupiter in seconds. Give your answer to 3 significant figures.
Calculate the orbital period in years. Give your answer to 3 significant figures. (Use either your unrounded answer or 3.77 × 108 s.)
A planet orbits the Sun at a larger radius than Jupiter. State and explain how its orbital speed compares with that of Jupiter.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 distance in one orbit = 2πr | ✓ 1 | |
| M1 T = 2πr / v = 2π × 7.8 × 1011 / 1.3 × 104 | ✓ 1 | |
| A1 T = 3.77 × 108 s | ✓ 1 | |
| Part (b) | ||
| M1 divide by 3.16 × 107 | ✓ 1 | |
| A1 period = 11.9 years | ✓ 1 | |
| Part (c) | ||
| Its orbital speed is smaller | ✓ 1 | |
| The Sun's gravitational force on it is weaker at larger distance, so less speed is needed to stay in orbit | ✓ 1 | |
Exam tip: Convert the period into years only at the end, and use the unrounded value from the earlier part.
Planet P orbits a star in a circular orbit of radius r with period T. Planet Q orbits the same star in a circular orbit of radius 4r with period 8T. What is the value of (orbital speed of Q) ÷ (orbital speed of P)?
| Marking point | Mark | Notes |
|---|---|---|
| Orbital speed is \(2\pi r/T\). For Q the radius is four times larger and the period eight times larger, so the speed is multiplied by \(4/8=0.50\). Using 8/4 gives 2.0, using only the radius ratio gives 4.0, and multiplying the two ratios gives 32. | ✓ 1 | Answer B |
Answer: B
Common error: Speed is distance divided by time, so a larger period reduces the speed, while a larger radius increases it.
Stars form, shine and eventually die. Many of the heavy elements found in the rocks of a volcano were made inside earlier stars.
Describe how a protostar forms from a cloud of gas and dust.
Explain why a star remains stable for most of its life.
A star with a mass much greater than the mass of the Sun reaches the end of its stable life. State three stages that follow, in the correct order.
Suggest how the heavy elements made in a star can end up in the rocks of a planet.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gravity pulls the gas and dust together | ✓ 1 | |
| gravitational potential energy is transferred to thermal energy so the temperature rises | ✓ 1 | |
| when the temperature is high enough, nuclear fusion of hydrogen begins and the star becomes stable | ✓ 1 | |
| Part (b) | ||
| the inward force of gravity is balanced by the outward pressure from the fusion energy | ✓ 1 | |
| the star neither expands nor contracts while hydrogen is available as fuel | ✓ 1 | |
| Part (c) | ||
| red supergiant | ✓ 1 | |
| supernova | ✓ 1 | |
| neutron star or black hole | ✓ 1 | |
| Part (d) | ||
| a supernova explosion scatters the elements into space, where they join a new cloud of gas and dust | ✓ 1 | |
Common error: A star is stable because forces are balanced, not because nothing is happening: fusion is releasing energy all the time.
Mars orbits the Sun in a path that can be treated as a circle of radius 2.3 × 108 km. One orbit takes 687 days. 1 day = 86 400 s.
What is the orbital speed of Mars in m/s?
| Marking point | Mark | Notes |
|---|---|---|
| Radius = 2.3 × 1011 m and period = 687 × 86 400 = 5.94 × 107 s. Speed = 2πr / T = 2π × 2.3 × 1011 / 5.94 × 107 = 2.4 × 104 m/s. 3.9 × 103 omits the factor 2π. 2.1 × 109 uses the period in days instead of seconds. 2.4 × 101 leaves the radius in km, which gives km/s, not m/s. | ✓ 1 | Answer C |
Answer: C
Exam tip: Convert the radius to metres and the period to seconds before using v = 2πr / T.
Observations give a Hubble constant of H0 = 2.4 × 10−18 s−1. Estimate the age of the Universe using age = 1 / H0. Use 1 year = 3.2 × 107 s.
What is the age in years?
| Marking point | Mark | Notes |
|---|---|---|
| Age = 1 / (2.4 × 10−18) = 4.17 × 1017 s. Dividing by 3.2 × 107 s per year gives 1.3 × 1010 years. 4.2 × 1017 is the answer left in seconds. 7.7 × 10−11 multiplies H0 by the seconds in a year instead of dividing. 4.8 × 1012 divides by the seconds in a day, which gives days, not years. | ✓ 1 | Answer B |
Answer: B
Exam tip: The unit of H₀ is s⁻¹, so 1 / H₀ is in seconds and must be converted to years.
A sports scientist models a vertical throw for an astronaut training programme. A ball is thrown vertically upwards at 4.0 m/s on the Moon, where the gravitational field strength is 1.6 N/kg. Ignore air resistance except where stated.
State why the ball on the Moon is not affected by air resistance.
Calculate the maximum height reached by the ball above the point of release on the Moon.
Calculate the time taken for the ball to reach its highest point on the Moon.
The same throw, at 4.0 m/s, is made on the Earth where g = 9.8 N/kg. Ignoring air resistance, calculate the maximum height reached. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the Moon has no atmosphere / there is no air | ✓ 1 | |
| Part (b) | ||
| M1 0 = 4.02 − 2 × 1.6 × h (or h = u2 / 2g) | ✓ 1 | |
| A1 5.0 m | ✓ 1 | |
| Part (c) | ||
| M1 t = u / g = 4.0 / 1.6 | ✓ 1 | |
| A1 2.5 s | ✓ 1 | |
| Part (d) | ||
| M1 h = 4.02 / (2 × 9.8) | ✓ 1 | |
| A1 0.82 m (0.816 m) | ✓ 1 | |
Exam tip: Use the equation v² = u² + 2as with v = 0 at the top and a = −g.
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