Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
Doppler shift, redshift and the universe
32 free, high-quality past-paper-style questions on Doppler shift, redshift and the universe for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.32 new this month · 114 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U5-9-A-01Unit 5 · A2Easy1 mark
An ambulance sounding its siren moves at constant speed directly towards a listener who is standing still. Compared with the sound emitted by the siren, which statement describes the sound received by the listener?
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Answer
Mark
Notes
The siren moves towards the listener between emitting one wavefront and the next, so the wavefronts arriving are closer together: a shorter wavelength. The speed of sound in air is unchanged, so by f = v/λ the frequency received is higher.
✓ 1
Answer D
Answer: D
Common error: The motion of the source changes the spacing of the wavefronts, not the speed of the wave. The speed of sound is set by the air.
A hydrogen line with a laboratory wavelength of 486.1 nm is observed at 488.5 nm in the spectrum of a galaxy. What is the speed of the galaxy along the line of sight?
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Answer
Mark
Notes
Δλ = 488.5 − 486.1 = 2.4 nm. v = cΔλ/λ = 3.00 × 108 × 2.4 ÷ 486.1 = 1.5 × 106 m s−1, away from the observer.
✓ 1
Answer A
Answer: A
Exam tip: Δλ and λ can both stay in nanometres: the units cancel in the ratio. Converting only one of them is a common slip.
Which observation is evidence for the existence of dark matter?
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Answer
Mark
Notes
If only the visible mass, concentrated near the centre, were present, orbital speeds would fall with distance (v = √(GM/r)). Speeds that stay roughly constant far out mean the mass inside the orbit keeps increasing: there is unseen mass — dark matter.
✓ 1
Answer D
Answer: D
Exam tip: Link the rotation curve to v = √(GM/r): a flat curve needs M to grow in proportion to r.
A train sounds its horn at a constant frequency as it travels at constant speed along a straight track towards a passenger standing on a platform. The train then passes the platform and moves away.
(a)
State how the frequency heard by the person on the platform compares with the frequency emitted while the train approaches.
(1)
(b)
Explain, in terms of wavefronts, why the wavelength of the sound reaching the platform is shorter while the train approaches.
(2)
(c)
A passenger sitting on the train hears the horn at the frequency emitted. Explain why.
(1)
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Answer
Mark
Notes
Part (a)
higher
✓ 1
Part (b)
the horn moves towards the platform between emitting one wavefront and the next / each wavefront is emitted from a point closer to the listener
✓ 1
so the wavefronts are closer together ahead of the train (the speed of sound is unchanged)
✓ 1
Part (c)
there is no relative motion between the horn and the passenger on the train
✓ 1
Model sketch
Figure: wavefronts from the moving horn (not to scale)
Common error: Writing that the sound travels faster towards the platform. The wave speed depends on the air, not on the motion of the horn.
A calcium absorption line has a laboratory wavelength of 393.4 nm. In the spectrum of a distant galaxy the same line is observed at 401.2 nm. The Hubble constant is 70 km s−1 Mpc−1.
Figure: position of the calcium line in the laboratory and in the galaxy's spectrum (not to scale)
(a)
State what is meant by redshift.
(1)
(b)
Calculate the redshift z of the galaxy.
(1)
(c)
Calculate the recession speed of the galaxy.
(1)
(d)
Calculate the distance to the galaxy in Mpc.
(2)
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Answer
Mark
Notes
Part (a)
an increase in the observed wavelength (decrease in frequency) of radiation from a source moving away from the observer
✓ 1
Part (b)
z = (401.2 − 393.4) ÷ 393.4 = 0.0198
✓ 1
Part (c)
v = zc = 0.01983 × 3.00 × 108 = 5.95 × 106 m s−1
✓ 1
Part (d)
v = 5950 km s−1
✓ 1
d = v/H0 = 5950 ÷ 70 = 85 Mpc
✓ 1
Exam tip: Keep v in km s−1 when H0 is in km s−1 Mpc−1; the distance then comes out directly in Mpc.
Figure 1 shows the wavefronts of sound from a source S at one instant. S moves at constant speed along the line joining two stationary observers, P and R. How do the frequencies heard by P and by R compare with the frequency emitted by S?
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Answer
Mark
Notes
Each wavefront is centred where S was when it was emitted. The centres are displaced towards P, so S is moving towards P: the wavefronts are closer together on P's side (shorter wavelength, higher frequency) and further apart on R's side (longer wavelength, lower frequency).
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Answer D
Answer: D
Exam tip: In a wavefront diagram the source is moving towards the side where the wavefronts are crowded together. Find the direction of motion first, then decide each observer's frequency.
Figure 1 shows the position of a hydrogen absorption line in a laboratory spectrum and in the spectrum of a distant galaxy. The laboratory wavelength of the line is 656.3 nm. What is the recession speed of the galaxy?
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Answer
Mark
Notes
The galaxy line is at 670.0 nm. Δλ = 670.0 − 656.3 = 13.7 nm, so z = 13.7 ÷ 656.3 = 0.0209. v = zc = 0.0209 × 3.00 × 108 = 6.3 × 106 m s−1.
✓ 1
Answer B
Answer: B
Common error: Using c = 3.00 × 105 (the value in km s−1) and then quoting the answer in m s−1. Keep the unit of c and the unit of the answer the same.
The bar chart shows the mean density of four model universes, W, X, Y and Z, compared with the critical density. All four are expanding now. Consider the effect of gravity alone. Which universe or universes will eventually stop expanding and then contract?
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Answer
Mark
Notes
Only a universe with a mean density greater than the critical density has enough gravity to stop the expansion and reverse it. That is Y only. X, at exactly the critical density, keeps expanding with its rate approaching zero; W and Z expand for ever.
✓ 1
Answer A
Answer: A
Exam tip: Compare each density with the critical density: above it, the expansion stops and reverses; equal to it, the expansion rate approaches zero but never reverses; below it, the expansion continues for ever.
Figure 1 shows wavefronts of sound spreading out from a stationary loudspeaker S. One grid square represents 0.20 m. A cyclist C rides directly towards S at a constant speed. The speed of sound in air is 340 m s−1.
(a)
Use Figure 1 to determine the frequency of the sound emitted by S.
(2)
(b)
S is stationary, so the spacing of the wavefronts in Figure 1 does not depend on where the listener is. Explain why the cyclist nevertheless hears a frequency higher than your answer to (a).
(2)
(c)
The cyclist rides past S and then away from it at the same speed. State how the frequency the cyclist hears now compares with your answer to (a). Give a reason.
(1)
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Answer
Mark
Notes
Part (a)
wavelength = 4 squares × 0.20 m = 0.80 m
✓ 1
f = 340 ÷ 0.80 = 425 Hz
✓ 1
Part (b)
the wavefronts move towards the cyclist while the cyclist moves towards them
✓ 1
so the cyclist meets more wavefronts each second than a stationary listener does / the time between wavefronts arriving is shorter
✓ 1
Part (c)
lower than 425 Hz, because the cyclist now meets fewer wavefronts each second (the wavefronts have to catch up with the cyclist)
✓ 1
Common error: Thinking the wavelength changes when the listener moves. A moving listener changes how often the wavefronts arrive, not their spacing; a moving source changes the spacing.
A satellite transmits radio waves of frequency 400 MHz. As it passes over a ground station, the station records the change Δf between the received frequency and 400 MHz. The graph shows Δf against time t; t = 0 is the moment of closest approach. The satellite moves at a constant speed of 7.5 km s−1.
(a)
Explain why Δf is positive before t = 0.
(1)
(b)
Use point P to calculate the component of the satellite's velocity along the line to the station at t = −5.0 minutes.
(2)
(c)
Explain why Δf = 0 at t = 0.
(1)
(d)
Calculate the angle between the satellite's velocity and the line to the station at t = −5.0 minutes.
(1)
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Answer
Mark
Notes
Part (a)
the satellite is moving towards the station, so the waves received are compressed / of shorter wavelength and higher frequency
at closest approach the satellite's velocity is at right angles to the line to the station, so it has no component towards or away from the station
✓ 1
Part (d)
cos θ = 6750 ÷ 7500 = 0.90, θ = 26° (allow 25°–26°, e.g. from 6.8 × 103 m s−1)
✓ 1
Exam tip: Only the component of velocity along the line of sight causes a Doppler shift. Motion at right angles to the line of sight gives no shift (at the speeds met in this course).
The Hubble constant is taken as 72 km s−1 Mpc−1. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s. Assuming the universe has always expanded at the same rate, what is its age?
Considering gravity alone, the mean density of a model universe is greater than the critical density. What is the ultimate fate of this universe?
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Answer
Mark
Notes
Above the critical density, gravity is strong enough to stop the expansion and reverse it; the universe then contracts. At exactly the critical density the expansion rate approaches zero after infinite time; below it, expansion continues for ever.
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Answer C
Answer: C
Exam tip: Learn the three cases as a set: density above, equal to and below the critical density give collapse, expansion slowing towards zero, and expansion for ever.
New measurements increase the accepted value of the Hubble constant by 10%. The age of the universe is estimated as 1/H0. By what factor does this estimate change?
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Answer
Mark
Notes
Age ∝ 1/H0, so the new estimate is 1 ÷ 1.10 = 0.91 of the old one: a faster expansion rate means the galaxies took less time to separate.
✓ 1
Answer B
Answer: B
Common error: Thinking a larger Hubble constant means an older universe. A larger H0 is a faster expansion, so less time was needed.
A student analyses data for five galaxies. The distances were found using standard candles and the recession speeds from redshifts.
Galaxy
A
B
C
D
E
distance / Mpc
25
52
80
112
140
recession speed / km s−1
1900
3700
5900
8100
10 200
She plots recession speed against distance and draws a best-fit straight line through the origin. The line passes through the point for galaxy E. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.Figure: recession speed against distance for the five galaxies
(a)
Describe how the recession speed of a galaxy is found from its spectrum.
(2)
(b)
Determine the Hubble constant in km s−1 Mpc−1.
(1)
(c)
Convert this value to s−1.
(1)
(d)
Estimate the age of the universe in years.
(2)
(e)
Suggest one reason why the points do not all lie exactly on the line.
(1)
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Answer
Mark
Notes
Part (a)
measure the wavelength of an identified spectral line and compare it with the laboratory wavelength of the same line
✓ 1
v = cΔλ/λ
✓ 1
Part (b)
gradient = 10 200 ÷ 140 = 72.9 km s−1 Mpc−1
✓ 1
Part (c)
72.9 × 103 ÷ 3.09 × 1022 = 2.36 × 10−18 s−1
✓ 1
Part (d)
1/H0 = 1 ÷ 2.36 × 10−18 = 4.24 × 1017 s
✓ 1
÷ 3.15 × 107 = 1.35 × 1010 years
✓ 1
Part (e)
the distances are uncertain because the luminosities of the standard candles are uncertain / galaxies have their own motions in addition to the expansion
✓ 1
Model sketch
Figure: best-fit line through the origin and galaxy E
Exam tip: Read the gradient of a Hubble graph from the best-fit line, not from a single data point, unless the question tells you that the point lies on the line.
Cold hydrogen gas in space emits radio waves of frequency 1420 MHz. A radio telescope receives this emission from a distant galaxy at a frequency of 1385 MHz. The Hubble constant is 68 km s−1 Mpc−1. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.
(a)
Calculate the redshift of the galaxy, using z ≈ Δf/f.
(1)
(b)
Calculate the recession speed of the galaxy.
(1)
(c)
Calculate the distance to the galaxy in Mpc.
(1)
(d)
Estimate how many years the radio waves have been travelling.
(2)
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Answer
Mark
Notes
Part (a)
z = (1420 − 1385) ÷ 1420 = 0.0246
✓ 1
Part (b)
v = 0.02465 × 3.00 × 108 = 7.39 × 106 m s−1
✓ 1
Part (c)
d = 7394 ÷ 68 = 109 Mpc (108.7)
✓ 1
Part (d)
d = 108.7 × 3.09 × 1022 = 3.36 × 1024 m
✓ 1
t = d/c = 1.12 × 1016 s = 3.6 × 108 years
✓ 1
Common error: Dividing Δf by the received frequency instead of the emitted one. For small z the difference is tiny, but the method should use the emitted value.
A star orbits a companion star. Its orbit is circular and is seen edge-on from the Earth. Over several weeks, the wavelength of a line in the star's spectrum is measured; the laboratory wavelength of this line is 600.00 nm. The observed wavelength varies smoothly between a minimum of 599.96 nm and a maximum of 600.08 nm and repeats every 6.0 days.
Figure: observed wavelength of the line minus 600.00 nm, against time (range 599.96 nm to 600.08 nm)
(a)
Explain why the observed wavelength varies with a period of 6.0 days.
(2)
(b)
Show that the pair of stars as a whole is moving away from the Earth at about 1 × 104 m s−1.
(2)
(c)
Calculate the orbital speed of the star.
(1)
(d)
Calculate the radius of the star's orbit.
(2)
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Answer
Mark
Notes
Part (a)
as the star orbits it moves alternately towards and away from the Earth, once every 6.0 days
✓ 1
the Doppler effect shortens the wavelength when it approaches and lengthens it when it recedes
The graph shows the recession speed v of nine galaxies plotted against their distance d, with a best-fit line through the origin. 1 Mpc = 3.09 × 1022 m. What is the Hubble constant, in s−1?
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Answer
Mark
Notes
Gradient = 12 000 ÷ 160 = 75 km s−1 Mpc−1. Convert: 75 km s−1 = 7.5 × 104 m s−1, and 1 Mpc = 3.09 × 1022 m, so H0 = 7.5 × 104 ÷ 3.09 × 1022 = 2.4 × 10−18 s−1.
✓ 1
Answer B
Answer: B
Exam tip: Take the gradient from two points on the best-fit line that are far apart and lie on gridline intersections, not from a single plotted galaxy.
A star moves in an orbit, so its velocity along the line of sight from the Earth changes as shown in the graph. A positive velocity means the star is moving away from the Earth. A sodium line in the star's spectrum has a laboratory wavelength of 589.00 nm. What is the shortest wavelength observed for this line?
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Answer
Mark
Notes
The shortest wavelength occurs when the star moves towards the Earth fastest: the minimum of the graph, v = −20 km s−1. Δλ = λv/c = 589.00 × 2.0 × 104 ÷ 3.00 × 108 = 0.039 nm, so λ = 589.00 − 0.04 = 588.96 nm.
✓ 1
Answer C
Answer: C
Exam tip: The whole system can also be moving: here the curve is centred on +20 km s−1, not on zero. Read the actual extreme value from the graph rather than using the amplitude.
A small loudspeaker emitting a note of constant frequency is fixed to the end of a rotating arm, so that it moves anticlockwise at constant speed around the circle shown in Figure 1. A listener L is a long way away, level with the centre of the circle. At which position is the loudspeaker when it emits the sound that L hears with the highest frequency?
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Answer
Mark
Notes
The frequency heard is highest when the source is moving directly towards L. Moving anticlockwise, the loudspeaker at the bottom of the circle moves to the right, straight towards L: position R. At P it moves directly away (lowest frequency); at Q and S it moves at right angles to the line to L, so there is no shift.
✓ 1
Answer C
Answer: C
Common error: Choosing the position nearest to the listener. The Doppler shift depends on the velocity towards or away from the listener, not on the distance.
Figure 1 is a scale drawing of the wavefronts of sound from a source S that is moving at constant speed along the horizontal line. The wavefronts were emitted at intervals of one period of the source. One grid square represents 5.0 cm. The speed of sound in air is 340 m s−1.
(a)
State the direction in which S is moving and explain how Figure 1 shows this.
(2)
(b)
Use Figure 1 to determine the frequency heard by a stationary observer on the line to the right of S.
(2)
(c)
Determine the frequency heard by a stationary observer on the line to the left of S.
(1)
(d)
S emits sound of frequency 680 Hz. Show that the speed of S is about 70 m s−1.
(2)
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Answer
Mark
Notes
Part (a)
to the right
✓ 1
the centres of the circles move to the right / the wavefronts are closer together on the right-hand side
✓ 1
Part (b)
wavelength = 8 squares × 5.0 cm = 0.40 m
✓ 1
f = 340 ÷ 0.40 = 850 Hz
✓ 1
Part (c)
λ = 12 × 5.0 cm = 0.60 m, f = 340 ÷ 0.60 = 570 Hz (567)
✓ 1
Part (d)
λ0 = 340 ÷ 680 = 0.50 m / period T = 1/680 s
✓ 1
in one period S moves 0.50 − 0.40 = 0.10 m (2 squares between centres), so v = 0.10 × 680 = 68 m s−1
✓ 1
Common error: Measuring the wavelength between the source and the first wavefront. Measure between neighbouring wavefronts along the line of motion.
Figure 1 shows part of the emission spectrum of oxygen gas in a laboratory, and the same part of the spectra of two distant galaxies, P and Q. The laboratory wavelength of the emission line is 500.7 nm. The Hubble constant is 71 km s−1 Mpc−1.
(a)
State which galaxy is further from the Earth. Give a reason for your answer.
(1)
(b)
Use Figure 1 to determine the redshift z of galaxy Q.
(2)
(c)
Calculate the ratio (distance to Q) ÷ (distance to P).
(2)
(d)
Calculate the distance to galaxy Q in Mpc.
(2)
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Answer
Mark
Notes
Part (a)
Q, because its line has the greater redshift, so it is receding faster and recession speed is proportional to distance
An astronomer measures the redshift z and the distance d of nine galaxies and plots the graph shown, with distance in metres. The best-fit line through the origin passes through the gridline intersection at 8.0 × 1024 m, z = 0.060. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.
(a)
Explain how the graph supports the idea that the universe is expanding.
(2)
(b)
Use the graph to show that the Hubble constant is about 2.3 × 10−18 s−1.
(2)
(c)
Estimate the age of the universe in years.
(2)
(d)
Express the Hubble constant in km s−1 Mpc−1.
(1)
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Answer
Mark
Notes
Part (a)
all the galaxies show a redshift, so all are moving away from us
✓ 1
z (and so recession speed) is proportional to distance, as expected if space itself is expanding uniformly
✓ 1
Part (b)
gradient = 0.060 ÷ 8.0 × 1024 = 7.5 × 10−27 m−1
✓ 1
v = zc, so H0 = c × gradient = 3.00 × 108 × 7.5 × 10−27 = 2.25 × 10−18 s−1
A galaxy is seen edge-on. A spectral line with a laboratory wavelength of 500.0 nm is observed at 503.2 nm from one edge of the galaxy and at 502.0 nm from the opposite edge. What is the orbital speed of the stars at the edges?
Figure: galaxy seen edge-on, with the wavelength observed from each edge (not to scale)
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Answer
Mark
Notes
The galaxy's own recession shifts both edges equally; rotation adds to one shift and subtracts from the other. Rotation shift = (503.2 − 502.0) ÷ 2 = 0.6 nm, so v = 3.00 × 108 × 0.6 ÷ 500.0 = 3.6 × 105 m s−1.
✓ 1
Answer A
Answer: A
Exam tip: For a rotating object, the average shift gives the motion of the whole object; half the difference between the two edges gives the rotation speed.
Astronomers measure the orbital speeds of gas clouds in a spiral galaxy from Doppler shifts. At 8.0 kpc from the centre the speed is 230 km s−1; at 25 kpc it is 225 km s−1. Almost all of the galaxy's visible stars lie within 8.0 kpc of the centre. Assume circular orbits. 1 kpc = 3.09 × 1019 m; G = 6.67 × 10−11 N m2 kg−2.
(a)
Show that the mass inside an orbit of radius 8.0 kpc is about 2 × 1041 kg.
(2)
(b)
Calculate the orbital speed expected at 25 kpc if there were no mass beyond 8.0 kpc.
(2)
(c)
Explain what the measured speed at 25 kpc suggests about the galaxy.
(2)
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Answer
Mark
Notes
Part (a)
GMm/r2 = mv2/r, so M = v2r/G
✓ 1
M = (2.30 × 105)2 × 8.0 × 3.09 × 1019 ÷ 6.67 × 10−11 = 1.96 × 1041 kg
Astronomers do not yet agree on the age of the universe or on its ultimate fate.
(a)
Discuss the reasons for this uncertainty.
(6)
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Answer
Mark
Notes
Indicative content (one mark for each point, linked reasoning expected):
—
the age is estimated from 1/H0
✓ 1
H0 depends on the distances to galaxies, which rely on standard candles of uncertain luminosity
✓ 1
different methods give different values of H0, and a larger H0 gives a younger universe
✓ 1
1/H0 assumes a constant rate of expansion; gravity slows the expansion, so the true age is less than 1/H0 by an amount that depends on the density
✓ 1
the fate depends on the mean density compared with the critical density: above it the universe eventually contracts, below it the universe expands for ever
✓ 1
the mean density is uncertain because dark matter can be detected only by its gravitational effects (for example on galaxy rotation curves)
✓ 1
Exam tip: In an extended answer, link each point to the next: measurement → why it is uncertain → what that uncertainty changes. Use the terms critical density and dark matter accurately.
Two independent methods of measuring the Hubble constant give 67 km s−1 Mpc−1 and 73 km s−1 Mpc−1. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.
(a)
Calculate the age of the universe given by 1/H0 for each value.
(2)
(b)
The expansion of the universe has been slowed by gravity. Explain why 1/H0 is then an overestimate of the age.
(2)
(c)
The oldest stars are estimated to be about 1.3 × 1010 years old. Use your answers to (a) and (b) to explain why the larger value of H0 causes a difficulty.
(1)
(d)
Suggest why it is difficult to measure H0 accurately.
(1)
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Answer
Mark
Notes
Part (a)
67: 1 ÷ (67 × 103 ÷ 3.09 × 1022) = 4.61 × 1017 s = 1.46 × 1010 years
✓ 1
73: 1.34 × 1010 years
✓ 1
Part (b)
the galaxies were moving apart faster in the past
✓ 1
so they reached their present separations in less time than d/v = 1/H0
✓ 1
Part (c)
with 73, the true age is less than 1.34 × 1010 years and could be less than the age of the oldest stars, which is impossible
✓ 1
Part (d)
distances to galaxies are uncertain (standard candles of uncertain luminosity) / galaxies have their own motions in addition to the expansion
✓ 1
Exam tip: When a question says “use your answers”, quote the numbers from the earlier parts in your explanation; a general statement without them may not earn the mark.
The graph shows the rotation curve of a spiral galaxy: the orbital speed v of stars and gas against distance r from the centre. Two points are marked on the curve. Assume circular orbits, so that the mass inside an orbit is M = v2r/G. What is the ratio (mass inside the orbit at 30 kpc) ÷ (mass inside the orbit at 10 kpc)?
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Answer
Mark
Notes
M ∝ v2r, so the ratio = (2202 × 30) ÷ (2002 × 10) = 1.21 × 3 = 3.6. The visible stars are concentrated near the centre, so most of this extra mass is unseen: evidence for dark matter.
✓ 1
Answer D
Answer: D
Common error: Squaring the distance as well as the speed, or forgetting to square the speed. From GMm/r2 = mv2/r, the mass is proportional to v2r.
The graph shows how the separation of two distant galaxy clusters has changed with time in a model of the universe, together with the tangent to the curve at the present time. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s. What is the Hubble constant at the present time in this model?
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Answer
Mark
Notes
H0 = v/d = (gradient of tangent) ÷ (separation now) = 1/(time at which the tangent meets the axis, measured back from now). So 1/H0 = 15 × 109 years = 4.725 × 1017 s, H0 = 2.12 × 10−18 s−1 × 3.09 × 1019 km Mpc−1 = 65 km s−1 Mpc−1.
✓ 1
Answer A
Answer: A
Exam tip: 1/H0 is the age only if the expansion rate has been constant. In a slowing universe the curve starts later than the tangent suggests, so the true age (here 10 × 109 years) is less than 1/H0.
The graph shows how the relative size of the universe changes with time in three models, X, Y and Z, considering gravity alone. In each model the universe is expanding at the same rate today. The dashed line shows a universe that has always expanded at this constant rate. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.
(a)
State which model has a mean density greater than the critical density. Describe how the graph shows the fate of this universe.
(2)
(b)
Use the dashed line to determine the Hubble constant in km s−1 Mpc−1.
(2)
(c)
Use the graph to estimate the age of universe Y. Explain why it is less than 1/H0.
(2)
(d)
Explain why the possible existence of dark matter makes it difficult to decide which model describes our universe.
(1)
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Answer
Mark
Notes
Part (a)
X
✓ 1
its size reaches a maximum and then decreases to zero: the expansion stops and the universe contracts
✓ 1
Part (b)
1/H0 = 14 × 109 years (allow 13.8–14.2 × 109) = 4.41 × 1017 s
gravity has slowed the expansion, so it was faster in the past and the universe reached its present size in less time than at a constant rate
✓ 1
Part (d)
the model depends on the mean density compared with the critical density, and dark matter can be detected only by its gravitational effects, so its contribution to the density is uncertain
✓ 1
Exam tip: On a size–time graph the gradient now is the same for every model with the same H0. The models differ in how the gradient changed in the past and how it will change.
The chart shows values of the Hubble constant H0 obtained by four different methods. Each value is shown with its uncertainty as an error bar. 1 Mpc = 3.09 × 1022 m. 1 year = 3.15 × 107 s.
(a)
Use the error bar for Method 2 to determine the range of values for the age of the universe, estimated as 1/H0, in years.
(3)
(b)
Explain why the results of Methods 1 and 2 cannot both be correct.
(1)
(c)
A galaxy is 160 Mpc away. Calculate the difference between its recession speeds predicted by the central values of Methods 1 and 2.
(2)
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Answer
Mark
Notes
Part (a)
range of H0 is 72 to 76 km s−1 Mpc−1
✓ 1
72: 1/H0 = 3.09 × 1022 ÷ 72 × 103 = 4.29 × 1017 s = 1.36 × 1010 years
✓ 1
76: 4.07 × 1017 s = 1.29 × 1010 years, so the range is 1.29 × 1010 to 1.36 × 1010 years
✓ 1
Part (b)
their error bars do not overlap (66–68 and 72–76), so no single value of H0 is consistent with both / at least one method has an unidentified systematic error
✓ 1
Part (c)
67 × 160 = 10 720 km s−1 and 74 × 160 = 11 840 km s−1
✓ 1
difference = 1.1 × 103 km s−1 (1120)
✓ 1
Exam tip: A larger Hubble constant gives a smaller value of 1/H0. Check that the top of an error bar gives the bottom of the age range.
A spiral galaxy is seen edge-on. The graph shows the observed wavelength of the hydrogen-alpha line, measured at different positions along the galaxy's major axis. The laboratory wavelength of the line is 656.30 nm. Three points are marked on the curve. Assume the stars and gas move in circular orbits. 1 kpc = 3.09 × 1019 m.
(a)
Use the graph to calculate the speed at which the galaxy as a whole is moving away from the Earth.
(2)
(b)
Determine the orbital speed of the stars 20 kpc from the centre.
(2)
(c)
Calculate the mass inside an orbit of radius 20 kpc.
(2)
(d)
The total mass of the visible stars and gas in the galaxy is estimated as 1.5 × 1041 kg, almost all of it within 10 kpc of the centre. Discuss what the measurements suggest.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
centre: λ = 661.00 nm, Δλ = 4.70 nm
✓ 1
v = 3.00 × 108 × 4.70 ÷ 656.30 = 2.15 × 106 m s−1 (allow 2.1–2.2 × 106)
✓ 1
Part (b)
shift due to rotation = (661.50 − 660.50) ÷ 2 = 0.50 nm
✓ 1
v = 3.00 × 108 × 0.50 ÷ 656.30 = 2.3 × 105 m s−1 (2.29 × 105; allow 2.2–2.3 × 105, e.g. 661.00 nm used for λ)
✓ 1
Part (c)
M = v2r/G
✓ 1
= (2.286 × 105)2 × 20 × 3.09 × 1019 ÷ 6.67 × 10−11 = 4.8 × 1041 kg (allow 4.7–4.9 × 1041, or ECF from (b))
✓ 1
Part (d)
the mass inside 20 kpc is about 3 times the visible mass: about 3.3 × 1041 kg is unseen
✓ 1
with no mass beyond 10 kpc the orbital speed would fall with distance, but the curve stays level, so there is unseen (dark) matter extending beyond the visible galaxy
✓ 1
Exam tip: On a position–wavelength graph for a rotating galaxy, the value at the centre gives the motion of the whole galaxy; half the difference between the two sides at the same distance gives the orbital speed there.
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