Topic 6: Space Physics -- Cambridge Challenge Level (Set 2)
1 hour 15 minutes
80
7
75:00
0625 / 0972
⚡ Cambridge Challenge Level
These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- The Weather Satellite
Total: 12 marks
The Met Office receives images from a polar-orbiting weather satellite that circles the Earth at an orbital radius of 7.2 × 10⁶ m (measured from the centre of the Earth), completing one orbit every 101 minutes. The Moon orbits at a radius of 3.84 × 10⁸ m with a period of 27.3 days.
(a)[3]
Calculate the orbital speed of the satellite in m/s.
Model Answer -- 1(a)
v = 2πr / T [1]
T = 101 × 60 = 6060 s [1]
v = 2π × 7.2 × 10⁶ / 6060 = 7470 m/s (≈ 7.5 km/s) [1]
⚠ If you missed marks here: Convert the period to seconds first: 101 min = 6060 s. Then v = 2πr/T = 7470 m/s — about 27 000 km/h. If you got 450 km/s you left T in minutes; if your answer looks slower than a car, check the powers of ten in r.
Mark 1 -- orbital speed equation 2 pi r over T used (1 mark)
Mark 2 -- period converted to 6060 seconds (1 mark)
Mark 3 -- speed about 7470 m per s calculated (1 mark)
(b)[3]
Name the force that keeps the satellite in orbit, state its direction, and explain why the satellite does not fly off in a straight line.
Model Answer -- 1(b)
Gravitational attraction (of the Earth on the satellite) [1]
Directed towards the centre of the Earth [1]
This force continuously pulls the satellite away from straight-line motion / changes its direction, keeping it on the (roughly) circular path [1]
⚠ If you missed marks here: Newton’s first law says the satellite WOULD travel in a straight line if no force acted. Gravity, pulling always towards the Earth’s centre, bends the path into a circle. There is no "force of the orbit" and nothing balancing gravity — the pull is unbalanced, which is exactly why the direction keeps changing.
Mark 1 -- gravitational force named (1 mark)
Mark 2 -- directed towards centre of Earth (1 mark)
Mark 3 -- force bends path from straight line into orbit (1 mark)
(c)[3]
The satellite travels at constant speed. Explain why its velocity is NOT constant, and why it is correct to say the satellite is accelerating.
Model Answer -- 1(c)
Velocity is speed in a stated DIRECTION (a vector) [1]
The direction of motion changes continuously around the orbit, so the velocity changes even though the speed does not [1]
Acceleration is change of velocity (per unit time), so a changing direction means the satellite is accelerating (towards the Earth’s centre) [1]
⚠ If you missed marks here: This is the vector trap: speed (scalar) constant, velocity (vector) changing because the DIRECTION changes. Since acceleration is any change of velocity, circular motion at steady speed is still accelerated motion — the acceleration points the same way as the force, towards the centre.
Mark 1 -- velocity defined as vector with direction (1 mark)
Mark 2 -- direction changes so velocity changes (1 mark)
Mark 3 -- changing velocity means acceleration towards centre (1 mark)
(d)[3]
Calculate the orbital speed of the Moon, and comment on how it compares with the satellite.
Model Answer -- 1(d)
T = 27.3 × 24 × 3600 = 2.36 × 10⁶ s [1]
v = 2π × 3.84 × 10⁸ / 2.36 × 10⁶ = 1020 m/s [1]
The Moon moves much more slowly than the satellite (about 1 km/s against 7.5 km/s) — the more distant orbiting body travels slower [1]
⚠ If you missed marks here: 27.3 days = 2.36 × 10⁶ s — the day-to-second conversion (×86 400) is where this question is won. The Moon’s 1020 m/s against the satellite’s 7470 m/s illustrates the general rule: the farther the orbit, the slower the orbital speed.
Mark 1 -- period converted to about 2.36 million seconds (1 mark)
Mark 2 -- Moon speed about 1020 m per s calculated (1 mark)
Mark 3 -- comparison farther orbit is slower stated (1 mark)
Question 2 -- Stargazing at Kielder
Total: 11 marks
Kielder Observatory in Northumberland has some of the darkest skies in England. Visitors observe the Sun (safely, by projection), the stars, and their nightly motion across the sky. Data: speed of light = 3.0 × 10⁸ m/s; Sun–Earth distance = 1.496 × 10¹¹ m; 1 light-year = 9.5 × 10¹⁵ m.
(a)[3]
Calculate the time taken for light to travel from the Sun to the Earth. Give your answer in seconds and in minutes.
Model Answer -- 2(a)
t = d / v = 1.496 × 10¹¹ / 3.0 × 10⁸ [1]
t = 499 s [1]
499 / 60 = 8.3 minutes [1]
⚠ If you missed marks here: Time = distance ÷ speed: 1.496 × 10¹¹ ÷ 3.0 × 10⁸ = 499 s ≈ 8.3 min. If you got 2 × 10¹⁹ you multiplied instead of dividing — sanity-check: light is fast, so the answer should be minutes, not years.
Mark 1 -- time equals distance over speed used (1 mark)
Mark 2 -- time 499 seconds calculated (1 mark)
Mark 3 -- converted to about 8.3 minutes (1 mark)
(b)[3]
The bright summer star Vega is 25 light-years away. Define the light-year, and calculate this distance in metres.
Model Answer -- 2(b)
A light-year is the DISTANCE travelled (in a vacuum) by light in one year [1]
d = 25 × 9.5 × 10¹⁵ [1]
d = 2.4 × 10¹⁷ m (2.375 × 10¹⁷ m) [1]
⚠ If you missed marks here: A light-year measures DISTANCE, not time — the single most common definition error. 25 × 9.5 × 10¹⁵ = 2.4 × 10¹⁷ m — and remember to re-normalise 237.5 × 10¹⁵ into standard form. Numbers this size are why astronomers invented the unit.
Mark 1 -- light-year defined as distance light travels in one year (1 mark)
Mark 2 -- multiplication 25 times 9.5e15 shown (1 mark)
Mark 3 -- distance 2.4 times 10 to the 17 m calculated (1 mark)
(c)[2]
A visitor says: "When you look at Vega you are looking into the past." Explain why this is true.
Model Answer -- 2(c)
The light now arriving left the star 25 years ago (light takes 25 years to travel the distance) [1]
So we see the star as it WAS 25 years ago, not as it is now [1]
⚠ If you missed marks here: Light is a messenger with a travel time: what reaches the eye tonight left Vega 25 years ago. For a galaxy millions of light-years away the image is millions of years out of date — telescopes are time machines pointing backwards.
Mark 1 -- light took 25 years to arrive (1 mark)
Mark 2 -- star seen as it was in the past (1 mark)
(d)[3]
During the night the stars appear to move across the sky from east to west, yet return to almost the same positions 24 hours later. Explain these observations.
Model Answer -- 2(d)
The Earth rotates on its axis (from west to east) [1]
This rotation makes the stars (and the Sun) APPEAR to move across the sky the opposite way, east to west [1]
One complete rotation takes (approximately) 24 hours, so the sky pattern repeats each day [1]
⚠ If you missed marks here: The stars are not orbiting the Earth nightly — the EARTH is turning beneath them. Spinning west-to-east makes everything in the sky drift east-to-west, and one full turn each 24 h brings the pattern back. The same logic explains day and night.
Mark 1 -- Earth rotates on its axis (1 mark)
Mark 2 -- apparent star motion east to west explained (1 mark)
Mark 3 -- 24 hour rotation period gives daily repeat (1 mark)
Question 3 -- The Fate of Antares
Total: 12 marks
Members of an astronomy society in Edinburgh follow Antares, a red supergiant star in Scorpius far more massive than the Sun. One day — possibly within the next million years — it will explode. The Sun, a much less massive star, faces a quieter future.
(a)[4]
Describe the remaining life stages of a star much more massive than the Sun, from red supergiant onwards.
Model Answer -- 3(a)
The red supergiant collapses when its (fusion) fuel runs out... [1]
...and explodes as a SUPERNOVA [1]
The core that remains becomes a neutron star... [1]
...or, for the most massive stars, a black hole [1]
⚠ If you missed marks here: The massive-star sequence is: red supergiant → supernova (the explosion) → neutron star OR black hole (the collapsed core, depending on mass). Mixing in "white dwarf" belongs to the Sun’s path, not this one.
Mark 1 -- fuel exhausted and collapse begins (1 mark)
Mark 2 -- supernova explosion named (1 mark)
Mark 3 -- neutron star as remnant (1 mark)
Mark 4 -- black hole for most massive stars (1 mark)
(b)[2]
Gold and uranium are found on Earth, yet ordinary stars cannot make elements heavier than iron by fusion. Explain how these heavy elements came to exist.
Model Answer -- 3(b)
Elements heavier than iron are formed IN a supernova (explosion) [1]
The explosion flings these elements into space, where they become part of the clouds (nebulae) that form new stars and planets — including the Earth [1]
⚠ If you missed marks here: Fusion in a stable star stops paying energy at iron; the heavier elements need a supernova’s violence. Every gold atom on Earth was made in an exploding star and scattered into the cloud that later became the Solar System.
Mark 1 -- heavier than iron elements made in supernova (1 mark)
Mark 2 -- explosion scatters elements into space for new systems (1 mark)
(c)[4]
Describe the remaining life stages of the Sun, from its present stable state onwards.
Model Answer -- 3(c)
The Sun remains a stable (main sequence) star until most of its hydrogen is used up [1]
It then expands to become a RED GIANT [1]
Its outer layers are ejected as a PLANETARY NEBULA [1]
The remaining core is a WHITE DWARF, which slowly cools (and fades) [1]
⚠ If you missed marks here: The low-mass path: stable star → red giant → planetary nebula (the shed outer layers) → white dwarf that cools forever. No supernova, no neutron star — the Sun is not massive enough. "Planetary nebula" has nothing to do with planets; the name is a historical accident.
Mark 1 -- stable until hydrogen exhausted (1 mark)
Mark 2 -- expands to red giant (1 mark)
Mark 3 -- outer layers ejected as planetary nebula (1 mark)
Mark 4 -- core left as cooling white dwarf (1 mark)
(d)[2]
Explain how a stable star like the present-day Sun maintains a constant size, in terms of the forces or pressures acting.
Model Answer -- 3(d)
Gravity pulls the star’s material inwards [1]
This is balanced by the outward pressure (of the hot gas / radiation produced by fusion in the core) — the two are in equilibrium, so the size stays constant [1]
⚠ If you missed marks here: A stable star is a tug-of-war in balance: gravity inwards against the outward pressure of the fusion-heated gas. When the fuel fails, the outward push weakens and gravity wins — which is what launches every later stage of the story.
Mark 1 -- gravity acts inwards on star material (1 mark)
Mark 2 -- balanced by outward pressure from fusion heated gas (1 mark)
Question 4 -- Measuring the Expanding Universe
Total: 11 marks
Astronomers at the Institute of Astronomy in Cambridge study the light from five distant galaxies. For each galaxy they measure its distance d and its speed of recession v (found from redshift). Their results are plotted below.
(a)[2]
The light from all five galaxies is redshifted. State what redshift is, and what it tells us about the motion of these galaxies.
Model Answer -- 4(a)
The light is shifted to longer wavelength (towards the red end of the spectrum) [1]
This shows the galaxies are moving AWAY from us (receding) [1]
⚠ If you missed marks here: Redshift = wavelength stretched longer, and its size grows with recession speed. It does not mean the galaxies look red — the whole spectrum (every line in it) is displaced towards longer wavelengths.
Mark 1 -- redshift is increase in observed wavelength (1 mark)
Mark 2 -- shows galaxies receding from us (1 mark)
(b)[3]
The points lie on a straight line through the origin. State the relationship this shows between v and d, and use the graph to calculate the Hubble constant H₀ = v/d in s⁻¹.
Model Answer -- 4(b)
v is (directly) proportional to d [1]
Gradient: e.g. H₀ = 2200 × 10³ m/s ÷ 1.0 × 10²⁴ m [1]
H₀ = 2.2 × 10⁻¹⁸ s⁻¹ [1]
⚠ If you missed marks here: Two unit conversions hide in the gradient: v is plotted in km/s (2200 km/s = 2.2 × 10⁶ m/s) and d in units of 10²⁴ m. H₀ = 2.2 × 10⁶ ÷ 10²⁴ = 2.2 × 10⁻¹⁸ s⁻¹. Any point on the line gives the same answer — that is what proportionality means.
Mark 1 -- v proportional to d stated (1 mark)
Mark 2 -- gradient method with correct unit conversion (1 mark)
Mark 3 -- Hubble constant 2.2e-18 per second calculated (1 mark)
(c)[3]
Assuming the galaxies have always moved at their present speeds, the age of the universe is 1/H₀. Calculate this age in seconds, and convert it to years (1 year ≈ 3.15 × 10⁷ s).
⚠ If you missed marks here: 1 ÷ 2.2 × 10⁻¹⁸ = 4.5 × 10¹⁷ s; dividing by 3.15 × 10⁷ s/year gives 1.4 × 10¹⁰ years. If your age came out at a few thousand years, the negative exponent flipped somewhere — 10⁻¹⁸ inverted must give a POSITIVE power of ten.
Mark 1 -- age as reciprocal of Hubble constant (1 mark)
Mark 2 -- age 4.5 times 10 to 17 seconds (1 mark)
Mark 3 -- converted to about 14 billion years (1 mark)
(d)[3]
Explain how the pattern in this graph supports the Big Bang theory.
Model Answer -- 4(d)
Every distant galaxy is receding, and the farther it is the faster it recedes — so space itself is expanding [1]
Running the expansion backwards, all the matter/energy of the universe was once together at a single (very small, very hot, very dense) point [1]
This origin, expanding ever since, is the Big Bang — and the measured age (1/H₀) dates it [1]
⚠ If you missed marks here: The argument runs the film backwards: if everything is flying apart now, it was closer together in the past, and at time 1/H₀ ago it was all in one place. The graph does not merely show movement — the PROPORTIONALITY is what makes a single common origin work from every galaxy’s point of view.
Mark 1 -- expansion of universe deduced from recession pattern (1 mark)
Mark 2 -- running back in time gives single dense hot origin (1 mark)
Mark 3 -- identified as Big Bang with age from reciprocal (1 mark)
Question 5 -- The Mars Rover from Stevenage
Total: 12 marks
The Rosalind Franklin Mars rover was built in Stevenage, Hertfordshire. The gravitational field strength at the surface of Mars is 3.7 N/kg, and the rover has a mass of 300 kg. At closest approach, Mars is 5.6 × 10¹⁰ m from Earth. Radio commands travel at the speed of light, 3.0 × 10⁸ m/s.
(a)[2]
Calculate the weight of the rover on Mars, and state what its weight would be on Earth (g = 10 N/kg).
Model Answer -- 5(a)
W = m g = 300 × 3.7 = 1110 N on Mars [1]
On Earth: W = 300 × 10 = 3000 N (mass unchanged at 300 kg) [1]
⚠ If you missed marks here: Weight changes with g; MASS does not. 300 kg × 3.7 = 1110 N on Mars, 3000 N on Earth — same rover, same 300 kg everywhere. If you divided by g you found the mass you already had.
Mark 1 -- Mars weight 1110 N calculated (1 mark)
Mark 2 -- Earth weight 3000 N with mass unchanged (1 mark)
(b)[2]
Suggest why the gravitational field strength at the surface of Mars is smaller than at the surface of the Earth.
Model Answer -- 5(b)
Mars has a (much) smaller mass than the Earth [1]
A smaller planetary mass produces a weaker gravitational field at its surface, so g is smaller (3.7 N/kg against 10 N/kg) [1]
⚠ If you missed marks here: Field strength at a planet’s surface is set by the planet: less mass, weaker pull. "Mars is farther from the Sun" is the wrong planet’s gravity — the SUN’s field has nothing to do with the rover’s weight on the Martian surface.
Mark 1 -- Mars has smaller mass than Earth (1 mark)
Mark 2 -- smaller mass gives weaker surface field (1 mark)
(c)[4]
Calculate the time for a radio command to travel from Earth to Mars at closest approach, and explain why the rover must be able to drive itself rather than being steered live from Earth.
Model Answer -- 5(c)
t = d / v [1]
t = 5.6 × 10¹⁰ / 3.0 × 10⁸ [1]
t = 187 s ≈ 3.1 minutes (and over 6 minutes for a round trip) [1]
By the time a driver on Earth saw a hazard and their command arrived back, minutes would have passed — far too slow to react, so the rover needs on-board (autonomous) control [1]
⚠ If you missed marks here: 5.6 × 10¹⁰ ÷ 3.0 × 10⁸ = 187 s — about 3.1 minutes EACH WAY, and that is at closest approach. A see-hazard-send-command loop takes over six minutes; a rover can drive off a rock in far less. The physics of signal delay is the entire reason for autonomy.
Mark 1 -- time distance over speed stated (1 mark)
Mark 2 -- powers of ten substituted correctly (1 mark)
Mark 3 -- one-way delay about 3.1 minutes calculated (1 mark)
Mark 4 -- round-trip delay makes live steering impossible (1 mark)
(d)[2]
The rover’s solar panels generate less power on Mars than identical panels would on Earth, even in full sunshine. Explain why.
Model Answer -- 5(d)
Mars is farther from the Sun than the Earth is [1]
The Sun’s radiation spreads out as it travels, so the intensity (power per unit area) reaching the panels is lower at Mars’s distance [1]
⚠ If you missed marks here: Sunlight thins with distance: the same radiated power spreads over an ever larger area, so each square metre of panel at Mars catches less than half what it would at Earth. Nothing is "blocked" — it is geometry, not obstruction.
Mark 1 -- Mars farther from Sun stated (1 mark)
Mark 2 -- radiation spread over larger area lower intensity (1 mark)
(e)[2]
State the force that keeps Mars in orbit around the Sun, and the direction in which it acts.
Model Answer -- 5(e)
The gravitational attraction between the Sun and Mars [1]
Acting on Mars towards the Sun (towards the centre of the orbit) [1]
⚠ If you missed marks here: One force, one direction: the Sun’s gravitational pull, directed from Mars towards the Sun. There is no outward force balancing it — an unbalanced centre-seeking force is precisely what circular motion requires.
Mark 1 -- gravitational attraction of Sun named (1 mark)
Mark 2 -- direction towards the Sun stated (1 mark)
Question 6 -- The Milky Way and Beyond
Total: 12 marks
On a clear night at Exmoor’s Dark Sky Reserve, the Milky Way is visible as a faint band across the sky. Through the society’s telescope, visitors view the Whirlpool galaxy, 3.0 × 10⁷ light-years away. (1 light-year = 9.5 × 10¹⁵ m.)
(a)[3]
Describe what the Milky Way is, and the Sun’s place within it.
Model Answer -- 6(a)
The Milky Way is a GALAXY: a collection of many billions of stars (held together by gravity) [1]
The Sun is one (ordinary) star within the Milky Way [1]
The universe contains many billions of such galaxies (the Whirlpool being another) [1]
⚠ If you missed marks here: Scale ladder: star → galaxy (billions of stars) → universe (billions of galaxies). The band across the sky is our own galaxy seen edge-on from inside. The Sun is not at the centre and not special — one star among hundreds of billions.
Mark 1 -- Milky Way is galaxy of billions of stars (1 mark)
Mark 2 -- Sun is one star within it (1 mark)
Mark 3 -- universe holds billions of galaxies (1 mark)
(b)[2]
Astronomical distances are given in light-years rather than metres. Define the light-year, and explain why it is a convenient unit for these distances.
Model Answer -- 6(b)
The distance light travels (in a vacuum) in one year [1]
Distances between stars/galaxies are so enormous that metres give unmanageably large numbers; light-years keep the numbers small and also show how long the light took [1]
⚠ If you missed marks here: Definition first (a DISTANCE, not a time), then the convenience: "2.5 million light-years" is both a manageable number and a built-in statement that the light left 2.5 million years ago.
Mark 1 -- distance light travels in one year (1 mark)
Mark 2 -- avoids huge numbers and shows light travel time (1 mark)
(c)[3]
Calculate the distance to the Whirlpool galaxy in metres. Give your answer in standard form.
⚠ If you missed marks here: Multiply the numbers (3.0 × 9.5 = 28.5) and ADD the powers (7 + 15 = 22), then re-normalise: 28.5 × 10²² = 2.85 × 10²³ ≈ 2.9 × 10²³ m. Dropping the re-normalisation step leaves 28.5 × 10²², which is not standard form.
Mark 1 -- 3.0e7 light years times 9.5e15 set up (1 mark)
Mark 2 -- powers of ten handled correctly (1 mark)
Mark 3 -- distance 2.9 times 10 to 23 m in standard form (1 mark)
(d)[4]
The cosmic microwave background radiation (CMBR) is often called the strongest single piece of evidence for the Big Bang. State what the CMBR is, where it comes from, and why its existence supports the Big Bang theory.
Model Answer -- 6(d)
Microwave radiation detected coming (almost equally) from EVERY direction in space [1]
It is radiation produced shortly after the Big Bang [1]
Its wavelength has been stretched (redshifted) into the microwave region by the expansion of the universe [1]
The Big Bang theory PREDICTED such left-over radiation; no rival theory explains an all-sky glow, so detecting it strongly supports the theory [1]
⚠ If you missed marks here: Four ideas: microwaves from ALL directions; origin just after the Big Bang; wavelength stretched by the expansion of space (it began as much shorter-wavelength radiation); and the clincher — the theory predicted it before it was found. An answer that never says "from every direction" has missed what makes it cosmic.
Mark 1 -- microwave radiation from all directions (1 mark)
Mark 2 -- produced shortly after Big Bang (1 mark)
Mark 3 -- wavelength stretched by expansion of universe (1 mark)
Mark 4 -- predicted by theory so detection supports it (1 mark)
Question 7 -- The Moons of Jupiter
Total: 10 marks
A school astronomy club in Kent observes Jupiter through a small telescope and sees four moons, first recorded by Galileo in 1610. The innermost large moon, Io, orbits at a radius of 4.22 × 10⁸ m with a period of 1.77 (Earth) days. Europa orbits farther out, at 6.71 × 10⁸ m.
(a)[3]
Calculate the orbital speed of Io.
Model Answer -- 7(a)
T = 1.77 × 24 × 3600 = 1.53 × 10⁵ s [1]
v = 2πr / T = 2π × 4.22 × 10⁸ / 1.53 × 10⁵ [1]
v = 17 300 m/s [1]
⚠ If you missed marks here: 1.77 days = 152 900 s. Then v = 2π × 4.22 × 10⁸ ÷ 1.53 × 10⁵ ≈ 17 300 m/s — more than twice a low-Earth satellite’s speed, because Jupiter’s pull at Io’s distance is ferocious.
Mark 1 -- period converted to about 153000 seconds (1 mark)
Mark 2 -- circumference over period substituted (1 mark)
Mark 3 -- orbital speed about 17300 m per s (1 mark)
(b)[2]
Without calculation, state and explain how Europa’s orbital speed compares with Io’s.
Model Answer -- 7(b)
Europa orbits more SLOWLY than Io [1]
It is farther from Jupiter, where Jupiter’s gravitational field is weaker — orbital speed decreases with orbital radius [1]
⚠ If you missed marks here: Same rule as the planets round the Sun: farther out, weaker field, slower orbit. No calculation needed — the comparison plus the field reason earns both marks.
Mark 1 -- Europa slower than Io stated (1 mark)
Mark 2 -- weaker field at greater radius explains it (1 mark)
(c)[2]
State the force that holds Io in its orbit, and the direction of Io’s acceleration.
Model Answer -- 7(c)
The gravitational attraction of Jupiter (on Io) [1]
Io’s acceleration is directed towards (the centre of) Jupiter [1]
⚠ If you missed marks here: Force and acceleration point the same way: towards Jupiter’s centre. Even at constant speed Io accelerates, because its direction changes continuously — the same argument as any orbiting satellite, in miniature.
Mark 1 -- Jupiter gravitational attraction named (1 mark)
Mark 2 -- acceleration towards Jupiter centre (1 mark)
(d)[3]
Jupiter takes almost 12 Earth-years to orbit the Sun, while the Earth takes 1 year. Give TWO reasons why Jupiter’s orbital period is so much longer.
Model Answer -- 7(d)
Jupiter’s orbit has a much larger radius, so the circumference (distance travelled per orbit) is much greater [1]
Jupiter also moves more slowly, because the Sun’s gravitational field is weaker at its distance [1]
Both effects lengthen the period: T = 2πr / v grows because r is larger AND v is smaller [1]
⚠ If you missed marks here: Two compounding reasons, and the question asks for both: a longer track (bigger circumference) covered at a lower speed (weaker solar gravity). Writing only "it is farther away" gives one idea — unpack it into distance AND speed via T = 2πr/v.
Mark 1 -- larger radius means greater circumference (1 mark)
Mark 2 -- slower speed due to weaker solar field (1 mark)
Mark 3 -- period formula links both effects (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
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