S6 Physics Practice Paper: Gravitation And Satellite Motion (12 Questions, 100 Marks, 75 Minutes)

S6 Physics Practice Paper: Gravitation and Satellite Motion (12 questions, 100 marks, 75 minutes)

2026-11-09

HKDSE Physics: Gravitation and Satellite Motion

S6 Physics paper: 12 questions, 100 marks, 75 minutes. Gravitation, orbital speed and period, geostationary orbits, orbit transfer energy, binary stars and escape speed, with answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show the formula used, the values substituted and the units. Give numerical answers to three significant figures.

In explanations, name the law used, for example the law of gravitation or conservation of energy.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A (short questions, 29 marks)

Answer all questions. Show the formula used and the values substituted.

#	Question	Answer and marking notes
1	Earth has mass 5.97 × 10^24 kg and radius 6.37 × 10^6 m; G = 6.67 × 10^-11 N m^2 kg^-2. (a) Find g at the surface. (b) Find g at a height of 1.0 × 10^7 m above the surface. (c) Find the ratio of the two values.	(a) g0 = GM/R^2 = (6.67e-11)(5.97e24)/(6.37e6)^2 = 9.81 m s^-2 (2 marks) (b) r = R + h = 1.637e7 m, g = 1.49 m s^-2 (2 marks) (c) ratio = 6.60 (2 marks) Marking: using r = R + h is required; using R alone scores 1.
2	The Moon orbits Earth with an orbital radius of about 3.84 × 10^8 m. (a) Treating the orbit as circular, find the orbital period in days. (b) Explain why the Moon does not fall to Earth. (c) If the Moon had twice the mass at the same radius, would the period change? Explain.	(a) GMm/r^2 = m(4 pi^2 r/T^2) so T = 2 pi sqrt(r^3/GM) = 2369336 s = 27.4 days (3 marks) (b) It has enough tangential speed: the distance it falls equals the curvature of Earth, so it keeps orbiting (2 marks) (c) No change (1 mark): the period depends only on the central mass and the radius, not on the orbiting mass (1 mark)
3	A satellite orbits at a height of 400 km above Earth (R = 6.37 × 10^6 m). (a) Find the orbital radius. (b) Find its speed. (c) Find its period in minutes. (d) How does the speed change if the mass of the satellite doubles?	(a) r = 6.37e6 + 4.0e5 = 6.77e6 m (1 mark) (b) v = sqrt(GM/r) = 7669 m/s (3 marks) (c) T = 2 pi r/v = 92.4 minutes (3 marks) (d) Unchanged (1 mark): the mass cancels.
4	(a) Use conservation of energy to derive the escape speed v = sqrt(2GM/R). (b) Evaluate it in km/s. (c) Explain why the escape speed does not depend on the mass of the projectile.	(a) 0.5 m v^2 – GMm/R = 0 (1 mark) so v = sqrt(2GM/R) (2 marks) (b) v = 11.18 km/s (3 marks) (c) Both kinetic and potential energy are proportional to m, so m cancels (2 marks) Marking: answer with no derivation scores at most 4.

Section B (calculation and explanation, 33 marks)

G = 6.67e-11 N m^2 kg^-2, mass of Earth 5.97e24 kg, radius of Earth 6.37e6 m.

#	Question	Answer and marking notes
5	Mars has mass 6.42 × 10^23 kg and radius 3.39 × 10^6 m. (a) Find g at its surface. (b) Compare it with Earth and give two reasons for the difference. (c) An instrument weighing 600 N on Earth is taken to Mars; find its weight there.	(a) g = GM/R^2 = 3.73 m s^-2 (3 marks) (b) Smaller (1 mark): the mass is about 0.107 of Earth while the radius is about 0.532; since g depends on M/R^2 the mass factor dominates (2 marks) (c) mass = 61.2 kg, weight = 228 N (2 marks)
6	A geostationary satellite has a period equal to the sidereal day, 86 164 s. (a) Find its orbital radius. (b) Find its height above the surface. (c) Explain why it must lie above the equator.	(a) r = (GMT^2/4 pi^2)^(1/3) = 4.21e7 m (3 marks) (b) height = 3.58e7 m (2 marks) (c) Only in the equatorial plane is the centripetal acceleration collinear with the gravitational pull; otherwise the pull has a component towards the equator and the orbit drifts (3 marks)
7	A 1200 kg satellite moves from a 400 km circular orbit to a geostationary orbit of radius 4.22 × 10^7 m. (a) Write the total mechanical energy of a circular orbit. (b) Find the energy needed per kilogram. (c) Find the total energy needed. (d) Where does this energy come from?	(a) E = -GMm/2r (2 marks) (b) dE/m = (GM/2)(1/r1 – 1/r2) = 2.47e7 J/kg (3 marks) (c) dE = 2.96e10 J (2 marks) (d) From the chemical energy of the propellant burned in the thruster (2 marks) Marking: using GMm(1/r1 – 1/r2) as total energy loses 1.
8	A binary system consists of two stars, each of 1.5 solar masses (1 M_sun = 1.99 × 10^30 kg), separated by 2.0 × 10^11 m and orbiting their common centre of mass. (a) Find each orbital radius. (b) Find their speed. (c) Find the orbital period in days. (d) How does the centre of mass move if the masses differ?	(a) radius = a/2 = 1.0e11 m (2 marks) (b) Gm^2/a^2 = m v^2/(a/2) so v = sqrt(Gm/2a) = 2.23e4 m/s (3 marks) (c) T = 2 pi (a/2)/v = 326 days (3 marks) (d) The centre of mass shifts towards the heavier star; the radii are in inverse ratio to the masses (1 mark)

Section C (integrated and applied, 38 marks)

Give numerical answers to three significant figures; in explanations name the physical law used.

#	Question	Answer and marking notes
9	A moon of an unknown planet is observed with orbital radius r = 7.00 × 10^6 m and period T = 5.82 × 10^3 s. (a) Find the mass of the planet. (b) State the formula used and explain why the mass of the moon is not needed. (c) If T were measured 2% too small, how would the calculated mass be affected?	(a) GM = 4 pi^2 r^3/T^2 = 3.98e14 so M = 5.99e24 kg (3 marks) (b) T^2 = 4 pi^2 r^3/(GM) (1 mark); the mass of the moon cancels and it acts only as a probe (2 marks) (c) T too small by 2% means T^2 too small by about 4%, so M comes out 4% too large (3 marks) Marking: (c) needs the 4% figure.
10	The Earth rotates once in 86 164 s. (a) Find the centripetal acceleration required at the equator. (b) Find the effective value of g at the equator. (c) Explain why g is smaller at the equator than at the poles. (d) Why does an astronaut in orbit feel weightless?	(a) a = w^2 R = 0.034 m s^-2 (3 marks) (b) g_eff = 9.81 – 0.034 = 9.78 m s^-2 (2 marks) (c) At the equator the radius of rotation and the centripetal term are largest, and the oblate Earth gives a larger polar radius effect on 1/R^2 (2 marks) (d) Astronaut and craft fall with the same acceleration, so the seat exerts no normal force and apparent weight is zero (2 marks)
11	A satellite in a 400 km circular orbit moves at 7.67 km/s. (a) Find the escape speed at that height. (b) Find the required change in speed. (c) Why is it more efficient to escape this way than to accelerate to 11.2 km/s at the ground? Answer in terms of energy.	(a) v_esc = sqrt(2) x 7.67 = 10.85 km/s (3 marks) (b) dv = 3.18 km/s (3 marks) (c) Accelerating from the ground must overcome the whole potential well and fight air drag; from orbit only the extra energy above that orbit is needed (4 marks) Marking: (c) needs air drag or the potential-difference argument, both for full marks.
12	A planet has surface gravity 3.73 m s^-2 and radius 3.39 × 10^6 m. (a) Find its mean density. (b) If a body of the same radius had twice the density, how would the surface gravity change? (c) How does a simple pendulum on that planet compare with one on Earth? (d) Suggest a practical way to check the density by observation.	(a) M = gR^2/G; rho = 3g/(4 pi G R) = 3934 kg m^-3 (4 marks) (b) g doubles to 7.45 m s^-2 (2 marks) (c) T = 2 pi sqrt(L/g), so the period is longer by a factor sqrt(9.81/3.73) = 1.62 (2 marks) (d) Track a probe in orbit around the planet: get M from T^2 = 4 pi^2 r^3/(GM), then combine with the radius (2 marks)

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