OxfordAQA International AS and A-level Physics: Gravitational potential and orbits of planets and satellites (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, exams from May/June 2020 onwards. They cover sections 3.7.3 (Gravitational potential) and 3.7.4 (Orbits of planets and satellites), which are International A-level only. Learn the content from the study guide and the revision notes, and see the course hub and checklist.

Data: G = 6.67 × 10⁻¹¹ N m² kg⁻². Masses: Earth 5.97 × 10²⁴ kg, Moon 7.35 × 10²² kg, Jupiter 1.90 × 10²⁷ kg, Sun 1.99 × 10³⁰ kg, Saturn 5.68 × 10²⁶ kg. Radii: Earth 6.37 × 10⁶ m, Moon 1.74 × 10⁶ m. Treat all bodies as spheres and ignore other bodies.

Questions

1. State what is meant by gravitational potential, and explain why it is negative at every point near a planet. [3]

2. A 900 kg probe is near the Earth.

(a) The probe moves through half of a circular path of radius 8.00 × 10⁶ m, centred on the Earth. State the work done by the gravitational force and give a reason. [2] (b) The probe is then moved radially from r = 8.00 × 10⁶ m to r = 2.40 × 10⁷ m. Calculate the work that must be done on it. [3]

3. At a distance of 6.0 × 10⁶ m from the centre of a planet, the gravitational potential is −1.5 × 10⁷ J kg⁻¹. Assume that V = −GM/r outside the planet.

(a) Calculate the gravitational field strength at this distance. [2] (b) Calculate the mass of the planet. [2] (c) Calculate the potential at a distance of 1.2 × 10⁷ m from the centre. [1]

4. Show that the speed of a satellite in a circular orbit of radius r about a body of mass M is given by v = √(GM/r). [3]

5. A spacecraft orbits the Moon in a circular path 100 km above its surface. Calculate its orbital speed and its period in minutes. [4]

6. A planet orbits the Sun in a circular orbit of radius 1.08 × 10¹¹ m.

(a) Calculate its orbital period in days. [3] (b) A second planet has an orbital radius four times as large. State its period, in days, and show how you obtained it. [1]

7. An 850 kg satellite is moved from a circular orbit of radius 7.20 × 10⁶ m to a geosynchronous orbit of radius 4.21 × 10⁷ m about the Earth.

(a) Calculate its kinetic energy in the first orbit. [2] (b) Calculate its kinetic energy in the geosynchronous orbit. [1] (c) Calculate the change in the total energy of the satellite. [2] (d) Explain why the satellite has less kinetic energy in the final orbit even though energy has been supplied to it. [1]

8. A 300 kg satellite is in a circular orbit of radius 7.30 × 10⁶ m about the Earth. Air resistance causes the orbit to shrink slowly to a radius of 7.10 × 10⁶ m.

(a) Calculate the speed of the satellite at each radius. [2] (b) Calculate the change in kinetic energy and the change in total energy. [2] (c) Explain why the speed increases although the satellite is losing mechanical energy. [1]

9. Jupiter rotates once in 9.93 hours.

(a) Calculate the orbital radius of a satellite that stays above the same point on Jupiter’s equator. [3] (b) State two conditions, other than the period, that a satellite must meet to stay above the same point on a planet’s surface. [2]

10. A planet has mass 4.50 × 10²⁴ kg and radius 8.00 × 10⁶ m. A 640 kg satellite is placed in a circular orbit of radius equal to twice the planet’s radius.

(a) Calculate the gravitational potential at the surface of the planet. [1] (b) Calculate the orbital speed of the satellite. [2] (c) Calculate the kinetic energy, potential energy and total energy of the satellite in orbit. [3] (d) Calculate the minimum energy needed to put the satellite into this orbit from rest on the surface (ignore rotation). [2] (e) Calculate the further energy needed to take it from this orbit to infinity. [1]

11. Titan orbits Saturn in a circular orbit of radius 1.22 × 10⁹ m.

(a) Calculate the orbital period of Titan in days. [3] (b) Calculate its orbital speed. [1] (c) Another moon orbits at a radius of 4.0 × 10⁸ m. Calculate its period in days using your answer to (a) and not the full formula. [2] (d) Describe the graph of gravitational potential against distance from the centre of Saturn, for distances beyond Saturn’s radius. [2]

Answers

1. The work done per unit mass [1] by an external agent moving a small test mass from infinity (where V = 0) to the point [1]. The field is attractive and does the work as the mass comes in, so the agent’s work is negative and V is below zero [1]. [3] Examiner insight: Include every key phrase: “per unit mass”, “from infinity” and the zero reference. Without “per unit mass” you describe potential energy instead.

2. (a) Zero [1]. The path lies on an equipotential (constant r), so ΔV = 0 and the force is perpendicular to the motion [1]. [2] (b) V(8.00e6) = −3.982e14/8.00e6 = −4.977 × 10⁷ J kg⁻¹ and V(2.40e7) = −1.659 × 10⁷ J kg⁻¹ [1]. ΔV = +3.318 × 10⁷ J kg⁻¹ [1]. ΔW = 900 × 3.318e7 = 2.99 × 10¹⁰ J [1]. [3] Examiner insight: In part (a), give the reason: force at right angles to the displacement, or ΔV = 0. “Zero because there is no force” is wrong, since a force does act.

3. (a) g = −ΔV/Δr, and for V = −GM/r this has magnitude |V|/r [1]. g = 1.5e7/6.0e6 = 2.5 N kg⁻¹ [1]. [2] (b) M = gr²/G = 2.5 × (6.0e6)²/6.67e-11 [1] = 1.35 × 10²⁴ kg [1]. [2] (c) V is inversely proportional to r, so doubling r halves V: −7.5 × 10⁶ J kg⁻¹ [1]. [1] Examiner insight: Show the substitution into the rearranged formula, so your method stays visible even if an arithmetic slip occurs. Check units: N kg⁻¹ for g, J kg⁻¹ for V.

4. Gravity provides the centripetal force [1]: GMm/r² = mv²/r [1]. Cancel m and one factor of r: v² = GM/r, so v = √(GM/r) [1]. [3] Examiner insight: In a “show that” answer, write the starting equation and each step, including the satellite mass cancelling.

5. r = 1.74e6 + 1.00e5 = 1.84 × 10⁶ m [1]. v = √(4.902e12/1.84e6) = 1.63 × 10³ m s⁻¹ [1]. T = 2πr/v = 2π × 1.84e6/1632 = 7.08 × 10³ s [1], which is 118 minutes [1]. [4] Examiner insight: A common slip is using the height 1.00 × 10⁵ m, or the Moon’s radius alone, as r. Orbit distances are measured from the centre.

6. (a) T = 2π√(r³/GM) with GM = 6.67e-11 × 1.99e30 = 1.327 × 10²⁰ [1]. T = 2π√((1.08e11)³/1.327e20) = 1.94 × 10⁷ s [1] = 224 days [1]. [3] (b) T² ∝ r³, so T increases by 4^1.5 = 8, giving 8 × 224 = 1.79 × 10³ days [1]. [1] Examiner insight: Stating “T² ∝ r³, so ×8” in part (b) is quicker and more reliable than recalculating.

7. (a) Ek = GMm/(2r) = 3.982e14 × 850/(2 × 7.20e6) [1] = 2.35 × 10¹⁰ J [1]. [2] (b) Ek = 3.982e14 × 850/(2 × 4.21e7) = 4.02 × 10⁹ J [1]. [1] (c) E = −Ek, so E(initial) = −2.35 × 10¹⁰ J and E(final) = −4.02 × 10⁹ J [1]. Change = +1.95 × 10¹⁰ J [1]. [2] (d) The energy supplied increases the potential energy, and the speed is lower at a larger radius [1]. [1] Examiner insight: Watch the sign of an energy change: a total energy that becomes less negative is an increase, so the energy supplied is positive.

8. (a) v = √(GM/r) = √(3.982e14/7.30e6) = 7.39 × 10³ m s⁻¹ [1]. At 7.10e6, v = 7.49 × 10³ m s⁻¹ [1]. [2] (b) Ek = GMm/(2r): ΔEk = 3.982e14 × 300/2 × (1/7.10e6 − 1/7.30e6) = +2.30 × 10⁸ J [1]. The total energy E = −Ek changes by −2.30 × 10⁸ J [1]. [2] (c) The satellite moves to a lower orbit, converting potential energy to kinetic energy: the potential energy falls by twice the kinetic energy gained and the surplus is dissipated [1]. [1] Examiner insight: Part (c) needs the energy transfer identified. “Drag speeds it up” is not an explanation, because drag opposes the motion.

9. (a) T = 9.93 × 3600 = 3.575 × 10⁴ s [1]. r³ = GMT²/(4π²) = 6.67e-11 × 1.90e27 × (3.575e4)²/(4π²) = 4.10 × 10²⁴ m³ [1]. r = 1.60 × 10⁸ m [1]. [3] (b) The orbit must be in the plane of the equator [1] and the satellite must move in the same direction as the planet’s rotation [1]. [2] Examiner insight: “Geosynchronous” and “geostationary” are different. The question asks for conditions beyond the period, so repeating the period gains nothing.

10. (a) V = −GM/R = −6.67e-11 × 4.50e24/8.00e6 = −3.75 × 10⁷ J kg⁻¹ [1]. [1] (b) r = 1.60e7 m. v = √(GM/r) = √(3.0015e14/1.60e7) [1] = 4.33 × 10³ m s⁻¹ [1]. [2] (c) Ek = ½mv² = 6.00 × 10⁹ J [1]. Ep = −2Ek = −1.20 × 10¹⁰ J [1]. E = Ek + Ep = −6.00 × 10⁹ J [1]. [3] (d) E(surface) = mV = 640 × (−3.75e7) = −2.40 × 10¹⁰ J [1]. Energy needed = −6.00e9 − (−2.40e10) = 1.80 × 10¹⁰ J [1]. [2] (e) The energy needed to reach infinity is −E = 6.00 × 10⁹ J [1]. [1] Examiner insight: In part (d) the energy on the ground is purely potential because the satellite is at rest, so do not add a kinetic term.

11. (a) GM = 6.67e-11 × 5.68e26 = 3.789 × 10¹⁶ [1]. T = 2π√((1.22e9)³/3.789e16) = 1.38 × 10⁶ s [1] = 15.9 days [1]. [3] (b) v = √(GM/r) = 5.57 × 10³ m s⁻¹ [1]. [1] (c) T² ∝ r³, so T = 15.9 × (4.0e8/1.22e9)^1.5 [1] = 2.99 days [1]. [2] (d) The potential is negative throughout [1]. It increases (becomes less negative) as the distance increases, with a gradient that decreases, approaching zero from below but never reaching it [1]. [2] Examiner insight: When describing a graph, state its sign, its direction of change and its end behaviour. Each is a separate point.

Where marks are usually lost

• Losing the minus sign of V, or subtracting initial from final the wrong way round.
• Using height above the surface as the orbital radius.
• Giving the change in total energy as the change in potential energy alone.
• Defining potential without “per unit mass” or “from infinity”.
• Calling any 24 hour orbit geostationary, and forgetting the equatorial plane and the direction of rotation.

Next steps

• Revisit the revision notes and the study guide.
• Tick off the checklist on the course hub.
• Try the all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.7.3 and 3.7.4.