A Level Physics: Gravitational Fields — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Gravitational Fields revision notes

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Questions

1. State Newton’s law of gravitation in words. [3]

2. Define gravitational field strength and gravitational potential. [4]

3. Explain why gravitational potential is always negative. [2]

4. A satellite of mass 850 kg orbits Earth at a height of 3.6 × 10⁵ m above the surface. (G = 6.67 × 10⁻¹¹; M_Earth = 5.97 × 10²⁴ kg; R_Earth = 6.37 × 10⁶ m)

(a) Calculate the orbital radius. [1] (b) Calculate the gravitational field strength at that radius. [3] (c) Calculate the orbital speed. [3] (d) Calculate the period of the orbit. [2]

5. A geostationary satellite must satisfy three conditions.

(a) State them. [3] (b) Explain why such an orbit must be equatorial. [2]

6. (Extension — beyond the 9702 syllabus, not examinable.) Explain what is meant by escape velocity, and derive an expression for it.

7. Explain why an astronaut in orbit appears weightless even though gravity acts on them. [3]

8. Two point masses are separated by a distance r. If the separation is doubled, state what happens to the gravitational force between them, and explain why. [2]

9. Distinguish between gravitational potential, ϕ, and gravitational potential energy, E, including their units and the equation relating them. [3]

10. Explain why g = GM/r² cannot be used to calculate the gravitational field strength at a point inside a uniform solid sphere. [2]

11. Using G = 6.67 × 10⁻¹¹ N m² kg⁻², M_Earth = 5.97 × 10²⁴ kg and R_Earth = 6.37 × 10⁶ m, show that the gravitational field strength at the Earth’s surface is approximately 9.8 m s⁻². [2]

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Answers

1. The gravitational force between two point masses is proportional to the product of their masses [1] and inversely proportional to the square of their separation [1], acting along the line joining them [1].

2. Field strength — the force per unit mass at a point [1], g = F ÷ m [1]. Potential — the work done per unit mass in bringing a small mass from infinity to that point [1], φ = W ÷ m [1].

3. Potential is defined as zero at infinity [1], and since gravity is attractive, work is released as a mass moves inwards, so the potential at any finite distance is less than zero [1].

4. (a) r = 6.37 × 10⁶ + 3.6 × 10⁵ = 6.73 × 10⁶ m [1]. (b) g = GM ÷ r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (6.73 × 10⁶)² [1] [1] = 3.982 × 10¹⁴ ÷ 4.529 × 10¹³ = 8.79 N kg⁻¹ [1]. (c) Gravitational force provides the centripetal force: v = √(GM ÷ r) [1] = √(3.982 × 10¹⁴ ÷ 6.73 × 10⁶) [1] = 7.69 × 10³ m s⁻¹ [1]. (d) T = 2πr ÷ v = (2π × 6.73 × 10⁶) ÷ 7.69 × 10³ [1] = 5498 s ≈ 92 minutes [1].

5. (a) Period exactly 24 hours [1]; orbit above the equator [1]; moving west to east, the same direction as the Earth’s rotation [1]. (b) For the satellite to remain above a fixed point, the centre of its orbit must coincide with the Earth’s centre [1]; only an equatorial orbit keeps it above the same longitude at all times [1].

6. (Extension, not part of the 9702 mark scheme.) The minimum speed needed to escape a gravitational field entirely without further propulsion. Equating kinetic energy to the magnitude of gravitational potential energy: ½mv² = GMm ÷ r, giving v = √(2GM ÷ r).

7. Both the astronaut and the spacecraft are in free fall — accelerating towards the Earth at the same rate [1]. There is therefore no contact force between the astronaut and the spacecraft [1], and it is that absence of a normal contact force, not the absence of gravity, that produces the sensation of weightlessness [1].

8. The force falls to one quarter of its original value [1]. This is because gravitational force follows an inverse square law, F = Gm₁m₂/r², so doubling r increases the denominator by a factor of 2² = 4 [1].

9. Gravitational potential, ϕ, is the work done per unit mass in bringing a small test mass from infinity to that point, measured in J kg⁻¹ [1]. Gravitational potential energy, E, is the work done in bringing an actual mass m from infinity, measured in joules [1], related by E = mϕ [1].

10. The formula g = GM/r² applies only outside a spherical mass, or for a genuine point mass [1]; inside a uniform solid sphere, only the mass enclosed within radius r contributes to the field at that point, so the simple point-mass formula does not apply [1].

11. g = GM/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (6.37 × 10⁶)² [1] ≈ 9.8 m s⁻² [1] — showing that g = 9.81 m s⁻² used throughout AS mechanics is not an arbitrary constant, but a direct consequence of Newton’s law of gravitation applied at the Earth’s surface.

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Where marks are usually lost

• Omitting the minus sign on gravitational potential.
• Using the height above the surface instead of the orbital radius.
• Saying astronauts are weightless because there is no gravity.
• Forgetting that gravitational force provides the centripetal force in an orbit.
• Confusing gravitational potential (J kg⁻¹) with gravitational potential energy (J) — check which the question asks for.
• Treating g as a fixed constant rather than a quantity that decreases with distance from a mass.