OCR A Level Physics A: Gravitational fields, Newton's law and planetary motion (H556) – 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 cover sections 5.4.1 to 5.4.3 of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² and, where needed, Earth mass 5.97 × 10²⁴ kg and Earth radius 6.37 × 10⁶ m.

Learn the content from the study guide and the revision notes. See also the OCR A Level Physics hub and the checklist.

Questions

1. (a) State the property of an object that gives rise to a gravitational field. [1] (b) Define gravitational field strength. [2]

2. Four moons orbit a planet. Their orbital radii r and periods T are given in the table.

r / 10⁸ m	1.2	2.0	3.5	5.0
T / days	0.676	1.45	3.37	5.75

(a) Show that T² ∝ r³ holds for the moons at 1.2 × 10⁸ m and 3.5 × 10⁸ m. [3] (b) Use your values to calculate the mass of the planet. [2]

3. (a) Describe the field lines around an isolated spherical planet and what their spacing shows. [2] (b) Describe the field lines close to the planet’s surface and explain why the field there is treated as uniform. [2]

4. Two uniform spheres of mass 6.0 kg and 4.0 kg have their centres 0.30 m apart.

(a) Calculate the gravitational force between them. [2] (b) State the force when the centres are 0.60 m apart. [1] (c) State the distance used in your calculation and why it is valid for spheres. [1]

5. Jupiter has mass 1.90 × 10²⁷ kg and radius 7.15 × 10⁷ m.

(a) Calculate g at the planet’s surface. [2] (b) Calculate the field strength at a point 2.0 radii above the surface. [2] (c) Calculate the gravitational force on a 3.0 kg probe at that point. [1]

6. A summit is 2000 m above sea level.

(a) Calculate the gravitational field strength at the summit, and the percentage decrease from the sea-level value. [2] (b) State what this shows about treating g as uniform near the surface. [1]

7. Derive the equation T² = (4π²/GM) r³ for a planet of mass m in a circular orbit of radius r around a star of mass M. [4]

8. (a) State Kepler’s three laws of planetary motion. [3] (b) An asteroid orbits the Sun. Its closest approach is 2.2 × 10¹¹ m, where its speed is 2.1 × 10⁴ m s⁻¹, and its furthest distance is 4.4 × 10¹¹ m. Calculate its speed at the furthest point. [2] (c) Calculate the area swept out per second by the line from the Sun to the asteroid at closest approach. [1]

9. A satellite orbits the Earth at a height of 800 km.

(a) State the orbital radius. [1] (b) Calculate its orbital period. [2] (c) Calculate its orbital speed. [2]

10. A star has two planets in circular orbits. The inner planet has orbital radius 2.0 × 10¹⁰ m and period 20 days. The outer planet has orbital radius 5.0 × 10¹⁰ m.

(a) Calculate the period of the outer planet. [2] (b) Calculate the mass of the star. [3] (c) Explain why the masses of the planets are not needed in part (b). [1] (d) State the condition that must be met for the comparison in part (a) to be valid. [1]

11. Saturn has mass 5.68 × 10²⁶ kg, equatorial radius 6.03 × 10⁷ m and a rotation period of 10.7 hours. A satellite is to stay above one point on Saturn’s equator.

(a) State two orbit conditions. [2] (b) Calculate the orbital radius needed. [3] (c) Calculate the height of the orbit above the surface. [1] (d) State one use and one disadvantage of geostationary satellites at the Earth. [2]

12. The Sun and the Earth have masses 1.99 × 10³⁰ kg and 5.97 × 10²⁴ kg. Their centres are 1.50 × 10¹¹ m apart. A point on the line between them has zero resultant gravitational field strength.

(a) State the condition for the resultant field to be zero. [1] (b) Calculate the distance of this point from the centre of the Earth. [4] (c) Calculate the magnitude of the Earth’s field strength at this point. [2]

Answers

1. (a) Mass [1]. [1] (b) The force per unit mass [1] on a small test mass at that point, g = F/m [1]. [2] Examiner insight: Define it with both “force” and “per unit mass”.

2. (a) For 1.2 × 10⁸ m: T = 5.84 × 10⁴ s, T²/r³ = 3.41 × 10⁹ / 1.728 × 10²⁴ = 1.97 × 10⁻¹⁵ s² m⁻³ [1]. For 3.5 × 10⁸ m: T = 2.91 × 10⁵ s, T²/r³ = 8.48 × 10¹⁰ / 4.2875 × 10²⁵ = 1.98 × 10⁻¹⁵ s² m⁻³ [1]. The ratios agree to within rounding, so T² ∝ r³ [1]. [3] (b) T²/r³ = 4π²/GM, so M = 4π² / (G × 1.98 × 10⁻¹⁵) [1] = 3.0 × 10²⁶ kg [1]. [2] Examiner insight: Convert days to seconds before squaring. One ratio alone proves nothing, so show both.

3. (a) Straight lines pointing radially inwards towards the centre, with arrows [1]. They are closer together nearer the planet, showing the field is stronger there [1]. [2] (b) The lines are parallel and equally spaced, vertically downwards [1], because over a small height range the distance from the centre barely changes, so g barely changes [1]. [2] Examiner insight: Mention both direction and spacing, and say what the spacing means.

4. (a) F = GMm/r² = 6.67 × 10⁻¹¹ × 6.0 × 4.0 / 0.30² [1] = 1.8 × 10⁻⁸ N (1.78 × 10⁻⁸ N) [1]. [2] (b) Distance doubled, so force is a quarter: 4.4 × 10⁻⁹ N [1]. [1] (c) The distance between the centres, 0.30 m [1]: a uniform sphere acts as if its mass is at its centre. [1] Examiner insight: Keep the power of ten with the unit throughout. Forces between small objects are tiny, so a large answer signals an error.

5. (a) g = GM/R² = 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / (7.15 × 10⁷)² [1] = 24.8 N kg⁻¹ [1]. [2] (b) The distance from the centre is R + 2.0R = 3.0R = 2.145 × 10⁸ m [1]. g = 24.8 / 3² = 2.75 N kg⁻¹ [1]. [2] (c) F = mg = 3.0 × 2.754 = 8.3 N [1]. [1] Examiner insight: “2.0 radii above the surface” means r = 3.0 R. State the distance you use at the start.

6. (a) r = 6.37 × 10⁶ + 2000 = 6.372 × 10⁶ m [1]. g = GM/r² = 9.807 N kg⁻¹ against 9.813 N kg⁻¹ at sea level, a decrease of 0.063 % [1]. [2] (b) The change is under 0.1 % over 2000 m, so g is effectively uniform near the surface [1]. [1] Examiner insight: Give the percentage, then a comparison. A bare number does not answer “what does this show”.

7. The gravitational force provides the centripetal force [1]: GMm/r² = mv²/r [1]. Substitute v = 2πr/T [1]: GM/r² = 4π²r/T². Rearranging: T² = (4π²/GM) r³ [1]. [4] Examiner insight: Write the starting statement in words and show m cancelling. A derivation that jumps to the answer leaves nothing to credit.

8. (a) Planets move in ellipses with the Sun at one focus [1]. The line joining a planet to the Sun sweeps equal areas in equal times [1]. T² is proportional to r³ [1]. [3] (b) Equal areas in equal times at the closest and furthest points gives v₁r₁ = v₂r₂ [1]. v₂ = 2.1 × 10⁴ × 2.2 × 10¹¹ / 4.4 × 10¹¹ = 1.05 × 10⁴ m s⁻¹ [1]. [2] (c) Area per second = ½rv = ½ × 2.2 × 10¹¹ × 2.1 × 10⁴ = 2.3 × 10¹⁵ m² s⁻¹ [1]. [1] Examiner insight: Say the relation holds at these two points because velocity is perpendicular to the radius there.

9. (a) r = 6.37 × 10⁶ + 8.0 × 10⁵ = 7.17 × 10⁶ m [1]. [1] (b) T = 2π √(r³/GM) [1] = 2π √((7.17 × 10⁶)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)) = 6.05 × 10³ s [1]. [2] (c) v = 2πr/T [1] = 2π × 7.17 × 10⁶ / 6045 = 7.45 × 10³ m s⁻¹ [1]. [2] Examiner insight: Writing “r = R + h” first avoids using the height as r. Cross-check with v = √(GM/r).

10. (a) T ∝ r^(3/2) so T = 20 × (5.0 / 2.0)^1.5 [1] = 79 days [1]. [2] (b) T = 20 × 86 400 = 1.728 × 10⁶ s [1]. M = 4π²r³/(GT²) = 4π² × (2.0 × 10¹⁰)³ / (6.67 × 10⁻¹¹ × (1.728 × 10⁶)²) [1] = 1.6 × 10³⁰ kg [1]. [3] (c) The planet’s mass appears in both the gravitational force and the centripetal force, so it cancels [1]. [1] (d) Both planets must orbit the same central star [1]. [1] Examiner insight: The ratio in part (a) needs no unit conversion, but part (b) needs seconds.

11. (a) Any two of: above the equator; same direction as Saturn rotates; circular; period equal to its rotation period [1] [1]. [2] (b) T = 10.7 × 3600 = 3.852 × 10⁴ s [1]. r³ = GMT²/4π² [1]. r = 1.13 × 10⁸ m [1]. [3] (c) h = 1.125 × 10⁸ − 6.03 × 10⁷ = 5.2 × 10⁷ m [1]. [1] (d) Use: television or telephone communications, or weather observation of a fixed region [1]. Disadvantage: a signal delay of about 0.24 s, or poor coverage at high latitudes [1]. [2] Examiner insight: “Geostationary” needs “above the equator” as well as “24 hours”. Missing the equator is the usual lost point.

12. (a) The fields from the two bodies are equal in magnitude and opposite in direction [1]. [1] (b) Let the distance from the Earth be x. GM_E/x² = GM_S/(d − x)² [1]. x/(d − x) = √(M_E/M_S) = 1.73 × 10⁻³ [1]. x = d × 1.73 × 10⁻³ / (1 + 1.73 × 10⁻³) [1] = 2.6 × 10⁸ m (2.59 × 10⁸ m) [1]. [4] (c) g = GM_E/x² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (2.59 × 10⁸)² [1] = 5.9 × 10⁻³ N kg⁻¹ [1]. [2] Examiner insight: Take the square root of both sides early to avoid a quadratic, then check that the point is much nearer the Earth.

Where marks are usually lost

• Substituting the height above the planet into GM/r² instead of the distance to its centre.
• Writing “2.0 radii above the surface” as r = 2.0R.
• Not squaring the power of ten when squaring r.
• Starting a derivation without saying gravity provides the centripetal force.
• Applying v₁r₁ = v₂r₂ away from the two closest and furthest points.
• Leaving out “above the equator” for a geostationary orbit.

Next steps

• Revisit the revision notes and study guide.
• Go back to the OCR A Level Physics hub and tick off the checklist.
• Take all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 5.4.1 Point and spherical masses, 5.4.2 Newton’s law of gravitation and 5.4.3 Planetary motion.