Physics 9702 · syllabus 13
Cambridge A Level Physics 9702 — Gravitational fields topical past papers
Gravity as a field of force: field strength, Newton's law of gravitation, gravitational potential and potential energy, and circular and geostationary orbits.
- Questions
- 103
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 30
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Gravitational fields covers
A gravitational field is a region in which a mass feels a force, and its strength g is the force per unit mass. Outside a uniform sphere such as a planet, the mass can be treated as a point at the centre, so Newton's law of gravitation, F = Gm₁m₂/r², gives the force between two bodies and g = GM/r² gives the field strength at a distance r from the centre. Near the Earth's surface, height changes are small compared with the radius, so g is close to constant.
Gravitational potential φ at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It is negative everywhere, φ = −GM/r, because gravity attracts and the zero is at infinity. Multiplying by a second mass gives the potential energy of the pair, Eₚ = −GMm/r, from which energy changes far from a planet's surface can be calculated.
For a satellite in a circular orbit, gravity supplies the centripetal force, which gives the orbital speed and the result that T² is proportional to r³. A geostationary satellite orbits directly above the Equator, from west to east, with a period of 24 hours. This is A Level content, examined on Paper 4, and its structure is mirrored closely by electric fields.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 13 Gravitational fields
- 13.1 Gravitational field
- 13.2 Gravitational force between point masses
- 13.3 Gravitational field of a point mass
- 13.4 Gravitational potential
What you should be able to do
- Define gravitational field strength and potential, and draw field lines for a point mass and near a planet's surface
- Use F = Gm₁m₂/r² and g = GM/r², and derive the second from the first
- Use φ = −GM/r and Eₚ = −GMm/r to calculate energy changes between heights or orbits
- Equate gravitational and centripetal force to find an orbital speed, period or radius
- State the conditions for a geostationary orbit and calculate its radius
- Explain why g is approximately constant near the Earth's surface
Key ideas
- Newton's law of gravitation F = Gm₁m₂/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻²
- Field strength g = F/m = GM/r² (N kg⁻¹)
- Potential φ = −GM/r (J kg⁻¹); potential energy of two masses Eₚ = −GMm/r
- Circular orbit: GMm/r² = mv²/r = mrω², which leads to T² = (4π²/GM)r³
How Gravitational fields has been examined
From 2022 to 2026, Gravitational fields was on Paper 4 · A Level Structured Questions in 10 of 11 exam series, worth about 10% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · A Level Structured Questions | 30 | 10% | 10 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 10.
Filed in Momentum under: Gravitational Fields.
A worked example
Calculate the radius of a geostationary orbit and its height above the Earth's surface. The Earth's mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m.
- Period T = 24 h = 24 × 3600 = 8.64 × 10⁴ s
- Gravity provides the centripetal force: GMm/r² = mr(2π/T)², so r³ = GMT²/4π²
- r³ = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (8.64 × 10⁴)² / 4π² = 7.53 × 10²² m³
- r = 4.22 × 10⁷ m
- Height above the surface = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.59 × 10⁷ m
Answer: Orbital radius 4.22 × 10⁷ m; height 3.59 × 10⁷ m
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Measuring r from the planet's surface rather than its centre — for a satellite at height h above a planet of radius R, r = R + h
- Dropping the minus sign in φ = −GM/r and then getting an energy change backwards — moving away from a planet makes the potential less negative, so potential energy increases
- Squaring r in the potential formula, or forgetting to square it in the field formula — φ goes as 1/r and g as 1/r²
- Using mgΔh for a change in height comparable to the planet's radius — g is no longer constant, so use the change in −GMm/r
- Giving only the 24-hour period for a geostationary orbit — the satellite must also be above the Equator and move in the same direction as the Earth rotates
Exam technique
- For a derivation or 'show that', start from the equation linking the forces, cancel the satellite's mass and say so, then rearrange step by step
- Keep at least three significant figures until the final answer, especially when a cube root is involved
- When defining potential, include 'work done per unit mass', 'small test mass' and 'from infinity'
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.