Gravitational force between point masses
Cambridge A-Level Physics (9702) · Unit 13: Gravitational fields · 7 flashcards
Gravitational force between point masses is topic 13.2 in the Cambridge A-Level Physics (9702) syllabus , positioned in Unit 13 — Gravitational fields , alongside Gravitational field, Gravitational field of a point mass and Gravitational potential. In one line: Newton's Law of Gravitation states that the gravitational force (F) between two point masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers: F = Gm1m2 / r².
Marked as A2 Level: examined at A Level in Paper 4 (A Level Structured Questions) and Paper 5 (Planning, Analysis and Evaluation). It is not tested on the AS-only papers (Papers 1, 2 and 3).
The deck below contains 7 flashcards — 2 definitions, 3 key concepts and 2 calculations — covering the precise wording mark schemes reward. Use the 2 definition cards to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.
Newton's Law of Gravitation
Newton's Law of Gravitation states that the gravitational force (F) between two point masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers: F = Gm1m2 / r²
What the Cambridge 9702 syllabus says
2025-2027 syllabus, in our words · A2 LevelWhat the syllabus expects you to be able to do in this topic, in plain English. For the exact wording, see the official syllabus (PDF).
- Understand that, at any point outside a uniform sphere, the sphere's mass can be treated as a point mass located at its centre.
- Recall Newton's law of gravitation, F = Gm₁m₂/r², giving the gravitational force that two point masses exert on each other, and use it.
- Analyse an object in a circular orbit within a gravitational field, linking its centripetal acceleration to the gravitational force that produces it.
- Understand that a geostationary satellite stays above the same point on the Earth's surface: it orbits directly above the Equator, from west to east, with a period of 24 hours.
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 9702 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Tips to avoid common mistakes in Gravitational force between point masses
- › Be precise with language; Newton’s law involves the product of masses and the inverse square of the separation between their centers.
Explain why we can treat a uniform sphere's mass as a point mass concentrated at its center when calculating gravitational force outside the sphere.
Gravitational field lines outside a uniform sphere are identical to those of a point mass at its center. This simplifies gravitational force calculations using Newton's Law of Gravitation, allowing us to treat the sphere as if all its mass is at a single point.
State Newton's Law of Gravitation.
Newton's Law of Gravitation states that the gravitational force (F) between two point masses (m1 and m2) is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers: F = Gm1m2 / r²
A satellite orbits a planet. How is the gravitational force related to the centripetal acceleration required for its orbit?
The gravitational force between the planet and the satellite provides the centripetal force needed for the circular orbit. So GMm/r² = mv²/r, where M is the planet's mass, m is the satellite's mass and r is the orbital radius measured from the planet's centre.
What are the defining characteristics of a geostationary orbit?
A geostationary orbit is a circular orbit where the satellite remains at a fixed point above the Earth's surface. This requires an orbital period of 24 hours, orbiting from west to east, directly above the Equator.
How would you calculate the orbital speed (v) of a satellite orbiting a planet of mass M at a distance r from its center?
Equate gravitational force to centripetal force (GMm/r^2 = mv^2/r), and solve for v. Therefore, v = sqrt(GM/r), where G is the gravitational constant, M is the planet's mass, and r is the orbital radius.
If the orbital radius of a satellite is doubled, how does this affect its orbital period?
From Kepler's Third Law (T² ∝ r³), if the orbital radius is doubled (r' = 2r), then the square of the new period (T'²) will be proportional to (2r)³. This means T'² = 8T², so the new period T' is sqrt(8) times the original period T.
A satellite orbits a planet at a distance 'r' with a velocity 'v'. If the distance is increased to '4r', what will be the new orbital velocity?
Since v = sqrt(GM/r), the velocity is inversely proportional to the square root of the radius. If the radius increases to 4r, the new velocity v' = sqrt(GM/4r) = v/2. The new velocity will be half of the original velocity.
Review the material
Read full revision notes on Gravitational force between point masses — definitions, equations, common mistakes, and exam tips.
Read NotesMore topics in Unit 13 — Gravitational fields
Gravitational force between point masses sits alongside these A-Level Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Key terms covered in this Gravitational force between point masses deck
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