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
Official 2025-2027 spec · A2 LevelThese are the exact learning outcomes Cambridge sets for this topic. The candidate is expected to be able to do each of these on the relevant paper.
- understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre
- recall and use Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point masses
- analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes
- understand that a satellite in a geostationary orbit remains at the same point above the Earth’s surface, with an orbital period of 24 hours, orbiting from west to east, directly above the Equator
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
- › Define gravitational potential as the work done *per unit mass* when moving a mass from infinity.
- › Definitions of potential always require the 'per unit mass' or 'per unit charge' component to be dimensionally correct.
- › Be precise with language; Newton’s law involves the product of masses and the inverse square of the separation between their centers.
- › State that force is proportional to the product of the masses and inversely proportional to the square of the separation between their centers.
- › Read the question carefully to distinguish between requests for field lines (radial with arrows) and equipotential lines (concentric circles).
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 necessary for the satellite's circular motion. Therefore, F_gravitational = F_centripetal, which means Gm1m2/r^2 = mv^2/r, where m is the satellite's mass.
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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