Learn9702Paper 4 · A Level Structured QuestionsGravitational fieldsGravitational force and field strength

Gravitational fields · Subtopic 1 of 2

Gravitational Force and Field Strength

Every pair of point masses attracts with a force $F=\dfrac{GMm}{r^2}$ , where $G=6.67\times10^{-11}$ N m² kg⁻². A uniform sphere acts as if its mass were at its centre. Gravitational field strength is force per unit mass, $g=\dfrac Fm=\dfrac{GM}{r^2}$ , an inverse-square law. Near Earth's surface, $g$ is nearly constant because height changes are tiny compared with Earth's radius.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define gravitational field strength as force per unit mass.
Use $F=\dfrac{GMm}{r^2}$ for point masses and uniform spheres.
Derive and use $g=\dfrac{GM}{r^2}$ .
Explain why $g$ is approximately constant near Earth's surface.

Pause and think

The International Space Station orbits about $400$ km up. Astronauts float, so is $g$ there nearly zero?

Section 2 of 14

Core idea

Start hereWhy the Moon doesn't fall

The Moon is falling towards Earth all the time; it just moves sideways so fast that it keeps missing. Newton realised the force that drops an apple is the same one that bends the Moon's path, weakening with the square of distance.

The big idea

$F=\frac{GMm}{r^2}$ ; $g=\frac{GM}{r^2}$ . Orbits: gravity provides $\frac{mv^2}r$ , giving $v=\sqrt{\frac{GM}r}$ and $T^2=\frac{4\pi^2}{GM}r^3$ .

1. Field strength

Field strength outside the Earth falls as 1/r²

Double the distance from the centre: a quarter of the field. Inside the Earth (not needed here) it would fall to zero at the centre.

Radial field lines around a planet; uniform close to the surface

Field lines point towards the mass and spread out with distance: the field weakens. Over a small region near the surface they are almost parallel.

F=\frac{GMm}{r^2}\qquad g=\frac{GM}{r^2}

$G$: $6.67\times10^{-11}$ N m² kg⁻².
$r$: distance between centres.

Worked exampleEarth's field

$M=5.97\times10^{24}$ kg, $R=6.37\times10^6$ m.

Goal: $g$ at the surface and at $2R$ .
Strategy: Formula; inverse square.

$g=\frac{6.67\times10^{-11}\times5.97\times10^{24}}{(6.37\times10^6)^2}=9.81$
Surface.
$\frac{9.81}{4}=2.45\text{ N kg}^{-1}$
Double distance, quarter field.

Answer

9.81

;

2.45

N kg

^{-1}

✓ CheckMatches measured

g

✓.

2. Orbits

Orbital speed and Kepler's third law

$\frac{GMm}{r^2}=\frac{mv^2}r$
Gravity is the centripetal force.
$v=\sqrt{\frac{GM}r}$
$m$ cancels.
$\left(\frac{2\pi r}T\right)^2=\frac{GM}r$
$v=2\pi r/T$ .
$T^2=\frac{4\pi^2}{GM}r^3$
Rearrange.

Worked exampleGeostationary orbit

Period 24 h, equatorial, same direction as Earth's spin.

Goal: Radius and speed.
Strategy: Kepler's third law.

$r^3=\frac{GMT^2}{4\pi^2}$
$T=86\,400$ s.
$r=4.22\times10^7\text{ m}$
Cube root: ~36 000 km above the surface.
$v=\sqrt{GM/r}=3.07\text{ km s}^{-1}$
Speed.

Answer

4.22\times10^7

3.07

km s

^{-1}

✓ Check

2\pi r/T=3.07

km s

^{-1}

✓.

Common misconceptionHeight vs radius Use distance from Earth's centre: add 6370 km to the height.

Summary

$F=GMm/r^2$ ; $g=GM/r^2$ .
Orbits: $T^2\propto r^3$ .

Gravitational field strength: Force per unit mass.
Geostationary: Orbit staying above one point on the equator.

HistoryWeighing the Earth. In

1798

Henry Cavendish measured the tiny attraction between lead spheres using a torsion balance in a sealed shed, watching through a telescope to avoid air currents. His result gives

G

, and with it the first measurement of Earth's mass: about

6\times10^{24}

kg.

Fun factYou pull the Earth up. When you jump, the Earth pulls you down with your weight, and you pull the Earth up with an equal force. The Earth's acceleration is about

10^{-22}

m s⁻², far smaller than the width of an atomic nucleus per second squared.

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