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Newton's law of gravitation

One equation covers the apple and the Moon. Every pair of masses attracts, with a force set by their product and by the inverse square of their separation. Divide out the test mass and the same law gives the field strength anywhere.

Year 13AQA 3.7.2.1, 3.7.2.2CIE 13.1.1, 13.2.1, 13.2.2, 13.3.1, 13.3.2OCR A 5.4.1(b), 5.4.1(d), 5.4.2(a), 5.4.2(b), 5.4.2(c)EDEXCEL 175, 176, 177

Builds on The field concept.

IN THIS TOPIC

  • Use Newton's law of gravitation for point masses.
  • Define g as force per unit mass and use g = F/m.
  • Use g = GM/r² in a radial field, with r measured from the centre.

COMMON MISCONCEPTION

There's no gravity in space.

The universal law

Gravity is a universal attractive force acting between all matter: every mass pulls every other. For point masses, and for spheres treated from outside as points at their centres, the magnitude is

F=Gm1m2r2F = \frac{Gm_{1}m_{2}}{r^{2}}ON THE AQA DATA SHEET

where G, the gravitational constant, is 6.67 × 10−11 N m2 kg−2, printed in the data booklet. That tiny number is why gravity between everyday objects goes unnoticed. Two people standing a metre apart attract with well under a millionth of a newton.

Gravity is mutual: the Earth pulls the apple and the apple pulls the Earth with exactly equal forcebig Msmall mF = Gm₁m₂/r², on each of themsame force; only the accelerations differ
FIG. 1The force acts on both masses equally: Newton's third law inside Newton's law of gravitation.

The force is mutual. The Earth pulls you and you pull the Earth, with exactly equal force; only the accelerations differ, because the masses do. Exam questions probe this precisely because intuition resists it.

WORKED EXAMPLE

Weighing the Moon's leash

Find the gravitational force between the Earth (5.97 × 1024 kg) and the Moon (7.35 × 1022 kg), separated by 3.84 × 108 m.

Both bodies are far enough apart to treat as point masses at their centres, which is what licenses the formula at all.

F = GMm/r2 = (6.67 × 10−11 × 5.97 × 1024 × 7.35 × 1022)/(3.84 × 108)2 = 2.0 × 1020 N.

The same 2.0 × 1020 N acts on each body, Newton's third law holding at planetary scale; the Moon simply answers with more acceleration for its smaller mass.

Inverse square

The inverse-square law: double the separation and the force falls to a quarterrFr2rFF/4double the distance, a quarter of the force
FIG. 2Double the separation and the force falls to a quarter: the inverse-square signature.

The r2 downstairs is the law's character. Double the distance, quarter the force; treble it, a ninth. Any question comparing forces at two separations is really asking you to square a ratio, and setting the ratio up before touching numbers is the fastest route through.

GUIDED PRACTICE

g at double the distance

The field strength at the Earth's surface is 9.81 N kg−1. Without touching G or M, find g at a height of one Earth radius above the surface.

Show the working

One radius above the surface is two radii from the centre, the distance the law actually uses.

Doubling r quarters the inverse-square field, so g = 9.81/4 = 2.5 N kg−1. Two classic traps live in that one line. Halving instead of quartering is the first, and measuring r from the surface instead of the centre is the second.

From force to field strength

The gravitational field strength at a point is the force per unit mass a body placed there would feel:

g=Fmg = \frac{F}{m}ON THE AQA DATA SHEET

measured in N kg−1, and that unit is identical to m s−2. Field strength and free-fall acceleration are one quantity under two names. Substitute Newton's law, watch the test mass cancel, and what remains is the field of a mass M anywhere in its radial region,

g=GMr2g = \frac{GM}{r^{2}}ON THE AQA DATA SHEET
Field strength above a planet: g is GM over r squared, measured from the centre, quartering by twice the radiusr (from centre)gR2Rsurface: 9.81a quarterr is measured from the planet's centre
FIG. 3g outside a planet: the surface value at radius R, one quarter at 2R, with r always measured from the centre.

One habit matters more than any other in this unit. Measure r from the centre, never from the surface. An orbit “400 km up” sits at r = 6.37 × 106 + 4.0 × 105 m, and forgetting to add the planet's own radius wrecks more answers here than anything else.

INDEPENDENT PRACTICE

The Moon's famous sixth

The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m. Find the field strength at its surface, and compare it with Earth's.

Show the working

g = GM/r2 = (6.67 × 10−11 × 7.35 × 1022)/(1.74 × 106)2 = 1.6 N kg−1.

That is the celebrated one sixth of Earth's 9.81, computed instead of quoted. Less mass pulls the value down, a smaller radius pushes it back up, and one sixth is where the contest settles.

ASSESSMENT FOCUS

  • The law applies to point masses, with spherical bodies treated as points at their centres. Stating that assumption is often a mark in itself.
  • In a ratio question, square the distance ratio first and reach for numbers second. Doubling r quarters both F and g.
  • The pull is mutual and equal on both bodies, however unequal the masses. “The Earth pulls harder on you than you pull on it” fails Newton's third law.
  • Add the planet's radius to any altitude before you square anything. r runs from the centre.
  • g at the International Space Station's altitude is about 89% of the surface value, and astronauts float because they are in free fall, not because gravity has stopped. Papers ask for the calculation and the explanation together.

CHECK YOURSELF

The ISS orbits about 400 km above the Earth's surface. Taking M = 5.97 × 1024 kg and RE = 6.37 × 106 m, calculate g at the station, and explain why astronauts float despite your answer.

Show a hint

Build r from the centre first, then ask what free fall feels like from inside.

Show the answer

r = 6.37 × 106 + 4.0 × 105 = 6.77 × 106 m.

g=GM/r2g = GM/r^{2} = (6.67 × 10−11 × 5.97 × 1024) / (6.77 × 106)2 = 8.7 N kg−1, about 89% of the surface value.

Astronauts float because station and crew are both in free fall, accelerating identically under that g while perpetually missing the ground. Gravity up there is nearly full strength; support forces are what vanished.

Every mass pulls every other, as the inverse square.

Field strength is force per unit mass.

Measure r from the centre, every single time.

WORKBOOK

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  • Use Newton's law of gravitation for point masses.
  • Define g as force per unit mass and use g = F/m.
  • Use g = GM/r² in a radial field, with r measured from the centre.

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