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Newton's law of gravitation

Newton's law of gravitation makes every pair of masses attract with a force Gm₁m₂/r², so doubling their separation quarters the pull. There is no repulsive case, because mass has one sign. Dividing out the test mass gives the gravitational field strength g = GM/r², with r measured from the centre of the body.

Newton's law of gravitation

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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.

Gravity follows an inverse square law and never reaches zero; at the ISS it is about 90% of surface g. Astronauts float because they and the station free-fall together.

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}}IN THE AQA DATA BOOKLET

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.

A large grey mass and a small cyan one, with equal-length force arrows pointing at each other: the pair share one force magnitude, however unequal their masses.
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 on you and you pull on the Earth with forces of equal magnitude. Only the accelerations differ, because the masses differ.

WORKED EXAMPLE

The force between the Earth and the Moon

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 the condition for using the formula.

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, as Newton's third law requires; the Moon has the larger acceleration because it has the smaller mass.

Inverse square

A steep inverse-square curve of force against separation, with dashed markers showing the force at distance r and the quarter-sized force at two r.
FIG. 2Double the separation and the force falls to a quarter: the inverse-square signature.

The r2 in the denominator sets the law's behaviour with distance: double the separation and the force falls to a quarter; treble it and the force falls to a ninth. A question comparing forces at two separations is asking for a squared ratio, so set the ratio up before substituting numbers.

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 points matter here: the field falls to a quarter, not a half, and r is measured from the centre, not the surface.

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}IN THE AQA DATA BOOKLET

measured in N kg−1, a unit identical to m s−2: field strength and free-fall acceleration are the same quantity. Substituting Newton's law, the test mass cancels, leaving the field of a mass M at any point in its radial region,

g=GMr2g = \frac{GM}{r^{2}}IN THE AQA DATA BOOKLET
The field strength curve outside a planet, starting at the surface value at radius R and quartering by two R, with an amber reminder that r is measured from the centre, never the surface.
FIG. 3g outside a planet: the surface value at radius R, one quarter at 2R, with r always measured from the centre.

Measure r from the centre of the body, not from its surface. An orbit “400 km up” is at r = 6.37 × 106 + 4.0 × 105 m, so add the planet's radius to any altitude before substituting.

INDEPENDENT PRACTICE

Surface gravity on the Moon

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 about one sixth of the Earth's 9.81 N kg−1. The Moon's smaller mass reduces g and its smaller radius increases it, and the two effects combine to give one sixth.

ASSESSMENT FOCUS

  • The law applies to point masses, with spherical bodies treated as points at their centres. State that assumption, since it carries credit.
  • In a ratio question, square the distance ratio first and reach for numbers second. Doubling r quarters both F and g.
  • The force is mutual and equal in magnitude on both bodies, whatever their masses. “The Earth pulls harder on you than you pull on it” contradicts Newton's third law.
  • Add the planet's radius to any altitude before squaring. r is measured 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

Calculate r from the centre first, then consider what free fall means for the astronauts.

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 the station and the crew are both in free fall, accelerating at the same rate under that g while moving fast enough tangentially to remain in orbit. The gravitational field strength there is nearly the surface value; what is absent is the contact force that would otherwise support them.

Every mass pulls every other with an inverse-square force.

Field strength is force per unit mass.

Measure r from the centre of the body, not from its surface.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

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Answer book

PDF · 57 KB · fully worked · navy
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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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