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Newton's law of gravitation and field strength

Physik A-level Updated Wed 7 Oct 2026

Every mass attracts every other mass. Newton's law gives the size of that force, and dividing by the mass of the object being pulled gives the field strength, which tells you how strong gravity is at any point.

Part 1 of 3: Learn it

In short

  1. F = Gm₁m₂ ÷ r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
  2. Gravitational field strength g = F ÷ m; near a planet, g = GM ÷ r².
  3. Doubling the distance between centres makes the force a quarter as big.

Where this is in your specification

Spec points: AQA 7408 3.7.2.1 and 3.7.2.2, OCR A H556 5.4.1 and 5.4.2, Edexcel 9PH0 Topic 12

BoardTopic: Gravitational and electric fields
AQA 74083.7.1 to 3.7.3
Edexcel 9PH0Topics 7 and 12
OCR H5565.4 and 6.2
Higher C857 76Our dynamic Universe: gravitation; Particles and waves: forces on charged particles

The law

F = G m₁ m₂ ÷ r²

r is measured between the centres of the masses. A planet or a sphere of even density behaves as if all its mass were at its centre. The force is always attractive and acts along the line joining the centres.

Field strength

g = F ÷ m
g = G M ÷ r²

g is the force per unit mass on a small test mass, in N kg⁻¹. Outside a planet it falls with the square of the distance from the centre. Near the surface, over small changes in height, it is almost constant, so the field is treated as uniform.

Field lines

Around a planet, field lines are radial, pointing inwards to the centre, and spread out with distance, showing the field getting weaker. Close to the surface they look parallel and evenly spaced, showing a uniform field.

Quick check

The distance between two masses is tripled. What happens to the force between them?

Show the answer

It becomes one ninth as big.

Part 2 of 3: See it worked

Worked examples

Example 1

Find the gravitational force between the Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) when their centres are 3.84 × 10⁸ m apart.

  1. F = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² ÷ (3.84 × 10⁸)²
  2. = 2.93 × 10³⁷ ÷ 1.47 × 10¹⁷

Answer: 2.0 × 10²⁰ N.

Example 2

Find g at a height of 400 km above the Earth's surface. (Earth: radius 6.37 × 10⁶ m, mass 5.97 × 10²⁴ kg)

  1. r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m
  2. g = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ (6.77 × 10⁶)²

Answer: 8.7 N kg⁻¹, so the International Space Station is not beyond gravity: it is in free fall.

Common mistakes

  • Using the height above the surface as r.
  • Forgetting to square r.
  • Saying astronauts in orbit float because there is no gravity.
  • Mixing up G (a universal constant) and g (field strength at a place).
Quick check

What are the units of gravitational field strength?

Show the answer

N kg⁻¹ (equivalent to m s⁻²).

Part 3 of 3: Test yourself

Check yourself

Answer each one in your head or on paper first, then open it to check.

The distance between two masses is tripled. What happens to the force between them?

It becomes one ninth as big.

What are the units of gravitational field strength?

N kg⁻¹ (equivalent to m s⁻²).

Describe the field lines around a planet.

Radial lines pointing towards the centre, further apart further from the planet.

Jobs that use this

Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).

Diese Lernzettel sind auf Englisch, weil sie britischen Prüfungslehrplänen folgen.

Full lessons and marked practice for this course are coming soon to Brainlag Learn. See courses