Newton’s law of gravitation

Every mass attracts every other mass. Newton’s law of gravitation says the force depends on the product of the two masses and gets weaker with the square of the distance between their centres. It explains the weight of objects, the value of g on other worlds and the orbits of moons and satellites.

Key facts

  • F = Gm₁m₂ / r², with G = 6.67 × 10^−11 N m²/kg² and r measured between centres.
  • Inverse square law: double the distance, a quarter of the force.
  • Gravitational field strength at distance r from a planet of mass M: g = GM / r².
  • For a circular orbit gravity provides the centripetal force: v = √(GM/r), T = 2π√(r³/GM).
  • Gravitational potential energy far from the surface: E = −GMm / r. Escape velocity v = √(2GM/R).
  • Gravity is always attractive, and the forces on the two masses are equal and opposite.

Worked examples

Example 1

Find the gravitational force between two 60 kg people 0.50 m apart.

  1. Each pulls the other: equal and opposite forces along the line between them. F = 6.67 × 10^−11 × 60 × 60 ÷ 0.50²
  2. Work it out: tiny! F = 9.6 × 10^−7 N ≈ 0.96 µN

Answer: 0.96 µN

Example 2

Find g at the surface of Mars (mass 6.42 × 10^23 kg, radius 3.39 × 10^6 m).

  1. Field strength at the surface points to the centre. g = 6.67 × 10^−11 × 6.42 × 10^23 ÷ (3.39 × 10^6)²
  2. Work it out. g = 3.73 N/kg

Answer: 3.73 N/kg

Example 3

A satellite orbits 400 km above Earth (M = 5.97 × 10^24 kg, R = 6.37 × 10^6 m). Find its speed and period.

  1. Measure r from the centre of the Earth. r = 6.37 × 10^6 + 0.40 × 10^6 = 6.77 × 10^6 m
  2. GMm/r² = mv²/r gives v = √(GM/r). v = 7669 m/s
  3. Time for one orbit: T = 2πr ÷ v = 5546 s ≈ 92.4 min

Answer: 7670 m/s; 92.4 min