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Newton's law of gravitation

Edexcel IAL Mathematics · WME03 Mechanics 3

Near the Earth's surface the weight mg is constant. Far from the surface it is not: the gravitational force falls off with the square of the distance from the centre. Pearson's specification says M3 problems may use this inverse square law.

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The inverse square law

The force of attraction on a particle of mass m at distance x from the centre of the Earth is F = k/x2 for some constant k. At the surface (x = R) the force is mg, so k = mgR2 and

F = mgR2/x2.

The acceleration towards the centre is gR2/x2.

Projected away from the Earth

Taking x outwards: v dv/dx = -gR2/x2. Integrating, ½v2 = gR2/x + c. With speed U at x = R:

v2 = U2 - 2gR + 2gR2/x.

The greatest distance is where v = 0.

Escaping

If U2 is at least 2gR, then v2 stays positive for every x and the particle never returns. The least such speed, sqrt(2gR), is the escape speed (ignoring air resistance).

Worked example

A particle is projected vertically upwards from the Earth's surface with speed sqrt(gR), where R is the Earth's radius. Find its greatest distance from the centre of the Earth.

  1. v2 = gR - 2gR + 2gR2/x = 2gR2/x - gR.
  2. v = 0: 2gR2/x = gR, so x = 2R.

Answer: 2R from the centre (a height R above the surface)

Key idea: F = mgR2/x2. With v dv/dx = -gR2/x2 you find speeds and greatest heights; escape needs U2 ≥ 2gR.

Check yourself

  1. Find the gravitational force on a 50 kg satellite at a height R above the Earth's surface, where R is the Earth's radius, to 3 significant figures.
    Show answer

    123 N. x = 2R, so F = mgR2/(4R2) = mg/4 = 123 N.

  2. At what distance from the centre of the Earth is a particle's weight one ninth of its weight at the surface? Give your answer as a multiple of R.
    Show answer

    x = 3 R. mgR2/x2 = mg/9 gives x2 = 9R2, x = 3R.

  3. A particle is projected vertically from the Earth's surface with speed U, where U2 = 1.5gR. Find its greatest distance from the centre, as a multiple of R.
    Show answer

    4 R. v = 0: 2gR2/x = 2gR - 1.5gR = 0.5gR, so x = 4R.

  4. Taking R = 6.4 × 106 m, find the escape speed from the Earth's surface, to 3 significant figures.
    Show answer

    11200 m/s. sqrt(2gR) = sqrt(2 × 9.8 × 6.4 × 106) = 11200 m/s.

  5. A particle is projected upwards from the surface with speed sqrt(2gR). Taking R = 6.4 × 106 m, find its speed when it is at distance 2R from the centre, to 3 significant figures.
    Show answer

    7920 m/s. v2 = 2gR - 2gR + 2gR2/(2R) = gR, so v = sqrt(9.8 × 6.4 × 106) = 7920 m/s.

  6. Which expression gives the gravitational force on a particle of mass m at distance x (x ≥ R) from the centre of the Earth?
    • mgR2/x2
    • mg
    • mgx2/R2
    • mgR/x
    Show answer

    mgR2/x2. F = k/x2 with F = mg at x = R.

  7. Why can suvat not be used for a rocket coasting far above the Earth?
    • Air resistance is too large
    • The mass of the rocket is zero
    • The gravitational force, and so the acceleration, changes with distance
    • g is exactly 9.8 everywhere
    Show answer

    The gravitational force, and so the acceleration, changes with distance. suvat needs constant acceleration, but gR2/x2 changes with x.

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Last updated 8 November 2026. Spotted a mistake? Tell us.