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    Topic 12: Gravitational Fields — Edexcel A-Level Physics

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    Topic 12: Gravitational Fields explained

    Start from Newton's law of gravitation: the force on a small test mass m at distance r from a point mass M is F = GMm/r².

    Read the full explanation

    Gravitational field strength is defined as force per unit mass, g = F/m, so dividing gives g = GM/r². The test mass cancels, which is why g depends only on the source mass and the distance from its centre. The field is radial, pointing towards M, and follows an inverse-square law: doubling r reduces g to one quarter. For a planet, use its mass and radius to find surface g, for example g = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶)² ≈ 9.8 N kg⁻¹. In multiple-choice questions, check whether the distance is measured from the centre, keep masses in kg and distances in m, and watch the squared denominator.

    Your focus

    1. Derive g = Gm/r² from Newton's law of gravitation and the definition of field strength.
    2. Apply g = Gm/r² to calculate field strength at a given distance from a point mass.
    3. Explain how g varies with distance according to an inverse-square relationship.

    Topic 12: Gravitational Fields exam tips

    Marking Points
    • States Newton's law of gravitation F = GMm/r² for the force between point masses.
    • Defines gravitational field strength as force per unit mass, g = F/m.
    • Divides F by m to cancel the test mass, giving g = GM/r².
    • Explains that g is radial, directed towards M, and obeys an inverse-square dependence on r.
    • Applies the equation correctly with SI units, measuring r from the centre of the source mass.
    Examiner Tips
    • 💡Write the derivation as two clear steps: state F = GMm/r², then divide by m to obtain g = GM/r².
    • 💡Convert all quantities to SI units before substituting, and keep r in metres from the centre.
    • 💡Use ratio reasoning for inverse-square changes, for example tripling r reduces g by a factor of 9.
    Common Mistakes
    • Using the diameter instead of the radius for r: the correction is to measure r from the centre of the mass, so use the radius for a surface value.
    • Forgetting to square r in the denominator: the correction is to apply the inverse-square law and square the distance.
    • Leaving the test mass m in the final expression: the correction is to recognise that m cancels when forming g = F/m.