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5 November, 22:34

The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2

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  1. 6 November, 00:19
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    the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)

    Explanation:

    according to newton's law of universal gravitation (we will neglect relativistic effects)

    F = G*m*M/d², G = constant, M = planet mass, m = mass of an object, d=distance between the object and the centre of mass of the planet

    if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³,

    since M = ρ * V = ρ * 4/3πR³, ρ = density

    F = G*m*M/R² = G*m*ρ * 4/3πR³/R² = G*ρ * 4/3πR

    from Newton's second law

    F = m*g = G*ρ*m * 4/3πR

    thus

    g = G*ρ * 4/3π*R = (4/3π*G*ρ) * R

    g ∝ R
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