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    Our moon has mass M, and radius R. The gravitational field strength at its surface is 1.64N{kg}^{-1}.

    Calculate the numerical value of g at an altitude R above the surface of the Moon.
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    0.41 N kg^-1

    Fm^-1 = GM/R^2=1.64
    R above is 2R from centre, so Fm^-1 = GM/(2R)^2 = GM/4R^2, so it is a quarter of the original value, which is 0.41 N kg^-1
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    (Original post by 'jay')
    0.41 N kg^-1

    Fm^-1 = GM/R^2=1.64
    R above is 2R from centre, so Fm^-1 = GM/(2R)^2 = GM/4R^2, so it is a quarter of the original value, which is 0.41 N kg^-1
    Thanks, for some reason I didn't think to realise that R is R (i.e. radius and altitude are the same). So as R doubles, g decreases 4x.

    Thanks again
 
 
 
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