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# Gravitational field strength question watch

1. Our moon has mass M, and radius R. The gravitational field strength at its surface is .

Calculate the numerical value of g at an altitude R above the surface of the Moon.
2. 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
3. (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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