What is diff in true weight standing at nearest + farthest positions on moon relative to center of earth?

Question

Here's the full question:

If you are on the far side of the moon you are farther away from the earth than if you are on the near side. a.) What is the difference in your true weight between standing at the nearest and farthest surface positions on the moon relative to the center of the earth?

Well I think that you need to use Fg=(Gm1m2)/r^2
but what do you do with that? Perhaps add the force of gravity of the moon and the earth together? I'm not sure what to do. Thanks for the help.

Answer 1

Your true weight on the moon?

None. The tidal forces are virtually the same. On the near side you are closer to the Earth compared to the cg of the moon by r and on the far side you are farther away by r.

Since the earth is accelerating the cg of the moon, on the far side the earth is pulling the moon away from you and on the near side the earth is pulling you away from the moon.

>>>>>>>>>>>>>>>>>>

I may have been a bit quick. There is a minor difference between each side. There is a second order effect. But I don't think that is what the problem wants. (It amounts to ~10^-8 m/sec^2) See http://en.wikipedia.org/wiki/Tidal_forces regarding tidal forces.