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Near the earth surface PE = mgh and PE = GMm/r is equal .Give an example and prove
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@texaschic101
Recall that g = GM/Re^2 at a distance from earth. Also U(Re) = -gmM/Re. Subbing these equations into each other you get U(Re) = -mgRe. For very small heights h above the earths surface where h<<Re U(Re + h) = -gmM/(Re + h) is approximately equal to -mg(Re +h). So, the change in potential energy of a falling mass would be \[\Delta U = Uf - Ui\] which is U = (-mgRe - (-mg(Re +h)) U = mgh Hope this helps :)
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