
Yefim S. answered 05/01/23
Math Tutor with Experience
F = GmM/(R + x)2; Elementary work dW = GmMdx/(R + x)2; W = ∫d∞GmMdx/(R + x)2 =
-GmM/(R + x)d∞ = GmM/(R + d);
On surface of Eaeth mg = GMm/R2. So, GM = gR2 . So, additional work W = gR2m/(R + h);
g = 9.8 m/s2, R = 6.371·106m; m = 100 kg; d = 107m
W = 9.8·(6.371·106)·100/(6.371·106 + 107) =2.43·107 J
Potential energy U = - GmM/(R + d) = - 2.43·107 J