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What is the gain in the potential energy of an object of mass $m$ raised from the surface of earth to a height equal to radius $R$ of the earth?

I tried this problem in two ways and got two answers.

Process 1

g'= g[R/(R+H)]^2 = g/4

So, gain in potential energy is = mg'h = (1/4)mgR

Process 2

Change in P.E = (GMm/R)-(GMm/2R) = GMm/2R = (1/2) mgR

Where am I doing wrong ? Another thing, are we taking P.E. to be zero at the surface in first proceess ?

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1 Answer 1

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In the first approach, you are trying to use the law $\Delta E_{\rm pot} = mgh$ for the change in potential energy. This law is applicable only in the approximation of homogeneous gravitational field; it is just a linearization of the relation $E_{\rm pot} = -G\frac{mM}{r}$. On the other hand, as you have shown, the linear approximation of the potential between $R$ (Earth surface) and $2R$ is very poor.

Note: In your "trick" of supplying $g'$ instead of $g$ into the $mgh$ law, you actually only choose another point around which you linearize the potential.

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