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Physics Question on Gravitational Potential Energy

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by,

A

−GMmR+h-\frac{GMm}{R + h}

B

GMmhR(R+h)\frac{GMmh}{R\left(R + h\right)}

C

mgh

D

GMmR+h\frac{GMm}{R + h}

Answer

GMmhR(R+h)\frac{GMmh}{R\left(R + h\right)}

Explanation

Solution

ΔU=−GMm[1rf−1ri]\Delta U = -GMm [ \frac{1}{r_f} - \frac{1}{r_i}]
=−GMm[1R+h−1R]= - GMm[\frac{1}{R+h} - \frac{1}{R}]
=GMmhR(R+h)= \frac{GMmh}{R(R+h)}