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(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth (mass $M$) is given by $U = -\frac{GMm

Vedclass · Accepted Answer

$-\frac{GMm}{R} + mgh$

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(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth (mass $M$) is given by $U = -\frac{GMm}{r}$.Here,the body is at a height $h$ above the surface of the Earth,so its distance from the center is $r = R + h$.Therefore,the potential energy is $U = -\frac{GMm}{R+h}$.Note: If $h$ is very small compared to $R$,we can approximate this using the Taylor expansion: $U = -\frac{GMm}{R(1 + h/R)} \approx -\frac{GMm}{R}(1 - h/R) = -\frac{GMm}{R} + \frac{GMmh}{R^2}$.Since $g = \frac{GM}{R^2}$,this becomes $U \approx -\frac{GMm}{R} + mgh$.

$A$ body of mass $m$ is kept at a small height $h$ above the ground. If the radius of the earth is $R$ and its mass is $M$,the potential energy of the body and earth system (with $h = \infty$ being the reference position) is

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