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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 $\mathrm{h}$ from the surface of earth (of radius $\mathrm{R}$ ) is given by,
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The correct answer is:
$\frac{G M m h}{R(R+h)}$
The gravitational potential energy of an object placed at earth's surface is
$$
\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}}
$$
where, $G=$ gravitational constant,
$M=$ Mass of earth, $m=$ mass of object and
$\mathrm{R}=$ radius of the earth
The negative sign in the above relation indicates that it is the work done in bringing the object from infinity to a distance $R$.
The gravitational potential energy of object at a height habove the surface of earth is
$$
\mathrm{U}_2=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}
$$
So, the change in potential energy is
$$
\begin{aligned}
\Delta U & =\mathrm{U}_2-\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right) \\
& =-\mathrm{GMm}\left[\frac{1}{(\mathrm{R}+\mathrm{h})}-\frac{1}{\mathrm{R}}\right]=-\mathrm{GMm}\left(-\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right) \\
& =\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}
\end{aligned}
$$
$$
\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}}
$$
where, $G=$ gravitational constant,
$M=$ Mass of earth, $m=$ mass of object and
$\mathrm{R}=$ radius of the earth
The negative sign in the above relation indicates that it is the work done in bringing the object from infinity to a distance $R$.
The gravitational potential energy of object at a height habove the surface of earth is
$$
\mathrm{U}_2=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}
$$
So, the change in potential energy is
$$
\begin{aligned}
\Delta U & =\mathrm{U}_2-\mathrm{U}_1=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right) \\
& =-\mathrm{GMm}\left[\frac{1}{(\mathrm{R}+\mathrm{h})}-\frac{1}{\mathrm{R}}\right]=-\mathrm{GMm}\left(-\frac{\mathrm{h}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right) \\
& =\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}
\end{aligned}
$$
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