Question
Derive an expression for work done against gravity.
✓
Answer
Potential energy of the body on the surface of the earth $=\frac{-G M m}{R}$
Potential energy at a height $h$ from the surface of the earth $=-\frac{G M m}{(R+h)}$
$
\begin{aligned}
& \text { Work done }=\left(-\frac{G M m}{R+h}\right)-\left(-\frac{G M m}{R}\right) \\
& =\frac{G M m}{R}-\frac{G M m}{R+h} \\
& =G M m\left(\frac{1}{R}-\frac{1}{R+h}\right) \\
& =\frac{G M m h}{R(R+h)}=\frac{M g R^2 h}{R(R+h)}\left[\because g=\frac{G M}{R^2}\right] \\
& =\frac{(M g h) R}{(R+h)}=\frac{M g h}{1+\frac{h}{R}}
\end{aligned}
$
Potential energy at a height $h$ from the surface of the earth $=-\frac{G M m}{(R+h)}$
$
\begin{aligned}
& \text { Work done }=\left(-\frac{G M m}{R+h}\right)-\left(-\frac{G M m}{R}\right) \\
& =\frac{G M m}{R}-\frac{G M m}{R+h} \\
& =G M m\left(\frac{1}{R}-\frac{1}{R+h}\right) \\
& =\frac{G M m h}{R(R+h)}=\frac{M g R^2 h}{R(R+h)}\left[\because g=\frac{G M}{R^2}\right] \\
& =\frac{(M g h) R}{(R+h)}=\frac{M g h}{1+\frac{h}{R}}
\end{aligned}
$
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