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The ratio of energy required to raise a satellite of mass ' $\mathrm{m}$ ' to height ' $h$ ' above the earth's surface to that required to put it into the orbit at same height is [ $\mathrm{R}=$ radius of earth $]$
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Verified Answer
The correct answer is:
$\frac{2 h}{R}$
Energy required to raise to satellite of $m$ to a height $h$ is equal to change in its potential energy.
$$
\therefore \mathrm{W}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMmh}}{(\mathrm{R}+\mathrm{h}) \mathrm{R}}
$$
The energy of a satellite moving in a circular orbit is given by
$$
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})} \\
& \therefore \frac{\mathrm{W}}{\mathrm{E}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}
$$
$$
\therefore \mathrm{W}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMmh}}{(\mathrm{R}+\mathrm{h}) \mathrm{R}}
$$
The energy of a satellite moving in a circular orbit is given by
$$
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})} \\
& \therefore \frac{\mathrm{W}}{\mathrm{E}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}
$$
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