The ratio of energy required to raise a satellite of mass ' $\mathrm{m}$ ' to height ' $h$ ' above the…
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 $]$
$\frac{\mathrm{h}}{\mathrm{R}}$
$\frac{2 \mathrm{~h}}{\mathrm{R}^2}$
$\frac{3 \mathrm{~h}}{\mathrm{R}^2}$
$\frac{2 h}{R}$
Solution
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
$\begin{aligned}
& \mathrm{E}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})} \\
& \therefore \frac{\mathrm{W}}{\mathrm{E}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}$