Question 345: Gravitation

Question

The ratio of energy required to raise a satellite of mass '$m$' 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]

Accepted Answer

Energy required to raise to satellite of $\mathrm{m}$ to a height $\mathrm{h}$ is equal to change in its potential energy.
$	herefore W=-\frac{G M m}{R+h}+\frac{G M m}{R}=\frac{G M m h}{(R+h) R}$ .... (1)
The energy of a satellite moving in a circular orbit is given by
$\begin{aligned}
E & =\frac{G M m}{2(R+h)} \quad 	ext{......(2)}\
	herefore \quad \frac{W}{E} & =\frac{2 h}{R}
\end{aligned}$

Answer

$\frac{h}{R}$

Answer

$\frac{2 h}{\mathrm{R}^2}$

Answer

$\frac{3 \mathrm{~h}}{\mathrm{R}^2}$

Answer

$\frac{2 \mathrm{~h}}{\mathrm{R}}$

Gravitation

The ratio of energy required to raise a satellite of mass '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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