Question 361: 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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