Question 316: Gravitation

Question

The ratio of energy required to raise a satellite to a height '$h$' above the earth's surface to that required to put it into the orbit at the same height is ($\mathrm{R}=$ radius of earth)

Accepted Answer

The formula for the energy required to raise a satellite to height $\mathrm{h}$ is
$\mathrm{E}_1=\Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}=\frac{\mathrm{mghR}}{\mathrm{R}+\mathrm{h}}$
The formula for the energy required to set the satellite in orbit is
$\begin{aligned}
\mathrm{E}_2 & =\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}+\frac{\mathrm{GMm}}{\mathrm{R}} \
& =\operatorname{mgR}\left[1-\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}ight)}ight]\left(\because \mathrm{GM}=\mathrm{gR}^2ight) \
	herefore \quad \mathrm{E}_2 & =\frac{\mathrm{mgR}\left(\frac{2 \mathrm{~h}}{\mathrm{R}}+1ight)}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}ight)} \
	herefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}} 	imes \frac{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}ight)}{\mathrm{mgR}}
\end{aligned}$
$=\frac{2 \mathrm{~h}}{\mathrm{R}} \quad\left(\because \mathrm{h}<\mathrm{R} \Rightarrow 1+\frac{2 \mathrm{~h}}{\mathrm{R}} \approx 0ight)$

Answer

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

Answer

$\frac{h}{R}$

Answer

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

Answer

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

Gravitation

The ratio of energy required to raise a satellite to a height 'h' above the earth's surface to that required to put it into the orbit at the same height is (\mathrm{R}= radius of earth)

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