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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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The correct answer is:
$\frac{2 \mathrm{~h}}{\mathrm{R}}$
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
$\mathrm{E}_2=\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}+\frac{\mathrm{GMm}}{\mathrm{R}}$
$=\operatorname{mgR}\left[1-\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}\right]\left(\because \mathrm{GM}=\mathrm{gR}^2\right)$
$\begin{aligned} \therefore \quad E_2 & =\frac{m g R\left(\frac{2 h}{R}+1\right)}{2\left(1+\frac{h}{R}\right)} \\ \therefore \quad \frac{E_1}{E_2} & =\frac{m g h}{1+\frac{h}{R}} \times \frac{2\left(1+\frac{h}{R}\right)}{m g R} \\ & =\frac{2 h}{R} \quad\left(\because h \ll R \Rightarrow 1+\frac{2 h}{R} \approx 0\right)\end{aligned}$
$\mathrm{E}_1=\Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}=\frac{\mathrm{mghR}}{\mathrm{R}+\mathrm{h}}$
The formula for the energy required to set the satellite in orbit is
$\mathrm{E}_2=\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}+\frac{\mathrm{GMm}}{\mathrm{R}}$
$=\operatorname{mgR}\left[1-\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}\right]\left(\because \mathrm{GM}=\mathrm{gR}^2\right)$
$\begin{aligned} \therefore \quad E_2 & =\frac{m g R\left(\frac{2 h}{R}+1\right)}{2\left(1+\frac{h}{R}\right)} \\ \therefore \quad \frac{E_1}{E_2} & =\frac{m g h}{1+\frac{h}{R}} \times \frac{2\left(1+\frac{h}{R}\right)}{m g R} \\ & =\frac{2 h}{R} \quad\left(\because h \ll R \Rightarrow 1+\frac{2 h}{R} \approx 0\right)\end{aligned}$
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