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Let E be the energy required to raise a satellite to height h above earth's surface and E' be the energy
required to put the same satellite into orbit at that height. Then, E/E' is equal to

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$\frac{2 h}{(2 R + 3 h)}$

$\frac{R}{R + h}$

$\frac{2 R}{2 h + R}$

$\frac{2 h}{(R + 2 h)}$

answer is A.

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Detailed Solution

$E = Δ U = \frac{m g h}{1 + h / R} = \frac{m g h R}{(R + h)}$

E' = Energy of satellite - Energy of satellite on surface of earth

$= - \frac{G M m}{2 (R + h)} - (- \frac{G M m}{R})$

$= m g R^{2} [\frac{1}{R} - \frac{1}{2 (R + h)}] = \frac{m g (R + 2 h) R}{2 (R + h)} (∵ G M = g R^{2})$

Now, $\frac{E}{E^{'}} = \frac{2 h}{(R + 2 h)}$

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