A satellite is moving in circular orbit around the earth with orbital speed v. If radius of earth is R and acceleration due to gravity on its surface is g, then height of the satellite above the surface of earth is

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$(\frac{2 g R^{2}}{V^{2}} - \frac{R}{2})$

$(\frac{g R^{2}}{V^{2}} - R)$

$(\frac{2 g R^{2}}{V^{2}} - R)$

$\frac{g R^{2}}{V^{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} O r b i t a l s p e e d V = \sqrt{\frac{G M}{r}} \\ N o w g = \frac{G M}{R^{2}} \Rightarrow G M = g R^{2} \\ A l s o, r = R + h \\ ∴ V = \sqrt{\frac{g R^{2}}{R + h}} \Rightarrow h = (\frac{g R^{2}}{V^{2}} - R) \end{array}$

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