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At What height from the Surface of the earth, the total energy of Satellite is equai to its potential energy at a height 2R from the surface of earth $(R \to Radius of earth)$

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$2 R$

$4 R$

$\frac{R}{4}$

$\frac{R}{2}$

answer is B.

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Detailed Solution

T . E at height ‘h’= $- \frac{G M_{m}}{2 [h + R]}$
P . E at height 2’R’ = $- \frac{G M_{m}}{[2 R + R]}$
According to Question

$- \frac{G M_{m}}{2 [h + R]} = - \frac{G M_{m}}{(2 R + R)} \Rightarrow h = \frac{R}{2}$

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