A satellite is revolving round the earth with orbital speed $v_{0}$ . If it stops suddenly, the speed with
which it will strike the surface of earth would be $\sqrt{v_{e}^{2} - {kv}_{0}^{2}}$ $(v_{e} =$ escape velocity of a particle on
earth’s surface). Find the value of K.

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Detailed Solution

$V_{0} = \sqrt{\frac{GM}{r}} & V_{e} = \sqrt{\frac{2 GM}{R}}; Apply LCE$

$\frac{- GMm}{r} + O = \frac{- GMm}{R} + \frac{1}{2} {mV}^{2}$

$\Rightarrow \frac{- GM}{r} = \frac{- GM}{R} + \frac{V^{2}}{2} \Rightarrow V^{2} = \frac{2 GM}{R} - \frac{2 GM}{r}$

$\Rightarrow V = \sqrt{V_{e}^{2} - 2 V_{0}^{2}}$

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