Orbital velocity of a satellite in its orbit (around earth) of radius r is v. It collides with another body in its orbit and comes to rest just after the collision. Taking the radius of earth as R, the speed with which it will fall on the surface of earth will be :

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$\frac{v}{\sqrt{2 (\frac{r}{R} + 1)}}$

$v \sqrt{2 (\frac{r}{R} - 1)}$

$v \sqrt{2 (\frac{r}{R} + 1)}$

$v \sqrt{(\frac{r}{R} - 1)}$

answer is B.

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Detailed Solution

$v = \sqrt{\frac{GM}{r}} . . . . . (1)$
$- \frac{GMm}{r} = - \frac{GMm}{R} + \frac{1}{2} {mv}^{' 2} . . . . (2)$
From (1) and (2) we have $v^{'} = v \sqrt{2 (\frac{r}{R} - 1)}$

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