A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing its mechanical energy due to small air resistance at the rate of K joule/sec. The time taken by the satellite to hit the surface of the earth is (M and R are the mass and radius of the earth)

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$\frac{2 G M m}{K} (\frac{1}{R} - \frac{1}{r})$

$\frac{G M m}{K} (\frac{1}{R} + \frac{1}{r})$

$\frac{G M m}{K} (\frac{1}{R} - \frac{1}{r})$

$\frac{G M m}{2 K} (\frac{1}{R} - \frac{1}{r})$

answer is B.

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Detailed Solution

$F . V = k \Rightarrow m . \frac{d v}{d t} . v = k \int_{0}^{1} d t = \frac{m}{k} (\frac{6 k}{R} - \frac{6 M}{r}) t = \frac{6 M m}{2 k} (\frac{1}{R} - \frac{1}{r})$

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