A satellite of mass m is moving in a circular orbit of radius R above the surface of a planet of mass M and radius R. The amount of work done to shift the satellite to a higher orbit of radius 2R is (here g is the acceleration due to gravity on planet's surface)

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$m g R$

$\frac{m M g R}{(M + m)}$

$\frac{m M g R}{6 (M + m)}$

$\frac{m g R}{6}$

answer is B.

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Detailed Solution

PE at a distance r from the centre of the planet

$= - \frac{G M m}{r}$

initial PE= $- \frac{G M m}{R + R} = - \frac{G M m}{2 R}$

Final PE = $- \frac{G M m}{R + 2 R} = - \frac{G m M}{3 R}$

Now, work down=increase in PE

= $\frac{G M m}{R} (\frac{1}{2} - \frac{1}{3}) = \frac{G M m}{6 R}$

= $\frac{1}{6} m g R$ $(∴ g = \frac{G M}{R^{2}})$

Hence the correct choice is (b ).

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