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Q.

The binding energy of a satellite of mass m in an orbit of radius r is (R = radius of earth, g = acceleration due to gravity on earth )

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a

${mgR}^{2}$

b

$\frac{{mgR}^{2}}{2 r}$

c

$- \frac{m g R^{2}}{r}$

d

$- \frac{{mgR}^{2}}{2 r}$

answer is B.

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Detailed Solution

Binding energy $= - (T . E)$
$= - (- \frac{GMm}{2 r}) = \frac{GMm}{2 r} = \frac{{mgR}^{2}}{2 r}$

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