A satellite of mass m is orbiting the earth in a circular orbit. Height of the satellite above the surface of earth is R.
Where R = radius of earth. If g be the accelaration due to gravity on the surface of earth, then total energy of the satellite is

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$- mgR$

$- 2 mgR$

$- \frac{1}{4} mgR$

$- \frac{1}{2} mgR$

answer is C.

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Detailed Solution

Total energy $E = - \frac{GMm}{2 r} = - \frac{G . M . m}{2 (R + R)} = - \frac{GMm}{4 R} = - m . (\frac{GM}{R^{2}}) . \frac{R}{4}$

$\Rightarrow E = - \frac{1}{4} mgR [∵ g = \frac{GM}{R^{2}}]$

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