The potential energy of a satellite of mass $m$ and revolving at a height $R_{e}$ above the surface of earth where $R_{e} =$ radius of earth, is

- m g R_{e}

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\frac{- m g R_{e}}{2}

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\frac{- m g R_{e}}{3}

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\frac{- m g R_{e}}{4}

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Solution

The correct option is A $\frac{- m g R_{e}}{2}$
At a height $h$ above the surface of earth the gravitational potential energy of the particle of mass $m$ is
$U_{h} = - \frac{G M_{e} m}{R_{e} + h}$
Where $M_{e}$ & $R_{e}$ are the mass & radius of earth respectively.
In this question, since $h = R_{e}$
So $U_{h = R_{e}} = - \frac{G M_{e} m}{2 R_{e}} = \frac{- m g R_{e}}{2}$

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The potential energy of a satellite of mass m and revolving at a height Re above the surface of earth where Re= radius of earth, is

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