What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

\frac{5 G m M}{6 R}

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\frac{2 G m M}{3 R}

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\frac{G m M}{2 R}

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\frac{G m M}{3 R}

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Solution

The correct option is A $\frac{5 G m M}{6 R}$
The kinetic energy at altitude $2 R = \frac{G M m}{6 R}$
The gravitational potential energy at altitude $2 R = \frac{G M m}{3 R}$
$∴$ Total energy $= k e + P E = \frac{G M m}{6 R}$
Potential energy at the surface is $- \frac{G M m}{R}$
$∴$ Req. kinetic energy $\frac{G M m}{R} - \frac{G M m}{6 R} = \frac{5 G M m}{6 R}$

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