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 $2 R$ ?

\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}{3 R}

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\frac{G m M}{2 R}

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Solution

The correct option is A $\frac{5 G m M}{6 R}$
Energy of the satellite on the surface of earth is only the potential energy given by,

$E_{i} = \frac{- G M m}{R}$

The energy of a satellite at a distance $(2 R + R) = 3 R$ , from the centre of Earth is,

$E_{f} = \frac{- G M m}{3 R} + \frac{1}{2} m v_{0}^{2}$

$= \frac{- G M m}{3 R} + \frac{1}{2} m \frac{G M}{3 R}$

$= \frac{- G M m}{6 R}$

Energy required to launch the satellite is, $E = E_{f} - E_{i}$

$E = \frac{- G M m}{6 R} + \frac{G M m}{R}$

$E = \frac{5 G M m}{6 R}$

Hence, option $(A)$ is correct.

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