The minimum energy required to launch a $m$ kg satellite from earth's surface in a circular orbit at an altitude of $2 R$ where $R$ is the radius of earth, will be:

3 m g R

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\frac{5}{6} m g R

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2 m g R

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\frac{1}{5} m g R

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Solution

The correct option is A $\frac{5}{6} m g R$

The kinetic energy at altitude $2 R$ is $= \frac{G M m}{6 R}$
The gravitational potential energy at altitude $2 R$ is $= \frac{- G M m}{3 R}$
Total energy $=$ Kinetic energy $+$ potential energy
$= \frac{G M m}{6 R} + \frac{- G M m}{3 R}$

Potential energy at the surfce is $= \frac{G M m}{R}$

Therefore, required kinetic energy $= - \frac{G M m}{6 R} + \frac{G M m}{R} = \frac{5 G M m}{6 R}$
hence the best possible answer is option B.

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