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(B) The total mechanical energy $E$ of a satellite of mass $m$ at an orbital radius $r$ is given by the sum of its kinetic energy and gravitational po

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$\frac{GMm}{12R}$

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(B) The total mechanical energy $E$ of a satellite of mass $m$ at an orbital radius $r$ is given by the sum of its kinetic energy and gravitational potential energy.$E = K.E. + P.E. = \frac{1}{2}mv^2 - \frac{GMm}{r}$Since the centripetal force is provided by gravity,$\frac{mv^2}{r} = \frac{GMm}{r^2}$,which implies $v^2 = \frac{GM}{r}$.Substituting this into the energy equation: $E = \frac{1}{2}m(\frac{GM}{r}) - \frac{GMm}{r} = -\frac{GMm}{2r}$.The work done $W$ is the change in total energy: $W = E_{final} - E_{initial}$.$W = \left(-\frac{GMm}{2(3R)}\right) - \left(-\frac{GMm}{2(2R)}\right) = \frac{GMm}{4R} - \frac{GMm}{6R}$.$W = \frac{GMm}{R} \left(\frac{3-2}{12}\right) = \frac{GMm}{12R}$.

$A$ satellite of earth of mass $m$ is moved from an orbital radius of $2R$ to $3R$. Calculate the minimum work done.

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