In order to shift a body of mass m from a cir | vedclass

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(B) The total mechanical energy $E$ of a body of mass $m$ in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initial energy at $r_1 =

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$\frac{1}{15}\frac{GMm}{R}$

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(B) The total mechanical energy $E$ of a body of mass $m$ in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initial energy at $r_1 = 3R$ is $E_1 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.Final energy at $r_2 = 5R$ is $E_2 = -\frac{GMm}{2(5R)} = -\frac{GMm}{10R}$.The work done $W$ is equal to the change in mechanical energy: $W = E_2 - E_1$.$W = -\frac{GMm}{10R} - (-\frac{GMm}{6R}) = GMm \left( \frac{1}{6R} - \frac{1}{10R} \right)$.$W = GMm \left( \frac{5 - 3}{30R} \right) = GMm \left( \frac{2}{30R} \right) = \frac{1}{15} \frac{GMm}{R}$.

In order to shift a body of mass $m$ from a circular orbit of radius $3R$ to a higher radius $5R$ around the earth,the work done is

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