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(B) The moment of inertia of a satellite of mass $m$ at a distance $R$ from the center of the Earth is $I = m R^2$.The kinetic energy $(K)$ of the sat

Vedclass · Accepted Answer

$K=\frac{J^2}{2 m R^2}, E=-\frac{J^2}{2 m R^2}$

Vedclass · Answer

(B) The moment of inertia of a satellite of mass $m$ at a distance $R$ from the center of the Earth is $I = m R^2$.The kinetic energy $(K)$ of the satellite in terms of angular momentum $(J)$ is given by $K = \frac{J^2}{2 I}$.Substituting $I = m R^2$, we get $K = \frac{J^2}{2 m R^2}$.For a satellite in a circular orbit, the total energy $(E)$ is related to the kinetic energy $(K)$ by the relation $E = -K$.Therefore, $E = -\frac{J^2}{2 m R^2}$.Thus, the correct option is $B$.

$A$ satellite of mass $m$ rotates around the Earth in a circular orbit of radius $R$. If the angular momentum of the satellite is $J$, then its kinetic energy $(K)$ and the total energy $(E)$ of the satellite are:

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