The kinetic energy of a revolving satellite o | vedclass

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Question

(A) The kinetic energy $(K.E.)$ of a satellite revolving at a distance $r = R + h$ from the center of the Earth is given by $K.E. = \frac{G M m}{2r}$.

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$\frac{m g R}{8}$

Vedclass · Answer

(A) The kinetic energy $(K.E.)$ of a satellite revolving at a distance $r = R + h$ from the center of the Earth is given by $K.E. = \frac{G M m}{2r}$.Given height $h = 3R$,the orbital radius is $r = R + 3R = 4R$.Substituting the value of $r$ into the formula: $K.E. = \frac{G M m}{2(4R)} = \frac{G M m}{8R}$.We know that the acceleration due to gravity at the Earth's surface is $g = \frac{G M}{R^2}$,which implies $G M = g R^2$.Substituting $G M = g R^2$ into the kinetic energy expression: $K.E. = \frac{(g R^2) m}{8R} = \frac{m g R}{8}$.

The kinetic energy of a revolving satellite of mass $m$ at a height equal to thrice the radius of the Earth $R$ is:

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