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(D) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M$) is given by $K.E. = \fra

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(D) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M$) is given by $K.E. = \frac{GMm}{2r}$.The potential energy $(P.E.)$ of the satellite at the same distance $r$ is given by $P.E. = -\frac{GMm}{r}$.We are asked to find the ratio of potential energy to kinetic energy. Note that the magnitude of potential energy is $|P.E.| = \frac{GMm}{r}$.The ratio is $\frac{P.E.}{K.E.} = \frac{-GMm/r}{GMm/2r} = -2$.Since the question asks for the ratio of potential energy to kinetic energy,and considering the sign,the value is $-2$. However,if the question implies the ratio of magnitudes,the answer is $2$. Given standard physics convention,the potential energy is negative,so the ratio is $-2$. If the options provided are positive,we select the magnitude $2$.

For a satellite orbiting around the earth in a circular orbit,the ratio of potential energy to kinetic energy at the same height is

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