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(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center is given by $K.E. = \frac{G

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$3$:$2$

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(B) The kinetic energy $(K.E.)$ of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center is given by $K.E. = \frac{GMm}{2r}$.Here,$r$ is the distance from the center of the Earth,so $r = R_{earth} + h$.For the first satellite at height $h_1 = R$,the orbital radius is $r_1 = R + R = 2R$.For the second satellite at height $h_2 = 2R$,the orbital radius is $r_2 = R + 2R = 3R$.Since both satellites have equal mass $m$,the kinetic energy is inversely proportional to the orbital radius: $K.E. \propto \frac{1}{r}$.Therefore,the ratio of their kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{r_2}{r_1} = \frac{3R}{2R} = \frac{3}{2}$.

Two satellites of equal mass are launched in circular orbits at heights $R$ and $2R$ above the surface of the Earth. The ratio of their kinetic energies is ($R =$ radius of the Earth).

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