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(A) The distance of a satellite from the center of the Earth is given by $r = R + h$,where $R$ is the radius of the Earth and $h$ is the height above

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$2$

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(A) The distance of a satellite from the center of the Earth is given by $r = R + h$,where $R$ is the radius of the Earth and $h$ is the height above the surface.For the first satellite at height $h_1 = R$,the distance is $r_1 = R + R = 2R$.For the second satellite at height $h_2 = 3R$,the distance is $r_2 = R + 3R = 4R$.The kinetic energy of a satellite in a circular orbit is given by $K = \frac{GMm}{2r}$,which implies $K \propto \frac{1}{r}$.Therefore,the ratio of the kinetic energies is $\frac{K_1}{K_2} = \frac{r_2}{r_1} = \frac{4R}{2R} = \frac{2}{1} = 2$.

What is the ratio of the kinetic energies of two satellites at heights $R$ and $3R$ from the surface of the Earth? ($R$ = radius of the Earth)

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