For a satellite moving in an orbit around the earth at height ' h ' the ratio of kinetic energy to potential energy is

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TestHub · Accepted Answer

The kinetic energy of a satellite revolving in an orbit of radius $r = R + h$ is given by $K = \frac{GM m}{2 r}$ .

The potential energy of the satellite is given by $U = - \frac{GM m}{r}$ .

Taking the ratio of their magnitudes, we get:

$\frac{∣ K ∣}{∣ U ∣} = \frac{\frac{GM m}{2 r}}{\frac{GM m}{r}} = \frac{1}{2}$

Thus, the ratio of kinetic energy to potential energy is $1 : 2$ .

Answer: $1 : 2$

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