For a satellite orbiting around the earth in a circular orbit, the ratio of potential energy to kinetic…
For a satellite orbiting around the earth in a circular orbit, the ratio of potential energy to kinetic energy at same height is
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2}$
- $\sqrt{2}$
- $2$
Solution
$\begin{aligned} & \text { K.E }=\frac{G M m}{2 r} \\ & \text { P.E }=-\frac{G M m}{r} \\ \therefore \quad & \frac{K \cdot E}{|P . E|}=\frac{G M m}{2 r} \times \frac{r}{G M m} \\ \therefore \quad & \frac{\text { P.E }}{\text { K.E }}=\frac{2}{1}=2\end{aligned}$
Asked in: MHT CET 2023 (10 May Shift 1)
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