MHT CET · Physics · Gravitation
For a satellite orbiting around the earth in a circular orbit, the ratio of potential energy to kinetic energy at same height is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C \(\sqrt{2}\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { K.E }=\frac{G M m}{2 r} \\ & \text { P.E }=-\frac{G M m}{r} \\ \therefore \quad & \frac{K \cdot E}{\mid P . E ~}=\frac{G M m}{2 r} \times \frac{r}{G M m} \\ \therefore \quad & \frac{\text { P.E }}{\text { K.E }}=\frac{2}{1}=2\end{aligned}\)
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