AP EAMCET · PHYSICS · Gravitation
If the time period of revolution of a satellite is \(T\). then its kinetic energy is proportional to
- A \(\mathrm{T}^{-1}\)
- B \(\mathrm{T}^{-2}\)
- C \(\mathrm{T}^{-3}\)
- D \(\mathrm{T}^{-2 / 3}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{T}^{-2 / 3}\)
Step-by-step Solution
Detailed explanation
The time period of satellite is \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}} \Rightarrow \mathrm{R} \propto \mathrm{~T}^{\frac{2}{3}}\) \(\therefore\) Kinetic energy,…
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