MHT CET · Physics · Gravitation
In a satellite, if the time of revolution is \(T\), then \(\mathrm{KE}\) is proportional to
- A \(\frac{1}{T}\)
- B \(\frac{1}{T^{2}}\)
- C \(\frac{1}{T^{3}}\)
- D \(T^{-2 / 3}\)
Answer & Solution
Correct Answer
(D) \(T^{-2 / 3}\)
Step-by-step Solution
Detailed explanation
Velocity of satellite \(v=\sqrt{\frac{G M}{r}}\)
\(\therefore\) \(\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}\)
\(\therefore \mathrm{KE} \propto T^{-2 / 3}\)
\(\therefore\) \(\mathrm{KE} \propto v^{2} \propto \frac{1}{r} \text { and } T^{2} \propto r^{3}\)
\(\therefore \mathrm{KE} \propto T^{-2 / 3}\)
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