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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2023
  • Medium
  • MCQ

The time period of a satellite, revolving above earth's surface at a height equal to \(R\) will be (Given \(g =\pi^2 m / s ^2, R =\) radius of earth)

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Correct Answer

(C) \(\sqrt{32 R}\)

Step-by-step Solution

Detailed explanation

\(\frac{ mv ^2}{2 R }=\frac{ GMm }{(2 R )^2} \Rightarrow v =\sqrt{\frac{ GM }{2 R }}=\sqrt{\frac{ Rg }{2}}\)

\(T =\frac{2 \pi(2 R )}{ v }=\frac{4 \pi R \sqrt{2}}{\sqrt{ Rg }}=\sqrt{32 R }\)

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