JEE Mains · Physics · STD 11 - 7. gravitation
Four identical particles of mass \(M\) are located at the corners of a square of side \(‘a’\). What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?

- A \(1.35\sqrt {\frac{{GM}}{a}} \)
- B \(1.16\sqrt {\frac{{GM}}{a}} \)
- C \(1.41\sqrt {\frac{{GM}}{a}} \)
- D \(1.21\sqrt {\frac{{GM}}{a}} \)
Answer & Solution
Correct Answer
(B) \(1.16\sqrt {\frac{{GM}}{a}} \)
Step-by-step Solution
Detailed explanation
Net force on particle towards center of circle is \({F_c} = \frac{{G{M^2}}}{{2{a^2}}} + \frac{{G{M^2}}}{{{a^2}}}\sqrt 2 \) \( = \frac{{G{M^2}}}{{{a^2}}}\left( {\frac{1}{2} + \sqrt 2 } \right)\) This force will act as centripetal force. Diatance of particle from center of circle…
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