ExamBro
ExamBro
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?

  1. A \(1.35\sqrt {\frac{{GM}}{a}} \)
  2. B \(1.16\sqrt {\frac{{GM}}{a}} \)
  3. C \(1.41\sqrt {\frac{{GM}}{a}} \)
  4. D \(1.21\sqrt {\frac{{GM}}{a}} \)
Verified Solution

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…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app