ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 7. gravitation

Four particles, each of mass \(M\) and equidistant from each other, move along a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is

  1. A \(\sqrt {2\sqrt 2 \frac{{GM}}{R}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\)
  2. B \(\;\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)
  3. C \(\frac{1}{2}\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)
  4. D \(\;\sqrt {\frac{{GM}}{R}} \)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2}\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)

Step-by-step Solution

Detailed explanation

\(2F\cos {45^{ \circ}} + F' = \frac{{M{v^2}}}{R}\,\,\,\,\left( {From\,figure} \right)\) \(where\,F = \frac{{G{M^2}}}{{{{\left( {\sqrt 2 R} \right)}^2}}}\,and\,F' = \frac{{G{M^2}}}{{4{R^2}}}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app