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
- A \(\sqrt {2\sqrt 2 \frac{{GM}}{R}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\)
- B \(\;\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)
- C \(\frac{1}{2}\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)
- D \(\;\sqrt {\frac{{GM}}{R}} \)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\sqrt {\frac{{GM}}{R}\left( {1 + 2\sqrt 2 } \right)} \)
Step-by-step Solution
Detailed explanation
\(2F\cos {45^{ \circ}} + F' = \frac{{M{v^2}}}{R}\,\,\,\,\left( {From\,figure} \right)\) \(where\,F = \frac{{G{M^2}}}{{{{\left( {\sqrt 2 R} \right)}^2}}}\,and\,F' = \frac{{G{M^2}}}{{4{R^2}}}\)…
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