Gravitation Question 299

Question: Two particles of equal mass ’m’ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

Options:

A) $ [\sqrt{\frac{Gm}{4R}}] $

B) $ [\sqrt{\frac{Gm}{3R}}] $

C) $ [\sqrt{\frac{Gm}{2R}}] $

D) $ [\sqrt{\frac{Gm}{R}}] $

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Answer:

Correct Answer: A

Solution:

  • Here, centripetal force will be given by the gravitational force between the two particles.

    $ [\frac{Gm^{2}}{{{(2R)}^{2}}}=m{{\omega }^{2}}R] $
    $ [\Rightarrow \frac{Gm}{4R^{3}}={{\omega }^{2}}\Rightarrow \omega =\sqrt{\frac{GM}{4R^{3}}}] $

    If the velocity of the two particles with respect to the centre of gravity is v then $ [v=\sqrt{\frac{Gm}{4R^{3}}}\times R=\sqrt{\frac{GM}{4R}}] $



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