Gravitation Question 282

Question: Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]

Options:

A)$ [v=\frac{1}{2R}\sqrt{\frac{1}{Gm}}] $

B)$ [v=\sqrt{\frac{Gm}{2R}}] $

C) $ [v=\frac{1}{2}\sqrt{\frac{Gm}{R}}] $

D) $ [v=\sqrt{\frac{4Gm}{R}}] $

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Answer:

Correct Answer: C

Solution:

Centripetal force provided by the gravitational force of attraction between two particles i.e. $ [\frac{mv^{2}}{R}=\frac{Gm\times m}{{{(2R)}^{2}}}] $ $ [\Rightarrow v=\frac{1}{2}\sqrt{\frac{Gm}{R}}] $



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