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:
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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}}] $
BETA
