Gravitation Question 324

Question: Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Then, velocity of a particle is given by

Options:

A) $ [{{\left[ \frac{Gm}{r}\left( \frac{1+2\sqrt{2}}{4} \right) \right]}^{\frac{1}{2}}}] $

B) $ [\sqrt{\frac{Gm}{r}}] $

C) $ [\sqrt{\frac{Gm}{r}}\left( 1+2\sqrt{2} \right)] $

D) $ [a{{\left[ \frac{1}{2}\frac{Gm}{r}\left( \frac{1+2\sqrt{2}}{2} \right) \right]}^{\frac{1}{2}}}] $

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Answer:

Correct Answer: A

Solution:

  • Centripetal force = net gravitational force

    $ [\Rightarrow \frac{mv_{0}^{2}}{r}=2F\cos {{45}^{o}}+F_{1}=\frac{2Gm^{2}}{{{\left( \sqrt{2} \right)}^{2}}}\frac{1}{\sqrt{2}}+\frac{Gm}{4r^{2}}] $

    $ [\frac{mv_{0}^{2}}{r}=\frac{Gm}{4r^{2}}\left[ 2\sqrt{2}+1 \right]] $ $ [\Rightarrow v_{0}={{\left[ \frac{Gm\left( 2\sqrt{2}+1 \right)}{4r} \right]}^{\frac{1}{2}}}] $



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