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:
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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}}}] $
BETA
