question_answer

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

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

Correct Answer: A

Solution :

[a] Centripetal force = net gravitational force \[\Rightarrow \frac{m{{v}_{0}}^{2}}{r}=2F\cos {{45}^{\operatorname{o}}}+{{F}_{1}}=\frac{2G{{m}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}\frac{1}{\sqrt{2}}+\frac{Gm}{4{{r}^{2}}}\]\[\frac{m{{v}_{0}}^{2}}{r}=\frac{Gm}{4{{r}^{2}}}\left[ 2\sqrt{2}+1 \right]\] \[\Rightarrow {{v}_{0}}={{\left[ \frac{Gm\left( 2\sqrt{2}+1 \right)}{4r} \right]}^{\frac{1}{2}}}\]