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Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be 

\(A) \space\sqrt{ (1+2\sqrt2)G\space \over2}\)  

\(B) \sqrt{ G(1+2\sqrt2)}\)

\(C) \sqrt { {G\over2} (2\sqrt2-1)}\)

\(D) \sqrt {{G\over2} (1+2\sqrt2)}\)




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1 Answer


The correct option is A)\(\space\sqrt{ (1+2\sqrt2)G\space \over2}\) 


24-feb-physics-shift-1-question{from -byzus} 

By resolving force F2, we get 

F1 + F2 cos 45° + F2 cos 45° = Fc

 F1 + 2F2 cos 45° = Fc

Fc = centripetal force = MV2 / R 

\({ GM^2\over(2R)^2}={ {2GM^2 \over(2R)^2} \cos 45^o}={ MV^2\over R}\) 

\({ GM^2\over 4R^2}+{ 2GM^2\over2\sqrt2R^2}={ MV^2\over R}\) 

\({GM^2\over 4R^2}+{GM\over\sqrt {2R}}=V^2\) 

\(V ={\sqrt{{ GM\over 4R}+{GM\over\sqrt{2R}}}}\) 

\(V={1\over2}{ \sqrt{{ GM\over R} [ {1+2\sqrt2}]}}\) 

now, mass=1kg and radius=1m  

\(⇒ V={1\over2}{\sqrt{ G(1+2\sqrt2)}}\)

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