Question

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 

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

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

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

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

 

 

 

Answer 1

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

explaination::

 

By resolving force F₂, we get 

F₁ + F₂ cos 45° + F₂ cos 45° = F_c

 F₁ + 2F₂ cos 45° = F_c

F_c = 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  

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