question_answer

Four particles, each of mass \[M\]and equidistant from each other, move along a circle of radius \[R\] under the action of their mutual gravitational attraction. The speed of each particle is

A) \[\sqrt{\frac{GM}{R}(1+2\sqrt{2})}\]

B) \[\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}\]

C) \[\sqrt{\frac{GM}{R}}\]            

D) \[\sqrt{2\sqrt{2}\frac{GM}{R}}\]

Correct Answer: B

Solution :

[b] Net force on any one particle \[=\frac{G{{M}^{2}}}{{{(2R)}^{2}}}+\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ +\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ \] \[=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1}{4}+\frac{1}{\sqrt{{}}} \right]\] This force will be equal to centripetal force so \[\frac{M{{u}^{2}}}{R}=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1+2\sqrt{2}}{4} \right]\] \[u=\sqrt{\frac{GM}{R}[1+2\sqrt{2}]}\] \[=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}\]