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Physics Question on Gravitation

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

GMR\sqrt{\frac{GM}{R}}

B

22GMR\sqrt{ 2 \sqrt 2 \frac{GM}{R}}

C

GMR(1+22)\sqrt{\frac{GM}{R} (1+2 \sqrt2)}

D

12GMR(1+22) \frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}

Answer

12GMR(1+22) \frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}

Explanation

Solution

Four particles, each of mass M and equidistant from each other, move along a circle of radius R

F2+F2+F′=Mv2R\frac{F}{\sqrt{2}}+\frac{F}{\sqrt{2}}+F'=\frac{M v^{2}}{R}

2×GM22(R2)2+GM24R2=Mv2R\frac{2 \times G M^{2}}{\sqrt{2}(R \sqrt{2})^{2}}+\frac{G M^{2}}{4 R^{2}}=\frac{M v^{2}}{R}

GM2R[14+12]=Mv2\frac{G M^{2}}{R}\left[\frac{1}{4}+\frac{1}{\sqrt{2}}\right]=M v^{2}

v=GmR(2+442)=12GmR(1+22)v=\sqrt{\frac{G m}{R}\left(\frac{\sqrt{2}+4}{4 \sqrt{2}}\right)}=\frac{1}{2} \sqrt{\frac{G m}{R}(1+2 \sqrt{2})}

The Correct Option is (D): 12GMR(1+22)\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}