Question

Four particles of equal mass $M$ move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is.

• A. $\frac{GM}{R}$
• B. $\sqrt{\left[2\sqrt{2}\frac{GM}{R}\right]}$
• C. $\sqrt{\left[\frac{GM}{R}(2\sqrt{2}+1)\right]}$
• D. $\sqrt{\left[\frac{GM}{R}\left(\frac{(2\sqrt{2}+1)}{4}\right)\right]}$

Answer 1

The correct option is D $\sqrt{\left[\frac{GM}{R}\left(\frac{(2\sqrt{2}+1)}{4}\right)\right]}$

Acceleration of mass A due to mass B=$\frac{GM}{(\sqrt{2}R{)}^2}=\frac{GM}{2R^2}$ along the line joining A and B.

Therefore acceleration of A due to B along line joining A and center=$\frac{GM}{2R^2}\cos {45}^{\circ }=\frac{GM}{2\sqrt{2}{R}^2}$

Total acceleration of mass A towards the center=Acceleration due to B+Acceleration due to C+Acceleration due to D

$\frac{GM}{2\sqrt{2}{R}^2}+\frac{GM}{2\sqrt{2}{R}^2}+\frac{GM}{(2R{)}^2}=\frac{v^2}{R}$=Centripetal acceleration

Therefore, $v=\sqrt{\frac{GM}{R}\left(\frac{2\sqrt{2}+1}{4}\right)}$