Discussion Forum

 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

Answer:

Explanation:

Net force acting on anyone particle M,

$=\frac{GM^{2}}{(2R)^{2}}+\frac{GM^{2}}{(R\sqrt{2})^{2}}\cos 45^{0}+\frac{GM^{2}}{(R\sqrt{2})^{2}}\cos 45^{0}$

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

 this force will equal to centripetal force.

 So,    $\frac{MV^{2}}{R}=\frac{GM^{2}}{R^{2}}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)$

            $v=\sqrt{\frac{GM^{}}{4R^{}}(1+2\sqrt{2})}$

                $=\frac{1}{2}\sqrt{\frac{GM^{}}{R^{}}(2\sqrt{2}+1)}$

 Hence, speed of each particle in a circular motion is 

           $\frac{1}{2}\sqrt{\frac{GM^{}}{R^{}}(2\sqrt{2}+1)}$