Question: Four particles, each of mass $\mathrm{M}$ and equidistant from each other, move along a circle of radius $\mathrm{R}$ under the action of their mutual gravitational attraction. The speed of each particle is :
$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
$\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$
$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$
$\sqrt{2 \sqrt{2} \frac{\mathrm{GM}}{\mathrm{R}}}$
Correct Option: , 2
Solution:
