**Question:**

**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:**

$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

$\sqrt{2 \sqrt{2} \frac{G M}{R}}$

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

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

Question of from chapter.

JEE Main Previous Year 2014

Correct Option: 4

**Solution:**