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:

• A. $\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$
• B. $\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$
• C. $\sqrt{\frac{GM}{R}}$
• D. $\sqrt{2\sqrt{2}\frac{GM}{R}}$

Answer 1

Answer: (B)

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

Net force on any one particle

$\frac{G{M}^2}{(2R{)}^2}+\frac{G{M}^2}{(R\sqrt{2}{)}^2}cos{45}^o+\frac{G{M}^2}{(R\sqrt{2}{)}^2}cos{45}^o$

$=\frac{G{M}^2}{R^2}[\frac{1}{4}+\frac{1}{\sqrt{2}}]$

This force will be equal to centripetal force so

$\frac{M{u}^2}{R}=\frac{G{M}^2}{R^2}\left[\frac{1+2\sqrt{2}}{4}\right]$

$u=\sqrt{\frac{GM}{4R}[1+2\sqrt{2}]}=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$