Question

Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the acing 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

The correct option is B $\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$

$\text{Solution}:$

Net force acting on any one particle M,

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

$=\frac{G{M}^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}})$

This force will equal to centripetal force.

So ,$\frac{M{v}^2}{R}=\frac{G{M}^2}{R^2}(\frac{1}{4}+\frac{1}{\sqrt{2}})$

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

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

Hence, speed of each particle in a circular motion is

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

$\text{Hence B is the correct option}$