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

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


Writing the equation for centripetal force for 4,
$\frac{m{v}^2}{r}=\overrightarrow{F_{14}}+\overrightarrow{F_{24}}+\overrightarrow{F_{34}}$
$=\frac{G{m}^2}{4r^2}+\frac{G{m}^2}{2r^2}\sqrt{2}$
$=\frac{G{m}^2}{2r^2}(\frac{1}{2}+\sqrt{2})$
$\frac{m{v}^2}{r}=\frac{G{m}^2}{2r^2}\left(\frac{1+2\sqrt{2}}{2}\right)$
$v=\sqrt{\frac{Gm}{4r}(1+2\sqrt{2})}$
$v=\frac{1}{2}\sqrt{\frac{Gm}{r}(1+2\sqrt{2})}$