Question

Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Answer 1

Force on M at c due to gravitational attraction
${\overrightarrow{F}}_{CB}=\frac{G{m}^2}{2R^2}\hat{j}{\overrightarrow{F}}_{CD}=\frac{G{m}^2}{4R^2}\hat{j}{\overrightarrow{F}}_{CA}=\frac{G{m}^2}{4R^2}\cos 45\hat{i}+\frac{G{m}^2}{4R^2}\sin 45\hat{j}\therefore {\overrightarrow{F}}_c={\overrightarrow{F}}_{CA}+{\overrightarrow{F}}_{CB}+{\overrightarrow{F}}_{CD}=\frac{-G{m}^2}{4R^2}(2+\frac{1}{2})\hat{i}+\frac{G{m}^2}{4R^2}(2+\frac{1}{2})\hat{j}$
So, Resultant force on c,
$F_C=\frac{G{m}^2}{4R^2}(\frac{2}{2}+1)$
For moving along the circle
$F=\frac{m{v}^2}{R}=V=\frac{GM}{R}\left(\frac{\frac{2}{2}+1}{4}\right)$