Question

Four particles, each of mass $M$, move along a circle of radius $R$, under the action of their mutual gravitational attraction. The speed of each particle is

• A. $\frac{GM}{R}$
• B. $\sqrt{2\sqrt{2}\frac{GM}{R}}$
• C. $\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}$
• D. $\sqrt{\frac{GM}{R}\left(\frac{2\sqrt{2}+1}{4}\right)}$

Answer 1

Answer: (D)

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

$\text{Gravitational force between two masses}=\frac{G{m}_1{m}_2}{d^2}$
$\text{So,}{F}_{12}=F_{14}=\frac{G{M}^2}{2R^2}$
$\text{The resultant of these two forces is}\frac{\sqrt{2}G{M}^2}{2R^2}$
$\text{Now,}{F}_{13}=\frac{G{M}^2}{4R^2}$

$\text{combined resultant of all the forces is}$
$F_{net}=\frac{\sqrt{2}G{M}^2}{2R^2}+\frac{G{M}^2}{4R^2}=\frac{G{M}^2}{R^2}[\frac{\sqrt{2}}{2}+\frac{1}{4}]$

$\text{Equating this with centripetal force,we get}$
$\frac{M{v}^2}{R}=\frac{G{M}^2}{R^2}\left[\frac{2\sqrt{2}+1}{4}\right]$
$v^2=\frac{GM}{R}\left[\frac{2\sqrt{2}+1}{4}\right]$
$v=\sqrt{\frac{GM}{R}\left[\frac{2\sqrt{2}+1}{4}\right]}$