Question

Four identical particles of mass $M$ are located at the corners of a square of side ' $a$ '. If each of them revolves under the influence of others gravitational field in a circular orbit circumscribing the square, their speed is found to be $Q$ times $\sqrt{\frac{G M}{a}}$. The value of $Q$ is_____.

Answer 1

Let $F _{1}, F _{2}$ and $F _{3}$ be the forces acting on particle at $A$ due to particles at $B$, C and D respectively.
Net force on $A$,
$F _{ A }= F _{1} \cos 45^{\circ}+ F _{2} \cos 45^{\circ}$
....[along $x$ - direction]
The $y$ - components of $F_{1}$ and $F_{2}$ cancels each other. Net force on A provides the centripetal force.
$\therefore \frac{ Mv ^{2}}{ r }=2 F _{1} \cos 45^{\circ}+ F _{3}$ ...(i)
for value of $r$,

From geometry
$r^{2}+r^{2}=a^{2}$
$\Rightarrow r=\frac{a}{\sqrt{2}}$
putting (ii) in (i)
$\frac{ Mv ^{2}}{\frac{ a }{\sqrt{2}}}=\frac{2 GMM }{ a ^{2}} \frac{1}{\sqrt{2}}+\frac{ GMM }{2 a ^{2}}$
$\Rightarrow v^{2}=\frac{G M}{a}\left(1+\frac{1}{2 \sqrt{2}}\right)$
$\Rightarrow v^{2}=\frac{2 \sqrt{2}+1}{2 \sqrt{2}} \frac{G M}{a}$
$\Rightarrow v=\sqrt{\frac{3.82}{2.82}} \sqrt{\frac{G M}{a}}$
$=1.16 \sqrt{\frac{G M}{a}}$