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 $\sqrt{\frac{GM}{a}}$. The value of $Q$ is_____.

Answer 1

Answer: 1.16

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.
$\therefore \frac{M{v}^2}{r}=2F_1\cos 45^{\circ }+F_3$ ...(i)
for value of $r$,

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