Question

The three particles, each having mass $m$, is kept at corners of an equilateral triangle of side length $a$ and are rotating under the effect of mutual gravitational force. Choose the correct alternative(s).

• A. The radius of circular path followed by the particles is $\frac{a}{2}$.
• B. Velocity of the particles is $\sqrt{\frac{Gm}{a}}$.
• C. Binding energy of the system is $\frac{1.5G{m}^2}{a}$.
• D. Time period of the particles is $\sqrt{\frac{\pi a^2}{2Gm}}$.

Answer 1

Answer: (B), (C)

Binding energy of the system is $\frac{1.5G{m}^2}{a}$.


From the $△AOD$,

$\cos {30}^{\circ }=\frac{a/2}{R}$

$\Rightarrow R=\frac{a}{\sqrt{3}}$

$\therefore$ Radius of circular path followed by the particles is $R=\frac{a}{\sqrt{3}}$.

Net mutual gravitational force,

$F_{net}=\sqrt{F^2+F^2+2F^2\cos {60}^{\circ }}=\sqrt{3}F=\frac{\sqrt{3}G{m}^2}{a^2}$

Net mutual gravitational force on the particles provides the necessary centripetal force.

Using circular dynamics,

$\frac{m{v}^2}{a/\sqrt{3}}=\frac{\sqrt{3}G{m}^2}{a^2}$

$\Rightarrow v=\sqrt{\frac{Gm}{a}}$

Now, the time period of rotation is,

$T=\frac{2\pi \left(a/\sqrt{3}\right)}{v}=2\pi \sqrt{\frac{a^3}{3Gm}}$

Further, total binding energy of the system is,

$E=|3\times \frac{-G{m}^2}{a}+3\times \frac{1}{2}m{\left(\sqrt{\frac{Gm}{a}}\right)}^2|$

$\Rightarrow E=\frac{1.5G{m}^2}{a}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, options $\left(B\right)$ and $\left(C\right)$ are correct.