Question

Three stars, each of mass $m$, are located at the vertices of an equilateral triangle of side $a$. They revolve in circular orbit under mutual gravitational force while preserving the equilateral triangle. Find the mechanical energy of the system.

• A. $+\frac{G{m}^2}{2a}$
• B. $-\frac{3G{m}^2}{2a}$
• C. $-\frac{G{m}^2}{2a}$
• D. $+\frac{3G{m}^2}{2a}$

Answer 1

The correct option is B $-\frac{3G{m}^2}{2a}$

Let us consider all the stars move in a common circle of radius $r$ about the centroid $O$ as shown in the figure.

Let us calculate the net force on any one of the three particles.


From figure,

$\sin {60}^{\circ }=\frac{a}{2r}$

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

Net force on the particle placed at point $\text{A}$:
$F_{net}=2F\cos {30}^{\circ }=\frac{2F\sqrt{3}}{2}=\sqrt{3}F$ (along $AO$)

Substituting the value of gravitational force $F$,

$\Rightarrow F_{net}=\frac{\sqrt{3}G\left(m\right)\left(m\right)}{a^2}=\frac{\sqrt{3}G{m}^2}{a^2}$

Since they revolve in circular orbit under mutual gravitational force, so we have

$F_{net}=F_c$

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

Putting the value of $r$,

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

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

Total mechanical energy : $E=K+U$

$\Rightarrow E=\frac{-3G{m}^2}{a}+3\left(\frac{1}{2}m{v}^2\right)$

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

$\therefore E=-\frac{3G{m}^2}{2a}$

Hence, option (b) is the correct answer.