Question

Three identical stars, each of mass $M$, form an equilateral triangle (stars are positioned at the corners) that rotates around the centre of the triangle. The system is isolated and edge length of the triangle is $L$. The amount of work done, that is required to dismantle the system, is:

• A. $\frac{3GM^2}{L}$
• B. $\frac{3}{2} \frac{GM^2}{L}$
• C. $\frac{3}{4} \frac{GM^2}{L}$
• D. $\frac{GM^2}{2L}$

Answer 1

Answer: (B)

Stars move around $COM$. Distance of each star from
$COM $ is $\frac{2}{3} \times L \,cos\,30^{\circ} = \frac{L}{\sqrt{3}}$
Force on each star $M$ due to the other two
$2\cdot \frac{G.M.M.}{L^2} cos\,30^{\circ} = \sqrt{3} \frac{GM^2}{L^2}$
This acts as centripetal force.
Therefore, $\sqrt{3} \frac{GM^2}{L^2} = \frac{mv^2}{(L/\sqrt{3})}$
$\Rightarrow V = \sqrt{\frac{GM}{L}}$
To dismantle the system, energy equal to total energy of the system must be provided.
i.e., $3\times \frac{1}{2} MV^2 + (3 \times \frac{GM^2}{L}) = -\frac{3}{2} \frac{GM^2}{L}$.