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. $\frac{3G{M}^2}{L}$
• B. $\frac{3}{2}\frac{G{M}^2}{L}$
• C. $\frac{3}{4}\frac{G{M}^2}{L}$
• D. $\frac{G{M}^2}{2L}$

Answer 1

Answer: (B)

Stars move around $COM$. Distance of each star from

$COM$ is $\frac{2}{3}\times Lcos30^{\circ }=\frac{L}{\sqrt{3}}$

Force on each star $M$ due to the other two

$2\cdot \frac{G.M.M.}{L^2}cos30^{\circ }=\sqrt{3}\frac{G{M}^2}{L^2}$

This acts as centripetal force.

Therefore, $\sqrt{3}\frac{G{M}^2}{L^2}=\frac{m{v}^2}{(L/\sqrt{3})}$

$\Rightarrow V=\sqrt{\frac{GM}{L}}$

To dismantle the system, energy equal to total energy of the system must be provided.

i.e., $3\times \frac{1}{2}M{V}^2+(3\times \frac{G{M}^2}{L})=-\frac{3}{2}\frac{G{M}^2}{L}$.