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# Three identical stars each of mass M form an equilateral triangle that rotates around the center of the triangle. The system is isolated and edge length of the triangle of the triangle is L. The amount of work that is required to dismantle the system is

$(a)\;\frac{3GM^2}{L} \quad (b)\;\frac{3GM^2}{2L} \quad (c)\;\frac{3GM^2}{4L} \quad (d)\;\frac{GM^2}{2L}$

Stars more around COM. Distance of each star from COM is $\large\frac{2}{3} \normalsize \times L \cos 30^{\circ}=\large\frac{L}{\sqrt 3}$
Force on each star M due to other two $=\large\frac{2GMM}{L^2}$$\cos 30^{\circ}=\sqrt 3 \large\frac{GM^2}{L^2} acting along the radius This acts as the centripetal force \therefore \sqrt 3 \large\frac{GM^2}{L^2}=\frac{Mv^2}{\bigg(\Large\frac{L}{\sqrt 3}\bigg)^2} v=\sqrt {\large\frac{GM}{L}} Total energy required to dismantle the system =Total KE +Total Gravitational energy \qquad= 3 \times \large\frac{1}{2}$$Mv^2+ \bigg[-3 \times \large\frac{GM^2}{L}\bigg]=\large\frac{3}{2} \frac{GM^2}{L}$
Hence a is the correct answer.
edited Jun 30, 2014