question_answer

Three identical bodies of mass M are located at the vertices of an equilateral triangle of side L. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is

A)  \[\sqrt{\frac{GM}{L}}\]

B)         \[\sqrt{\frac{3GM}{2L}}\]           

C)         \[\sqrt{\frac{3GM}{L}}\]             

D)         \[\sqrt{\frac{2GM}{3L}}\]

E)  \[\sqrt{\frac{GM}{3L}}\]

Correct Answer: A

Solution :

Given\[{{F}_{1}}={{F}_{2}}=F\]and\[\theta ={{60}^{o}}\] Resultant force\[=\sqrt{3}F\] \[\therefore \]Force on mass at A due to mass at B and C                 \[=\sqrt{3}\left( \frac{G{{M}^{2}}}{{{L}^{2}}} \right)\] Centripetal force for circumscribing the triangle in a circular orbit is provided by mutual gravitational interaction. ie, \[\frac{M{{v}^{2}}}{(L/\sqrt{3})}-=\sqrt{3}\left( \frac{G{{M}^{2}}}{{{L}^{2}}} \right)\]or \[v=\sqrt{\frac{GM}{L}}\]