question_answer

Three particles of equal mass m are situated at the vertices of an equilateral triangle of side L. What should be the velocity of each particle so that they move on a circular path without changing L?

A) Ö (GM/2L)                  

B) Ö (GM/L)

C) Ö (2GM/L)                  

D) Ö (GM/3L)

Correct Answer: B

Solution :

The resultant gravitational force on each particle provides the necessary centripetal force. \[\therefore \]\[m{{v}^{2}}/r=\]Ö\[({{F}^{2}}+{{F}^{2}}+2{{F}^{2}}\cos {{60}^{o}})=\]Ö3F But (Ö3L/2)x (2/3)=1/Ö3 \[\therefore \]\[m{{v}^{2}}\]Ö3/L=Ö3GM2/L2\[\Rightarrow v=\]Ö(GM/L)