Question

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,

Answer 1

force between A and B , $F_1=\frac{GM² ^2}{L^2}$
force between C and B, $F_2=\frac{GM^2}{L^2}$
here, $F_1=F_2=\frac{GM^2}{L^2}$

now Resultant force act on B due to A and C.
while angle ABC = 60° [ as ∆ is equilateral]
so, $F_{net}=\sqrt{F_1^2+F_2^2+2F_1F_2cos60^{\circ}}$
so, $F_{net}=\sqrt{3}\frac{GM^2}{L^2}$

now , net force due to gravitation is balanced by centripetal force .

so, $\frac{Mv^2}{\frac{L}{\sqrt{3}}}=\frac{\sqrt{3}GM^2}{L^2}$

$v=\sqrt{\frac{GM}{L}}$

Answer 2

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