Question

Three bodies of identical mass M from the vertices of an equilateral triangle of side L and rotate in circular orbits about the centre of the triangle. They are held in place by their mutual gravitation. What is the speed of each ?

Answer 1

Let speed of each particle is $v$

Given side of an equilateral triangle $L$ radius of orbit is $\frac{L}{\sqrt{3}}$

So, $F_c=\frac{m{v}^2}{L/\sqrt{3}}=\frac{\sqrt{3}m{v}^2}{L}$

And $F_g=\frac{G{m}^2}{L^2}$

Now, net gravitational force $F_{net}=2Fcos{30}^{\circ }=\frac{\sqrt{3}G{m}^2}{L^2}$

Now, $F_{net}=F_c$

$or,\frac{\sqrt{3}G{m}^2}{L^2}=\frac{\sqrt{3}m{v}^2}{L}$

$\therefore v=\sqrt{\frac{Gm}{L}}$