Question

Three particles each of mass $m=1kg$, are located at the vertices of an equilateral triangle of side $a=4m$. The speed at which they must move to revolve in a circular orbit, under the influence of mutual gravitational force, circumscribing and preserving the equilateral triangle is $v=n\sqrt{G}$, then the value of $n$ is

Answer 1

Answer: 0.50

The net gravitational force on any particle due to the other two is
${\Gamma }_g=\frac{2G{m}^2}{a^2}\cos \left(30^{\circ }\right)=\frac{\sqrt{3}G{m}^2}{a^2}$
Now, $r=\frac{a}{\sqrt{3}}$
For circular motion,
$F_g=\frac{\sqrt{3}G{m}^2}{a^2}=\frac{m{v}^2}{r}$
Solving it we obtain,
$v=0.50\sqrt{G}$