Question

Three identical bodies, each of mass m, are located at the vertices of an equilateral triangle with side r. At what speed must they move if they all revolve under the influence of one another's gravitation in a circular orbit circumscribing the triangle while still preserving the equilateral triangle?

Answer 1

From figure, stars orbit about a point O inside equilateral triangle,

The distance from the centre of its orbit is given by

$r^{\prime }=\frac{r}{2\cos 30}$

From Newton law of universal gravitation,

$F_1=\frac{G{M}^2}{r^2}\cos 30$

The net force of gravity on one star due to other two is

$f=2f_1$

$=\frac{2G{M}^2}{r^2}\cos {30}^{\circ }$

The centripetal force around triangle is

$f_c=\frac{M{v}^2}{r^{\prime }}$

Here speed of stars is v substitute $\frac{r}{2\cos {30}^{\circ }}$ for $r^{\prime }$

is above equation

$f_c=\frac{M{v}^2}{\left(\frac{r}{2\cos 30}\right)}=\frac{2M{v}^2\cos 30}{L}$

$f=f_c$ (now necessary centripetal force is provided by net force)

$\frac{2G{M}^2\cos {30}^{\circ }}{r^2}=\frac{2M{v}^2\cos {30}^{\circ }}{r}$

$\frac{2G{M}^2\times \sqrt{3}}{2r^2}=\frac{2M{v}^2\sqrt{3}}{2r}$

$\frac{GM}{r}=v^2$

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