Question

Three bodies of same masses are kept at the top of an equilateral triangle with side $a$. What speed would the three bodies be moved in a circle so that the triangle moves in the circular orbit and there should be no change in the side of the triangle?

Answer 1

The gravitational force acting between each couples of masses

$F=\frac{GMM}{a^2}=\frac{G{M}^2}{a^2}$

Each internal angle of equilateral triangle is of ${60}^{\circ }$.

Therefore the resultant of forces acting on each mass will be directed towards the center O of the circle and the same will the role of centripetal force for circular motion of the whole triangle.

Centripetal force = Resultant force

or $F_C=F_R=\sqrt{F^2+F^2+2F^2\cos {60}^{\circ }}$

or $\frac{M{v}^2}{r}=F\sqrt{3}$

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

or $\frac{v^2}{r}=\sqrt{3}\frac{GM}{a^2}$ ..............(1)

Now from right angled triangle ABD,

$A{D}^2+B{D}^2=A{B}^2$

or $A{D}^2+\frac{a^2}{4}=a^2$

or $A{D}^2=a^2-\frac{a^2}{4}=\frac{3}{4}{a}^2$

$\therefore$ $AD=\frac{a\sqrt{3}}{2}$

According to geometrical property of median of a equilateral triangle.

$r=AO=\frac{2}{3}AD=\frac{2}{3}\times \frac{a}{2}\sqrt{3}=\frac{a}{\sqrt{3}}$

$r=AO=\frac{2}{3}AD=\frac{2}{3}\times \frac{a}{2}\sqrt{3}=\frac{a}{\sqrt{3}}$

Therefore , from equation (1), we have

$\frac{v^2}{r}=\sqrt{3}\frac{GM}{a^2}\Rightarrow v^2=r\sqrt{3}\frac{GM}{a^2}$

or $v^2=\frac{a}{\sqrt{3}}\times \sqrt{3}\frac{GM}{a^2}=\frac{GM}{a}$

$\therefore$ $v=\sqrt{\frac{GM}{a}}$