Question

Three partieles, each of mass $m$ are situated at the vertices of an equilateral triangle of side $a$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation $a$. Find the initial velocity that should be given to each particle and also the time period of the circular motion. $(F=\frac{G{m}_1{m}_2}{r^2})$

• A. $v=\sqrt{\frac{Gm}{a}}$
• B. $v=\sqrt{\frac{Gm}{2a}}$
• C. $T=2\pi \sqrt{\frac{a^3}{3Gm}}$
• D. $T=2\pi \sqrt{\frac{a^3}{Gm}}$

Answer 1

Answer: (A), (C)

The correct options are
A $v=\sqrt{\frac{Gm}{a}}$
C $T=2\pi \sqrt{\frac{a^3}{3Gm}}$

ABC is an equililateral triangle of side $a$.

From figure, $\frac{a/2}{OB}=cos{30}^o=\frac{\sqrt{3}}{2}$ $⟹OB=\frac{a}{\sqrt{3}}$

Also $F_1=F_2=\frac{G{m}^2}{a^2}$

Resultant force $F_R=\sqrt{F_1^2+F_2^2+2F_1{F}_{2}cos{60}^o}$

$\therefore$ $F_R=\sqrt{F^2+F^2+2\left(F\right)\left(F\right)\times 0.5}=\sqrt{3}F$

Also $m{a}_r=F_R$

OR $\frac{m{v}^2}{OB}=\sqrt{3}F$

OR $\frac{m{v}^2}{a/\sqrt{3}}=\sqrt{3}\times \frac{G{m}^2}{a^2}$ $⟹v=\sqrt{\frac{Gm}{a}}$

Time period $T=\frac{2\pi \left(OB\right)}{v}=\frac{2\pi a/\sqrt{3}}{\sqrt{\frac{Gm}{a}}}=2\pi \sqrt{\frac{a^3}{3Gm}}$