question_answer

Three particles, 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 move 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.

Answer:

                In Fig. 5(HT).8, gravitational force between any two particles is    \[F=\frac{Gm\,m}{{{a}^{2}}}\] Resultant force on each particle due to the other two particles is   \[R=\sqrt{{{F}^{2}}+{{F}^{2}}+2F\,F\cos {{60}^{\text{o}}}}=\sqrt{{{F}^{2}}+{{F}^{2}}+{{F}^{2}}}=F\sqrt{3}\] \[R=\frac{\sqrt{3}G{{m}^{2}}}{{{a}^{2}}}\] If particles were at rest, each particle would move under the action of resultant force R (on each) and meet at the centroid O of the triangle. Let each particle be given a tangential velocity \[\upsilon \] so that R acts as the centripetal force, they would move m a circle of radius \[=\text{OA}=\text{OB}=\text{OC}=\text{r}=\frac{2}{3}a\,\sin {{60}^{\text{o}}}=\frac{2}{3}a\frac{\sqrt{3}}{2}=a/\sqrt{3}\] The original mutual separation will be maintained. As           \[R=\frac{m{{\upsilon }^{2}}}{r}\] \[\therefore \]  \[\upsilon =\sqrt{\frac{Rr}{m}}=\sqrt{\frac{\sqrt{3}G{{m}^{2}}}{{{a}^{2}}}.\frac{a}{\sqrt{3}m}}\] or                \[\upsilon =\sqrt{Gm/a}\] Time period of circular motion \[T=\frac{2\pi r}{\upsilon }=2\pi \frac{a}{\sqrt{3}}\sqrt{\frac{a}{Gm}}=2\pi \sqrt{\frac{{{a}^{3}}}{3Gm}}\]