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
\[T=\frac{2\pi r}{\upsilon
}=2\pi \frac{a}{\sqrt{3}}\sqrt{\frac{a}{Gm}}=2\pi
\sqrt{\frac{{{a}^{3}}}{3Gm}}\]
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