Question

A certain triple-star system consists of two stars, each of mass $m$, revolving about a central star of mass $M$, in the same circular orbit. The two stars stay at opposite ends of a diameter of the circular orbit. Then the expression for the period of revolution $\left(T\right)$ of the stars is (radius of the orbit is $r$)

• A. $T=\frac{6\pi r^{3/2}}{\sqrt{G(4M+m)}}$
• B. $T=\frac{4\pi r^{3/2}}{\sqrt{G(4M+m)}}$
• C. $T=\frac{4\pi r^{3/2}}{\sqrt{G(2M+m)}}$
• D. $T=\frac{4\pi r^{3/4}}{\sqrt{G(4M+m)}}$

Answer 1

The correct option is B $T=\frac{4\pi r^{3/2}}{\sqrt{G(4M+m)}}$

Time period
$T=\frac{2\pi r}{v}$ .....(1)

A centripetal force is working here, so that both bodies would stay in circular orbit.


so the net force:
$F_c=\frac{G{m}^2}{(2r{)}^2}+\frac{GMm}{(r{)}^2}=\frac{m{v}^2}{r}\frac{Gm}{\left(4r\right)}+\frac{GM}{\left(r\right)}=v^2\Rightarrow v=\sqrt{\frac{Gm}{\left(4r\right)}+\frac{GM}{\left(r\right)}}$

So, total time period,
$T=\frac{2\pi r}{\sqrt{\frac{Gm}{\left(4r\right)}+\frac{GM}{\left(r\right)}}}$
$T=\frac{4\pi r^{3/2}}{\sqrt{G(4M+m)}}$