Question

An isolated triple star system consists of two identical stars, each of mass m and a fixed star of mass M. The identical masses revolve around the central star in the same circular orbit of radius r. The two orbiting stars are always at opposite ends of a diameter of the orbit. The time period of revolution of each star around the fixed star is equal to :

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

Answer 1

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

The net gravitational force on a given star provides the necessary centripetal force.
$\frac{m{v}^2}{r}=\frac{GMm}{r^2}+\frac{G\left(m\right)\left(m\right)}{(2r{)}^2}$
$\Rightarrow v=\frac{\sqrt{G(4M+m)}}{2\sqrt{r}}$
The time period is,
$T=\frac{2\pi r}{v}$
$=\frac{4\pi r^{\frac{3}{2}}}{\sqrt{G(4M+m)}}$