Six stars of equal mass m are moving about the centre of mass of the system such that they are always on the vertices of a regular hexagon of side length a. Their common time period of revolution around center will be

 We know that,  $F=\frac{G{m}_1{m}_2}{r^2}$
 From the geometry of hexagon 
$$\begin{gathered} F_1=F_5=\frac{G{m}^2}{a^2}=F\left(let\right) \\ {F}_2=F_4=\frac{G{m}^2}{{\left(\sqrt{3}a\right)}^2}=\frac{G{m}^2}{3a^2}=\frac{F}{3} \\ {F}_3=\frac{G{m}^2}{{\left(2a\right)}^2}=\frac{G{m}^2}{4a^2}=\frac{F}{4} \end{gathered}$$

Resultant gravitation Force, 
 $F\text{'}=\sqrt{F_1^2+F_5^2+2F_1{F}_5\cos 120°}+\sqrt{F_2^2+F_4^2+2F_2{F}_4\cos 60°}+\frac{F}{4}$
 $=F+\frac{F}{\sqrt{3}}+\frac{F}{4}=(\frac{5}{4}+\frac{1}{\sqrt{3}})F=(\frac{5}{4}+\frac{1}{\sqrt{3}})\frac{G{m}^2}{a^2}$
Since, Centripetal Force = Gravitational Force
$m{\omega }^{2}a=\frac{G{m}^2}{a^2}(\frac{5}{4}+\frac{1}{\sqrt{3}})$

$\omega =\sqrt{\frac{Gm}{a^3}(\frac{5}{4}+\frac{1}{\sqrt{3}})}\Rightarrow T=2\pi \sqrt{\frac{4\sqrt{3}{a}^3}{Gm(5\sqrt{3}+4)}}$