A double-star, possessing masses m₁ and m₂ , are rotating with constant angular speed. When the maximum distance of separation between them is R, what is the angular speed of revolution of the stars:

$\sqrt{\frac{G\left(m_1+m_2\right)}{R^3}}$

Each star moves under the gravitational force exerted by the other star. Suppose they rotate with constant angular speed ω in the same sense.

Gravitational force provides necessary centripetal acceleration to keep the stars in their respective orbits of revolution.

$\Rightarrow F_{g_1}=F_{cp_1}=m_1 r_1 \omega^2$  and  $F_{g_2}=F_{cp_2}=m_2 r_2 \omega^2$

$F_{gr_1}=F_{gr_2}=G \frac{m_1 m_2}{R^2} \Rightarrow m_1 r_1 \omega^2=m_2 r_2 \omega^2$

$\Rightarrow m_1 r_1=m_2 r_2$

That means, the system rotates about its C.M., O.

$\frac{r_1}{r_2}=\frac{m_2}{m_1} \Rightarrow \quad \frac{r_1}{r_1+r_2}=\frac{m_2}{m_1+m_2}$

$\Rightarrow r_1=\frac{m_2}{m_1+m_2}$ R (since $r_1+r_2$ = R)

$F_{c p_1}=m_1 r_1 \omega^2$  and  $F_{gr_1}=\frac{G m_1 m_2}{R^2}$

Since $F_{cp_1}=F_{g_1}$

$\Rightarrow m_1 r_1 \omega^2=\frac{G m_1 m_2}{R^2} \Rightarrow r_1 \omega^2=\frac{G m_2}{R^2}$

Putting  $r_1=\frac{m_2 R}{m_1+m_2}$,  we obtain,

$\left(\frac{m_2 R}{m_1+m_2}\right) \omega^2=\frac{G m_2}{R^2}$

$\Rightarrow \omega=\sqrt{\frac{G\left(m_1+m_2\right)}{R^3}}$