Solution

The position of the center of mass (point $O$) could be found as

$$R_1=R\frac{m_2}{m_1+m_2}$$

$$R_2=R-R_1=R\frac{m_1}{m_1+m_2}$$

Since the bodies are stationary, their accelerations are equal, and therefore, according to Newton's first law, the sum of external forces is zero

$$F_{c1}=F_{21} \quad F_{c2}=F_{12}\tag{1}$$

By definition of centripetal acceleration

$$F_{c1}=m_1\omega_1^2R_1\tag{2.1}$$

$$F_{c2}=m_2\omega_2^2R_2\tag{2.2}$$

Since the body $m_1$ and $m_2$ rotate along the same straight line

$$\omega_1=\omega_2=\omega$$

On the other hand, the force of their gravitational attraction is equal to

$$F_{12}=F_{21}=G\frac{m_1m_2}{R^2}\tag{4}$$

Given $(2)$, we equate $(1)$ and $(3)$

$$G\frac{m_1m_2}{R^2}=m_1\omega^2R\frac{m_2}{m_1+m_2}$$

From where

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

Whence the circulation period is equal to

$$\boxed{T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{R^3}{G(m_1+m_2)}}}$$