$2.6.14^*.$ Two stars of mass $m_1$ and $m_2$ form a binary system with a constant distance between the stars $R$. What is the period of rotation of the stars around the common center of mass?
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 $$F_{c1}=F_{21} \quad F_{c2}=F_{12}\quad(1)$$ By definition of centripetal acceleration $$F_{c1}=m_1\omega_1^2R_1\quad(2.1)$$ $$F_{c2}=m_2\omega_2^2R_2\quad(2.2)$$ 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 $$F_{12}=F_{21}=G\frac{m_1m_2}{R^2}\quad(4)$$ 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 $$\boxed{T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{R^3}{G(m_1+m_2)}}}$$
$$T=2\pi\sqrt{R^3/\gamma(m_1+m_2)}$$
Almaskhan Arsen