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?

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    <h3> Statement </h3>
    <p>
        $2.6.14^*.$ 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?
    </p>

    <h3>Solution</h3>
    <p>

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

    <h4>Answer</h4>
    <p>
        $$T=2\pi\sqrt{R^3/\gamma(m_1+m_2)}$$
    </p>

    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>

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