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?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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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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