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+ <meta name="description" content="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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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3>
+
+ <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>
+
+</p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="260" />
+ <figcaption>
+ Forces acting on the system
+ </figcaption>
+ </figure>
+</center>
+<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
+$$\boxed{T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{R^3}{G(m_1+m_2)}}}$$
+ </p>
+
+ <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>
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