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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"
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			<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>
		@@ -88,7 +88,7 @@
			<p>
			$$T=2\pi\sqrt{R^3/\gamma(m_1+m_2)}$$
			</p>
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			<p style="text-align: right; font-style: italic; font-size: 14;">
			Almaskhan Arsen<br>
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	<h2>Solutions of Savchenko Problems in Physics</h2>		<h2>Solutions of Savchenko Problems in Physics</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3>

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

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="draw.png"		<img src="draw.png"
	loading="lazy" width="260" />		loading="lazy" width="260" />
	<figcaption>		<figcaption>
	Forces acting on the system		Forces acting on the system
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>

	The position of the center of mass (point $O$) could be found as		The position of the center of mass (point $O$) could be found as
	$$R_1=R\frac{m_2}{m_1+m_2}$$		$$R_1=R\frac{m_2}{m_1+m_2}$$
	$$R_2=R-R_1=R\frac{m_1}{m_1+m_2}$$		$$R_2=R-R_1=R\frac{m_1}{m_1+m_2}$$
	Since the bodies are stationary, their accelerations are equal, and therefore, according to Newton's first law, the sum of external forces is zero		Since the bodies are stationary, their accelerations are equal, and therefore, according to Newton's first law, the sum of external forces is zero
	$$F_{c1}=F_{21} \quad F_{c2}=F_{12}\quad(1)$$		$$F_{c1}=F_{21} \quad F_{c2}=F_{12}\quad(1)$$
	By definition of centripetal acceleration		By definition of centripetal acceleration
	$$F_{c1}=m_1\omega_1^2R_1\quad(2.1)$$		$$F_{c1}=m_1\omega_1^2R_1\quad(2.1)$$
	$$F_{c2}=m_2\omega_2^2R_2\quad(2.2)$$		$$F_{c2}=m_2\omega_2^2R_2\quad(2.2)$$
	Since the body $m_1$ and $m_2$ rotate along the same straight line		Since the body $m_1$ and $m_2$ rotate along the same straight line
	$$\omega_1=\omega_2=\omega$$		$$\omega_1=\omega_2=\omega$$
	On the other hand, the force of their gravitational attraction is equal to		On the other hand, the force of their gravitational attraction is equal to
	$$F_{12}=F_{21}=G\frac{m_1m_2}{R^2}\quad(4)$$		$$F_{12}=F_{21}=G\frac{m_1m_2}{R^2}\quad(4)$$
	Given $(2)$, we equate $(1)$ and $(3)$		Given $(2)$, we equate $(1)$ and $(3)$
	$$G\frac{m_1m_2}{R^2}=m_1\omega^2R\frac{m_2}{m_1+m_2}$$		$$G\frac{m_1m_2}{R^2}=m_1\omega^2R\frac{m_2}{m_1+m_2}$$
	From where		From where
	$$\omega=\sqrt{\frac{G(m_1+m_2)}{R^3}}$$		$$\omega=\sqrt{\frac{G(m_1+m_2)}{R^3}}$$
	Whence the circulation period is equal to		Whence the circulation period is equal to
	$$\boxed{T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{R^3}{G(m_1+m_2)}}}$$		$$\boxed{T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{R^3}{G(m_1+m_2)}}}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	@@ -88,7 +88,7 @@
	<p>		<p>
	$$T=2\pi\sqrt{R^3/\gamma(m_1+m_2)}$$		$$T=2\pi\sqrt{R^3/\gamma(m_1+m_2)}$$
	</p>		</p>

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