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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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@@ -14,9 +14,9 @@
<meta property="og: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?">
<title>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?</title>
$2.6.14^*.$ 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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
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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?">
<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?">
<meta property="og:title" 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?">
<meta property="og:title" 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?">
<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og: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?">
<meta property="og: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?">
<title>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?</title>
<title>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?</title>
$2.6.14^*.$ 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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
