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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> |
| | | |
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| <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> |
