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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="../../#2.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.2.23.$ To create artificial gravity, two compartments of the orbital station (mass ratio $1 : 2$) were separated by a distance $R$ from each other and spun around their common center of mass. Determine the time of complete rotation of the compartments if, in a more massive compartment, the artificial gravity is half the force of gravity on the Ground.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.2.23_1.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Velocities of the system
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+$$\sum \vec{R}_\text{external} = \vec{0}$$
+$F_3$ - The force of gravity on the ground acting on an object of mass $m$<br>
+$F_2$ - The force of gravity on a massive compartment<br>
+$F_1$ - The force of gravity on the less massive compartment <br>
+From the statement
+$$\frac{F_3}{F_2} = 2; \frac{m_xg}{m_xg_2}=2 \Rightarrow g_2 = \frac{g}{2}$$
+$$\frac{m_1}{m_2} = \frac{1}{2}$$
+$$\frac{F_1}{F_2} = \frac{m_1g_1}{m_2g_2} = \frac{g_1}{2g_2} = \frac{g_1}{g}$$
+Since centrifugal forces $F_1$ and $F_2$ are internal forces, they are equal to
+$$F_1 = F_2 \Rightarrow g_1 = g$$
+Centripetal acceleration could be found as
+$$g_1 = \frac{v_1^2}{x_1}; \quad g_2 = \frac{v_2^2}{x_2}$$
+From where
+$$x_1 = \frac{2mR}{3m}=\frac{2}{3}R; \quad x_2 = R-x_1 = \frac{1}{3}R$$
+Alternatively
+$$v_1 = \sqrt{\frac{2gR}{3}}; \quad v_2 = \sqrt{\frac{gR}{6}}$$
+Then we find the rotation period as the ratio of the trajectory circle length to the velocity of the first body
+$$ T = \frac{2\pi x_1}{v_1} $$
+$$ T = 2\pi \cdot \frac{2}{3}R \cdot \sqrt{\frac{3}{2gR}}$$
+$$\boxed{T=2\pi\sqrt{\frac{2R}{3g}}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$T=2\pi\sqrt{2R/3g}$$
+ </p>
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