$2.2.23.$ 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.
$$\sum \vec{R}_\text{external} = \vec{0}$$
$F_3$ - The force of gravity on the ground acting on an object of mass $m$
$F_2$ - The force of gravity on a massive compartment
$F_1$ - The force of gravity on the less massive compartment
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
$$ T = \frac{2\pi x_1}{v_1} $$
$$ T = 2\pi \cdot \frac{2}{3}R \cdot \sqrt{\frac{3}{2gR}}$$
$$\boxed{T=2\pi\sqrt{\frac{2R}{3g}}}$$
$$T=2\pi\sqrt{2R/3g}$$
Almaskhan Arsen