Four point bodies, each of mass $m$, attract each other due to the interaction gravitational. Such bodies are at the vertices of a $L$-side square rotating around their center with constant angular velocity. $G$ being the universal gravitational constant, the period of this rotation is given by:
What I tried:
$ F_R = F '+ F_ {DB} $
vector
so, $ F_{DB}= \frac{GM^2}{(L\sqrt{2})^2}$
answer: $2\pi \sqrt{\frac{L^3}{GM}(\frac{4-\sqrt{2}}{7})}$
The net attraction force on each mass is the sum of the gravitational attractions from the other three masses and points to the center of the square,
$$F=\frac{Gm^2}{(\sqrt2 L)^2}+2\cdot\cos 45\cdot\frac{Gm^2}{L^2}=\left(\frac12+\sqrt2\right)\frac{Gm^2}{L^2}\tag{1}$$
Additionally, from the Newton's law for the angular rotation,
$$F=mr\omega^2=\frac{\sqrt2}{2}mL\omega^2\tag{2}$$
where $r=\frac{\sqrt2}{2}L$ is the radius of the rotation. Combine (1) and (2) to obtain the period of the rotation,
$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L^3}{Gm}\left(\frac{4-\sqrt{2}}{7}\right)}$$