Many-body problem on gravitation I encountered the problem given below, and have been pondering on it for a while now

Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed, if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

I know how to solve it by taking the components of the gravitational force directed towards the center and equating it to the centripetal force.
Rather I'm on the lookout for a quick mathematical trick by reducing the four-body problem to a one body problem.
Is this possible?
Are there any other tricks?
 A: Easiest way is to do what you said. It's rather simple and quick.
There are however some tricks within that method you can use to simplify the calculations.

Namely, realize that the each mass will experience the same net force. This is essentially reducing the problem to a 1-body problem.
The solution is exactly what you pointed out:

I'll focus on the bottom mass. The forces exerted by the masses to its left and right the same in magnitude, and notice they're both applied at $45^\circ$ from the center line, so their projections along the center line will also be the same. You know the angle between the force and the center line is $45^\circ$ and that $\sin45=\dfrac {\sqrt{2}}2$. You can easily calculate their projections.
The force by the top mass is also found rather quickly with Newton's Law of Gravitation.
Then, as you said, equate that sum to the centripetal force and solve for $v$.

Other methods (like equations of motion) are tedious and filled with much more algebra. This method is the easiest and fastest. The only tricks there are pretty obvious.
A: You could use the virial theorem result for a system interacting solely via a $\frac{1}{r}$ interaction that it has an average kinetic energy equal to half the negative of the average potential energy. Since the relative positions and velocities don't change, the averages in this case are the same as the instantaneous values. There are 2 pairs a distance $a$ apart and 4 pairs $\frac{a}{\sqrt{2}}$ apart. So equate the total kinetic energy to the negative of half the potential energy,
$4 \frac{1}{2}mv^2 = \frac{1}{2}\frac{Gm^2}{a}\left [ 2+4\sqrt{2} \right]$, and then solve for $v$.
A: Perhaps look at the equation of motion for the scalar length(-squared) between any two of the bodies. This is the quantity you are directly interested in but I can't promise the algebra will be significantly easier.
