## Three-Way

I’ve been told that when you and your sweetheart get accepted to different grad schools, you’ve encountered the “Two-Body Problem”. It’s an inapt analogy, because the two-body problem can be solved elegantly. What’s really difficult is the three-body problem.

If we have three massive bodies interacting through Newton’s gravitational law, with arbitrary initial positions and velocities, it turns out that exact analytic solutions are extremely difficult to find. That’s not to say there is no solution – the massive bodies have no trouble finding it. But if we want to predict their motion we’ll need to do some numerical integration.

However, we might wonder if there are some particularly simple or intriguing solutions, perhaps in very symmetrical situations. There are. Quite a few of them are cited in the Scholarpedia article. Here are some:

I’m not sure how these solutions were discovered. But if we consider the problem a moment it’s clear the center of mass cannot accelerate, and that energy and angular momentum must be preserved. An especially simple way to do this is to have the angular momentum and energy of each individual mass be constant by making them all rotate around the center of mass at a fixed frequency.

To see if this works, measure the positions of the three masses by vectors $\vec{r_1}, \vec{r_2}, \vec{r_3}$ from the axis of rotation, which is through the center of mass and perpendicular to the plane of the bodies. Let the masses of the bodies be $m_1, m_2, m_3$. Then the equation for the center of mass states

$\vec{r_1}m_1 + \vec{r_2}m_2 + \vec{r_3}m_3 = 0$.

If we look at a reference frame comoving with the center of mass and corotating with the masses, the masses will be stationary, so there must be no force on them. Let’s take $m_1$ as an example. The centrifugal force on it is

$F_c = \omega^2 m_1 \vec{r_1}$.

The gravitational force on it is

$F_g = G m_1 \left(m_2\frac{\vec{r_2} - \vec{r_1}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + m_3\frac{\vec{r_3} - \vec{r_1}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$

If those forces are going to add to zero to give $m_1$ zero acceleration, they better at least point the same direction. It’s not clear that they do. The centrifugal force points only in the direction of $\vec{r_1}$, while the gravitational force has all three position vectors in there. So we need the amounts of $\vec{r_2}$ and $\vec{r_3}$ to be such that they add up to point towards $\vec{r_1}$.

From the equation for the center of mass we get,

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

whence we know the proportions in which the $\vec{r_2}$ and $\vec{r_3}$ must appear in the gravitational force. We conclude

$m_2\vec{r_2} + m_3\vec{r_3} \propto \left(\frac{m_2\vec{r_2}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + \frac{m_3 \vec{r_3}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$.

This will work perfectly iff the denominators on the right hand side of that equation are equal, meaning mass $m_1$ must be equidistant from masses $m_2$ and $m_3$. Repeating the argument while writing out forces on $m_2$ this time, the masses must be in an equilateral triangle.

We still don’t know if such a solution exists, only that it’s possible to get the centrifugal force to point opposite the gravitational force for arbitrary masses, as long as we put them in an equilateral triangle. Let’s continue, setting the centrifugal and gravitational forces equal for $m_1$ and see what the resulting rotation rate is. Also let’s let the sides of the triangle be length $l$

$F_c = \omega^2 m_1 \vec{r_1} = F_g = \frac{G m_1}{l^3} \left( m_2(\vec{r_2} - \vec{r_1}) + m_3 (\vec{r_3} - \vec{r_1}) \right)$

Applying the center of mass equation

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

one more time and setting the total mass equal to $M$, this simplifies to

$\omega^2 = \frac{G M}{l^3}$.

A clean result that is the same for all three masses, meaning these orbits indeed solve the equations of motion.