This answer to Are the orbits of all triple star systems at least technically unstable? mentions:

There are known solutions to the gravitational three body problem that can be shown to be stable. Lagrange found a three body solution for general masses where all three orbit the common center of mass in an equilateral triangular formation. Gascheau proved in 1843 that this solution is stable if the component masses satisfy

$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} < 1/27$$

When the smallest mass approaches zero the three masses are at the vertices of an equilateral triangle. In a realistic solar system this means Trojan asteroids are generally found in orbits of massive planets like Jupiter about 60 degrees ahead-of and behind it.

But if the smallest mass is large but the inequality above is still satisfied, what can we say about the triangle formed by the three bodies in a circular restricted three-body problem orbit?

Is it still known to be an equilateral triangle, but they spin around a point which is not the center of the triangle, but is weighted toward the heavier object?

  • If so, can this be shown by citing a math-based reference or shown here mathematically or computationally?
  • If not, is there an expression for two angles of the triangle as a function of to mass ratios?

The inner Solar System, from the Sun to Jupiter, including asteroid belt (Hildas, Trojans and NEOs Source click for full size


1 Answer 1


These questions are answered by the same references listed in my previous answer.

In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses. (However, it is stable only if the inequality in the previous answer is satisfied.). This equilateral configuration orbits the center of mass, which is generally not in the center of the equilateral triangle. Each of the bodies follows a circular orbit with radius (see e.g. 1212.0754

$$ r_1 = a\sqrt{\nu_2^2+\nu_2\nu_3+\nu_3^2}$$ $$ r_2 = a\sqrt{\nu_1^2+\nu_1\nu_3+\nu_3^2}$$ $$ r_3 = a\sqrt{\nu_1^2+\nu_1\nu_2+\nu_2^2}$$

and frequency

$$\omega = \sqrt{M/a^3}, $$

where $a$ is the length of the sides of the equilateral triangle, $M$ the total mass, and $\nu_i = m_i/M$.

The situation changes when one accounts for relativisitic effects. When one accounts for the leading (post-Newtonian) corrections, then a circular restricted three body solution still exists for general masses (with a smaller region of stability than in the Newtonian case.). However, the triangular configuration is no-longer circular (unless all three mass are equal or two masses are 0). Keeping the distances to the center of mass $r_i$ as in the Newtonian case, the sides of the triangle are now given by (again see 1212.0754)

$$ r_{ij} = a(1+\frac{M}{a}\epsilon_{ij}+ O(\tfrac{M^2}{a^2}) )$$


$$\epsilon_{12} = \frac{1}{24}[(\nu_1-\nu_3)(5-3\nu_2)+(\nu_2-\nu_3)(5-3\nu_1)]$$ $$\epsilon_{23} = \frac{1}{24}[(\nu_2-\nu_1)(5-3\nu_3)+(\nu_3-\nu_1)(5-3\nu_2)]$$ $$\epsilon_{31} = \frac{1}{24}[(\nu_3-\nu_2)(5-3\nu_1)+(\nu_1-\nu_2)(5-3\nu_3)].$$

The angles of the triangle can be calculated from these lengths if one is so inclined.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .