<p>These questions are answered by the same references listed in <a href="https://astronomy.stackexchange.com/a/33949/7982">my previous answer</a>.</p>

<p>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. <a href="https://arxiv.org/abs/1212.0754" rel="nofollow noreferrer">1212.0754</a></p>

<p><span class="math-container">$$ r_1 = a\sqrt{\nu_2^2+\nu_2\nu_3+\nu_3^2}$$</span>
<span class="math-container">$$ r_2 = a\sqrt{\nu_1^2+\nu_1\nu_3+\nu_3^2}$$</span>
<span class="math-container">$$ r_3 = a\sqrt{\nu_1^2+\nu_1\nu_2+\nu_2^2}$$</span></p>

<p>and frequency</p>

<p><span class="math-container">$$\omega = \sqrt{M/a^3}, $$</span></p>

<p>where <span class="math-container">$a$</span> is the length of the sides of the equilateral triangle, <span class="math-container">$M$</span> the total mass, and <span class="math-container">$\nu_i = m_i/M$</span>.</p>

<p>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 <span class="math-container">$r_i$</span> as in the Newtonian case, the sides of the triangle are now given by (again see  <a href="https://arxiv.org/abs/1212.0754" rel="nofollow noreferrer">1212.0754</a>)</p>

<p><span class="math-container">$$ r_{ij} = a(1+\frac{M}{a}\epsilon_{ij}+ O(\tfrac{M^2}{a^2}) )$$</span></p>

<p>with</p>

<p><span class="math-container">$$\epsilon_{12} = \frac{1}{24}[(\nu_1-\nu_3)(5-3\nu_2)+(\nu_2-\nu_3)(5-3\nu_1)]$$</span>
<span class="math-container">$$\epsilon_{23} = \frac{1}{24}[(\nu_2-\nu_1)(5-3\nu_3)+(\nu_3-\nu_1)(5-3\nu_2)]$$</span>
<span class="math-container">$$\epsilon_{31} = \frac{1}{24}[(\nu_3-\nu_2)(5-3\nu_1)+(\nu_1-\nu_2)(5-3\nu_3)].$$</span></p>

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

CC BY-SA 4.0
2020-05-11T13:14:30.060Z
"In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses." This is amazing, thank you!
uhoh
CC BY-SA 4.0
2020-05-11T13:29:32.990Z
I've just asked [Where can I find Lagrange’s equilateral triangular solution for arbitrary masses?](https://hsm.stackexchange.com/q/11781/3869) in HSM SE.
uhoh
CC BY-SA 4.0
2020-05-11T12:15:40.720Z
2020-05-11T13:14:09.690Z
2020-05-11T13:14:09.690Z
TimRias
<p><a href="https://astronomy.stackexchange.com/a/33949/7982">This answer</a> to <em>Are the orbits of all triple star systems at least technically unstable?</em> mentions:</p>
<blockquote>
<p>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</p>
<p><span class="math-container">$$ \frac{m_1 m_2+ m_1 m_3 + m_2 m_3}{(m_1+m_2+m_3)^2} &lt; 1/27$$</span></p>
</blockquote>
<p>When the smallest mass approaches zero the three masses are at the vertices of an equilateral triangle. In a <em>realistic</em> solar system this means <a href="https://en.wikipedia.org/wiki/Jupiter_trojan" rel="nofollow noreferrer">Trojan asteroids</a> are generally found in orbits of massive planets like Jupiter about 60 degrees ahead-of and behind it.</p>
<p>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 <a href="https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem" rel="nofollow noreferrer">circular restricted three-body problem</a> orbit?</p>
<p>Is it still known to be an <a href="https://astronomy.stackexchange.com/questions/33947/are-the-orbits-of-all-triple-star-systems-at-least-technically-unstable#comment64686_33949">equilateral triangle, but they spin around a point which is not the center of the triangle, but is weighted toward the heavier object</a>?</p>
<ul>
<li>If so, can this be shown by citing a math-based reference or shown here mathematically or computationally?</li>
<li>If not, is there an expression for two angles of the triangle as a function of to mass ratios?</li>
</ul>
<p><a href="https://i.sstatic.net/iN0op.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iN0opm.png" alt="The inner Solar System, from the Sun to Jupiter, including asteroid belt (Hildas, Trojans and NEOs" /></a> <a href="https://en.wikipedia.org/wiki/File:InnerSolarSystem-en.png" rel="nofollow noreferrer">Source</a> click for full size</p>

<orbital-mechanics><mathematics><n-body-simulations>
What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?
uhoh
# What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?

(Asked by: username_0 on 2020-05-10)

[This answer](https://astronomy.stackexchange.com/a/33949/7982) 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](https://en.wikipedia.org/wiki/Jupiter_trojan) 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](https://en.wikipedia.org/wiki/Three-body_problem#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](https://astronomy.stackexchange.com/questions/33947/are-the-orbits-of-all-triple-star-systems-at-least-technically-unstable#comment64686_33949)?

* 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](https://i.sstatic.net/iN0opm.png)](https://i.sstatic.net/iN0op.png) [Source](https://en.wikipedia.org/wiki/File:InnerSolarSystem-en.png) click for full size

**Tags:** orbital-mechanics, mathematics, n-body-simulations


---

## 1 answers

### Answer by username_1 on 2020-05-11. Score: 3 | Accepted answer

These questions are answered by the same references listed in [my previous answer](https://astronomy.stackexchange.com/a/33949/7982).

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](https://arxiv.org/abs/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](https://arxiv.org/abs/1212.0754))

$$ r\_{ij} = a(1+\frac{M}{a}\epsilon\_{ij}+ O(\tfrac{M^2}{a^2}) )$$

with

$$\epsilon\_{12} = \frac{1}{24}[(\nu\_1-\nu\_3)(5-3\nu\_2)+(\nu\_2-\nu\_3)(5-3\nu\_1)]$$
$$\epsilon\_{23} = \frac{1}{24}[(\nu\_2-\nu\_1)(5-3\nu\_3)+(\nu\_3-\nu\_1)(5-3\nu\_2)]$$
$$\epsilon\_{31} = \frac{1}{24}[(\nu\_3-\nu\_2)(5-3\nu\_1)+(\nu\_1-\nu\_2)(5-3\nu\_3)].$$

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

> **Comments:**
> **username_0**: "In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses." This is amazing, thank you!  
> **username_0**: I've just asked [Where can I find Lagrange’s equilateral triangular solution for arbitrary masses?](https://hsm.stackexchange.com/q/11781/3869) in HSM SE.  

---

