# Lagrange points (regions) for $n$-body systems

The title may be a bit misleading. They may not be points (small areas) at all in this case, but extended regions. Mathematically speaking:

Consider the origin for measurements at $O$. Let $R(t)$ = {$r_i ; i \in [1,n]$} denote the time varying set of radius vectors of the $n$ bodies, and let $M$ = {$m_i; i\in [1,n]$} denote their masses. Is it theoretically possible to determine the set of Lagrange points (or regions) $L(t) = {L_k}$ such that a satellite placed at one of these $L_k$ is fixed w.r.t. some subset of three bodies among the $n$ bodies? I can attempt a brute force programming computation for simulation purposes, but a mathematical approach is appreciated. Even proving the existence of such points will be considered a great help. I honestly have no idea how to proceed with a generalised $n$-body system.

• No certitude, but I have the feeling that if one could theoretically determine the set of Lagrange points for $n$ bodies, one wouldn't be very far of determining the $n$ bodies trajectories, which we know to be impossible in the general case. For the existence, did you take a look at the various saddle point existence theorems? (I'm not telling that all Lagrange points are saddle points)
– user130529
Nov 1, 2016 at 18:39
• I'll recheck them, but i don't recall one related that can be paricularly useful here. Any suggestions in particular which to see? Nov 2, 2016 at 9:07

## 1 Answer

1) Regarding determination. No certitude, but I have the feeling that if one could theoretically determine the set of Lagrange points for $n$ bodies, one wouldn't be very far of determining the $n$ bodies trajectories, which we know to be impossible in the general case.

2) Regarding existence. for fixed $t$, the potential $\Omega(r)$ issued from gravity + centrifugal force reads $$\Omega(r) = -\sum_{i=1}^{n} \frac{Gm_i}{|r-r_i|}-\frac{1}{2}|r|^2 \omega^2$$ in a suitable frame rotating with the $n$ bodies with a non zero angular velocity $\omega$.

To prove existence, it is enough to prove that $\Omega$ has a maximum somewhere. Take an arbitrary point $r_a$ different from the $r_i$'s, and let $\Omega_a=\Omega(r_a)$. Let $$K = \{r; \ \Omega(r) \geq \Omega_a-1 \}.$$ Because the function $\Omega(r)$ tends to $-\infty$ when $|r| \to \infty$ or $r \to r_i$, $K$ is compact and $\Omega$ is continuous on $K$, hence there exists an $r_L \in K$ where $\Omega$ reaches its maximum. Moreover, as $\Omega(r) = \Omega_a-1$ on the boundary of $K$, and $\Omega(r_L) \geq \Omega(r_a) > \Omega_a-1$, $r_L$ belongs to the interior of $K$. Hence $r_L$ is a local maximum of the function $\Omega$ on the whole space minus the $r_i$'s. Hence it is a Lagrange point (of type L$_4$ - L$_5$). Using the regularity of $\Omega$ and the implicit functions theorem, you can follow this point with the time for a certain time if it is not degenerate.

• Just a doubt: by your definition of $K$, i can choose $r_a$ to be arbitrarily close to some $r_i$, this would mean that the set K can be made to consist of arbitrarily small regions. Dosent this show that an arbitrarily small region of space must contain a maxima? Nov 3, 2016 at 15:32
• It would be rather the contrary: the closer $r_a$ to some $r_i$, the larger would usually become $K$. Indeed, with evident notation, if $r_b$ is closer to some $r_i$ than $r_a$ and $\Omega(r_b) < \Omega(r_a)$ (which would usually be expected, although not necessary), then by definition $K_a \subset K_b$, and $K_b$ is larger.
– user130529
Nov 3, 2016 at 16:10