{"id":2254,"date":"2026-05-29T22:00:00","date_gmt":"2026-05-30T02:00:00","guid":{"rendered":"https:\/\/underthehood.blogwyrm.com\/?p=2254"},"modified":"2026-05-26T21:26:33","modified_gmt":"2026-05-27T01:26:33","slug":"circular-restricted-three-body-problem-part-5-finding-the-equilibrium-points","status":"publish","type":"post","link":"https:\/\/underthehood.blogwyrm.com\/?p=2254","title":{"rendered":"Circular Restricted Three Body Problem \u2013 Part 5: Finding the Equilibrium Points"},"content":{"rendered":"\n

In the last two posts, we established the general structure of the pseudopotential and the fact that the allowed motions are restricted to certain spatial regions depending on the value of the Jacobi constant $C_J$.  In addition, we qualitatively argued for the existence of five equilibrium points without giving a hard-and-fast proof that they exist or a constructive method for finding them.  While having a heuristic understanding of why the points arise is important, ultimately not knowing how to find them is a serious gap.  This post remedies that deficiency by showing definitively how to find these equilibrium points.<\/p>\n\n\n\n

The condition of equilibrium means that if a test mass is placed at such a point with zero velocity (i.e., ${\\dot x} = 0$, ${\\dot y} = 0$, and ${\\dot z}=0$) then it must remain there.  This condition, in turn, implies that there are no forces on the test mass and that ${\\ddot x} = 0$, ${\\ddot y} = 0$, and ${\\ddot z} = 0$. <\/p>\n\n\n\n

Substituting these requirements into the equations of motion yields:<\/p>\n\n\n\n

\\[ \\frac{\\partial U}{\\partial x} \\equiv U_{,x} = 0 \\; , \\]<\/p>\n\n\n\n

\\[ \\frac{\\partial U}{\\partial y} \\equiv U_{,y} = 0 \\; , \\]<\/p>\n\n\n\n

and<\/p>\n\n\n\n

\\[ \\frac{\\partial U}{\\partial z} \\equiv U_{,z} = 0 \\; . \\]<\/p>\n\n\n\n

Thus, the process of finding the equilibrium points within a dynamical system transforms into a geometry problem of finding the critical points of the function $U(x,y,z)$ of three variables.<\/p>\n\n\n\n

The first equation to start with is $U_{,z} = 0$, which, when written in full, becomes<\/p>\n\n\n\n

\\[ z \\left[ \\frac{1-\\mu}{d_1^3} + \\frac{\\mu}{d_2^3} \\right] = 0 \\; .\\]<\/p>\n\n\n\n

Since $d_1$, $d_2$, and $\\mu \\leq \\frac{1}{2}$, are all positive quantities.  Note that we ignore the important but almost impossible to achieve condition that either $d_1$ or $d_2$ equal zero because when such a \u2018collision\u2019 occurs, the equations of motion are no longer well-defined.  Of course, this is only a mathematically possibility since a real particle will encounter a planetary surface or stellar atmosphere long before getting to the \u2018singularity\u2019 of $d_1=0$ or $d_2 = 0$.  Thus, we can conclude that $z = 0$, a result consistent with our qualitative argument for placing the equilibrium points in the $x$-$y$ plane.<\/p>\n\n\n\n

We now turn to the in-plane equations starting with the $y$ equation first to get:<\/p>\n\n\n\n

\\[ y \\left[ 1 – \\frac{1-\\mu}{d_1^3} – \\frac{\\mu}{d_2^3} \\right] = 0 \\; . \\]<\/p>\n\n\n\n

There are two choices: the quantity within the braces $[\\cdots] = 0$ or $y=0$.\u00a0 Let\u2019s examine them in order.<\/p>\n\n\n\n

Next, let\u2019s factor the $x$ equation to get<\/p>\n\n\n\n

\\[  x \\left[ 1 – \\frac{1-\\mu}{d_1^3} – \\frac{\\mu}{d_2^3} \\right] – \\mu \\frac{1-\\mu}{d_1^3} + \\mu \\frac{1-\\mu}{d_2^3} = 0 \\; . \\]<\/p>\n\n\n\n

By our assumptions, the first set of terms are equal to zero leaving us to conclude<\/p>\n\n\n\n

\\[ \\frac{1}{d_1^3} = \\frac{1}{d_2^3} \\equiv \\frac{1}{d^3} ; .\\]<\/p>\n\n\n\n

Returning to the $y$-equation and using this result in the bracketed-term, we find<\/p>\n\n\n\n

\\[ \\left( 1  – \\frac{1-\\mu}{d^3} – \\frac{\\mu}{d^3} \\right) = \\left( 1 – \\frac{1-\\mu+\\mu}{d^3} \\right) = 1 – \\frac{1}{d^3} = 0 \\; . \\]<\/p>\n\n\n\n

The only way to satisfy this condition is to set $d = 1$, which, when used in the definition of $d_1$ gives<\/p>\n\n\n\n

\\[ (x + \\mu)^2 + y^2 = 1 \\; \\]<\/p>\n\n\n\n

Likewise, the definition of $d_2$ gives<\/p>\n\n\n\n

\\[ (x+\\mu-1)^2 + y^2 = 1 \\; .\\]<\/p>\n\n\n\n

Combining gives<\/p>\n\n\n\n

\\[ (x+\\mu)^2 = (x + \\mu- 1)^2 \\; , \\]<\/p>\n\n\n\n

which only has a consistent solution if, when we take the square root of both sides, we pick the positive root on one and the negative root on the other.  Doing so immediately gives<\/p>\n\n\n\n

\\[ x = \\frac{1}{2} – \\mu \\; \\]<\/p>\n\n\n\n

and<\/p>\n\n\n\n

\\[ y = \\pm \\frac{\\sqrt{3}}{2} \\; .\\]<\/p>\n\n\n\n

These are the equilateral points L4 and L5 whose locations are off of the line joining the massives forming a equilateral triangle.<\/p>\n\n\n\n

The remaining three points come from going back to the choice we had to make after requiring $z=0$ and this time requiring $y=0$ as well.  All we now have to work with is the master $x$-equation that reads<\/p>\n\n\n\n

\\[ x – \\frac{(1-\\mu)(x+\\mu)}{|x+\\mu|^3} + \\frac{\\mu(1-\\mu-x)}{|x+\\mu-1|^3} \\; ,\\]<\/p>\n\n\n\n

where $d_1$ simplifies to $|x+\\mu|$ and $d_2$ to $|x+\\mu-1|$.\u00a0 \u00a0The master equation is actually three-fold based on the value of $x$; we get three different, specific equations for the three different regions shown in the figure below<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n

and defined by:<\/p>\n\n\n\n