$$\def\|{&}$$

## 1. The Positions

If two celestial bodies orbit around a common center of gravity, then there are five points that move with the bodies and in which the gravity of the two bodies and the centripetal force of the orbit of the point around the center of gravity are exactly balanced. Those five points are called the Lagrange points, after Mr. Lagrange who first calculated them. If a small and light (not massive) spacecraft is exactly in such a point and has exactly the right speed in just the right direction, then such a spacecraft can stay at that point without using its engine, so the Lagrange points can be interesting locations for satellites that must have a fixed orientation relative to the two celestial bodies.

The five Lagrange points can be divided into two groups. The first three points (L₁ through L₃) lie on the straight line that passes through the center of the two celestial bodies. The last two points (L₄ and L₅) form equilateral triangles with the two celestial bodies, and are equally far from each other and from each of the two celestial bodies. Here we derive the location of the first three Lagrange points in the case where the object in such a point has a negligible mass and the orbits of the two celestial bodies are circles.

The positions are determined by the solutions of the equation

$$x - \frac{(1 - μ)(x - μ)}{|x - μ|^3} - \frac{μ(x + 1 - μ)}{|x + 1 - μ|^3} = 0 \label{eq:hoofd}$$

where $$x$$ is the position on the line through the two celestial bodies in units of the separation between the two bodies, and $$μ$$ (mu) is the mass of the secondary (less massive) component relative to the total mass in the system. $$μ$$ is always between 0 and ½. The primary (most massive) object is at $$x = μ$$, and the secondary object is at $$x₂ = μ - 1$$.

We deal with the absolute values by replacing them by the original expressions multiplied by a factor which is either +1 or −1, depending on the sign of the expression. For instance, $$|x|$$ would be replaced by $$s×x$$, and $$s$$ can afterward be assigned the value −1 if $$x$$ is negative, or +1 if $$x$$ is positive or zero. Such substitutions lead to an equivalent equation which is of fifth order in $$x$$ and fourth order in $$μ$$, after discarding the common denominator (which is only zero at the positions of either of the two heavy bodies, and those positions are excluded anyway on physical grounds):

\begin{align} \| -s_2 - 2 s_2 x - s_2 x^2 + s_1 s_2 x^3 + 2 s_1 s_2 x^4 + s_1 s_2 x^5 \notag \\ \| + μ (3 s_2 + 4 s_2 x + (-s_1 + s_2 - 2 s_1 s_2) x^2 - 6 s_1 s_2 x^3 - 4 s_1 s_2 x^4) \notag \\ \| + μ^2 (−3 s_2 + (2 s_1 - 2 s_2 + s_1 s_2) x + 6 s_1 s_2 x^2 + 6 s_1 s_2 x^3) \notag \\ \| + μ^3 (-s_1 + s_2 - 2 s_1 s_2 x - 4 s_1 s_2 x^2) + μ^4 s_1 s_2 x = 0 \end{align}

where $$s_1$$ is the sign of $$x - μ$$ and $$s_2$$ is the sign of $$x + 1 - μ$$.

We now proceed to find the approximate solutions for $$x$$ to this fifth-order equation by assuming that $$μ$$ is small compared to 1. In this case, the zeroth-order term (i.e., without any $$x$$) $$-s_2$$ is the dominant one and it must be compensated if the whole equation is to remain balanced. A shift in origin of $$x$$ may do the trick.

If we replace $$x$$ by $$a + ε$$, and gather all zeroth-order terms in both $$ε$$ (epsilon) and $$μ$$ in the result, then we find:

$$-s_2 (a + 1)^2 + a^3 s_1 s_2 (a + 1)^2 = 0$$

which has solutions $$a = −1$$, and $$a = +s_1$$, so solutions to our problem are found for $$x$$ near $$a = −1$$ and $$a = +1$$ (the latter only if $$s_1 = +1$$, i.e., if $$x \gt μ$$).

First, we investigate the solutions near $$x = +1$$. We replace $$x$$ by $$1 + ε$$ in equation \ref{eq:hoofd} and assume that $$ε$$ is much smaller than 1. In that case, both $$s_1$$ and $$s_2$$ are equal to +1. Because both $$ε$$ and $$μ$$ are assumed to be numerically smaller than 1, the main contribution to the result comes from the terms which have the lowest orders in $$μ$$ and $$ε$$ (i.e., which have the fewest factors of $$μ$$ and $$ε$$). When we retain only the dominant terms in $$μ$$ and $$ε$$, we find $$ε = \frac{5}{12} μ$$, which satisfies the condition on $$x$$ that was found earlier ($$x \gt μ$$), so our first approximate position of a Lagrange point (actually the third Lagrange point, L₃) is $$x = 1 + \frac{5}{12} μ$$.

We now assume that the solution is of the form $$x = 1 + a_1 μ + a_2 μ^2 + a_3 μ^3 + a_4 μ^4 + \text{smaller terms}$$, substitute this into equation \ref{eq:hoofd} (with $$s_1 = s_2 = +1$$), set the coefficients of each order of $$μ$$ equal to zero, and solve for $$a_1$$ through $$a_4$$. We then find $$a_1 = \frac{5}{12}$$, $$a_2 = 0$$, $$a_3 = -\frac{1127}{20736}$$, $$a_4 = -\frac{7889}{248832}$$. Because the primary object is at $$x = μ$$, the position of L₃ relative to the primary is (to fourth order in $$μ$$) at

$$L_3 = 1 - \frac{7}{12} μ - \frac{1127}{20736} μ^3 - \frac{7889}{248832} μ^4$$

Next, we turn our attention to the solutions near $$x = −1$$. The solution $$x = −1$$ was obtained for $$μ = ε = 0$$, and coincides with the position of the secondary component. We are interested in the position of the Lagrange points relative to the secondary component so we replace $$x$$ by $$−1 + μ + ε$$ in equation \ref{eq:hoofd}. This substitution is consistent with $$x = −1$$ for $$μ = ε = 0$$. By retaining only the dominant terms involving $$μ$$ and $$ε$$ we discover the approximate solution $$ε = \left( \frac{μ}{3} \right)^{1/3}$$. We define $$z = \left( \frac{μ}{3} \right)^{1/3}$$, rewrite $$μ = 3 z^3$$, and expand $$ε$$ in terms of $$z$$ as $$ε = a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 + \text{smaller terms}$$. Substituting this in equation \ref{eq:hoofd} leads to an equation involving the $$a$$-coefficients and $$z$$ up to 6th order. By equating the coefficients of each order of $$z$$ with zero and solving for the $$a$$-coefficients we find that, for $$s_2 = 1$$, the only solution with all real values is that $$a_1 = 1$$, $$a_2 = -\frac{1}{3}$$, $$a_3 = -\frac{1}{9}$$, $$a_4 = \frac{58}{81}$$. For $$s_2 = −1$$, we find $$a_1 = −1$$, $$a_2 = -\frac{1}{3}$$, $$a_3 = \frac{1}{9}$$, $$a_4 = -\frac{50}{81}$$. These values are consistent with the conditions for $$s_2 = 1$$ and $$s_2 = −1$$, respectively, so the distances of the first and second Lagrange points to the secondary component, are

\begin{align} L_1 \| = z - \frac{1}{3} z^2 - \frac{1}{9} z^3 + \frac{58}{81} z^4 \\ L_2 \| = z + \frac{1}{3} z^2 - \frac{1}{9} z^3 + \frac{50}{81} z^4 \end{align}

with

$$z = \left( \frac{μ}{3} \right)^{1/3}$$

## 2. The Stability

What happens if you move a small distance away from a Lagrange point? If you then yet (without using propulsion) remain in the neighborhood of the Lagrange point for all eternity, then the Lagrange point is table. If you eventually get arbitrarily far away from the Lagrange point, then the Lagrange point is unstable. To be able to determine the stability of Lagrange points even to a first (linear) approximation, you need to use differential calculus and complex numbers, and it is too much work to explain those here. I'll provide the main argument and give the results.

The calculation goes roughly like this: First you determine what forces act if you move a small distance away from the Lagrange point in an arbitrary direction, and you ignore all terms that are of second or higher order in the distance to the Lagrange point. Then you try a standard solution with a separate starting amplitude for each coordinate but with the same exponential growth speed for all coordinates (which may be a complex number).

If the growth speed turns out to have an imaginary part, then the solution orbits around the Lagrange point. If the growth speed has a negative real part, then the distance to the Lagrange point decreases and the solution is stable. if the growth speed has a positive real part, then the Lagrange point is unstable. If the growth speed has a real part equal to zero, then the distance to the Lagrange point remains limited and then the Lagrange point is stable.

### 2.1. L₄ and L₅

For Lagrange points L₄ and L₅, the motion perpendicular to the orbital plane is an oscillation with a period equal to the period of the Kepler orbits of the two objects and the Lagrange point. For the motion within the orbital plane, the possible growth speeds $$λ$$ are solutions of

$$λ^4 + λ^2 + \frac{27}{4} μ(1 - μ) = 0$$

This has complex solutions that are unstable, and purely imaginary solutions that are stable (according to the linear analysis). The boundary is at $$μ = μ_0 = \frac{1}{2} - \sqrt{\frac{23}{108}} ≈ 0,0385209 ≈ \frac{1}{25.95993}$$: For greater values of $$μ$$ L₄ and L₅ are unstable, and for smaller values they are stable. The next table shows solutions for a few values of $$μ$$. $$t_2$$ is the time during which the distance to the Lagrange point doubles, in units of the period of the system. $$t_+$$ and $$t_-$$ are the two orbital periods that belong to the solution, also in units of the period of the system. The other two columns show certain combinations of the times and $$μ$$ which for very small values of $$μ$$ approach limiting values. Where $$t_2$$ is missing its value is infinite; that means that the solution is stable (in the linear analysis).

Table 1: Lagrange Point Periods

$${μ}$$$${t_2}$$$${t_+}$$$${t_+^2μ}$$$${t_-}$$$${(1-t_-)/μ}$$
0.5 2.282 1.054 0.5559 1.054 0.1087
0.4 2.321 1.062 0.4513 1.062 0.1554
0.3 2.455 1.088 0.3549 1.088 0.2922
0.2 2.778 1.140 0.2599 1.140 0.6995
0.15 3.120 1.184 0.2101 1.184 1.224 Pluto - Charon
0.1 3.860 1.250 0.1563 1.250 2.503
0.05 7.928 1.370 0.09379 1.370 7.392
0.04 21.368 1.408 0.07928 1.408 10.20
0.039 37.410 1.412 0.07777 1.412 10.57
0.0385209 1.414 0.07704 1.414 10.75
0.035 1.686 0.09954 1.242 6.911
0.03 1.930 0.1117 1.169 5.641
0.02 2.524 0.1274 1.089 4.455
0.0123 3.331 0.1365 1.048 3.932 Earth - Moon
0.01 3.727 0.1389 1.038 3.807
0.001 12.136 0.1473 1.003 3.412
0.000955 12.421 0.1473 1.003 3.410 Sun - Jupiter
0.0001 38.479 0.1481 1.000 3.379
10−05 121.713 0.1481 1.000 3.375
3 × 10−06 222.220 0.1481 1.000 3.375 Sun - Earth
0
       ∞
4/27 1 −27/8

The $$t_+$$ and $$t_-$$ for $$μ = μ_0$$ are equal to $$\sqrt{2}$$.

For example, the Moon has 0.0123 times as much mass as the Earth, so in the Earth-Moon system $$μ$$ is equal to 0.0123. This is less than $$μ_0$$, so L₄ and L₅ in the Earth-Moon system are stable. The orbital periods near those Lagrange points are equal to 3.331 and 1.048 times the period of the system, i.e., that many sidereal months. At a small distance from such a Lagrange point, you'll float around that point with a combination of both periods. In our Solar System, only the Pluto-Charon system has a mass ratio that is so great that their L₄ and L₅ are linearly unstable.

The linear stability analysis assumes that a small change in the initial situation yields similarly small changes in the final situation. This is not necessarily the case if the one orbital period is an exact multiple of the other one, because then you can get a "swing effect". You can make somebody on a swing swing well only if you given her a push just as she starts swinging forward again, i.e., if your push period is an exact multiple of her swing period. Even though you only provide a small push every time, the swing eventually swings very far, because the effect of all of the small pushes adds up. If instead you sometimes give a push when the swing goes forward, and sometimes when the swing goes backward, then the effect of all of those pushes will average out to zero. That happens when your push period is not an exact multiple of the swing period.

The results of the linear stability analysis can therefore not be trusted if the one orbital period is an exact multiple of the other one. A non-linear stability analysis shows that the Lagrange points are yet unstable if the ratio of the orbital periods is 2 or 3, and that is when $$μ$$ is equal to $$\frac{1}{2} - \frac{1}{90}\sqrt{1833} ≈ 0.0243$$ or $$\frac{1}{2} - \frac{1}{30}\sqrt{213} ≈ 0.0135$$.