# How are the Lagrange points determined?

According to Hyper Physics, there are 5 equilibrium, or Lagrange points of the Earth-Moon system and only 2 of them are said to represent stable equilibrium points.

This made me think if there is an equation that describes this system, and from which Physics Laws was it derived?

Sketched proof of all possible Lagrange points:

1. Consider first the 2-body problem. Deduce that possible Lagrange points must lie in the orbital plane (because a probe will always be gravitationally attracted towards the orbital plane). So from now on we restrict attention to the orbital plane, which we identify with the complex plane $$\mathbb{C}$$.

2. Consider the 2-body problem with circular orbits for simplicity. Let $$R$$ be the fixed distance between the 2 point masses $$m_1$$ and $$m_2$$. Go to the rotating center-of-mass coordinate system, where the point masses $$m_1$$ and $$m_2$$ are fixed at positions $$r_1~=~-\epsilon_2 R ~<~0\qquad\text{and}\qquad r_2~=~\epsilon_1 R~>~0 \tag{1}$$ along the real axis, where $$\epsilon_1~:=~ \frac{m_1}{m_1+m_2}~>~0,\qquad \epsilon_2~:=~ \frac{m_2}{m_1+m_2}~>~0,$$ $$\epsilon_1+\epsilon_2~=~1.\tag{2}$$

3. Deduce that that a test mass at position $$z\in\mathbb{C}$$ experiences an acceleration $$a~=~- \frac{Gm_1}{z^{\ast}_1|z_1|} - \frac{Gm_2}{z^{\ast}_2|z_2|} +\Omega^2 z,\tag{3}$$ from gravity and the centrifugal force, where $$z_1~:=~ z-r_1~\neq~0,\qquad z_2~:=~ z-r_2~\neq~0,$$ $$\Omega^2~:=~\frac{G(m_1+m_2)}{R^3}.\tag{4}$$

4. Deduce that the equation $$a=0$$ for Lagrange points is $$\frac{z}{R^3}~=~\frac{\epsilon_1}{z^{\ast}_1|z_1|} +\frac{\epsilon_2}{z^{\ast}_2|z_2|}~\equiv~\frac{\epsilon_1z_1}{|z_1|^3} +\frac{\epsilon_2z_2}{|z_2|^3} ,\tag{5}$$ or equivalently, $$z\underbrace{\left(\frac{\epsilon_1}{|z_1|^3}+ \frac{\epsilon_2}{|z_2|^3}-\frac{1}{R^3}\right)}_{\in~\mathbb{R}} ~\stackrel{(5)}{=}~ \underbrace{\epsilon_1\epsilon_2 \left(\frac{1}{|z_2|^3}-\frac{1}{|z_1|^3} \right)}_{\in~\mathbb{R}}. \tag{6}$$

5. The only way that $$z$$ on the lhs. of eq. (6) could be a non-real number is if the two parentheses in eq. (6) are both zero. This is the condition that the 3 bodies form an equilateral triangle $$|z_1|~=~R~=~|z_2|. \tag{7}$$ Eq. (7) has 2 solutions, namely the Lagrange points $$L_4$$ and $$L_5$$: $$\begin{array}{rcccl} z_1~&=&~R\exp\left\{\pm \frac{i\pi}{3} \right\} ~&=&~\frac{R}{2}\pm\frac{\sqrt{3}iR}{2}, \cr z_2~&=&~-R\exp\left\{\mp \frac{i\pi}{3} \right\} ~&=&~-\frac{R}{2}\pm\frac{\sqrt{3}iR}{2} . \end{array}\tag{8}$$

6. Hence we may (and will) assume from now on that $$z\in\mathbb{R}$$ is real, i.e. that the 3 bodies are collinear. Then eq. (5) becomes a 5th order equation, whose roots generically have no closed exact formula. Since the derivative $$\frac{da}{dz}~\stackrel{(3)}{=}~\frac{2Gm_1}{|z_1|^3} + \frac{2Gm_2}{|z_2|^3}+\Omega^2 ~>~0\tag{9}$$ is positive for $$z\in\mathbb{R}\backslash\{r_1,r_2\}$$, there can at most be one root in each of the continuous intervals $$]-\infty,r_1[, \qquad ]r_1,r_2[ \qquad\text{and}\qquad ]r_2,\infty[.\tag{10}$$ Hence the equation $$a=0$$ has at most 3 real roots. Further analysis reveals that the equation $$a=0$$ has exactly 3 real roots $$L_1$$, $$L_2$$ & $$L_3$$. See e.g. Ref. 1 and Wikipedia for further details.

7. For the stability question, see e.g. this and this Phys.SE posts. $$\Box$$

References:

1. J. Binney & S. Tremaine, Galactic Dynamics, 2nd edition (2008); p. 676.
• Can you tell me what is happening in eq6? where does the rhs come from? – Coolcrab Mar 1 at 22:36
• I updated the answer. – Qmechanic Mar 1 at 23:05