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:
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 $\mathbb{C}$.
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 (CM) 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 $$\begin{align} \epsilon_1~:=~&\frac{m_1}{m_1+m_2}~>~0,\cr \epsilon_2~:=~& \frac{m_2}{m_1+m_2}~>~0, \cr \epsilon_1+\epsilon_2~=~&1.\end{align}\tag{2}$$
m_1 CM m_2
------|-------------|----------------|-----------> z
r_1 0 r_2
| |
|<--------------R------------->|
$\uparrow $ Fig. 1: The positions $r_1$ and $r_2$ of the masses $m_1$ and $m_2$.
The gravitational force on $m_2$ must cancel the centrifugal force on $m_2$: $$ \frac{Gm_1m_2}{R^2} ~=~ m_2\Omega^2 r_2 \qquad\Rightarrow\qquad \Omega^2~=~\frac{G(m_1+m_2)}{R^3}.\tag{3}$$ This determines the angular velocity $\Omega$ of the coordinate system.
Deduce that that a test mass at position $z\in\mathbb{C}$ experiences an acceleration $$ a~=~ -\frac{Gm_1z_1}{|z_1|^3} -\frac{Gm_2z_2}{|z_2|^3}+\Omega^2 z,\tag{4}$$ from gravity and the centrifugal force, where we defined the relative positions $$z_1~:=~ z-r_1~\neq~0\qquad\text{and}\qquad z_2~:=~ z-r_2~\neq~0.\tag{5}$$
Deduce that the equation $$a~=~0\tag{6}$$ for Lagrange points is $$ \frac{z}{R^3} ~\stackrel{(2)+(3)+(4)+(6)}{=}~\frac{\epsilon_1z_1}{|z_1|^3} +\frac{\epsilon_2z_2}{|z_2|^3} ,\tag{7}$$ or equivalently, $$\begin{align}z~&\overbrace{\left(\frac{\epsilon_1}{|z_1|^3}+ \frac{\epsilon_2}{|z_2|^3}-\frac{1}{R^3}\right)}^{\in~\mathbb{R}} \cr ~\stackrel{(1)+(5)+(7)}{=}& \underbrace{\epsilon_1\epsilon_2R \left(\frac{1}{|z_2|^3}-\frac{1}{|z_1|^3} \right)}_{\in~\mathbb{R}}. \end{align}\tag{8}$$
The only way that $z$ on the lhs. of eq. (8) could be a non-real number is if the two parentheses in eq. (8) are both zero. This is the condition that the 3 bodies form an equilateral triangle $$ |z_1|~=~R~=~|z_2|. \tag{9} $$ Eq. (9) has 2 solutions, namely the Lagrange points $L_4$ and $L_5$: $$ \begin{array}{rcccl} z_1~&=&~R\exp\left\{\pm \frac{i\pi}{3} \right\} ~&=&~\frac{R}{2}\pm\frac{\sqrt{3}iR}{2}, \cr z_2~&=&~-R\exp\left\{\mp \frac{i\pi}{3} \right\} ~&=&~-\frac{R}{2}\pm\frac{\sqrt{3}iR}{2} . \end{array}\tag{10} $$
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. (7) becomes a 5th order equation, whose roots generically have no closed exact formula. Since the derivative $$\frac{da}{dz}~\stackrel{(4)}{=}~\frac{2Gm_1}{|z_1|^3} + \frac{2Gm_2}{|z_2|^3}+\Omega^2 ~>~0\tag{11}$$ 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 $$]-\infty,r_1[, \qquad ]r_1,r_2[ \qquad\text{and}\qquad ]r_2,\infty[.\tag{12} $$ Hence the equation $a=0$ has at most 3 real roots. The behavior of the function $a$ near the singularities $z\in\{-\infty,r_1,r_2,\infty\}$ reveals that the equation $a=0$ has exactly 3 real roots $L_1$, $L_2$ & $L_3$, cf. Fig. 2. See e.g. Ref. 1 and Wikipedia for further details.
$\uparrow $ Fig. 2: An example of the acceleration $a$ as a function (4) of the position $z$. The function $a$ has singularities at the positions $z\in\{-\infty,r_1,r_2,\infty\}$. The slope (11) is positive everywhere. There are always exactly 3 real roots $L_1$, $L_2$ & $L_3$.
For the stability question, see e.g. this and this Phys.SE posts. $\Box$
References:
The Lagrange points are positions where another object can orbit the sun with the same period as the earth. (L1 would be a good place to park an asteroid to block some of the heat from the sun.) Lets assume that the earth and moon act like a single combined mass at the center of mass. One might assume that an object at L1 (a smaller orbit around the sun) would move faster than the earth, but as long as it stays in line with the earth, the gravity from the earth offsets the extra pull from the sun. Similarly, at L2 and L3, the pull of the earth works with that from the sun. At L4 and L5, it is the vector sum of the two forces which determines the orbit. (See the answer from Qmechanic for formulas.)
