Simulation of the three body problem
Using the equations $r_{ij}$ for the distances and $a_{direction, i}$ for acceleration.
$$\begin{matrix}
r_{12} = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \\\
r_{13} = \sqrt{(x_1-x_3)^2+(y_1-y_3)^2} \\\
r_{23} = \sqrt{(x_2-x_3)^2+(y_2-y_3)^2} \\\
a_{x1} = -\frac{m_2}{r_{12}^3}(x_1-x_2)-\frac{m_3}{r_{13}^3}(x_1-x_3) \\\
a_{y1} = -\frac{m_2}{r_{12}^3}(y_1-y_2)-\frac{m_3}{r_{13}^3}(y_1-y_3) \\\
a_{x2} = -\frac{m_1}{r_{12}^3}(x_2-x_1)-\frac{m_3}{r_{23}^3}(x_2-x_3) \\\
a_{y2} = -\frac{m_1}{r_{12}^3}(y_2-y_1)-\frac{m_3}{r_{23}^3}(y_2-y_3) \\\
a_{x3} = -\frac{m_1}{r_{13}^3}(x_3-x_1)-\frac{m_2}{r_{23}^3}(x_3-x_2) \\\
a_{y3} = -\frac{m_1}{r_{13}^3}(y_3-y_1)-\frac{m_2}{r_{23}^3}(y_3-y_2) \\\
\end{matrix}$$
with the inital conditions:
$m_1=10$ and $m_2=m_3=1$
$$
\begin{bmatrix}
x \\
y \\
\end{bmatrix} \in \left\lbrace
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix},
\begin{bmatrix}
-0.55 \\
0 \\
\end{bmatrix},
\begin{bmatrix}
0.5 \\
0 \\
\end{bmatrix} \right\rbrace
$$
$$
\begin{bmatrix}
v_x \\
v_y \\
\end{bmatrix} \in \left\lbrace
\begin{bmatrix}
0 \\
-0.6 \\
\end{bmatrix},
\begin{bmatrix}
0.6 \\
0.6 \\
\end{bmatrix},
\begin{bmatrix}
-0.6 \\
0.6 \\
\end{bmatrix}\right\rbrace
$$
