Another Perturbed Orbit

Position of particle 1 at a general time is $\vec{r}_1=(x_1,y_1)$ and velocity is $\vec{v}_1=(\dot{x}_1,\dot{y}_1)$ . Position of particle 2 at a general time is $\vec{r}_2=(x_2,y_2)$ and velocity is $\vec{v}_2=(\dot{x}_2,\dot{y}_2)$ . Potential energy of the system is:

$V = -Gm_1m_2\left(\frac{1}{\lvert \vec{r}_2-\vec{r}_1\rvert} \right)$

The third term can be ignored as well as that is a constant. Kinetic energy of the system is:

$T = \frac{m_1}{2}\left(\dot{x}_1^2 + \dot{y}_1^2\right) + \frac{m_2}{2}\left(\dot{x}_2^2 + \dot{y}_2^2\right)$

Applying Lagrangian mechanics:

$\ddot{x}_1 = -\frac{1}{m_1}\frac{\partial V}{\partial x_1}$ $\ddot{y}_1 = -\frac{1}{m_1}\frac{\partial V}{\partial y_1}$ $\ddot{x}_2 = -\frac{1}{m_2}\frac{\partial V}{\partial x_2}$ $\ddot{y}_2 = -\frac{1}{m_2}\frac{\partial V}{\partial y_2}$

This system is solved numerically without explicitly crunching any derivatives, as demonstrated in the previous version of this problem of perturbed orbits.

The trajectory of the particle is:

The plot compared the trajectory if $m_1$ is fixed to the case where the orbit is perturbed. So, if the mass of $m_1 >> m_2$ the mass $m_1$ can be approximated to be fixed.