Logbook

2/28/2025

After re-reading the assignment details, I realized that the three body motion was meant to be modeled after an orbital system with a star, planet and moon, not 3 similiarly massive objects. I adjusted the initial masses appropriately, and solved for orbital motion using the following relation due to the large difference in masses.

$$v=\sqrt{\frac{GM}{r}}$$

Placing the star in a stationary position in the middle of the system, we can solve for the orbital speed of the planet and moon. The values in the briefing consist of the following:

$$\{\begin{array}{ll}{r}_{12}& :1\\ r_{23}& :0.0025\\ m_1& :1\\ m_2& :3\times 10^{-6}\\ m_3& :3.6\times 10^{-8}\end{array}$$

The simplest method to calculate the velocities would consist of placing all three bodies in a vertical line, and calculating the relative velocities. This yields the following:

$$v_{12}=\sqrt{\frac{1\times 1}{1}}=1$$ $$v_{23}=\sqrt{\frac{1\times 3\times 10^{-6}}{0.0025}}\approx 0.03464$$

And adding them together we get the total orbital velocity of body 3 to be 1.03464. These values yielded a stable 3 body orbital motion.

$${\vec{\mathbf{F}}}_{12}=\frac{G{m}_1{m}_2}{|{\mathbf{r}}_{12}{|}^3}{\mathbf{r}}_{12}$$ $$y_{n+1}=y_n+\Delta t\frac{dy}{dt}f(t_n,y_n)$$ $$F=ma$$ $$a=\frac{dv}{dt}=\frac{d{x}^2}{d^{2}t};v=\frac{dx}{dt}$$ $$x(t+\Delta t)=x(t)+\Delta t\cdot v(t)+\Delta t^2\frac{F(t)}{2m}$$ $$v(t+\Delta t)=v(t)+\Delta t\frac{F(t)+F(t+\Delta t)}{2m}$$