# Creating a Gravitational 2 body simulation

Gold Member
I am trying to create a simulation for a gravitational 2 body problem.
But I am kind of having trouble to define the equations that can be solve numerically. From an inertial frame I defined the position of the two objects as the ##\vec{r_1}## and ##\vec{r_2}## with masses ##m_1## and ##m_2##.

Let the ##\vec{R}_{CM}## be the position of the CM of the objects. Now from the perspective of the CM, we can write position vectors of the objects in terms of ##\vec{r'}_1## and ##\vec{r'}_2##.

$$\vec{r'}_1 = \frac{-m_2}{m_1 + m_2} \vec{r}~~(1)$$

and $$\vec{r'}_2 = \frac{m_1}{m_1 + m_2} \vec{r}~~(2)$$where

##\vec{r}= \vec{r'}_2 - \vec{r'}_1##

Now in this case we can use the reduced mass and define the force on this mass. So we have,

##\vec{F} = \mu \ddot{\hat{r}} = -\frac{Gm_1m_2}{r^2} \hat{r}##

Now I need to solve this equation and put back into the (1) and (2) right ?

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Mentor
If you want to keep track of both position vectors separately then using the real gravitational force is more useful. The reduced mass is great if you want to treat it as one-body problem.

Gold Member
If you want to keep track of both position vectors separately then using the real gravitational force is more useful.
$$m_1\ddot{\mathbf{r}}_1=-\frac{Gm_1m_2(\mathbf{r}_1-\mathbf{r}_2)}{|\mathbf{r}_1-\mathbf{r}_2|^3}\tag{1}$$

$$m_2\ddot{\mathbf{r}}_2=-\frac{Gm_1m_2(\mathbf{r}_2-\mathbf{r}_1)}{|\mathbf{r}_2-\mathbf{r}_1|^3}\tag{2}.$$

I am new at this topic and in the above equations the left side has ##r1## but right has ##r2## and ##r1##, so It seemed harder for me to solve it in this way. Thats kind of why I tried to use reduced mass.

DrStupid
I am new at this topic and in the above equations the left side has ##r1## but right has ##r2## and ##r1##, so It seemed harder for me to solve it in this way.

It seems but it isn't. Even using the reduced mass you actually have three equations - one for each component of the displacement vector. In each of these equations you have one component on the left side but all three components on the right side. With separate positions you have 6 instead of three equations but the basic principle doesn't change.

• Arman777
Gold Member
It seems but it isn't. Even using the reduced mass you actually have three equations - one for each component of the displacement vector. In each of these equations you have one component on the left side but all three components on the right side. With separate positions you have 6 instead of three equations but the basic principle doesn't change.
I see...I guess before jumping into these topics I should focus on the computational physics part, solving DE equations on computer.

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