Deriving position $\mathbf{r}$ in two-body problem with composite In classical mechanics, we know that accelerations are oppositely directed and inversely proportional to the masses:
$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2.$$
Let's say that we have a three-body system, where none of the masses of the bodies are equal. If two of the bodies (say, body 2 and 3) form a composite, then, since $m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2$ must remain true, how do we know what the position $\mathbf{r}$ of the composite is?
I am told that it must be $\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}$, but I don't understand how this can be derived from the basic equations of classical mechanics.
If I had to guess, I'd say that, since acceleration is the second derivative of position, we get something like
$$m_1 \ddot{\mathbf{r}} = -(m_2 + m_3) \ddot{\mathbf{r}} \Rightarrow -\dfrac{m_1}{m_2 + m_3}\ddot{\mathbf{r}} = \ddot{\mathbf{r}}.$$
This looks similar to the correct derivation, so I'm guessing that I'm somewhat on the right track.
I would greatly appreciate it if people would please take the time to explain how this is derived from the basic equations of classical mechanics.
 A: The two composite bodies, can be reduced down to one body with combined mass, and the location where forces are applied being the barycenter (the center of mass). Then we have the following two body problem.
Consider two free massive bodies located at $\boldsymbol{r}_1$ and $\boldsymbol{r}_2$ at some moment in time. A central force exists $\boldsymbol{F}(r)$ that is applied in equal and opposite measure to each body, that depends only on the distance $r$. The equations of motion are
$$ \begin{aligned}
 m_1 \ddot{\boldsymbol{r}}_1 & = -\boldsymbol{F}(r) \\
 m_2  \ddot{\boldsymbol{r}}_2 & = \boldsymbol{F}(r)
\end{aligned} \tag{1}$$
where $r = \sqrt{ (\boldsymbol{r}_2-\boldsymbol{r}_1) \cdot ( \boldsymbol{r}_2 - \boldsymbol{r}_1)} \tag{2} $
This is a hard problem to solve, as $r$ depends on the two positions.
Now let us do a change of variables. I am going to pick two vector quantities, and then I am going to show how this specific choice diagonalizes the equations of motion and makes them solvable.
$$\begin{aligned}
  \boldsymbol{\Delta r} & = \boldsymbol{r}_2 - \boldsymbol{r}_1 \\
  \boldsymbol{r}_C & = \frac{m_1 \boldsymbol{r}_1 + m_2 \boldsymbol{r}_2}{m_1 +m_2}
\end{aligned} \tag{3} $$
The first is the vector separation and the second is the center of mass. You can immediately restate the distance as $r=\sqrt{ \boldsymbol{\Delta r} \cdot \boldsymbol{\Delta r} }$.
Now use (3) to solve for the position vectors
$$ \begin{aligned} 
 \boldsymbol{r}_1 & = \boldsymbol{r}_C + \frac{m_1}{m_1+m_2} \boldsymbol{\Delta r} \\
 \boldsymbol{r}_2 &= \boldsymbol{r}_C - \frac{m_1}{m_1+m_2} \boldsymbol{\Delta r}
\end{aligned} \tag{4}$$
Now consider the accelerations
$$ \begin{aligned} 
 \ddot{\boldsymbol{r}}_1 & = \ddot{\boldsymbol{r}}_C + \frac{m_1}{m_1+m_2} \ddot{\boldsymbol{\Delta r}} \\
 \ddot{\boldsymbol{r}}_2 &= \ddot{\boldsymbol{r}}_C - \frac{m_1}{m_1+m_2} \ddot{\boldsymbol{\Delta r}} \end{aligned} \tag{5} $$
Use (5) in (1) and solve for $\ddot{\boldsymbol{r}}_C$ and $\dot{\boldsymbol{\Delta r}}$. The solution is
$$ \begin{aligned} \ddot{ \boldsymbol{r}}_C & = 0  \\
 \ddot{\boldsymbol{\Delta r}} &= \left( \tfrac{1}{m_1} + \tfrac{1}{m_2} \right) F(r)
\end{aligned} \tag{6}$$
So half the solution is the center of mass moves with constant velocity. $\dot{\boldsymbol{r}}_C = \text{(const.)}$. We just found the special point in space which helps us solve this problem.
The rest of the solution is really an equation only in terms of $\boldsymbol{\Delta r}$ which can be solved for special cases of $\boldsymbol{F}(r)$.
