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.

  • $\begingroup$ The title mentions a 3 body problem, but the question is about the 2 body problem. Please edit and clarify. $\endgroup$ – John Alexiou Jun 12 '20 at 21:51
  • $\begingroup$ @ja72 oh, is it? It was referred to as a three-body problem. Does it become a two-body problem due to the composite? $\endgroup$ – The Pointer Jun 12 '20 at 21:52
  • $\begingroup$ Then it is just a two-body problem. You took away the degrees of freedom between two of the three bodies. It just confuses the issue here. Just ask about how the center of mass arises from the equations of motion on the two-body problem. $\endgroup$ – John Alexiou Jun 12 '20 at 21:58
  • $\begingroup$ @ja72 Doing further research on two-body problems, I found this en.wikipedia.org/wiki/… ${\displaystyle {\ddot {\mathbf {R} }}\equiv {\frac {m_{1}{\ddot {\mathbf {x} }}_{1}+m_{2}{\ddot {\mathbf {x} }}_{2}}{m_{1}+m_{2}}}.}$ Clearly, this is the equation in question. However, the Wikipedia article (1) assumes it by definition, which is not clarifying for us, and (2) for some reason has the acceleration $\ddot{\mathbf{r}}$ instead of the position $\mathbf{r}$, which I find confusing. $\endgroup$ – The Pointer Jun 12 '20 at 22:51
  • $\begingroup$ @ja72 I found it here as well: en.wikipedia.org/wiki/Center_of_mass#Barycentric_coordinates , hyperphysics.phy-astr.gsu.edu/hbase/cm.html $\endgroup$ – The Pointer Jun 12 '20 at 23:30

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 $\boldsymbol{F}(r)$.

  • $\begingroup$ But here you're assuming $\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}$ in the first place, right? My question was about how to derive this in the first place, using the basic force equations of classical mechanics. $\endgroup$ – The Pointer Jun 12 '20 at 23:15
  • $\begingroup$ The logic is this: If you use $\boldsymbol{r}_C$ then the problem becomes solvable. therefore it is an important point in space. But as I said, it is all a trick. It is a special coordinate system where $$\sum_i m_i \boldsymbol{r}_i = 0$$ is true. You can reverse the logic and start from the mass balance above and find $\boldsymbol{r}_C$, if that is what you wanted. $\endgroup$ – John Alexiou Jun 13 '20 at 0:47
  • $\begingroup$ I am trying to carefully follow your reasoning. Can you please explain how (4) is valid? $\endgroup$ – The Pointer Jun 16 '20 at 2:49
  • $\begingroup$ Take the two equations in (3) and solve for $r_1$ and $r_2$. Use $r_2 = r_1 + \Delta r$ in the first expression to get $r_c = r_1 + \Delta r \frac{m_2}{m_1+m_2}$. Solve for $r_1$ and then back substitute to get $r_2$. $\endgroup$ – John Alexiou Jun 21 '20 at 1:08

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