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I am interested in solving a differential equation from the two-body problem. This is not homework, and I just thought of this on my own during physics class one day. However, after many attempts, I still cannot find a way to go at this. I know Wikipedia does something with the center of mass of the system but that is not a solution that satisfies me or one that I understand. That's why I'm asking in Math.SE and not Physics, which would just tell me to look at the com. So, if the following is solvable (general solution), please, let me know how.

For two masses $m_1$ and $m_2$, and constant $G$, we know that $$ m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_1-\vec r_2)}{|\vec r_1-\vec r_2|^3} $$And that $$ m_1\frac{d^2\vec r_1}{dt^2}=-m_2\frac{d^2\vec r_2}{dt^2} $$ My approach:

Let $\vec v_n$ correspond to $\frac{d\vec r_n}{dt}$ $$ \int_0^t\int_0^tm_1\frac{d^2\vec r_1}{dt^2}dtdt=\int_0^t\int_0^t-m_2\frac{d^2\vec r_2}{dt^2}dtdt $$ $$ \int_0^tm_1(\vec v_1(t)-\vec v_1(0))dt=\int_0^t-m_2(\vec v_2(t)-\vec v_2(0))dt $$ $$ m_1(\vec r_1(t)-r_1(0)-t\vec v_1(0))=-m_2(\vec r_2(t)-r_2(0)-t\vec v_2(0)) $$ $$ \frac{-m_1}{m_2}(\vec r_1(t)-\vec r_1(0)-t\vec v_1(0))+\vec r_2(0)+t\vec v_2(0)=\vec r_2(t) $$ Now that $\vec r_2(t)$ can be expressed with $\vec r_1(t)$ and constants, I replace it in the DE. For brevity consider that for a function $f$, $\vec r_2 = f(\vec r_1)$ $$ m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_1-f(\vec r_1))}{|\vec r_1-f(\vec r_1)|^3} $$ Now, convert to $x_1$ and $y_1$ $$ m_1\frac{d^2x_1}{dt^2}=G\frac{m_1 m_2 (x_1-f(x_1))}{D} $$ $$ m_1\frac{d^2y_1}{dt^2}=G\frac{m_1 m_2 (y_1-f(y_1))}{D} $$ D is the denominator: $$ D=|\vec r_1-f(\vec r_1)|^3=((x_1+\frac{m_1}{m_2}x_1 +c_1 )^2+(y_1+\frac{m_1}{m_2}y_1 +c_2 )^2)^{3/2} $$ Where $c_1$ and $c_2$ in the above are constants dependent on the initial conditions.

I can't even imagine beginning to take the Laplace transform of those equations, let alone using it to solve for $x_1$ and $y_1$.

Any suggestions?

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I don't see what is so problematic with the center of mass coordinates. It's just a change of variables with $\vec{R}=w_1\vec{r}_1+w_2\vec{r}_2$ and $\vec{r}=\vec{r}_1-\vec{r}_2$ and it is very useful. ($w_i=m_i/(m_1+m_2)$) –  Raskolnikov Aug 22 '12 at 21:46
@Raskolnikov Right, that's what Wikipedia did. But the site hand-waves finding $\vec r(t)$ and $\vec R(t)$, and I'd appreciate a way of finding $\vec r_1(t)$ and $\vec r_2(t)$, which are dependant on r(t) and R(t). –  VF1 Aug 22 '12 at 21:51
What do you mean "handwaves"? I just gave them to you. –  Raskolnikov Aug 22 '12 at 21:52
Only thing I can say for now is that you'd really benefit by reading the wikipage thoroughly. Because the way you go about it is not just wrong. Your first integration is wrong from a formal point of view. Your introduction of the functions $f$ and $D$ are useless, certainly compared to the slick trick of a change of coordinates. Do yourself a favor and take some time to read diligently about the two-body problem. It's not that trivial that you can just solve it with your level of knowledge. Trust me on that. But you can understand it, if you make the effort to read about it. –  Raskolnikov Aug 22 '12 at 22:03
Once again: what you label as abuse is the will to help. –  Did Aug 22 '12 at 22:32

1 Answer 1

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Question has been answered in comments to question. Apparently the two body system cannot be solved without shifting the reference point.

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