# Solving Lagrangian equations of motion for two point-bodies with gravitational interaction

I would like to solve the equations of motion with the Lagrangian function for two point-bodies that interact gravitationally via the potential $$V= {-Gm_1m_2 \over r_{12}}$$ where $$r_{12} = **r_1 - r_2**$$ the distance between them with the beginning of the axes on a point different than the bodies.

I work on polar coordinates.With $$L=T-V$$and $$T= {{1} \over {2}} m_1 (x_1 '+y_1') + {{1} \over {2}} m_2 (x_2 '+y_2') .$$

I have $$x_1=r_1 \cos u_1 ~~y_1=r_1\sin u_1$$ and respectively for the second body. Also, for convenience $$θ=u_1 - u_2$$

If I haven't any mistakes then the Lagrangian is $$L={1 \over 2 } m_1 ( {r_1 '}^2 + {r_1}^2 {u_1 '}^2 ) + {1 \over 2 } m_2 ( {r_2 '}^2 + {r_2}^2 {u_2 '}^2 ) + {Gm_1 m_2 \over [{r_1}^2 +{r_2}^2 + 2r_1 r_2 \cos θ ]^{1/2} }$$

If the above is correct, then for the first body we have $$r_1 '' - r_1 {u_1 ' } ^2 + {Gm_2 (r_1 + r_2 cosθ) \over [{r_1}^2 + {r_2}^2 + 2r_1 r_2 cosθ]^{3/2}}=0 ~~~~(1)$$ and $$r_1 ' u_1 ' +{r_1}^2 u_1 '' - {Gm_2 sinθ \over 4[{r_1}^2 + {r_2}^2 + 2r_1 r_2 \cos θ]^{3/2}}=0 ~~~ (2)$$Respectively we can find the equation for the second body

Even with the assumption that $u_i '' =0$ (which eliminates the second term in (2) ) the system is very difficult.

The question is this (I don't ask the solution of this problem):

1. Is there an analytical solution to this system? If yes:

2. Is there a way to make it simpler? (without changing the beginning of the axons)

3. Shouldn’t this system have at least angular momentum as a constant of motion?

I'm dealing with this problem because I would like to understand not only in principle or theoretically, but to see , through the equations how the motion is dependable on the reference frame(for example that the earth doesn't move around the sun but that in another coordinate system - like that of the example- we have a different but equivalent understanding of the system's motion - like a geocentric system in another example).

Note: The tones are for time derivatives

• You'll save yourself a lot of time and energy by replacing ${\vec r_{1}}$ and ${\vec r_{2}}$ with ${\vec r} = {\vec r_{1}} - {\vec r_{2}}$ and ${\vec R} = \frac{1}{m_{1}+m_{2}}\left(m_{1}{\vec r_{1}} + m_{2}{\vec r_{2}}\right)$ – Jerry Schirmer Apr 17 '15 at 16:58
• If you do this, then you'll see that your answers are yes, yes, and yes – Jerry Schirmer Apr 17 '15 at 16:58
• @Jerry Schirmer I should do this before deriving the equations or is it irrelevant? – Constantine Black Apr 17 '15 at 17:09
• It will make deriving the equations a lot easier if you do the substitution first. – Jerry Schirmer Apr 17 '15 at 17:38

The two body problem can be made equivalent to the one body problem. Say you have a mass $m_1$ at $\vec{r}_1$ and a mass $m_2$ at $\vec{r}_2$ interacting gravitationally. Now focus your attention on the center of mass $\vec{r}$ and the difference vector $\vec{r}_{12}=\vec{r}_2-\vec{r}_1$. Then the kinetic energy of the system can be written as $$K=\frac{1}{2}(m_1+m_2)\dot{\vec{r}}^2+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\dot{\vec{r}_{12}}^2$$. Since the potential energy doesn't depend on the the center of mass, the $\vec{r}$ coordinates will be cyclic and you can drop it from the Lagrangian. Then you will only have a Lagrangian in terms of $\vec{r}_{12}$ which is the same as that of a body under a central potential with mass $\frac{m_1m_2}{m_1+m_2}$. See the treatment given in Goldstein's Classical Mechanics