# Question about Lagrange's solution for the three-body problem

The original text in my textbook, written in short:

"By Newton's second law, $$\ddot{\mathbf{x}}_1=-Gm_{2}\frac{\mathbf{x}_1-\mathbf{x}_2}{|\mathbf{x}_1-\mathbf{x}_2|^3}-Gm_{3}\frac{\mathbf{x}_1-\mathbf{x}_3}{|\mathbf{x}_1-\mathbf{x}_3|^3}$$ and analogously for the other two masses. If we make use of the relative-position vectors $$\mathbf{s}_i$$, then clearly $$\mathbf{s}_1+\mathbf{s}_2+\mathbf{s}_3=0$$ After a little algebra, the equations of motion assume the symmetrical form $$\ddot{\mathbf{s}}_{i}=-mG\frac{\mathbf{s}_i}{s_{i}^3}+m_{i}\mathbf{G}$$ where $$m=m_1+m_2+m_3$$ and $$\mathbf{G}=G\left ( \frac{\mathbf{s}_1}{s_{1}^3}+\frac{\mathbf{s}_2}{s_{2}^3}+\frac{\mathbf{s}_3}{s_{3}^3}\right )$$. If the vector $$\mathbf{G}=0$$, the equations of motion decouple, and they reduce to the two-body form of the Kepler problem, $$\ddot{\mathbf{s}}_{i}=-mG\frac{\mathbf{s}_{i}}{s_{i}^3}$$ with each mass moving along an elliptical orbit lying in the same plane with the same focal point and the same period. This decoupling occurs when the three masses are at the vertices of an equilateral triangle. As the motion proceeds, the equations remain uncoupled so the equilateral triangle condition continues to be satisfied, but the triangle changes in size and orientation."

The author has proved that the separation vectors $$\mathbf{s}_i$$ describe ellipses, but how can the author conclude that the trajectories $$\mathbf{x}_i$$ of the three masses are also ellipses? I have not been able to directly prove that those are ellipses.

My attempts:

If each $$\mathbf{x}_i$$ satisfies

$$\ddot{\mathbf{x}_{i}}=-k_{i}\frac{\mathbf{x}_{i}}{x_{i}^3}$$

for some constant $$k_i$$, then the proof is completed. I striaghtly substituted these into $$\mathbf{s}_{i}=-mG\frac{\mathbf{s}_i}{s_{i}^3}$$, and get hardly any results. Also, I later thought that this wouldn't be even a 'proof' in usual sense.

So I tried another attempt. I used the relation $$m_{1}\mathbf{x}_{1}+m_{2}\mathbf{x}_{2}+m_{3}\mathbf{x}_{3}=0$$ and the fact that $$s_{1}=s_{2}=s_{3}=s$$. Still, I couldn't figure out any clues for identifying their trajectories. How should I prove it? Does $$\ddot{\mathbf{s}}_{i}=-mG\frac{\mathbf{s}_{i}}{s_{i}^3}$$ itself already prove the fact?

• What is the definition of ${\bf s}_i$ in terms of ${\bf r}_i$ ? Jan 14 at 9:04
• @GiorgioP-DoomsdayClockIsAt-90 You mean $\mathbf{x}_i$? Like in the figure, $\mathbf{s}_1=\mathbf{x}_3-\mathbf{x}_2$ for instance. I also tried to switch $\mathbf{s}_i$'s in terms of $\mathbf{x}_i$'s. But then it only tells us mass 2 draws elliptic orbit in mass 1's frame(Am I right?).
– q q
Jan 14 at 9:51
• Voting to reopen. Once again, the catch-all "homework-like" reason is being used to incorrectly close a question that is obviously not homework, and where the questioner is asking about a concept, not a specific calculation. Jan 16 at 10:58

It seems like nothing can directly proves that those <the trajectories> are ellipses.

The trajectories of the three masses are not in general ellipses. They are only ellipses if

$$\displaystyle \frac {\vec {s_1}}{s_1^3} + \frac {\vec {s_2}}{s_2^3} + \frac {\vec {s_3}}{s_3^3}=0.$$

If this condition holds then the differential equations for the three vectors $$\vec {s_i}$$ are decoupled from one another, and each one is

$$\displaystyle \ddot {\vec {s_i}} = -mG \frac {\vec {s_i}}{s_i^3}.$$

The solutions of this differential equation are conic sections - circles, ellipses or hyperbolas. I think the author must assume that the orbits are bounded when he infers that they are all ellipses (of which the circle is a special case).

• Then do $\mathbf{x}_i$'s draw conic sections, too?
– q q
Jan 14 at 10:22