Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand three body problem in Newtonian space. I want to make formulation of differential equations for known initial conditions for the case with:

  1. Identical three masses
  2. Periodic solution
  3. Zero Angular momentum

The equations in vector form are:

$$\frac{d^2 \overset{\rightharpoonup }{r}_1}{\text{dt}^2}=\frac{k^2 m_2 \left(\overset{\rightharpoonup }{r}_2-\overset{\rightharpoonup }{r}_1\right)}{r_{12}^3}+\frac{k^2 m_3 \left(\overset{\rightharpoonup }{r}_3-\overset{\rightharpoonup }{r}_1\right)}{r_{13}^3}$$

$$\frac{d^2 \overset{\rightharpoonup }{r}_2}{\text{dt}^2}=\frac{k^2 m_1 \left(\overset{\rightharpoonup }{r}_1-\overset{\rightharpoonup }{r}_2\right)}{r_{12}^3}+\frac{k^2 m_3 \left(\overset{\rightharpoonup }{r}_3-\overset{\rightharpoonup }{r}_2\right)}{r_{23}^3}$$

$$\frac{d^2 \overset{\rightharpoonup }{r}_3}{\text{dt}^2}=\frac{k^2 m_1 \left(\overset{\rightharpoonup }{r}_1-\overset{\rightharpoonup }{r}_3\right)}{r_{13}^3}+\frac{k^2 m_2 \left(\overset{\rightharpoonup }{r}_2-\overset{\rightharpoonup }{r}_3\right)}{r_{23}^3}$$

where $k$ denotes the gravity constant.

I want to formulate (not solve) this problem properly.

I've found this reference can help understand zero angular momentum. I don't understand it well.

share|cite|improve this question
It is not clear what you're looking for. You have the equations of motion, the only other piece is initial conditions. – DilithiumMatrix Mar 16 '13 at 17:30
ok, initial conditions but how to set them to satisfy periodic solution and angular momentum to be equal to zero – Pipe Mar 16 '13 at 17:58
hope that it is clear what I am looking for. explain mathematically condition of angular momentum to be zero and periodic solution. for two body problem is solvable without these conditions – Pipe Mar 16 '13 at 18:11
Finding periodic solution to the three-body problem is extremely difficult. Very few such solutions are known, despite people working on the problem for hundreds of years. In the last few weeks a recent paper came out with the largest set of such orbits, but its not something you can trivially solve. – DilithiumMatrix Mar 16 '13 at 19:12
@ zhermes thank you for new comments. So the problem of two bodies is nonlinear in basis and then solvable analytically, am I right? I think that you are not right with momentum, this is not a strict condition because rotation of the center of mass should be equal to zero? – Pipe Mar 16 '13 at 19:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.