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Two particles are interacting through gravitational forces. How to find their positions in function of time?

Given the initial positions and velocities $r_1(0), r_2(0), v_1(0), v_2(0)$ of 2 particles interacting by a inverse square law force $F(r) = k/r^2$, what is their positions as a function of time?

I've spent my entire day trying to solve this. Every site seems to be ending up with a "solution" like:

$$r''(t) = k / r(t)²$$

But this is not a formula it is the same differential equation I started with. How I am supposed to actually evaluate this?

I've posted a similar question and I've been linked to this "duplicate". Unfortunatelly the formula provided there finds the time an object falling will hit the ground, which is quite different from the position of particles interacting in function of time. I don't know how to adapt.

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MaiaVictor
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  • Comments from the floor are welcome. Have I been hasty in closing this? – dmckee --- ex-moderator kitten May 17 '12 at 23:06
  • Pardon, I don't understand you. What "comments from the floor" means? And it was not you who closed the question. – MaiaVictor May 17 '12 at 23:11
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    Stack Exchange sites are user driven. Unilateral actions by a moderator, like this one, should be a matter for discussion. I am inviting people to tell me I'm wrong and explain why. If I'm convinced I'll re-open it. Or more likely your first instance, because you should not respond to a closing by re-posting. The other two moderator might also jump in if they think I'm in the wrong. We've all been reversed from time to time. – dmckee --- ex-moderator kitten May 17 '12 at 23:15
  • I believe this question should be reopenned because I still didn't get the answer I'm looking for. "Unfortunatelly the formula provided there finds the time an object falling will hit the ground, which is quite different from the position of particles interacting in function of time. I don't know how to adapt." Please reopen if possible. – MaiaVictor May 17 '12 at 23:18
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    The solution Lagrangian mechanics is at the wikipedia page for Kepler problem, which was linked off of the page for the two body problem. It is also done in every undergraduate mechanics book, which may or may not constitute "general reference". That's one of the things I'm interested in opinions on. – dmckee --- ex-moderator kitten May 17 '12 at 23:19
  • I can't understand some of the math on this article as it is above my current level. I can't see any explicit formula the way I asked it. Can you help me? – MaiaVictor May 17 '12 at 23:22
  • dmckee the problem is that you closed my question and I still don't have my answer! I know you are doing your job but think about it. It is not because the answer is somewhere on that article that everyone is supposed to understand it. This is why this is a site of questions. I just need this function to continue my work. – MaiaVictor May 17 '12 at 23:30
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    @Dokkat: The "explicit formula" doesn't exist with elementary function for a stupid reason--- the area of a section of an ellipse cut out by two lines eminating from a focus is not an elementary function of the angle between the lines. The solution is first to make it a one-body problem by using center of mass coordinates, and then to use conservation of angular momentum to reduce it to a one-dimensional problem, then solve for the orbit $r(\theta)$, using a different nonuniform time coordinate. To map to the original time coordinate, you need the aformentioned special function. – Ron Maimon May 17 '12 at 23:58
  • @dmckee: It's not a duplicate--- he is looking for the solution to the non-radial, ellipse, Kepler problem in time. This requires elliptic functions, and is not usually given in books. The solution in books is for the orbit shape, not the position as a function of time. One can explain how to transform this solution to a solution in time using the equal area law. The previous question is for radial free fall. – Ron Maimon May 18 '12 at 00:00
  • @Ron Re-opening the original version. Thanks. – dmckee --- ex-moderator kitten May 18 '12 at 00:19
  • Thank you both...! @Ron Maimon good explanation. This complicates the thing a lot; but why is there not an explicit formula in function the eliptical area function you mentioned? This way I can just find a way to implement that eliptical function on my program and it will work. – MaiaVictor May 18 '12 at 01:34
  • @Dokkat: It's a known special function, it's just not one I personally work with. Its not usually provided off-the-shelf in a numerical program. You can called it "A(a,\theta)", it's the indefinite integral $\int {d\theta \over (1+a \cos(θ))^2}$ , and you need its inverse function. It is very easy to write down a Fourier series for this (away from parabolic or zero area degenerations) by expanding the denominator in a power series, and integrating term by term. If you want a numerical approximation, this is sufficient, although numerical recipes might have a better algorithm for this integral. – Ron Maimon May 18 '12 at 05:11

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