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I was playing a game on my smartphone whose goal is to draw certain orbit in presence of certain central gravitational potential. I noticed that when there are two center of force is possible to have a eight(or lemniscate)-shaped orbit. I would like to find analytical solution to that problem in the most symmetric situation: I have a potential such as

$$ V=-\frac{1}{d_1}-\frac{1}{d_2} $$

where

$d_1=\sqrt{(x-1)^2 + y^2}$ and $d_2=\sqrt{(x+1)^2+y^2}$.

I would like to write down the equation of orbit in the having as initial conditions: $x(0)=0$, $y(0)=0$, $x'(0)=v\cos{θ}$, $y'(0)=v\sin{θ}$.

Of course in euclidean coordinates the square roots makes the problem difficult (Mathematica doesn't succede in solving them analytically). I was thinking what is the most clever system of coordinates but I'm a bit doubtful.

Do you know any book or paper that solves this problem that seems very simple?

MaPo
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  • Related: http://physics.stackexchange.com/q/31201/50583; Note that this is a three-body problem and thus it's not clear that you should expect an analytical solution to exist. – ACuriousMind Jan 17 '17 at 13:36
  • I read that post but it's not discussing about how to solve it. – MaPo Jan 17 '17 at 13:48
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    The solution, if it exists, is unstable - that is, a small perturbation over time will change the orbit. But presumably you already discovered this when you played your game. – Floris Jan 17 '17 at 13:56
  • I noticed that the solution is unstable. My question was simply if anyone knows a way to write down the equation of the orbit, exactly with the above initial condition, disregarding stability issues – MaPo Jan 17 '17 at 14:02
  • This is not a true 3-body problem because the two force centers are fixed. – Lewis Miller Jan 17 '17 at 15:26
  • @LewisMiller This is what I suspected, but I'm not an expert of the definitions. – MaPo Jan 17 '17 at 15:28
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    This is the problem of two fixed centers, first studied by Euler. Papers continue to be published on this subject to this day. – David Hammen Jan 20 '17 at 18:00

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