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Is the radial two-body problem solvable, i.e. does the differential equation $$r''(t) = 1/r(t)^2$$ have an analytic solution? If it does, what is the solution? And if not why does everyone say that it is, in the words of Britannica, 'the completely solvable two-body problem'*? *https://www.britannica.com/science/celestial-mechanics-physics/The-three-body-problem

Qmechanic
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DrWill
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1 Answers1

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Yes, the two-body problem for an inverse-square force is considered completely solvable.

The equation you wrote is the radial infall equation. It has an analytic solution for $t(r)$. See Wikipedia.

If you actually meant $r$ to be a vector, there is also an analytic solution. A bound orbit is an ellipse, with a circle as a special case of zero eccentricity. Wikipedia gives the position on the ellipse in terms of a parameter related to time, although computing this parameter from the time involves solving a transcendental equation. Unbound trajectories are parabolas or hyperbolas. Note that all these trajectories are conic sections!

The Wikipedia articles on the general two-body problem and the more specific gravitational two-body problem have still more information, as does any textbook on classical mechanics, since this was one of the greatest triumphs of Newtonian physics three centuries ago.

G. Smith
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  • From wiki: "However, solving for E (i.e. position) when M (i.e. time) is given can be considerably more challenging. There is no closed-form solution. One can write an infinite series expression for the solution to Kepler's equation using Lagrange inversion, but the series does not converge for all combinations of e and M (see below)........... Therefore, this solution is a formal definition of the inverse Kepler equation. However, E is not an entire function of M at a given non-zero e." So, the 'solution' doesn't converge for some values of e and M? – DrWill Nov 22 '19 at 16:35
  • As the article goes on to explain, “Alternatively, Kepler's equation can be solved numerically.” I don’t believe there is any problem getting $E$ from $M$ to any desired degree of accuracy. – G. Smith Nov 22 '19 at 18:28
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    The problem is considered analytically solvable because one can obtain a simple analytic parameterization of the trajectory, $x(M)$, $y(M)$, $t(M)$. The fact that inverting $t(M)$ may require numerical techniques is not considered to make the two-body problem unsolvable. Parameterizations of trajectories are common; for example, in relativity we usually use the proper time as a parameter and try to get expressions for $x,y,z,t$ in terms of $\tau$. – G. Smith Nov 22 '19 at 18:40
  • Wiki - " It is often claimed that Kepler's equation "cannot be solved analytically"; see for example here. Whether this is true or not depends on whether one considers an infinite series (or one which does not always converge) to be an analytical solution." This implies to me that 1) - a solution, not a parameterization of the curve, is required, and 2) that wiki considers a series that doesn't converge to be a solution ! I cannot find an explicit statement of what 'solvable' means. – DrWill Nov 22 '19 at 21:51
  • That’s because it doesn’t have a precise and widely accepted meaning among physicists. I have explained why it is commonly said that the two-body inverse-square problem is exactly solvable. After 300 years, I don’t think you are likely to convince many physicists that they should call it unsolvable just because you don’t think an analytic parameterized trajectory should count. – G. Smith Nov 22 '19 at 22:11
  • In any case, solvability has no physical significance whatsoever, and it does not deserve further discussion. – G. Smith Nov 22 '19 at 22:14