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Recently I was surprised to discover that no exact solution for the position of a planet as a function of time exists. I am referring to the two-body problem in a gravitational field where Newtons law of gravity holds.

Well known are proofs that the planet will move in an ellipse, Keplers laws can be derived fairly easy etc., but for the exact position of the planet on the ellipse as a function of time, no formula exists, only numerical approximations.

Is this correct?
Can anybody elaborate on the deeper reason(s) that this relatively simple case cannot be solved?

Qmechanic
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Gerard
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  • Are you sure that there is not exact solution? Maybe under extra conditions. Otherwise, you have http://en.wikipedia.org/wiki/Two-body_problem#Displacement_vector_motion_.282nd_one-body_problem.29 – r_31415 Nov 08 '10 at 15:21
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    See Wikipedia: The problem is not deriving an equation that involves time, but that the equation is transcendal. But I think this "insolvability" only occurs for three and more bodies – Tobias Kienzler Nov 08 '10 at 15:25
  • Aren't elliptic functions (see mbq's answer) more evil than transcendal functions such as sin, exp, log? – Gerard Nov 08 '10 at 15:45
  • Nothing "evil" about transcendental functions like elliptic functions. One actually ought to be glad that the solution to this problem is expressible in terms of well-studied functions! –  Nov 08 '10 at 20:59
  • Ok transcendental. Log, exp, sin, cos are transcendental functions, but a solution in such terms would be "exact" from my point of view. Isn't there a fundamental difference between log, exp, sin, cos and elliptic functions?! – Gerard Nov 08 '10 at 21:16
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    Nope, not as far as I know. It's just that we happen to be less familiar with elliptic functions than, say, trigonometric ones. Though that might be something to ask at the math SE site. – David Z Nov 08 '10 at 21:29
  • Gerard: This might be of interest. :) The point is that elliptic functions are at least as useful as trigonometric functions, only less popular (which I think is a shame, since studying them is no more difficult than studying trigonometric functions). –  Nov 08 '10 at 22:02

3 Answers3

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This is not that there is no exact solution, only the exact solutions for $x(t)$ and $y(t)$ use elliptic functions. The problem whether elliptic functions (which are defined by inverse of some integrals) are "good" functions is a bit philosophical one; one can on one hand state that sine is not a real function because one must integrate or sum a infinite series to calculate it, and on the other that even Lorentz attractor solution can be called three Lorentz chaotic functions with 4 parameters $a$, $b$, $c$ and $t$ and tabularized.

  • I accepted this as the solution, however I stil feel as if these elliptic functions somehow demonstrate that the solution is not "exact".. – Gerard Nov 08 '10 at 21:13
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    Then why are things expressed in terms of sines and cosines "exact"? Not all of them are expressible as algebraic numbers like $\frac{\sqrt{2}}{2}$. –  Nov 08 '10 at 22:04
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    @Gerard As I tried to express, exactness is not something well defined, so you can call it however you wish (-; On the other hand, this problem is usually considered very exact, since the path equation in radial coordinates ($r(\phi)$) is quite simple. –  Dec 03 '10 at 23:05
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    @Gerard: An elliptic function is just as exact as the sine functions. The only difference is that latter is taught, while the former is barely used in college. What does one have that the other doesn't? – Malabarba Dec 22 '10 at 01:13
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    Agreed: sines and cosines are merely solutions to a differential equation y'' + cy = 0, which is the eigenvalue equation for the operator ''. This is the Laplacian operator in 1d. Study this in higher dimensions with rectilinear boundary conditions and you still have sines. But with cylindrical/spherical boundary conditions, you immediately discover Bessel functions and spherical harmonics. You learn you don't really "know" anything much about sine except that it is familiar. Likewise for many other kinds of solutions to ODE's: they appear naturally and should be given some respect! – Eric Zaslow Dec 22 '10 at 17:09
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Considering the ellipse case, the Kepler equation $$ E-\epsilon\sin E=M $$ can be solved, where $M$ is time $t$ dependent(actually linear dependent), and $r,\theta$ in polar system can be expressed in terms of $E$ since we've already known the polar equation(containing initial conditions). So if we can solve the equation "exactly", the position with time parameter is given out. We could firstly write the solution as $$ E=M+ \frac2\pi \sum_{n=1}^\infty \frac{\sin(nM)}n \int_0^\pi \cos(n\phi-n\epsilon \sin\phi) \ \mathrm d\phi, $$ using Fourier series. Then we can compute the sum and give $$ \begin{align} E&=M+\frac1\pi\int_0^\pi \bigg(\tan^{-1}\cot \frac{\phi -\epsilon \sin \phi+M}{2} \\ &\qquad\qquad\qquad-\tan^{-1}\cot \frac{\phi-\epsilon \sin \phi-M}{2}\bigg) \,\mathrm{d}\phi, \end{align} $$ namely $$ E=\int_0^\pi \left(\left\lfloor{\phi-\epsilon\sin\phi+M\over2\pi}\right\rfloor-\left\lfloor{\phi-\epsilon\sin\phi-M\over2\pi}\right\rfloor\right)\text d\phi. $$ The $\lfloor\cdot\rfloor$ here means Floor function. Although the solution cannot be written more "obviously", it's still exact. Other cases are similarly solved. Details of solving methods can be easily found in some relative papers online.

Here's a desmos project link where you can watch the motion: https://www.desmos.com/calculator/z6kxsolfox

Cynthia Xia
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I think i have recently learned a solution of that.
I hope Kepler equation can be helpful; that is:

$M=\frac{2\pi t}{p}=E-e\sin E$

where $E$ is defined by:

$\tan \frac{E}{2}=\sqrt{\frac{1-e}{1+e}}\tan \frac{\theta}{2}$

and $e$ & $\theta$ are eccentricity of the orbit and true anomaly, respectively.

fibonatic
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astronomy olympiadist
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