How about an exact solution for the position of a planet as a function of time?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.