The eccentricity of Earths orbit follows a bounded random walk-like pattern, see this chart. I presume most other planets are similar. One could think of eccentricity and argument of periapsis as "polar coordinates"; the centroid of the ellipse would be roughly analogous to "Cartesian coordinates". Either way the orbital parameters are chaotic but bounded.

What causes this behavior? When the orbit drifts away from it's "equilibrium point" what effect "kicks in" to push it back rather than let it drift even further away? Would this picture be qualitatively different under Newtonian gravity (these drifts are very slow so small relativistic corrections may matter)?

My hypothesis is that, due to the smallness of these perturbations, the system is almost linear. i.e Venus and Jupiter will contribute additively to the perturbations (additively in the "Cartesian coordinates" at least), and said contribution is insensitive to the small perturbations for other planets. So we are looking at a superposition of multiple periodic functions; one for each of important enough planet and maybe one for corrections from general relativity itself. Is this line of reasoning valid?

  • 1
    $\begingroup$ A sum of sines and cosines is nothing like a random walk. $\endgroup$ – G. Smith Feb 23 '20 at 2:55
  • $\begingroup$ There is a deterministic prediction, so I don’t think it’s a random walk. However do we have enough experimental data to verify? This talk of changes in long periods of time (“thousands of years”). $\endgroup$ – Superfast Jellyfish Feb 23 '20 at 3:35
  • $\begingroup$ @Fellow Traveller: Numerical modeling does wonders here (there is an efficient post-newtonian model to use for very high accuracy). But there should be a simpler way to understand whats going on, namely why it is simply just "adding a bunch of sinewaves" together. $\endgroup$ – Kevin Kostlan Feb 23 '20 at 3:47
  • $\begingroup$ @KevinKostlan I see. Can you point me to a good source regarding the same? $\endgroup$ – Superfast Jellyfish Feb 23 '20 at 3:50
  • $\begingroup$ @Fellow Traveller: I think the following is enough to brute-force orbits around the sun (for moons tidal friction is far more important!): The Einstein–Infeld–Hoffmann equations are accurate to O(v^4/c^4) ~10^(-16) (Newtonian is only accurate to O(v^2/c^2)). Add the solar quadrupole (bulging from spin) moment (planet quadrupoles are too small I think to matter). Also solar mass loss (stellar evolution). Add the galactic tide as an F=(matrix)*x forcefield, which is important for orbits far beyond Neptune. Finally, throw other stars past the sun "randomly" that can occasionally perturb orbits. $\endgroup$ – Kevin Kostlan Feb 25 '20 at 18:58

It is been already well-explained and I am just going to translate it to the language of the Hamiltonian Chaos. There is a theory called KAM Theory which is roughly a classical counterpart of the quantum degenerate perturbation theory.
Simply it says that:

For a Hamiltonian $H=H_0 + \lambda H^\prime$, where $H_0$ is the integrable part (2-body problem of earth and sun) and $H^\prime$ is the non-integrable part (3rd body interaction like jupyter gravity), for small $\lambda$, the trajectories (tori is its jargon) of $H_0$ remains periodic but slightly deformed.\

Deformation of a tori in many-body phase space (18-dimension phase space for the 3-body problem) leads to a bounded variation of constants of motions, in this case, the eccentricity of Earth's orbit.

And at the end, your assumption:

So we are looking at a superposition of multiple periodic functions; one for each of important enough planet and maybe one for corrections from general relativity itself. Is this line of reasoning valid?

is almost true. Almost true because $R_{earth}$ is a linear superposition of $R^{0}_{earth}+\lambda R^{1}_{earth}+\lambda^2 R^{2}_{earth}+ \dots$, but $R^{i}_{earth}$ are not the periodic trajectory of earth around the $i^{th}$ planet. As an intuition, recall the quantum pertutbation theory.


It is my understanding that the solar system is not inherently stable.

The giant planets didn't form at their current distance to the Sun, in a distant past there has been a period during which there was significant migration of the giant planets.

That said: as far as we can tell the current solar system configuration has been like it is now for the past billions of years. By far most objects are in regular orbit. An example of a rere exception to that is Oterma

My understanding is that the the current long term continuity of the solar system should be regarded as a lucky coincidence.

By lucky coincidence: the perturbations of orbits average out over time, such that the effects do not accumulate.

If memory serves me: long duration computer simulations show the following feature: orbits tend to become circularized. If that is indeed the case it would be interesting to explore why that is so.

If orbits tend to become circularized it is unlikely for accumulation of eccentricity of orbit to ocur.

What remains is migration of a celestial object to a higher or lower orbit around the Sun. Such a migration requires a lot of energy. To migrate to a lower orbit you have to lose a lot of energy, to migrate to a higher orbit you have to gain a lot of energy. Such energy transfer is possible: as I mentioned, current thinking is that there has been a period during which there was significant migration of the giant planets. But for such massive transfer you need accumulation, and clearly for the past billions of years the conditions for such accumulation have not been there.

  • $\begingroup$ "To migrate to a lower orbit you have to lose a lot of energy", that's true, but it can be a little confusing because total orbital energy of a bound orbit is negative (and inversely proportional to the mean orbit radius). See en.wikipedia.org/wiki/Specific_orbital_energy Specific energy is energy/mass. For a circular orbit, $-E_K = E_T = \frac12 E_P$, where $E_K, E_T, E_P$ are kinetic, total, and potential energies. $\endgroup$ – PM 2Ring Nov 25 '20 at 15:26

In the short term (read "many millions of years"), changes are bounded, as you observed. The reason for that is essentially, as it's been remarked, that:

An orbit isn't a fragile balance, where some small deviation can knock an object out of orbit. For an object at a given distance from a central body (the sun for the Earth, the Earth for an Earth satellite), any velocity less than the escape velocity results in an elliptical orbit

On top of that, the Solar System is quite old, so any strong instabilities that might have existed have done their damage long ago. Thus only mild (slow) instabilities are left - in a system whose time scales are already very long.

However, in time scales hundreds of millions of years we can't make predictions and dramatic changes can't be ruled out, though extreme events such as collisions between planets shouldn't happen in the next few billion years (see, e.g., this paper).

See also this answer and the Wikipedia entry it refers to for more on the stability of the Solar System.

  • $\begingroup$ A single orbit is stable, but the effects of long-term perturbations don't have cancel themselves out on 10myr scales. The moon has/is slowly receding, so it is an example where small perturbations to the ideal two-body system add up over time. But for planet orbits, the system keeps coming back to (nearly) where it came from. $\endgroup$ – Kevin Kostlan Feb 25 '20 at 18:44
  • $\begingroup$ @KevinKostlan No one said anything about canceling out in 10 My - it's precisely the combined, cummulative effect of the known factors that doesn't allow us to predict precisely much further. $\endgroup$ – stafusa Feb 25 '20 at 20:54
  • $\begingroup$ The chart of Earths orbit over time-scales up to 10 My and maybe further shows no cumulative effect, things just bounce back and forth. However, the same with the moon's orbit would, it would be drifting steadily farther away over the period. $\endgroup$ – Kevin Kostlan Feb 25 '20 at 21:49
  • 1
    $\begingroup$ @KevinKostlan With the the Lyapunov times for the Solar System being of the order of 10 My (see also this presentation), predictions much further should be considered only statistically; but, yes, the Earth is supposed to stay put until the Sun becomes a red giant, due to the stability our old, settled-down Solar System displays, as I write in the answer - though other planet might face some adventure. $\endgroup$ – stafusa Feb 25 '20 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.