Kepler's first two laws tell us that in a two body orbit in which one of the bodies is much more massive than the other, that:

  1. The smaller body orbiting the larger body has an orbital path that traces an ellipse.
  2. A segment between the two bodies sweeps out equal area in equal time.

Given these simplifications, it appears that such an orbit is time-symmetric. That is, an orbit propagator that predicts the future orbit could also reproduce the prior orbit if we input a position with negative orbital velocity.

Is this also true for simplified parabolic and hyperbolic orbits? How about for n-body orbital simulations? It can't be true for mutually orbiting black holes which radiate energy in the form of gravitational waves and eventually merge, can it?

Which leads me to the question: Just how time-symmetric is orbital mechanics?

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    $\begingroup$ Any N-Body system interacting via conservative forces is time-reversible. This also applies to gas dynamics, you only run into the problem of finding specific initial conditions for getting back to your starting point. If you talk about time-symmetric simulation codes, then you have to apply a specific class of them, so-called symplectic codes, otherwise your solutions are not reversible due to numerical dissipation. $\endgroup$ Apr 29 at 9:20
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    $\begingroup$ It's completely time-symmetric up to but not including entropic events such as collisions and drag from gas/dust. $\endgroup$ Apr 29 at 14:05
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    $\begingroup$ All classical physics is time-symmetric, isn't it? $\endgroup$
    – Barmar
    Apr 29 at 14:56
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    $\begingroup$ @Barmar friction and other losses seems to be an explicit part of classical mechanics (at least in that article) so perhaps energy losses in one direction (of time) have to switch to (very carefully implemented) energy increases going the other way? Would that still be considered time-symmetric? $\endgroup$
    – uhoh
    Apr 29 at 19:33

3 Answers 3


2-body or finite-number-body Newton/Kepler-based models are as symmetric as possible. This is how we know where 1I/Omuamua (in a hyperbolic orbit around the Sun) came from.

Multiple-body systems are chaotic - a small deviation in the initial conditions diverge with time, but the divergence is the same in both directions of time. I.e. they are symmetrically chaotic.

In regard to real-world orbits of objects having a finite size (e.g. ability to collide), finite rigidity (tides), less than perfect vacuum around them (aerodynamic resistance), their own mass changing with time, other bodies perturbing them and the gravity not being exactly Newtonian:

Modelling their orbits is symmetrically hard and inaccurate in either direction.

  • $\begingroup$ Viscous tidal dissipation in the bodies is basically 100% of the non-time symmetric part of solar system evolution. There is no appreciable gravitational waves so general relativity is still reversible. $\endgroup$ Apr 29 at 2:10
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    $\begingroup$ en.wikipedia.org/wiki/Gravitational_wave#Binaries has an approximation of gravitational radiation energy loss in a binary system of modest mass. Eg, the Earth-Sun system radiates ~200 watts, which is insignificant relative to the other factors: it would take over 30 trillion times the current age of the universe to drop the Earth into the Sun. :) $\endgroup$
    – PM 2Ring
    Apr 29 at 8:11
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    $\begingroup$ @KevinKostlan even with gravitational waves the situation is reversible, you'd just have to also include the (now) incoming gravitational wavefronts. Obviously these wavefronts couldn't have any reasonable physical source, but that doesn't affect the ability to reverse the mechanics $\endgroup$
    – Tristan
    Apr 29 at 9:40
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    $\begingroup$ How is the fact that a multiple-body system is chaotic relevant to the time-symmetry of the situation? Chaotic systems are still time symmetric. $\endgroup$
    – zephyr
    Apr 29 at 13:42
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    $\begingroup$ @zephyr, chaotic systems are not necessarily symmetrically chaotic. For example, weather, a notoriously chaotic system, can be forecasted several days out, but can only be back-casted a few hours at best, because effects that are dissipative with time running forward instead magnify things with time running backwards. $\endgroup$
    – Mark
    Apr 29 at 22:04

It seems that a source of time-asymmetry is becoming appreciable in increasingly high-precision accounts of the moon's orbit.

The latest JPL lunar and planetary ephemeris comes in two varieties, DE440 and DE441, because it is easier in forward-going integrations to model the pattern of excitation and damping between the lunar mantle and core, than in integrations going back. There is something random and not fully time-symmetrical about these excitations and dampings. The phenomenon is not modeled in version DE441, which has a long historical backwards extension, thus:

The ephemerides DE440 and DE441 are fit to the same data set, but DE441 assumes no damping between the lunar liquid core and the solid mantle, which avoids a divergence when integrated backward in time. Therefore, DE441 is less accurate than DE440 for the current century, but covers a much longer duration of years −13,200 to +17,191, compared to DE440 covering years 1550–2650.

The quote is from "The JPL Planetary and Lunar Ephemerides DE440 and DE441", Ryan S Park et al., Astronomical Journal, 161:105 (15pp), 2021.


It's hard to give a definitive answer to this question. For the spherical-cow, Physics 101 answer, the answer is yes, the dynamics of orbital mechanics are time-reversible. However, as I'll get into later in my answer, for pretty much all practical scenarios that move beyond any simplified picture, the answer is no.

I think AtmosphericPrisonEscape really hit the nail on the head with their comment, so I can't take all the credit for this part of my answer. When one sets up the dynamical equations for orbital mechanics, and tries to solve them (be it with a Hamiltonian, Lagrangian, or something else), fundamentally they've built in the force governing the motion as the (Newtonian) force of gravity. I'd suggest taking a look at this question and answer for an example of setting up the equations of motion. In anycase, because we know the force of gravity is conservative (since it only depends on position and not on anything else, such as velocity), then the the orbital mechanics equations that are governed by the force of gravity are time-reversible. A conservative force allows one to perform an operation and then undo that operation and return to the same state. This statement is independent of how many bodies we have or what types of orbits they may have.

Now let's move beyond the spherical cow. Reality is not accurately modeled by the 101 orbital mechanics equations. Reality is more complex. There's all sorts of other processes at work such as gravitational radiation, yarkovsky effects, Poytnint-Roberston effects, drag within an accretion disk/solar wind/gas/dust/etc. (including drag from magnetic fields!), and many, many other effects that can cause orbits to deviate from the standard orbit one would calculate with the naive orbital dynamics as described above. So in that consideration, Kepler's laws, while generally true (by that I mean, it's often the "leading order term" in the list of all things effecting motions of a body), they are not exclusively true and don't 100% account for any given object's orbits.

However, it gets even worse! It may be that even in the spherical-cow, physics 101 case, the orbital motion equations may not be time symmetric. One often finds that in order to accurately predict orbital motions, they are relegated to using simulations (since it is known that anything more than a 2-body problem has no closed-form, analytic solution). Simulations come with their own issues. Computers do not have perfect precision or infinite computing power, which means concessions must be made. We must, necessarily compute with limited precision (that is, we may only be able to calculate numbers out to 15 digits) or with limited computing power (that is, we may only be able to compute the equations every 0.0001 seconds, but not to an accuracy better than that in time). What's more, different numerical techniques may have varying numerical stability. This leads to the concept of numerical dissipation or diffusion wherein, over time, small errors introduced by imprecision of the simulation results in a building inaccuracy of the simulation. This can cause your physics simulations to "bleed away energy" (hence the use of "dissipation" in the term above). This steady, stochastic loss of energy represents a time asymmetry meaning the simulation of orbital mechanics is always asymmetric in time.

I bring all of that up, as an interlude to the concept of Planck time and Planck length. Planck time and lengths are often seen as the smallest fundamental units of the universe. By that I mean, it doesn't make physical sense to talk about events happening on time scales shorter than a Planck time or on lengths smaller than a Planck length. This brings me to Boekhalt et al. (2020) who set out simulate three orbiting black holes with extreme precision (that is, they invested considerable computing resources to minimize the numerical dissipation as much as possible). The punch line of their research was that they managed to reduce their imprecision to below the Planck length and still found that a selection of their orbits were time asymmetric. If it is true that the Planck length is the smallest length at which physics makes sense, then they've shown that 100% time symmetry of orbital equations would require a precision greater than the universe would seem to allow.

So all of that is summarized by saying that yes, orbital dynamics are time symmetric if you don't consider all the things that can make them time asymmetric.


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