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I was thinking about orbits a few days ago, and realized that orbits shift/precess naturally. Given that a two-body problem with a star and a planet, if the planet has an eccentric orbit that precesses, would we still call it a Keplerian orbit?

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    $\begingroup$ There is no such thing in nature as a Keplerian orbit. The concept remains useful because (a) many orbits are approximately Keplerian, and (b) the deviations from a true Keplerian orbit can be modeled rather nicely in terms of things like a precession rate. $\endgroup$ Commented Jan 18, 2021 at 10:40
  • $\begingroup$ Kepler is a first approximation. In the real world you need to add virial coefficients: en.wikipedia.org/wiki/Virial_coefficient It gets ugly. $\endgroup$ Commented Feb 7, 2021 at 17:15

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No orbits are strictly Keplerian, unless you really monkey about with your coordinate systems. However, a precessing orbit is pretty darned Keplerian.

enter image description here

The relationship between the various orbital elements can be a little complicated, and precession is traditionally thought of as the rotation of the periapsis of the orbit around the more massive body. If we look at the orbital elements, we can think of it instead as a changing term to the longitude of the ascending node. So we could capture the longitude of the ascending node, $\Omega_0$ at some initial time $t_0$. Then, at any time t, the longitude of the ascending node would be $\Omega=\Omega_0+(t-t_0)\omega_p$, where $\omega_p$ is the rate of precession. The rest of the orbital elements can be treated as constant.

enter image description here

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    $\begingroup$ Love the diagram! +1 Why is a binary star system not d̶a̶r̶n̶e̶d̶ Keplerian even though they are just 2 bodies with e<1? $\endgroup$
    – WarpPrime
    Commented Jan 18, 2021 at 19:05
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    $\begingroup$ @fasterthanlight A binary star system is just a little Keplerian because Kepler didn't know about the barycenter. Kepler would have had a hard time rectifying the movement of both bodies around an empty focus of an ellipse. Of course, which orbit goes in which column is highly subjective. $\endgroup$
    – Connor Garcia
    Commented Jan 18, 2021 at 22:33
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    $\begingroup$ as soon as I saw "monkey about with your coordinate systems" I knew who the author was going to be :-) $\endgroup$
    – uhoh
    Commented Feb 6, 2021 at 5:37
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    $\begingroup$ @uhoh Guilty as charged! I can't resist a good orbit or coordinate system question! $\endgroup$
    – Connor Garcia
    Commented Feb 6, 2021 at 20:10
  • $\begingroup$ If two bodies are in elliptical orbits about the barycenter they also are in elliptical orbits about one another, and the Moon's orbit about the Sun doesn't look anything like that $\endgroup$ Commented Mar 30 at 8:18
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Strictly speaking, pure Keplerian orbits can only occur with two bodies obeying Newton's law of gravitation. If there are more than two bodies then the orbits will be perturbed, to some degree.

Kepler orbits are a reasonable first approximation to the orbits of the planets in the Solar System because the Sun is so massive compared to the planets, and the distances between the planets are relatively large, so the planets don't perturb each other's orbits very much. In units of $10^{-6}$ solar masses, the masses of the larger planets are approximately:

Planet Mass
Jupiter 954.8
Saturn 285.9
Uranus 43.66
Neptune 51.51

The Earth-Moon system weighs in at around 3.04.

Kepler was fortunate that he had sufficiently accurate data to see the patterns that led to his 3 laws of planetary motion, but not so accurate that he could see those laws were only a rough approximation. ;) Fortunately, Newton came along a few decades later and showed how Kepler's laws were a consequence of the laws of mechanics and gravitation applied to two bodies. However, Newton also realised that the Moon's orbit around the Earth was significantly perturbed by the Sun. In fact, the Moon's motion is quite complex and its orbit precesses in various ways on rather short time scales.

This complexity is (mostly) because the Moon's mass is quite large relative to the mass of the Earth, so the Earth-Moon barycentre is a significant distance from the centre of the Earth (on average 4,671 km or 0.75 of the Earth's radius). So the Moon's orbit is not well-approximated by an ellipse with the centre of the Earth at one focus. Also, most moons have orbits that are close to their primary's equatorial plane, but the Moon's orbital plane is near the ecliptic plane (with a mean inclination ~5.1°), like a planet's orbit. If the Moon had a small mass it probably would have been pulled into an equatorial orbit long ago. The torque caused by the Earth trying to pull the Moon into an equatorial orbit causes the nodal precession.

Under Newton's laws, it's impossible for the orbit of a perfectly spherical single planet (without moons) orbiting a perfectly spherical single star to precess. The only possible (closed) orbits are nice Keplerian ellipses (including circles), with a fixed orientation and eccentricity. To be precise, the star and planet will move in complementary elliptical trajectories around their barycentre, always remaining in the same plane, and collinear with the barycentre. If the bodies aren't perfect spheres, with spherically symmetric density profiles, then they cannot be modelled as points via the Shell theorem and nodal precession is possible.

However, under the laws of General Relativity, such a planet with non-zero eccentricity must precess (and the greater the eccentricity, the greater the rate of precession), but the amount of precession is very small unless the star is very massive. Mercury's orbit precesses by about 532 arc-seconds per century, Newtonian gravity accounts for all but 43 of those arc-seconds.

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  • $\begingroup$ "because the Moon's mass is relatively large for a satellite its motion is quite complex and its orbit precesses in various ways on rather short time scales." Wouldn't apsidal and nodal precession take place at nearly the same rate even if the Moon were 100x smaller? There could be a difference of a percent or so due to center-of-mass shift, but I don't see why these are otherwise related to the mass of the Moon. Cool table formatting btw :-) $\endgroup$
    – uhoh
    Commented Feb 6, 2021 at 23:02
  • $\begingroup$ @uhoh Fair points. However, if the Moon's mass were small then it would have been pulled into an equatorial orbit fairly quickly, so the nodal precession would be much smaller. I think the apsidal precession would be much smaller if the Earth-Moon barycentre were closer to the centre of the Earth, but I'd need to calculate (or simulate) that to be certain. ;) $\endgroup$
    – PM 2Ring
    Commented Feb 7, 2021 at 5:00
  • $\begingroup$ This is very interesting, the first apsidal precession equations we can find are for a massless particle in a central field, I don't know how they are modified for two more equivalent bodies orbiting a barycenter. But why then would a smaller moon be pulled into an equatorial orbit fairly quickly? All gravitational accelerations of the center of mass of body are independent of the body's mass. Perhaps I can ask this as a new question? "Does the equatorialization rate of inclinded Moons depend on the Moon's mass?" $\endgroup$
    – uhoh
    Commented Feb 7, 2021 at 5:28
  • $\begingroup$ I'd like to see a reference or mathematical treatment, I'm not good at prose explanations, they don't necessarily obey conservation laws. If I asked that proposed title as a question could you author a well-supported SE answer to it? $\endgroup$
    – uhoh
    Commented Feb 7, 2021 at 5:39
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – PM 2Ring
    Commented Feb 7, 2021 at 9:02
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To the good and thoughtful answers given above, I would add some general considerations.

While a Keplerian orbit exists only within a two-body problem, in a perturbed setting an actual orbit is modelled with a continuous sequence of instantaneous Keplerian orbits sharing a focus.

Usually (not always!) the instantaneous Keplerian orbits are chosen to be osculating (a term coined by Lagrange). This means that the actual perturbed orbit is, at each point, tangential to the corresponding instantaneous Keplerian orbit (and, as terry rightly pointed out in a comment below, has the same instantaneous velocity).

This approach enables us to describe the actual orbit using the parameters of a Keplerian orbit (eccentricity, semimajor axis, etc) and endowing them with time-dependence. The motion of a body is thus split into two components: motion along an instantaneous Keplerian orbit, and non-stop switching from one instantaneous orbit to another. Generally, this splitting is ambiguous. It becomes unambiguous, e.g., when the condition of osculation is imposed.

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    $\begingroup$ Yes. And the osculating orbit is not only tangential, it also has the same velocity at the point of tangency. $\endgroup$
    – terry-s
    Commented Aug 18, 2023 at 11:24
  • $\begingroup$ @terry-s Yes, very true. Will amend. Thank you. $\endgroup$ Commented Aug 18, 2023 at 12:51
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It depends how you define 'Keplerian orbit'. A precessing orbit can obviously not be represented by a stationary ellipse anymore, but it is essentially still the same ellipse, merely uniformly rotating about its focus. The main features of the orbit, like semi-major axis and period of revolution (as defined by the time interval between two perihelion passages) are still the same. So Kepler's laws still apply if interpreted in this sense. If you go to the Horizons Website you can check for yourself that the planetary ephemerides, even though in practice calculated numerically in terms of $(x,y,z)$, $(v_x,v_y,v_z)$ coordinates, are still representable in terms of Kepler's laws (although, since there are more than 2 bodies involved, the orbital elements vary with time and are only treated as 'osculating' elements). At least for orbits that are merely moderately disturbed 2-body orbits, this tends to give better insights in the orbital features (like precession) than just looking at tables of Cartesian coordinates.

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  • $\begingroup$ The JPL ephemerides in the Horizons system aren't computed using Keplerian elements. They're constructed by integrating "the n-body problem, in effect putting the entire Solar System into motion in the computer's memory, accounting for all relevant physical laws." OTOH, Horizons can give output (& take input) in the form of osculating elements, and they're used to store data for small bodies, which are integrated on demand, see ssd.jpl.nasa.gov/horizons/manual.html#intdisp $\endgroup$
    – PM 2Ring
    Commented Aug 18, 2023 at 3:10
  • $\begingroup$ Also see The JPL Planetary and Lunar Ephemerides DE440 and DE441 doi.org/10.3847/1538-3881/abd414 $\endgroup$
    – PM 2Ring
    Commented Aug 18, 2023 at 3:11
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    $\begingroup$ @PM2Ring I wasn't suggesting that Kepler's laws are used to calculate the ephemerides, but the latter are still often formulated in terms of Kepler's laws using osculating elements. They tend to give better insights into orbits than just looking at tables of Cartesian coordinates. I have clarified my answer in this respect. $\endgroup$
    – Thomas
    Commented Aug 19, 2023 at 17:29
  • $\begingroup$ Ah, ok. But I'll leave my previous comments, since they contain useful links. $\endgroup$
    – PM 2Ring
    Commented Aug 19, 2023 at 18:18
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Kepler didn't prescribe that the axis of his elliptical orbits should be either stationary or not.

But the numbers in his tables of mean elements in his Rudolphine Tables of 1627 (https://library.oarcloud.noaa.gov/docs.lib/htdocs/rescue/Rarebook_treasures/QB41K431627.PDF) and their annual/centurial variations show that he accepted/believed it to be an observational fact that the known elliptical orbits do have slowly rotating axes.

A few recent authors have adopted a way of describing an (otherwise?) Keplerian elliptical orbit where the axis rotates, by calling it 'pseudo-Keplerian'. But it is only by the well-known analysis of two-body motion using Newtonian gravitational mechanics that one reaches the result, that in the absence of perturbing forces, the direction of the axis should be unchanging. So this feature does not appear to have anything to do with Kepler.

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