# What are “non-Keplerian” orbits? What are some familiar examples in our solar system, and can some still be closed?

This excellent answer to Forms of stellar orbits around the galactic center invokes the following concepts:

• non-Keplerian orbits
• closed orbits

I have a fairly good idea what these mean and so might many of us, but our ideas may not overlap completely.

### For the sake of future stack exchangers and orbiteers, how can we best define these?

Question(s):

1. What exactly are "non-Keplerian" orbits?
2. What are some familiar examples of distinctly non-Keplerian orbits in our solar system?
3. What are closed orbits?
4. What are some familiar examples of distinctly non-closed orbits in our solar system?
5. Can some non-Keplerian orbits still be closed?

Included should be the understanding that we're talking about soft limits here; no orbit is exactly Keplerian because gravity goes everywhere (there are no "gravitational dielectrics" or opposite charges for field lines to terminate). But for practical purposes we can treat many orbits as Keplerian for many purposes, and some we can't at all.

• Strictly speaking, all orbits are non-Keplerian and no orbits are closed. – David Hammen Nov 3 '20 at 2:49
• @DavidHammen I saw that coming! "...no orbit is exactly Keplerian because gravity goes everywhere (there are no "gravitational dielectrics" or opposite charges for field lines to terminate). " I suppose that non-closure is a near-corollary but a mathematical proof in all cases might be elusive. – uhoh Nov 3 '20 at 2:50
• @uhoh I hope you enjoyed my little table of cartoon orbits below. I hope I don't disappoint with my other answer to this question. I think you will make a number of people unhappy if you end up choosing either of my answers! – Connor Garcia Nov 16 '20 at 1:25
• @uhoh I will post a second answer by Tuesday of next week. – Connor Garcia Dec 3 '20 at 5:57
• @ConnorGarcia the universe is 13.8 billion years old and the age of Earth's orbit is 4.5 billion; The Tuesday after next Tuesday is only a billion milliseconds from now, so that sounds fine to me! – uhoh Dec 3 '20 at 6:02

What exactly are "non-Keplerian" orbits?

Orbits that don't follow Kepler's laws.

Strictly speaking, all orbits are non-Keplerian. In practice, one can model some orbits as basically Keplerian, but with perturbations. Sun synchronous satellites are one example of orbits that are close to Keplerian, but not quite so. The Earth's equatorial bulge makes satellites fail to orbit in a plane. Sun synchronous satellites take advantage of the precession induced by the Earth's equatorial bulge so that their orbits precess by 360° in one year. Connor Garcia's example of satellites in pseudo-orbits about the Sun-Earth linear Lagrange points form another set of examples.

What are some familiar examples of distinctly non-closed orbits in our solar system?

Every planet, every moon, every asteroid, ...

What are closed orbits?

Bound orbits are orbits that stay bound to some central object. Parabolic and hyperbolic trajectories are not bound. Closed orbits are bound orbits that repeat their path. Any attractive central force law can result in circular orbits. These are trivially closed. There are only two attractive central force laws that can result in closed noncircular orbits: An inverse square law (e.g., Newtonian gravitation), and a linear force law (e.g., a Hookean spring). This is Bertrand's Theorem.

What are some familiar examples of distinctly non-Keplerian orbits in our solar system?

Every planet, every moon, every asteroid, ...

Can some non-Keplerian orbits still be closed?

Other than a Hookean spring, no.

• Do orbits in tunnels embedded within spherically symmetric bodies count as Hookean? – uhoh Nov 3 '20 at 3:22
• @uhoh In the case of an object that is spherically symmetric, has a uniform density, is the only object in the universe, and is not subject to general relativity, yes. By Newton's shell theorem, only the portion of the central object that is closer to the center of mass than the "orbiting" object counts. Newton's law of gravitation inside the central object reduces to $F=-kr$, which is Hooke's Law, where $k$ is $\frac43 \pi G \rho$. – David Hammen Nov 3 '20 at 3:31
1. What exactly are "non-Keplerian" orbits?

Strictly speaking, no orbits are in perfect accordance with Kepler’s laws. Kepler’s laws aren’t really “laws” in terms of physical laws, but are instead trends that Kepler noticed and calculated using astronomical observations of the planets. Kepler’s laws are very accurate for planetary orbits since he used very accurate (for the time) planetary observations. I think Wikipedia states Kepler’s laws quite nicely:

1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its orbit.

Newton proved that Kepler’s laws would perfectly describe planetary motion using an inverse r-squared law for gravitational force if the planet’s mass was negligible with respect to the sun’s mass, if the only gravity the planet was subjected to was the sun, and if the planet hadn’t achieved escape velocity.

Does that mean we should stop using Kepler’s laws? Heck no! They are super useful for all kinds of things. As an example, I give the only rigorous answer to this question using Kepler’s 2nd and 3rd laws (in the paragraph just before my notes): If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth? . This question was around for over a year, and many people had taken a crack at answering and failed (in my opinion). My solution is much less complicated than another clever physics answer that assumes gravitational force is constant only to get a lower bound!

A "non-Keplerian" orbit is an orbit in which Kepler's laws lack predictive and descriptive power. If a question about an orbit requiring a specified accuracy can’t be answered with the required accuracy using Kepler’s laws, the orbit is ‘Non-Keplerian’ in the context of that question. This is probably not a very satisfying answer, especially since the same orbit could be considered Keplerian for one question, but not for another. Unfortunately, this is typically how engineering gets done. We may make some assumptions to get to an answer and then refine that answer by reversing those assumptions or using a more sophisticated model. When Kepler’s laws fail to give us required accuracy, we can move to orbit propagation simulations.

Kepler’s laws were written specifically for the motion of planets around the sun! So one could argue that a moon orbit around a planet is not a Kepler orbit, even though Kepler’s laws may still be very accurate if you replace “planet” with “moon” and then replace “sun” with “planet”. I would rather argue that any orbital system is “Keplerian” if Kepler’s laws still accurately describe the objects’ motions.

Kepler’s laws don’t include the concept of a barycenter (system center of mass). So they degrade for 2-body orbits when there isn’t very much difference between their mass. A comment below says that a binary star system is an ‘almost perfect example of Keplerian behavior’, but I disagree. One has to generalize Kepler’s laws to accurately describe binary star motion:

a) Orbiting bodies move in elliptical orbits around the system barycenter.

b) A line segment between the barycenter and a body sweeps out equal areas in equal time.

c) The square of a body’s orbital period is proportional to the cube of its mean distance from the barycenter.

Here is a table I made to accompany my definition of non-Keplerian orbits:

1. What are some familiar examples of distinctly non-Keplerian our solar system?

I think the easiest example of a mostly non-Keplerian orbit is the James Webb telescope soon to be at Earth's L2 point: https://en.wikipedia.org/wiki/Lagrange_point . Kepler's 3rd law states that the ratio of the cube of the orbit SMA (semi-major axis) to the square of the orbital period is constant for everything orbiting the same massive body. But something in Earth's L2 point has the same orbital period as Earth, but a way larger SMA. Kepler's 3rd law is violated, hence the James Webb telescope will be in a non-Keplerian orbit. If you want a more exotic orbit and a naturally mostly non-Keplerian orbit, take a look at this weird orbit for a retrograde Jupiter coorbital asteroid: http://www.astro.uwo.ca/~wiegert/2015BZ509/ This is an example of a 3-body problem orbit.

1. What are closed orbits?

A closed orbit is one that repeats itself in an inertial coordinate system with the origin at the center of mass of the most massive body. If you don't specify a coordinate system, I can say that the orbit of any object is closed by specifying a coordinate system with that object always at the origin.

1. Can some non-Keplerian orbits still be closed?

Yes, the two orbit examples given above are closed enough. Oumuamua is an example of a "non-closed" "non-Keplerian" solar orbit. https://en.wikipedia.org/wiki/%CA%BBOumuamua

• You are thinking of bound orbits rather than closed orbits. A closed orbit is one that repeats itself. There are only two central force power laws that result in closed non-circular orbits, an inverse square force such as Newtonian gravity, and a linear force such as a spring. This is Bertrand's theorem. – David Hammen Nov 3 '20 at 2:54
• David Hammen - How embarrassing! You are absolutely correct, I changed my answer accordingly above. – Connor Garcia Nov 3 '20 at 3:03
• uhoh - I will take a look at the ALMA questions, but mostly what I worked on was the control software, so I may not be able to answer them. – Connor Garcia Nov 3 '20 at 3:06
• 'I think if a prediction based on Kepler's laws fails to within the second or third significant digit in an orbit calculation, then that orbit is "non-Keplerian".' This isn't a good criterion. First, we don't in general have fixed criteria to say whether something is or is not a good approximation. That's just not how science works. Also, it matters whether the error has a secular trend. If the error has a secular trend, then the Keplerian approximation can be excellent at short times and bad at long times. – Ben Crowell Nov 3 '20 at 14:08
• @BenCrowell Rather than the "that's just not how science works" lecture, double-check how I've written the question itself: "Included should be the understanding that we're talking about soft limits here; no orbit is exactly Keplerian because gravity goes everywhere (there are no "gravitational dielectrics" or opposite charges for field lines to terminate). But for practical purposes we can treat many orbits as Keplerian for many purposes, and some we can't at all." Question asks for an arbitrary line in the sand, and one has been provided. – uhoh Nov 3 '20 at 23:56

Preface:

In all the existing coordinate systems I am aware of, no orbits are strictly Keplerian. But perhaps in an Earth Centered coordinate system, one could say that the Earth's orbit is closed, since the Earth, by definition, is unmoving at [0,0,0].

Throughout history, CS (Coordinate Systems) have been a huge bone of contention. Einstein said:

Can we formulate physical laws so that they are valid for all CS?... The struggle, so violent in the early days of science, between the views of Ptolemy and Copernicus would then be quite meaningless. Either CS could be used with equal justification. The two sentences, 'the sun is at rest and the Earth moves', or 'the sun moves and the Earth is at rest', would simply mean two different conventions concerning two different CS.

On a practical level, I think it is convenient to treat some orbits as Keplerian, since we can answer a lot of questions quickly and accurately off the 'back of the napkin' using Kepler's laws. As inspiration from Ptolemy, Kepler, and Einstein, I wondered if a coordinate system could be imagined in which every orbit is perfectly Keplerian and closed. The following is the result:

The Moon's orbit around the Earth is Keplerian and closed.

In order to track and quantify the motion of celestial objects, we define coordinate systems. Our choice of coordinate system is typically made to ease the calculation complexity for a particular problem. For example, Earth satellite calculations are often performed in an ECI (Earth Centered Inertial) coordinate system. That means the origin of the system is the center of the Earth, and the Earth rotates in place about the origin. This is a convenient system to study the motion of Earth bound satellites, since it is consistent with Kepler’s laws for Earth satellites, including the Moon. ECI:

In some cases, it's convenient to use an ECEF (Earth Centered Earth Fixed) coordinate system. This coordinate system fixes the rotation of the Earth, so the axes don't change with respect to the Earth’s surface. This is a convenient system for space launch since the coordinates of Earthed based sensors don’t change. ECEF:

We can define a coordinate system called ECMF (Earth Centered Moon Fixed). In this coordinate system, we set the x-axis to be coincident with the vector from the Earth to the Moon. As the Moon rotates around the Earth, the whole coordinate system moves with it. The z and y axes are offset by 90 degrees and lie in the plane orthogonal to the vector from Earth to the Moon.

In order to ‘fix the Moon’ in our ECMF coordinate system, we have to account for the variations in lunar distance due to the eccentricity of the orbit. If we switch from cartesian coordinates to polar coordinates, we see that we can set r, the distance from the Earth to the Moon to be equal to $$k=500,000km$$. ECMF (not to scale):

The coordinate transformation from ECEF to ECMF is dependent on the moon’s polar coordinates in ECEF at time $$t$$: [$$\lambda$$, $$\phi$$, $$r$$]. To translate a point $$P = \alpha, \beta, d$$ from ECEF to ECMF, $$\alpha’=\alpha-\lambda$$, $$\beta’=\beta-\phi$$, and $$d’=d*k/r$$. Note that the moon’s position [$$\lambda$$, $$\phi$$, $$r$$] in ECEF always becomes $$[0,0,k]$$ in ECMF.

The ECMF coordinate system has some really bad qualities. It non-uniformly stretches the rest of the universe based on time. Depending on the direction, light no longer travels in a straight line! Regular shapes in ECEF become irregular in ECMF. The z-axis becomes irregularly offset from the Earth's rotation axis within the range of the Moon's inclination from the equatorial plane. All kinds of bad stuff happens in ECMF.

From the ECMF coordinate system, we can make another coordinate system called a ECMFDR (Earth Centered, Moon Fixed Distance, Rotating) system. This coordinate system just rotates the ECMF system around the z’-axis so that a full revolution takes 1 year, or $$p$$. To translate a point P = $$\alpha', \beta', d'$$ from ECMF to ECMFDR, $$\alpha’’=\alpha’$$, $$\beta’’=\beta’+2\pi(t-t_0)/p$$, and $$d’’=d’$$. ECMFDR (not to scale):

In our usual ECEF system, the moon’s orbit is not quite Keplerian. Instead, it's perturbed by the irregular shape of the earth, it's perturbed by other gravitational bodies in the solar system, and it's slowly spiraling outward, away from the Earth. In the ECMFDR system, the moon is, by definition and construction, in a perfectly circular orbit which is both a Kepler orbit and a closed orbit.

If we can create this coordinate system for the Earth and the Moon, then we can generalize it to any pair of orbiting bodies.

Every orbit is Keplerian: Consider a body $$b_2$$ in orbit around a body $$b_1$$. For a particular time $$t_0$$, define a coordinate system with an origin at the center of mass of $$b_1$$, with the center of mass of $$b_2$$ at $$[0,\sin(2\pi (t-t_0)/p),k]$$, where $$p$$ is one year, and $$k$$ is one AU.

Then $$b_2$$’s orbit around $$b_1$$ is Keplerian because it complies with Kepler’s 3 laws,

1. $$b_2$$’s orbit traces out an ellipse (since it traces out a circle and a circle is an ellipse).

2. A segment from $$b_1$$ to $$b_2$$ sweeps out equal area in equal time (since distance between $$b_1$$ and $$b_2$$ is always 1 AU, $$b_1$$ is fixed, and $$b_2$$ maintains constant speed)

3. All orbits around $$b_1$$ have the same ratio of square of the SMA divided by the cube of the period, since all orbits have an SMA of 1 AU and a period of one year.

Every orbit is closed. Keplerian orbits are closed orbits because a Keplerian orbit traces out a closed shape (the ellipse). Since every orbit is a Keplerian orbit, every orbit is a closed orbit.

• This was certainly waiting for, now it will take some time to figure out exactly what this does and doesn't say. Because gravity is what it is everything effects everything and usually people end up saying no orbit is Keplerian and no orbit is closed. You've said the opposite which is in dissonance to what people usually say, but I have a hunch you've made some interesting points here that need to be considered first. Hmm... :-) – uhoh Dec 5 '20 at 0:17
• @uhoh, I've done something pretty taboo with this answer, I think. I am happy to chat with you about it whenever you like, though perhaps there is a better forum than the comments. – Connor Garcia Dec 5 '20 at 0:21
• I don't really go off-site but here's a new chat room What are "non-Keplerian" orbits? For some reason it got created in Space SE's area instead of Astronomy SE's area (I think that that's related to this). – uhoh Dec 5 '20 at 0:30
• @copy that, I think I am in your chat and just said hello. – Connor Garcia Dec 5 '20 at 0:34