<blockquote>
<ol>
<li>What exactly are &quot;non-Keplerian&quot; orbits?</li>
</ol>
</blockquote>
<p>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:</p>
<blockquote>
<ol>
<li>The orbit of a planet is an ellipse with the Sun at one of the two foci.</li>
<li>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</li>
<li>The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its orbit.</li>
</ol>
</blockquote>
<p>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.</p>
<p>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): <a href="https://astronomy.stackexchange.com/questions/32859/if-the-moon-were-impacted-by-a-suitably-sized-meteor-how-long-would-it-take-to/39558#39558">If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth?</a> .  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!</p>
<p>A &quot;non-Keplerian&quot; orbit is an orbit in which Kepler's laws lack predictive and descriptive power. <strong>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.</strong>  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.</p>
<p>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.</p>
<p>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:</p>
<p>a)       Orbiting bodies move in elliptical orbits around the system barycenter.</p>
<p>b)      A line segment between the barycenter and a body sweeps out equal areas in equal time.</p>
<p>c)       The square of a body’s orbital period is proportional to the cube of its mean distance from the barycenter.</p>
<p>Here is a table I made to accompany my definition of non-Keplerian orbits:<a href="https://i.sstatic.net/UwpoC.png" rel="noreferrer"><img src="https://i.sstatic.net/UwpoC.png" alt="enter image description here" /></a></p>
<blockquote>
<ol start="2">
<li>What are some familiar examples of distinctly non-Keplerian our solar system?</li>
</ol>
</blockquote>
<p>I think the easiest example of a mostly non-Keplerian orbit is the James Webb telescope soon to be at Earth's L2 point: <a href="https://en.wikipedia.org/wiki/Lagrange_point" rel="noreferrer">https://en.wikipedia.org/wiki/Lagrange_point</a> . 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: <a href="http://www.astro.uwo.ca/%7Ewiegert/2015BZ509/" rel="noreferrer">http://www.astro.uwo.ca/~wiegert/2015BZ509/</a> This is an example of a 3-body problem orbit.</p>
<blockquote>
<ol start="3">
<li>What are closed orbits?</li>
</ol>
</blockquote>
<p>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.</p>
<blockquote>
<ol start="4">
<li>Can some non-Keplerian orbits still be closed?</li>
</ol>
</blockquote>
<p>Yes, the two orbit examples given above are closed enough. Oumuamua is an example of a &quot;non-closed&quot; &quot;non-Keplerian&quot; solar orbit. <a href="https://en.wikipedia.org/wiki/%CA%BBOumuamua" rel="noreferrer">https://en.wikipedia.org/wiki/%CA%BBOumuamua</a></p>

CC BY-SA 4.0
2020-11-03T02:52:51.707Z
Thanks for your answer! DSCOVR and SOHO are also good examples, and unlike JWST they are actually in space :-)
uhoh
CC BY-SA 4.0
2020-11-03T02:54:18.367Z
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](https://en.wikipedia.org/wiki/Bertrand%27s_theorem).
David Hammen
CC BY-SA 4.0
2020-11-03T02:55:43.977Z
Welcome to Stack Exchange! I saw your profile; you may be able to track down answers to the following: [How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?](https://astronomy.stackexchange.com/q/28666) and [Why are the ALMA receivers' ADCs only 3-bits?](https://astronomy.stackexchange.com/q/18052)
uhoh
CC BY-SA 4.0
2020-11-03T03:03:10.960Z
David Hammen - How embarrassing! You are absolutely correct, I changed my answer accordingly above.
Connor Garcia
CC BY-SA 4.0
2020-11-03T03:06:07.010Z
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
CC BY-SA 4.0
2020-11-03T23:56:07.193Z
@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
CC BY-SA 4.0
2020-11-04T19:19:55.567Z
As @BenCrowell, notes, among other things surely "secular terms" are definitely non-Keplerian, but in some way should count as "non-exotic"?
paul garrett
CC BY-SA 4.0
2023-08-18T07:33:54.717Z
The square of the period is proportional to the cube of the mean separation, not the distance to the barycentre (which is smaller). Two body orbits of point masses are Keplerian.
ProfRob
CC BY-SA 4.0
2020-11-03T02:16:56.220Z
2020-11-06T01:52:11.893Z
2020-11-06T01:52:11.893Z
Connor Garcia
<blockquote>
<p>What exactly are &quot;non-Keplerian&quot; orbits?</p>
</blockquote>
<p>Orbits that don't follow Kepler's laws.</p>
<p>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.</p>
<blockquote>
<p>What are some familiar examples of distinctly non-closed orbits in our solar system?</p>
</blockquote>
<p>Every planet, every moon, every asteroid, ...</p>
<blockquote>
<p>What are closed orbits?</p>
</blockquote>
<p>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 <a href="https://en.wikipedia.org/wiki/Bertrand%27s_theorem" rel="noreferrer">Bertrand's Theorem</a>.</p>
<blockquote>
<p>What are some familiar examples of distinctly non-Keplerian orbits in our solar system?</p>
</blockquote>
<p>Every planet, every moon, every asteroid, ...</p>
<blockquote>
<p>Can some non-Keplerian orbits still be closed?</p>
</blockquote>
<p>Other than a Hookean spring, no.</p>

CC BY-SA 4.0
2020-11-03T03:22:24.557Z
Do orbits in tunnels embedded within spherically symmetric bodies count as Hookean?
uhoh
CC BY-SA 4.0
2020-11-03T03:31:40.930Z
@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
CC BY-SA 4.0
2020-11-03T03:19:36.980Z
2020-11-04T08:12:19.607Z
2020-11-04T08:12:19.607Z
David Hammen
<p><strong>Preface:</strong></p>
<p>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].</p>
<p>Throughout history, CS (Coordinate Systems) have been a huge bone of contention.  Einstein said:</p>
<blockquote>
<p>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.</p>
</blockquote>
<p>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:</p>
<p><strong>The Moon's orbit around the Earth is Keplerian and closed.</strong></p>
<p>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:</p>
<p><a href="https://i.sstatic.net/QWL9s.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QWL9s.gif" alt="enter image description here" /></a></p>
<p>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:</p>
<p><a href="https://i.sstatic.net/9g01x.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9g01x.jpg" alt="enter image description here" /></a></p>
<p>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.</p>
<p>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 <span class="math-container">$k=500,000km$</span>.  ECMF (not to scale):</p>
<p><a href="https://i.sstatic.net/7rNOB.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7rNOB.gif" alt="enter image description here" /></a></p>
<p>The coordinate transformation from ECEF to ECMF is dependent on the moon’s polar coordinates in ECEF at time <span class="math-container">$t$</span>: [<span class="math-container">$\lambda$</span>, <span class="math-container">$\phi$</span>, <span class="math-container">$r$</span>].  To translate a point <span class="math-container">$P = \alpha, \beta, d$</span> from ECEF to ECMF, <span class="math-container">$\alpha’=\alpha-\lambda$</span>, <span class="math-container">$\beta’=\beta-\phi$</span>, and <span class="math-container">$d’=d*k/r$</span>.  Note that the moon’s position [<span class="math-container">$\lambda$</span>, <span class="math-container">$\phi$</span>, <span class="math-container">$r$</span>] in ECEF always becomes <span class="math-container">$[0,0,k]$</span> in ECMF.</p>
<p>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.</p>
<p>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 <span class="math-container">$p$</span>.  To translate a point P = <span class="math-container">$\alpha', \beta', d'$</span> from ECMF to ECMFDR, <span class="math-container">$\alpha’’=\alpha’$</span>, <span class="math-container">$\beta’’=\beta’+2\pi(t-t_0)/p$</span>, and <span class="math-container">$d’’=d’$</span>. ECMFDR (not to scale):</p>
<p><a href="https://i.sstatic.net/ulWjj.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ulWjj.gif" alt="enter image description here" /></a></p>
<p>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.</p>
<p>If we can create this coordinate system for the Earth and the Moon, then we can generalize it to any pair of orbiting bodies.</p>
<p><strong>Every orbit is Keplerian:</strong> Consider a body <span class="math-container">$b_2$</span> in orbit around a body <span class="math-container">$b_1$</span>. For a particular time <span class="math-container">$t_0$</span>, define a coordinate system with an origin at the center of mass of <span class="math-container">$b_1$</span>, with the center of mass of <span class="math-container">$b_2$</span> at <span class="math-container">$[0,\sin(2\pi (t-t_0)/p),k]$</span>, where <span class="math-container">$p$</span> is one year, and <span class="math-container">$k$</span> is one AU.</p>
<p>Then <span class="math-container">$b_2$</span>’s orbit around <span class="math-container">$b_1$</span> is Keplerian because it complies with Kepler’s 3 laws,</p>
<ol>
<li><p><span class="math-container">$b_2$</span>’s orbit traces out an ellipse (since it traces out a circle and a circle is an ellipse).</p>
</li>
<li><p>A segment from <span class="math-container">$b_1$</span> to <span class="math-container">$b_2$</span> sweeps out equal area in equal time (since distance between <span class="math-container">$b_1$</span> and <span class="math-container">$b_2$</span> is always 1 AU, <span class="math-container">$b_1$</span> is fixed, and <span class="math-container">$b_2$</span> maintains constant speed)</p>
</li>
<li><p>All orbits around <span class="math-container">$b_1$</span> 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.</p>
</li>
</ol>
<p><strong>Every orbit is closed.</strong> 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.</p>

CC BY-SA 4.0
2020-12-05T00:08:54.217Z
2020-12-05T01:38:21.303Z
2020-12-05T01:38:21.303Z
Connor Garcia
<blockquote>
<h2>What exactly are &quot;non-Keplerian&quot; orbits?</h2>
</blockquote>
<p>Let's first start with Keplerian orbits. A Keplerian orbit is the motion of one body relative to another body, and takes the form of an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space according to <a href="https://en.wikipedia.org/wiki/Kepler_orbit" rel="nofollow noreferrer">Wikipedia</a>. Keplerian orbits also neglect atmospheric drag, solar radiation pressure, non-spherical central bodies, and other perturbations. Now, you could say &quot;Oh but if Keplerian orbits neglect perturbations, then perturbations related to gravitational interactions therefore 'says' that a spacecraft orbiting Earth is non-Keplerian!&quot; It <em>technically</em> is... but for the sake of avoiding doubt in this answer, let's just assume that orbits do <em>not</em> account for gravity perturbations, as historically the term &quot;non-Keplerian&quot; has been used in reference to specific orbits where a perturbing or propulsive acceleration acts on the orbit in addition to the gravitational interactions. So we will assume that other perturbations such as solar pressure will be formally termed as &quot;Non-Keplerian orbits&quot; (in this answer and in <a href="https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion" rel="nofollow noreferrer">this paper - Survey of Highly Non-Keplerian Orbits with Low Thrust Propulsion</a>). A special subset of the non-Keplerian orbit set would then be the <strong>highly</strong> non-Keplerian orbits, where the term “highly” will refer to the magnitude of the perturbing force.</p>
<p><em>I am using this paper that describes Highly Non-Keplerian orbits as for some reason research papers on Non-Keplerian orbits are <strong>extremely</strong> hard to find.</em></p>
<p>In order to have a non-Keplerian object, some additional accelerating/propelling force (beyond just gravity from a point mass) is required. For a highly non-Keplerian orbit, the time-average of this acceleration over a full orbit is considered to be of at least equal or greater magnitude than the sum of the gravitational and centripetal accelerations experienced by the object (as viewed in a reference frame rotating at a given angular velocity relative to an inertial reference frame).</p>
<p>Now let's introduce a parameter <span class="math-container">$\lambda$</span> to represent the ratio, therefore <span class="math-container">$\lambda = \Vert \mathbf{a}_{av} \Vert / \Vert \nabla V \Vert \geq 1$</span> represents highly non-Keplerian orbits, where <span class="math-container">$\Vert \mathbf{a}_{av} \Vert$</span> is the time-average of the magnitude of the propulsive acceleration over one complete highly non-Keplerian orbit, and <span class="math-container">$\Vert \nabla V \Vert$</span> is the magnitude of the gravitational and centripetal accelerations experienced by the object (This information will be needed to explain non-Keplerian orbits later on).</p>
<p><strong>Warning</strong> from the <a href="https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion" rel="nofollow noreferrer">research paper</a>:</p>
<blockquote>
<p>Even with such a careful definition, the authors do not consider this definition to be <strong>&quot;fool-proof&quot;</strong> - by which it is meant that, in the course of this paper orbits with <span class="math-container">$\lambda \ggg 1$</span> are not considered (if such orbits even exist). For the purposes of this review, <span class="math-container">$\lambda$</span> may be greater than exactly 1 but is still approximately of order 1.</p>
</blockquote>
<p>Now let's go through how <span class="math-container">$\lambda$</span> represents the different types of orbits.   Orbits with <span class="math-container">$\lambda = 0$</span> represent simple Keplerian orbits - in essence the large subset of classical celestial orbital mechanics without the addition of a non-conservative force. Orbits with <span class="math-container">$\lambda \ll 1$</span> (but nevertheless finite) are, in effect, weakly-perturbed Keplerian orbits, and may be referred to as non-Keplerian orbits.</p>
<p>The figure below shows the classification of Keplerian orbits, non-Keplerian orbits and highly non-Keplerian orbits, where the box represents the set of all types of orbits.</p>
<p><a href="https://i.sstatic.net/wY7uk.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wY7uk.png" alt="Set of orbits" /></a></p>
<h2>Examples of non-Keplerian orbits</h2>
<blockquote>
<p>There are numerous examples of such orbits: for instance, there is a considerable body of work which has considered the use of rather small solar sails to artificially precess Earth-centered elliptical orbits for space physics mission applications. Macdonald and McInnes showed that an ellipse will be precessed if the sail is Sun-facing, leaving the orbit averaged semi-major axis and eccentricity unchanged. By an appropriate choice of sail loading, the elliptical orbit can be forced to precess at a Sun-synchronous rate, maintaining a science payload permanently within the geomagnetic tail, a concept utilized by the GeoSail mission. An even more recent example of such orbits is the extended Sun-synchronous orbits proposed by Macdonald et al.. A separate example of non-Keplerian orbits would be those of the open spiral trajectories used for spacecraft transfers. It is worth remembering, however, that although these examples are of, in strict terms, non-Keplerian orbits, they are a quite separate subset of orbits to those considered in this review.</p>
</blockquote>
<h3>Sources:</h3>
<ul>
<li><a href="https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion" rel="nofollow noreferrer">Survey of Highly Non-Keplerian Orbits with Low Thrust Propulsion</a>)</li>
</ul>

CC BY-SA 4.0
2023-08-20T20:41:38.553Z
2023-08-20T20:41:38.553Z
DialFrost
<p><a href="https://astronomy.stackexchange.com/a/39621/7982">This excellent answer</a> to <em>Forms of stellar orbits around the galactic center</em> invokes the following concepts:</p>
<ul>
<li>non-Keplerian orbits</li>
<li>closed orbits</li>
</ul>
<p>I have a fairly good idea what these mean and so might many of us, but our ideas may not overlap completely.</p>
<h3>For the sake of future stack exchangers and orbiteers, how can we best define these?</h3>
<p><strong>Question(s):</strong></p>
<ol>
<li>What exactly are &quot;non-Keplerian&quot; orbits?</li>
<li>What are some familiar examples of distinctly non-Keplerian orbits in our solar system?</li>
<li>What are closed orbits?</li>
<li>What are some familiar examples of distinctly non-closed orbits in our solar system?</li>
<li>Can some non-Keplerian orbits still be closed?</li>
</ol>
<p>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 &quot;gravitational dielectrics&quot; 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.</p>

CC BY-SA 4.0
2020-11-03T02:49:16.103Z
Strictly speaking, all orbits are non-Keplerian and no orbits are closed.
David Hammen
CC BY-SA 4.0
2020-11-03T02:50:12.887Z
@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
CC BY-SA 4.0
2020-11-15T05:11:39.310Z
@uhoh Would you consider giving another week or two before you choose an answer on this question? I've been contemplating giving a third answer, but it is fiendishly difficult.
Connor Garcia
CC BY-SA 4.0
2020-11-16T00:09:21.673Z
@ConnorGarcia I was doing a weekend review of my posts and noticed all the additional goodies in your current answer and momentarily clicked accept, then noticed your comment above. Sure I'm happy to wait. I"m always game for reading something that a mathematician describes as "fiendishly difficult" :-)
uhoh
CC BY-SA 4.0
2020-11-16T01:25:55.617Z
@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
CC BY-SA 4.0
2020-12-03T05:57:44.973Z
@uhoh I will post a second answer by Tuesday of next week.
Connor Garcia
CC BY-SA 4.0
2020-12-03T06:02:47.920Z
@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
CC BY-SA 4.0
2022-01-19T00:39:14.093Z
@uhoh I don't plan to add any more answers to the question at this point, so please feel free to choose an answer if you wish!
Connor Garcia
CC BY-SA 4.0
2022-01-19T01:27:54.950Z
@ConnorGarcia thanks for calling this to my attention! I'm going to add it to my "to do today" list. SE doesn't have built-in reminder alarms unfortunately so wish me luck!
uhoh
CC BY-SA 4.0
2023-08-17T16:26:21.020Z
I don't mean to say these answers aren't complete, they're awesome :), but if you need any extra info/answers for this question @uhoh I'll try my best to answer :D
DialFrost
CC BY-SA 4.0
2023-08-17T20:08:46.633Z
@DialFrost  you've piqued my curiosity; *go for it!*
uhoh
CC BY-SA 4.0
2023-08-20T19:27:54.883Z
Is there any specific reason why it's so hard to find anything about Non-Keplerian orbits on the internet :P
DialFrost
<orbit><gravity><orbital-mechanics><orbital-elements><terminology>
What are "non-Keplerian" orbits? What are some familiar examples in our solar system, and can some still be closed?
uhoh
# What are "non-Keplerian" orbits? What are some familiar examples in our solar system, and can some still be closed?

(Asked by: username_0 on 2020-11-03)

[This excellent answer](https://astronomy.stackexchange.com/a/39621/7982) 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.

**Tags:** orbit, gravity, orbital-mechanics, orbital-elements, terminology
> **Comments:**
> **username_1**: Strictly speaking, all orbits are non-Keplerian and no orbits are closed.  
> **username_0**: @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.  
> **username_2**: @uhoh Would you consider giving another week or two before you choose an answer on this question? I've been contemplating giving a third answer, but it is fiendishly difficult.  
> **username_0**: @ConnorGarcia I was doing a weekend review of my posts and noticed all the additional goodies in your current answer and momentarily clicked accept, then noticed your comment above. Sure I'm happy to wait. I"m always game for reading something that a mathematician describes as "fiendishly difficult" :-)  
> **username_2**: @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!  
> **username_2**: @uhoh I will post a second answer by Tuesday of next week.  
> **username_0**: @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!  
> **username_2**: @uhoh I don't plan to add any more answers to the question at this point, so please feel free to choose an answer if you wish!  
> **username_0**: @ConnorGarcia thanks for calling this to my attention! I'm going to add it to my "to do today" list. SE doesn't have built-in reminder alarms unfortunately so wish me luck!  
> **username_3**: I don't mean to say these answers aren't complete, they're awesome :), but if you need any extra info/answers for this question @uhoh I'll try my best to answer :D  
> **username_0**: @DialFrost  you've piqued my curiosity; *go for it!*  
> **username_3**: Is there any specific reason why it's so hard to find anything about Non-Keplerian orbits on the internet :P  

---

## 4 answers

### Answer by username_2 on 2020-11-03. Score: 16

> 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?](https://astronomy.stackexchange.com/questions/32859/if-the-moon-were-impacted-by-a-suitably-sized-meteor-how-long-would-it-take-to/39558#39558) . 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:[![enter image description here](https://i.sstatic.net/UwpoC.png)](https://i.sstatic.net/UwpoC.png)

> 2. 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/](http://www.astro.uwo.ca/%7Ewiegert/2015BZ509/) This is an example of a 3-body problem orbit.

> 3. 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.

> 4. 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>

> **Comments:**
> **username_0**: Thanks for your answer! DSCOVR and SOHO are also good examples, and unlike JWST they are actually in space :-)  
> **username_1**: 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](https://en.wikipedia.org/wiki/Bertrand%27s_theorem).  
> **username_0**: Welcome to Stack Exchange! I saw your profile; you may be able to track down answers to the following: [How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?](https://astronomy.stackexchange.com/q/28666) and [Why are the ALMA receivers' ADCs only 3-bits?](https://astronomy.stackexchange.com/q/18052)  
> **username_2**: David Hammen - How embarrassing! You are absolutely correct, I changed my answer accordingly above.  
> **username_2**: 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.  
> **username_0**: @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.  
> **username_4**: As @BenCrowell, notes, among other things surely "secular terms" are definitely non-Keplerian, but in some way should count as "non-exotic"?  
> **username_5**: The square of the period is proportional to the cube of the mean separation, not the distance to the barycentre (which is smaller). Two body orbits of point masses are Keplerian.  

---

### Answer by username_1 on 2020-11-03. Score: 15

> 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](https://en.wikipedia.org/wiki/Bertrand%27s_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.

> **Comments:**
> **username_0**: Do orbits in tunnels embedded within spherically symmetric bodies count as Hookean?  
> **username_1**: @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 $\frac43 \pi G \rho$.  

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### Answer by username_2 on 2020-12-05. Score: 3

**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:

[![enter image description here](https://i.sstatic.net/QWL9s.gif)](https://i.sstatic.net/QWL9s.gif)

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:

[![enter image description here](https://i.sstatic.net/9g01x.jpg)](https://i.sstatic.net/9g01x.jpg)

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):

[![enter image description here](https://i.sstatic.net/7rNOB.gif)](https://i.sstatic.net/7rNOB.gif)

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):

[![enter image description here](https://i.sstatic.net/ulWjj.gif)](https://i.sstatic.net/ulWjj.gif)

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.


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### Answer by username_3 on 2023-08-20. Score: 2

> ## What exactly are "non-Keplerian" orbits?

Let's first start with Keplerian orbits. A Keplerian orbit is the motion of one body relative to another body, and takes the form of an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space according to [Wikipedia](https://en.wikipedia.org/wiki/Kepler_orbit). Keplerian orbits also neglect atmospheric drag, solar radiation pressure, non-spherical central bodies, and other perturbations. Now, you could say "Oh but if Keplerian orbits neglect perturbations, then perturbations related to gravitational interactions therefore 'says' that a spacecraft orbiting Earth is non-Keplerian!" It *technically* is... but for the sake of avoiding doubt in this answer, let's just assume that orbits do *not* account for gravity perturbations, as historically the term "non-Keplerian" has been used in reference to specific orbits where a perturbing or propulsive acceleration acts on the orbit in addition to the gravitational interactions. So we will assume that other perturbations such as solar pressure will be formally termed as "Non-Keplerian orbits" (in this answer and in [this paper - Survey of Highly Non-Keplerian Orbits with Low Thrust Propulsion](https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion)). A special subset of the non-Keplerian orbit set would then be the **highly** non-Keplerian orbits, where the term “highly” will refer to the magnitude of the perturbing force.

*I am using this paper that describes Highly Non-Keplerian orbits as for some reason research papers on Non-Keplerian orbits are **extremely** hard to find.*

In order to have a non-Keplerian object, some additional accelerating/propelling force (beyond just gravity from a point mass) is required. For a highly non-Keplerian orbit, the time-average of this acceleration over a full orbit is considered to be of at least equal or greater magnitude than the sum of the gravitational and centripetal accelerations experienced by the object (as viewed in a reference frame rotating at a given angular velocity relative to an inertial reference frame).

Now let's introduce a parameter $\lambda$ to represent the ratio, therefore $\lambda = \Vert \mathbf{a}\_{av} \Vert / \Vert \nabla V \Vert \geq 1$ represents highly non-Keplerian orbits, where $\Vert \mathbf{a}\_{av} \Vert$ is the time-average of the magnitude of the propulsive acceleration over one complete highly non-Keplerian orbit, and $\Vert \nabla V \Vert$ is the magnitude of the gravitational and centripetal accelerations experienced by the object (This information will be needed to explain non-Keplerian orbits later on).

**Warning** from the [research paper](https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion):

> Even with such a careful definition, the authors do not consider this definition to be **"fool-proof"** - by which it is meant that, in the course of this paper orbits with $\lambda \ggg 1$ are not considered (if such orbits even exist). For the purposes of this review, $\lambda$ may be greater than exactly 1 but is still approximately of order 1.

Now let's go through how $\lambda$ represents the different types of orbits. Orbits with $\lambda = 0$ represent simple Keplerian orbits - in essence the large subset of classical celestial orbital mechanics without the addition of a non-conservative force. Orbits with $\lambda \ll 1$ (but nevertheless finite) are, in effect, weakly-perturbed Keplerian orbits, and may be referred to as non-Keplerian orbits.

The figure below shows the classification of Keplerian orbits, non-Keplerian orbits and highly non-Keplerian orbits, where the box represents the set of all types of orbits.

[![Set of orbits](https://i.sstatic.net/wY7uk.png)](https://i.sstatic.net/wY7uk.png)

## Examples of non-Keplerian orbits

> There are numerous examples of such orbits: for instance, there is a considerable body of work which has considered the use of rather small solar sails to artificially precess Earth-centered elliptical orbits for space physics mission applications. Macdonald and McInnes showed that an ellipse will be precessed if the sail is Sun-facing, leaving the orbit averaged semi-major axis and eccentricity unchanged. By an appropriate choice of sail loading, the elliptical orbit can be forced to precess at a Sun-synchronous rate, maintaining a science payload permanently within the geomagnetic tail, a concept utilized by the GeoSail mission. An even more recent example of such orbits is the extended Sun-synchronous orbits proposed by Macdonald et al.. A separate example of non-Keplerian orbits would be those of the open spiral trajectories used for spacecraft transfers. It is worth remembering, however, that although these examples are of, in strict terms, non-Keplerian orbits, they are a quite separate subset of orbits to those considered in this review.

### Sources:

* [Survey of Highly Non-Keplerian Orbits with Low Thrust Propulsion](https://www.researchgate.net/publication/270030742_Survey_of_Highly_Non-Keplerian_Orbits_with_Low-Thrust_Propulsion))


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