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Halo Orbits for Dummies: can halo orbits be usefully approximated by Simple Harmonic Motion?

I am an undergraduate in the Kerbal Academy of Astrodynamics. Patched conics have given me an intuitive understanding of 2-body orbital mechanics. But 3-body problems (like halo orbits) are as mysterious as quantum physics.

I’m hoping that, for small perturbations, an established L2 halo orbit can be treated as a simple harmonic oscillator. This would allow “orbital rules of thumb” that resolve some of the 3-body mystery.

To qualify as a radial harmonic oscillator, we only need to show the restoring force (towards L2) is linearly proportional to displacement from L2.

To demonstrate, consider an idealized 3-body system of Earth, Sun and JWST orbiting at L2. No Moon or Jupiter is present to make 4-body complications.

enter image description here

Our rotating frame of reference has the Earth and Sun stationary. Since the frame of reference is rotating, we deal with centrifugal force (which we know is nonexistent in inertial frames.) Centrifugal acceleration acts parallel to the Sun-Earth axis (at right angle to the frame’s rotational axis). Centrifugal acceleration is equal to the sum of Sun and Earth gravitational accelerations because that is the definition of a Lagrange point.

enter image description here

To simplify, we will only consider Earth’s gravity since similar results arise when the Sun’s gravity is included. The example uses the Z/X plane, but similar results arise from other planes of rotation around the X axis.

The centrifugal force acts parallel to the X axis, but gravitational attraction acts radially towards the Earth’s center. This generates a radial restoring force R towards the L2 point. R is proportional to sin a. For small values of angle a, R is proportional to angle a.

enter image description here

A restoring force proportional to displacement is the central requirement for Simple Harmonic Motion, and for a 2 dimensional Radial Harmonic Oscillator as described by Bertrand's Theorem.

The condition of proportional radial restoring force only holds for the Y/Z plane at L2. Deviations closer (P1) or further (P3) from Earth create a destabilizing component to the R vector along the X axis as in the sketch below

enter image description here

If this Simple Harmonic Oscillator model is valid, the following rules of thumb should apply to small, established halo orbits:

  1. All halo orbits at a given L2 will have the same orbital period, analogous to all swing periods of a pendulum being equal.
  2. Any circular halo orbit can be transferred to any other halo orbit by two tangential burns separated by 90* orbital phase angle. This is analogous to Hohmann Transfers, but in Hohmann Transfers the burns are 180* apart.
  3. Z and Y deviations are reasonably stable
  4. X deviations are destabilizing, particularly deviations towards earth since the instability is driven by an inverse square function.

Question: Can Simple Harmonic Motion usefully model established L2 Halo Orbits?

BrendanLuke15
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Woody
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  • different but related: What sort of orbital elements are used to describe halo orbits? and What is the period of halo/Lissajous orbits? and especially What is the "mass" of a Lagrange point? – uhoh Jan 09 '22 at 05:05
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    May I just say that this is a fantastic way of asking questions? I really enjoyed reading it. You clearly stated the premise, the theory, your conjecture and included illustrations to drive your point. Bravo! Your approach seems legit to me but people who much more knowledgeable would better describe the facts and fallacies. I have seen the orbit being represented by sinusoidal wave in an inertial frame of reference (please correct me if I'm wrong). Btw, is the enrollment for Kerbal Academy of Astrodynamics open for Fall 2022? How do I apply for their graduate program? – OrangeDurito Jan 09 '22 at 05:22
  • I have a problem understanding "The centrifugal force acts parallel to the X axis,...". I agree that this is the case for any point on the Z-X plane. But how can it be called centrifugal (running away from the center) if it is always parallel to X? – Ng Ph Jan 09 '22 at 22:21
  • @uhoh, answering "What sort of orbital elements are used to describe halo orbits?" amounts to telling Newton, Poincaré and scores of mathematicians that the 3-body problem has been solved analytically, for some (but infinite) particular cases. In other terms, with a finite set of parameters it would be possible to uniquely describe the periodic trajectory of a mass point, due to gravity from only 2 other primary mass points. My current "take away" is, todate there is no equivalent to Kepler's discovery of the law of the planets for Halo orbits. Would be delighted if somebody can dispute this. – Ng Ph Jan 09 '22 at 22:54
  • @NgPh I think you can post that as an answer there if you expand a bit. – uhoh Jan 09 '22 at 23:04
  • @Ng Ph Re: “centrifugal force parallel to X axis” I was referring to the plane of the second diagram only, not the rest of 3D space in the revolving frame of reference. I was attempting to make the point that the centrifugal vector is not parallel to the gravitational vector, which is aligned with the barycenter. It is this misalignment which produces the resultant R. – Woody Jan 09 '22 at 23:06
  • Then you should explicit in the sentence that it is true only in the plane of your 2nd diagram. But then how to interpret your sentence just above " ... similar results arise from other planes of rotation around the X axis"? How can a reader follow this generalization? – Ng Ph Jan 10 '22 at 09:02
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    @uhoh, it was not intended as an answer, but a warning. For my part, I am trying to undertsand what Newton, Lagrange, Alembert, Poincaré wrote themselves. Currently, I think that's an undertaking for the rest of my life. – Ng Ph Jan 10 '22 at 13:20
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    Related: https://space.stackexchange.com/questions/20590/how-to-best-think-of-the-state-transition-matrix-and-how-to-use-it-to-find-peri/ – ChrisR Jan 10 '22 at 18:50
  • "To simplify, we will only consider Earth’s gravity since similar results arise when the Sun’s gravity is included". I would thought it is the converse! At L2, according to my quick calculation, the Sun's pull is approximately 70 times that of Earth's pull. So, you need to demonstrate your theory with the Sun gravity alone first (which, I think would not work since L2 would not exist without Earth, but for the sake of simplification, why not?) – Ng Ph Jan 11 '22 at 10:31
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    @NgPh --- You are correct. The gravitational acceleration of the Sun at L2 is many times greater than the Earth's. However, the Sun's included angle "a" is much smaller (I calculate 0.3* vs 28* for JWST orbit). As a result, in the present example (Sun, Earth, JWST) the restoring force R has a much larger component from Earth's gravity than the Sun's. The Sun's destabilizing effects of X-axis movement are much lower than Earth's as well. – Woody Jan 11 '22 at 16:30
  • I think I now get what you are modelling. But I think your explanations are misleading. My guess is that you observe that the Sun pull is almost colinear to the centrifugal force. Around L2, it's mainly the Sun pull that almost cancels the centrifugal force. It's the residual centrifugal force that you showed in your 2nd diagram (you did not ignore, you substract). It's angle with the (small) Earth pull makes an angle that results in a net (small) force towards L2. It remains to show this in actual numbers. – Ng Ph Jan 12 '22 at 14:14
  • @NgPh ---- Correct. Note Roger Wood's comment following his answer. He calculated the simple harmonic motion halo orbital period as 7 months using only the Earth's gravity. But adding in the Sun's gravity brought the answer to 6 months. This is a reasonable approximation of JWST's L2 orbit. – Woody Jan 12 '22 at 15:41

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Nice descriptive question with the illustrations! I'll look forward to a more educated answer, but here are my two-cents:

[Edited in response to @Woody's comments]

Looking at the definition of a Halo orbit, it seems to be quite specific and involves all three dimensions and is not a small perturbation: "In 1973 Farquhar and Ahmed Kamel found that when the in-plane amplitude of a Lissajous orbit was large enough there would be a corresponding out-of-plane amplitude that would have the same period."

So it's out of the linearized regime that the paper by Neal Cornish describes. Also that paper describes only in-plane orbits.

However, there are small in-plane orbits for which the harmonic oscillator analogy presumably does apply. In the linearized system in the rotating frame, in plane, two pairs of eigenmodes can be excited. For L2/3, one pair of modes is oscilliatory at a frequency of $\sqrt{2 \sqrt7-1}$ times the orbital frequency (i.e. a period just under six months). This doesn't seem to depend on the mass of the secondary body (in contrast to the Trojans at L4/5). One of the other two modes corresponds to an exponentially increasing drift away from the Lagrange point with a time constant of $1/\sqrt{2 \sqrt7+1}$ or about 5 months.

For L4/5, small coplanar Lagrange orbits are ellipses with a 2:1 aspect ratio elongated along the direction of the planetary orbit. Presumably the orbits around L2/L3 are also elliptical but I don't know how to figure the eccentricity or orientation. In illustrations, the in-plane projection also looks to have about a 2:1 aspect ratio.

Cornish's linearized analysis shows (inevitably) that the period is independent of magnitude for small perturbations. This is certainly indicative of simple harmonic motion in the presence of a linear central force. However, the Coriolis force is a dominant factor in determining the shape of the orbit. So I don't think you have full freedom to set the eccentricity and orientation of the orbit as you would for a simple linear central force (questionable statement).

Again, this is just for very small orbits and ignores their inherent instability.

It seems like the pendulum analogy should be valid for small motions purely along the z-axis and avoiding any Coriolis effects. If I use the pendulum formula on your second figure, $T = 2 \pi \sqrt{L/g}$ with $L = 150 \times 10^9$ and $g = 9.8\times(6.4 \times 10^6/150 \times 10^9)^2$, I get a period of seven months, which doesn't quite match the quoted "about six months" for JWST.

Perhaps the halo orbit can be thought of as an orbit large enough that the in-plane orbital period and z-axis 'pendulum' period match each other with neither being in their linear harmonic motion regime?

Roger Wood
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  • @ Roger Wood ------The analysis in this post only applies to halo orbits perpendicular to the X axis (Sun-Earth) since there is no Coriolis effect (unless x-axis deviations are being considered). --------I don’t think there is a restriction on ellipse eccentricity for L2 orbits in this model. This model similar to Bertrand’s Radial Harmonic Oscillator (except restricted to 2 dimensions): orthogonal axis must have the same dV/dr restoring force but can have different amplitudes. ------- I believe the eccentricity of JWST halo was chosen to avoid Lunar eclipses. – Woody Jan 09 '22 at 16:37
  • @Woody I guess (from Wikipedia) a Halo orbit is not 'small' and specifially includes motions in all three axes: "In 1973 Farquhar and Ahmed Kamel found that when the in-plane amplitude of a Lissajous orbit was large enough there would be a corresponding out-of-plane amplitude that would have the same period". – Roger Wood Jan 09 '22 at 18:53
  • @Woody the Coriolis force acts in two dimensions and orbits are usually in two dimensions, so it's difficult to avoid Coriolis except for a straight-line 'orbit' along the z-axis - for which the pendulum formula should work? Maybe a Halo orbit is where the pendulum formula matches the period of a suitably large in-plane orbit? – Roger Wood Jan 09 '22 at 19:00
  • I think the Coriolis Force only acts on the X axis component of an object moving in a rotating plane of reference. it manifests as a force in the Y axis, but doesn't act on objects moving only in the Y axis. – Woody Jan 09 '22 at 20:45
  • @Woody I don't think Gaspard-Gustave de Coriolis would agree. It's omega cross velocity where omega and velocity are both vectors. So it doesn't affect motion in the z (omega) direction, but it does have an effect on the x and y axes in equal measure. – Roger Wood Jan 10 '22 at 02:11
  • It’s been a long time since school let out for me, so I’m groping with this. The rotation vector is parallel to the axis of rotation (Z axis). The object’s velocity is measured in the rotating reference frame (X axis). The cross product of these two vectors is at right angles to them both, (Y axis). Am I on the right page? – Woody Jan 10 '22 at 02:54
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    @Woody I'm equally rusty, if not more so. I rely heavily on Wikipedia. Velocity is a vector in any 3D direction but only the component in the X-Y plane (perpendicular to the rotation axis) gets involved. The force acts at 90 degrees to that velocity and also acts in the X-Y plane. – Roger Wood Jan 10 '22 at 03:01
  • That clears it up. – Woody Jan 10 '22 at 03:44
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    Looking at various examples of actual Halo orbits, in particular the "representative" one for JWST, I don't see any of them being confined to the YZ plane (orthogonal to X). So I agree with you that Halo orbits are 3D and would question Woody's method of analysis. I understand that Woody needs a simplifying assumption, but if this simplifying assumption is contrary to the main conditions of the problem, what's the point? – Ng Ph Jan 10 '22 at 13:34
  • @RogerWood ----- In your answer, you calculated a simple harmonic period of seven months. Did you include the restoring force R from the sun? The Sun's Gravitational field is 33 times as strong as the earth's at L2, but the angle "a" is much smaller than that measured at the Earth. In my posted example I ignored the Sun's effect to simplify (and hopefully clarify! ) the example. – Woody Jan 11 '22 at 17:02
  • @Woody good point! I did think about it but assumed it was insignificant. It's 333,000 times heavier but about 100 times further away. So that's the 33 times stronger. But the length is also 100 times longer, so the angle is 100 times smaller, but that still leaves 0.33 which sounds like a good correction. That would make the period 1/sqrt(1+.33) = shorter i.e. almost exactly 6 months! – Roger Wood Jan 11 '22 at 20:11
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    See Eqs. 19 and 20 in Farquhar & Kamel 1973 Quasi-Periodic Orbits about the Translunar Libration Point For the first-order linearized solution they get $\omega_{xy}$ = 1.865 The CR3BP frame rotates at $\omega$ = 1 (period T = $2 \pi$) so that's 12/ 1.865 = 6.4 months. That paper plus Howell 1983 Three-Dimensional Periodic Halo Orbits have all the halo goodies. – uhoh Jan 18 '22 at 11:34
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    @uhoh I missed this earlier comment. That Farquhar paper does sound like the definitive paper and especially the amazing thesis on which it's based (and is a bit easier to understand) https://ntrs.nasa.gov/api/citations/19710000821/downloads/19710000821.pdf The thesis Table 2.5 quotes the Earth-Sun L2 as 2.057. I gather Cornish's sqrt(2.sqrt(7)-1) would be for infinite ratio of Sun/Earth mass. – Roger Wood Feb 05 '22 at 02:13