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

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.

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.

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.

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

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?

• 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? Jan 9 at 5:22
• @Orange Durito This is definitely a simplistic model which gives only qualitative predictions. But approximating an answer first helps recognize colossal computation errors later. However, I don’t think it will get me a job at JPL. Can you come up with any more “rules of thumb” from this model? Jan 9 at 5:47
• @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. Jan 10 at 13:20
• Jan 10 at 18:50
• @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. Jan 11 at 16:30

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 ------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. Jan 9 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". Jan 9 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? Jan 9 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. Jan 9 at 20:45
• @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. Jan 10 at 3:01