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L2 Halo orbits are often portrayed as a pair of “North” and “South” orbits. Could the difference in “tilt” around the Y axis be explained by the direction of rotation and the Coriolis Effect?

enter image description here https://www.semanticscholar.org/paper/Maintenance-of-earth-moon-halo-orbit-Liu-Hu/6fa14df4563cd71841a53b0017ff44264ac45cee

Below is a sketch of the Sun-Earth L2 and a closer view of two hypothetical counter-rotating halo orbits

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Since this is a rotating frame of reference, the Coriolis Force will affect objects moving in the X-Y plane in relation to the frame of reference. Movement along the Z axis will experience no Coriolis Effect.

In sketch a) below, the green, “counterclockwise” orbit is isolated. The blue arrows indicate the Z-axis which is the rotational axis of the frame of reference. The dark green arrows represent the velocity vectors at various points on the orbit. The orange arrows represent the Coriolis Force acting at that point in the orbit. In sketch b), extraneous vectors have been removed. The dashed green line shows the expected displacement of the orbit around the Y axis due to the Coriolis Force.

Sketch c) shows similar effect on the counter-rotating red orbit, but with the tilt in the opposite direction as expected. Sketch d) shows the relative tilt of the two orbits together.

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The final sketch e) shows a slightly rotated view

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I cannot source any information on the orbital direction of “North” and “South” halo orbits. Is Coriolis Force a reasonable model for the difference in their orientation?

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    $\begingroup$ the caption on figure 5 in citeseerx.ist.psu.edu/viewdoc/… seems to emphasize the relative independence of the in-plane (ecliptic) motion and the out-of-plane (polar axis) motion, which is what your good SHM question led me to believe earlier. $\endgroup$
    – Roger Wood
    Jan 19, 2022 at 7:48
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    $\begingroup$ I have a hard time reading your own plots - there are not enough visual cues for my eyes to see how they are supposed to be oriented in 3D. $\endgroup$
    – asdfex
    Jan 19, 2022 at 9:53
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    $\begingroup$ @asdfex I wish we could somehow easily generate and embed WebGL in our SE posts. What are ways that we can show 3D orbits and other 3D things in posts? $\endgroup$
    – uhoh
    Jan 19, 2022 at 10:05
  • $\begingroup$ If I get the drawings right, the orange arrows are not Coriolis, but residual centrifugal forces. $\endgroup$
    – asdfex
    Jan 19, 2022 at 10:27
  • $\begingroup$ @RogerWood ---- but the caption says "The amplitudes can be chosen such that the frequencies become equal.." In other words, there are a range of diameters for halo orbits. Outside this range, the "orbit" breaks down. Very small halo orbits don't exist. $\endgroup$
    – Woody
    Jan 19, 2022 at 17:32

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halo orbit 'tilt' has nothing to do with coriolis but is caused by the object speeding up and slowing down as it orbits the sun.

In a Sun-Earth L2 halo orbit the object is ALWAYS orbiting the sun in the same direction as the earth at an AVERAGE orbital velocity equal to earth's. HOWEVER... the object in halo speeds up and slows down a tiny bit so that it is always in a cycle of apparently overtaking the earth or falling behind the earth. simple 'orbital velocity about Sun' calculation results in an orbit closer to sun when object is faster (overtaking earth) and orbit farther from sun when slower (falling back from earth).

Because of this behavior it is impossible to get a halo orbit with no tilt. It is possible to get a halo with 90 degree tilt (halo plane in same plane as earth orbit).

EDIT: earth gravity causes speedup and slowdown. When object is 'behind' earth the earth gravity works to pull object along, increasing its speed. In same way, when object is 'ahead' of earth the earth gravity slows down the object. Max object velocity at 12 oclock position gives minimum radius sun orbit, minimum velocity at 6 oclock gives maximum radius sun orbit for "north" halo.

Edit 2-24-2022: I need to acknowledge that I am in way over my head here. I no longer 'have the chops' to support my assertions in a scientific/mathematical way. I would like to invite ANYONE to provide another answer. Just because I can no longer 'math out' the answer does not automatically mean my assertion is wrong. Its just not supported.

THIS IS AN INTERESTING ASPECT OF ORBITAL MECHAINCS. I WISH BUZZ ALDRIN COULD HELP HERE!

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  • $\begingroup$ I think you've captured the idea that the orbit around L2 has to be retrograde. I'm not sure exactly what the definition of a halo orbit is, but I think it refers to a 3D closed orbit that includes a weak definition of stability such that small disturbances shift the orbit but don't necessarily lead to a runaway instability. I don't think this is true for the small elliptical orbits in the plane of the ecliptic or for small linear oscillations along the z-axis $\endgroup$
    – Roger Wood
    Jan 22, 2022 at 20:57
  • $\begingroup$ @BradV ---- An object at SEL2 (or a halo orbit around it) does NOT have the same average orbital velocity as the Earth. It has the same angular velocity as Earth, but because it is 1% further from the Sun, its average orbital velocity must be 1% greater that the Earth's. The object is not in a Keplerian orbit around the Sun, so you cannot use 2-body formulas for radius/speed/period. $\endgroup$
    – Woody
    Jan 22, 2022 at 22:01
  • $\begingroup$ @BradV Woody is correct. The Earth's field gradient dominates at L2, so it's better to say closer or further from the Earth. I do like the recent edit about the object lagging or leading the Earth (though I'm not sure where 12 o'clock and 6 o'clock are) $\endgroup$
    – Roger Wood
    Jan 23, 2022 at 6:48
  • $\begingroup$ @Woody.. good catch! Thanks! and Roger Wood... I wish I could recall the acronym for the non-inertial reference frame used for looking at halo orbits. Basically, X axis is through system barycenter and +x direction passes thru second body. The +Y axis is thru the L point being discussed in the direction of planet orbit. The +Z axis points in direction of system north. SO... 12 o'clock when viewing a halo orbit from this reference frame would be when object has +z, zero y location, and 6 o'clock would be at -z, zero y location. $\endgroup$
    – BradV
    Jan 24, 2022 at 4:31
  • $\begingroup$ @roger wood I'm still not sure if I buy into the idea that a halo orbit can be prograde or retrograde. The terms I've read are 'north' which is also known as class 1 and 'south' or class 2. However... I have no idea what these definitions mean in terms of direction. I'm guessing that north (class 1) means object is north side (+Z) when prograde. $\endgroup$
    – BradV
    Jan 24, 2022 at 6:49

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