After plotting the JWST halo orbit to a flat plane, and finding that the halo appears to closely approximate an ellipse (yes, the halo has some 'potato chip' warping) I began to wonder. A planetary orbit sweeps out an equal area for a given time period. Would a halo orbit do the same? Since there is no equivalent gravitational center for the halo,(EDIT: should have said 'gravitational center AT the halo') I'd guess 'NO'. Since I have no means to crunch the numbers in a supportable manner I'd like the community here to look at this!

Question: In a Rotating Libration Point framework, would a "swept area per time unit" be the same?

enter image description here

came across this today. enter image description here


1 Answer 1


I don't know the exact answer, but my guess is that it is approximately true for JWST for two reasons:

  1. For small perturbations from L2, there are linear restoring forces that result in simple harmonic motion. This is true along the axis of rotation (z-axis) and also in the direction of rotation (y-axis). In the ecliptic plane (x-y plane), the Coriolis force acts to automatically bend the resulting simple harmonic motion into an ellipse.

  2. So, at least for small perturbations, there is an apparent linear central restoring force that gives rise to the elliptical orbits. Orbits arising from a central force conserve angular momentum which means the orbit sweeps out equal areas in equal times. This is true whether it's an inverse-square law (Kepler) or a linear law (Hooke). Orbits with Hooke's law correspond to simple harmonic motion.

Obviously the motions for JWST are not small and the family of halo orbits does incude some very extreme examples. (These include highly elliptical orbits that pass close to the secondary body and are not centered very close to the Lagrange point.)

However, the JWST orbit seems relatively well-behaved. It is fairly well centered on L2 and it looks fairly elliptical. Admittedly it's not a very small orbit and it's a bit warped, but I would still guess that "equal areas in equal times" works pretty well.

Here is a plot of the triangular areas for each 1-week interval and also the velocity.

area swept per week enter image description here

It's interesting that the velocity is highest at the inner edge of the orbit (closest to Earth). But the main feature is the twice-around variation with the velocity lowest at each y-axis extremum. The velocity variation is almost 2:1, whereas the area per unit time variation is only 4:5. So most of the velocity variation is cancelled by the radial variation - though not to the extent I was expecting.

To clarify: the first graph shows the area of the triangle $(0,0,0),r_1,r_2$ where $r_1$ is the 3D vector position at the beginning of a week and $r_2$ at the end of the week. The second graph shows the speed $|r_2-r_1|$. The vectors include x,y, and z.

[update per @BradV comment] Moving the center can get rid of the once-around component in the swept-area graph:

reduced once-around graph

[update per @BradV comment #2] Moving the center to the centroid of the area enclosed by the orbit doesn't eliminate the once-around component quite as well. It doesn't take into account the velocity round the orbit (faster on the near-earth side).

enter image description here

  • $\begingroup$ "Orbits arising from a central force conserve angular momentum" might need some qualifications; exclude velocity dependent forces and perhaps those with other than $r^{-2}$ dependence? I wonder if "conservative potential" is helpful vocabulary? $\endgroup$
    – uhoh
    Commented Feb 3, 2022 at 3:38
  • 4
    $\begingroup$ @uhoh it's true for any central force as long as the angular momentum is measured around that central point. (but, as you say, only r and 1/r^2 give closed orbits) $\endgroup$
    – Roger Wood
    Commented Feb 3, 2022 at 5:07
  • $\begingroup$ @Roger Wood: I wonder what your 'area' graph represents. Is it a calculation of the warped potato chip surface or a planar surface. What was used as center? RLP 0,0,0 or point on 'chip'? ALSO... is speed graph the full Vx,Vy,Vz or just Vy? $\endgroup$
    – BradV
    Commented Feb 3, 2022 at 15:09
  • 1
    $\begingroup$ @Roger Wood: thank you so much for being so helpful to me. I am new to this and trying to integrate so many different concepts into a useful whole. Here I find myself on the embarrassing end of a comment I've used on other engineers... "I've given you an answer that is 100% correct but is of no use to you because you've asked the wrong question." One question in particular I got wrong was about your 'speed' graph. After trying to make some things work out it dawned on me that the graph is nothing more than distance/time and has no 'direction' other than making assumption $\endgroup$
    – BradV
    Commented Feb 20, 2022 at 21:06
  • 1
    $\begingroup$ that max speed happens at max Z, prograde (and in that moment) parallel to earth orbit and the lower speed peak is at min Z ... also prograde and parallel to earth orbit. Do I have it now? I meant dist/time V years $\endgroup$
    – BradV
    Commented Feb 20, 2022 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.