7
$\begingroup$

In this answer I've generated some oscillations of eccentricity by being somewhat naughty and propagating a TLE for +/-250 days from epoch. I believe it's okay for the purposes of demonstrating gross eccentricity oscillations.

This does tend to reproduce Tiangong-1's 50 day oscillation in eccentricity, but I'd like to see if I can confirm this behavior, and the way it varies with inclination, with some analytical approximation.

These oscillations are a known effect, but I don't know much about it. I think there is a name for oscillations in eccentricity due to non-spherical gravity terms and may have used it in the past, but I can not find that now, nor a simple expression. It may be a 2nd order effect connected to or driven by apsidal precession, note the apparent nulling between 60 and 65 degrees, consistent with the Molniya orbit inclination of 63.4 degrees.

You can also see that the semi-major axis oscillates as well, the apoapsis hits a plateau at the same time the periapsis makes sharp "u-turn" at it's minimum.

Tiangong-1 rough simulation

Tiangong-1 rough simulation

$\endgroup$
6
  • 2
    $\begingroup$ I don't know the answer but the plots are op art, frame-worthy! $\endgroup$ Oct 9, 2018 at 12:25
  • $\begingroup$ @uhoh This source defines the reason for the phenomenon: "The potential generated by the non-spherical Earth causes periodic variations in all the orbital elements". So the term itself might be "geopotential orbit variations"? $\endgroup$ Dec 26, 2019 at 5:02
  • $\begingroup$ @uhoh In that link there's also analytical derivation for Molniya orbit inclination, that goes inline with your results. $\endgroup$ Dec 26, 2019 at 5:24
  • $\begingroup$ @LeoS Thanks but I'm looking for a specific term for this specific perturbation that probably depends primarily on the $J_2$ term of the geopotential. $\endgroup$
    – uhoh
    Dec 26, 2019 at 5:34
  • 1
    $\begingroup$ @uhoh "J2 perturbation" is used in some scientific articles 1, 2 etc. $\endgroup$ Dec 26, 2019 at 5:42

1 Answer 1

4
+200
$\begingroup$

As the comments on the original question have stated, this is due to the spherical harmonics of the Earth. This answer is adding more context.

In the two-body problem, one assumes that the central body (in this case Earth) is a perfect sphere with evenly distributed mass. If this is true, its force due to gravity can be modeled as a point source at its geometric center. Of course in real life this isn't true. Since the Earth is rotating, its equatorial radius is larger than its polar radius. This is the J2 perturbation, which is a first order approximation of Earth's spherical harmonics. Also note that the farther you get away from the Earth, the weaker this perturbation gets, since Earth looks more like a point mass from far away.

In the cases of sun synchronous orbits, this is actually extremely useful. One can set up their altitude and inclination such that the J2 perturbation will cause a drift in right ascension that matches the rate of change of Earth's true anomaly around the sun. Here is a plot of the Keplerian orbital elements for a sun synchronous orbit for one year:

enter image description here

Since this sun synchronous orbit is also in LEO, it shows the same oscillations in semi-major axis and eccentricity. The secular decrease in the semi-major axis is numerical error, since we know that gravity is a conservative force, and semi-major axis is inversely proportional to (absolute value) of orbital energy. This plot was made from the results of integration with a RK4/5 solver. Other solvers may have more or less energy drift for these equations of motion

Also note J2 is only a first order estimation. Spherical harmonic models can get much more complex. For example, the GRAIL missions around the Moon produced a 1500x1500 gravity field for the Moon (JGGRX model).

It is possible to represent the exact variation in these orbital elements using the Gaussian Variation of Parameters (VOPs), from Vallado 4th edition, pages 636. Please refer to the ten prior pages for the exact derivation. Note that general astrodynamics tool (like GMAT or Nyx) do not use the Gaussian VOP for their propagation as Keplerian orbital elements are singular (e.g. one cannot use these equations for near-circular or near-equatorial orbits).

enter image description here

$\endgroup$
5
  • $\begingroup$ I know this arises due specifically $J_2$ because that's precisely the perturbation I used to generate these plots as linked in the first sentence of my question. And so to my actual question: "What is this perturbation effect called, and what would be an analytical expression for the resulting eccentricity oscillations?" is there some way to address it directly? $\endgroup$
    – uhoh
    Mar 13, 2021 at 19:56
  • 1
    $\begingroup$ Yes. The equations are referred to as the variation of parameters (or sometimes Gauss' planetary equations), which describe the change in all the orbital elements with respect to time given a perturbation. Its another way to propagate orbits besides in cartesian coordinates. This PDF from MUT goes over specifically how the elements drift due to J2, but these equations can also be generalized to any perturbation: ocw.mit.edu/courses/aeronautics-and-astronautics/… edit: forgot to add link $\endgroup$ Mar 13, 2021 at 20:21
  • $\begingroup$ If the link and the terms are part of the answer to the question, then when you get a chance it's best to add it to the answer itself. In Stack Exchange comments are considered temporary and can be cleaned up at any time, and future readers may not dig down into comments to find the answer. Thanks! $\endgroup$
    – uhoh
    Mar 14, 2021 at 2:03
  • 1
    $\begingroup$ I don't quite know why this answer was downvoted... Alfonso is correct: these are due to the spherical harmonics. He's also correct that the Gauss VSOPs allow that modeling, but I'll add a few notes. First, the formulation is a singular formulation of an orbit, and therefore is not used in any high fidelity modeling. Second, the high fidelity formulation of this usually uses Pine's equations, and those are represented in Cartesian form (to be applied in a non-singular representation of an orbit). $\endgroup$
    – ChrisR
    Mar 16, 2021 at 3:50
  • $\begingroup$ @uhoh, I've updated the answer with the exact equations. $\endgroup$
    – ChrisR
    Mar 16, 2021 at 4:01

Your Answer

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

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