4
$\begingroup$

This answer provides quite interesting plots of apogee and perigee for objects 41332 and 41333.

The primary (sine wave -like) oscillations are probably due to perturbations caused by potential generated by non-spherical Earth, as described in this answer (at least period of few months is in the same order of magnitude with apogee and perigee oscillations for Tiangong-1, as shown in a plot in the question body to the answer).

But, along with primary oscillation, there are quite strong secondary oscillations of irregular(?) nature shown on the plots: enter image description here

What does look interesting is the symmetry of these otherwise irregular secondary oscillations along vertical line connecting max/min values.

What is/are the reason(s) of these secondary oscillations?

Could it be a measurement noise/error or some weird processing artifacts, or is it indeed the actual behaviour of the orbiting bodies?

$\endgroup$
3
  • $\begingroup$ @LeoS - I suspect you have selected the wrong answer. The plot in Cristiano's answer shows orbital radius - 6371 km. That is not how altitude is defined. Altitude is typical computed as height above the reference ellipsoid. $\endgroup$ Dec 30, 2019 at 19:35
  • $\begingroup$ @DavidHammen from the both answers I took it as both answerers (Including the author of the original plot) agreed the oscillations are artifacts rather than real behaviour. With the absence of other options, I made a selection. I would appreciate if you could post your answer that sheds more light on this effect, and then I can un-accept the current answer and accept the more correct one. $\endgroup$ Dec 30, 2019 at 22:05

2 Answers 2

3
$\begingroup$

They are some weird processing artifacts.

Here is the correct plot for KMS-4 (ID 41332) obtained from the TLEs downloaded from www.space-track.org and processed with the CSpOC library (downloadable from the same link):

KMS4 perigee, apogee and semi-major axis

the shape is incredibly smooth and there are no evident secondary perturbations (surely there are several small components of perturbations, but they are not visible with this scale; I'll add a plot to show the perturbations).


ADDENDUM: Perturbations Since this addendum seems to generate more confusion than clarity, I preferred to delete it, also because I think the first part of my post answers the question.

$\endgroup$
13
  • 2
    $\begingroup$ For whatever reason, they show up with STK, which is generally a good tool to do such analysis. Will have to look more in to this... $\endgroup$
    – PearsonArtPhoto
    Dec 27, 2019 at 11:01
  • 1
    $\begingroup$ @PearsonArtPhoto I can confirm that using 3,470 Space-Track TLEs for 41332 and 1,782 TLEs for 41333 that the lines are smooth (quick-n-dirty calc, SGP4 propagate each for 120 minutes (in Skyfield) and grab the max and min distance) i.stack.imgur.com/pepJ5.png and pastebin.com/p6JFppEk Did you use the same TLEs for your plots? I wonder if you used more sophisticated tools and/or had access to a different/better set of data? $\endgroup$
    – uhoh
    Dec 27, 2019 at 11:07
  • $\begingroup$ I deleted "54895 points": my first version of the "Addendum" included the Sma and the osculating inclination, but since the question was about the perigee and the apogee, I deleted the graph of the inclination and I added the one about the perigee, but I forgot to delete the 54895 points used to sample the inclination. Thanks to @uhoh. $\endgroup$
    – Cristiano
    Dec 29, 2019 at 11:32
  • 1
    $\begingroup$ @Cristiano I'll have a look again tomorrow, thanks for the speedy replies! (other question deleted again) $\endgroup$
    – uhoh
    Dec 29, 2019 at 11:39
  • 1
    $\begingroup$ @LeoS I take the length of the position vectors (distanced from the Geocenter) by squaring, and then subtract 6378.137 km to get altitude. It's not altitude above the moving sub-satellite point, it's the standard recipe for satellite altitude that everyone uses. $\endgroup$
    – uhoh
    Dec 31, 2019 at 2:34
3
$\begingroup$

I looked more carefully at my source, which is plots using STK, and here is a recent closely zoomed in plot. Note that for some reason it uses a different perigee and apogee on a regular basis, which seems a bit odd, to say the least... I'm using the classical orbital parameters, mean of epoch.

enter image description here

I switched this from the "Apogee Altitude" to the "Apogee Radius". It showed a similar chart.

I then moved to a "True of Epoch" time to see if that changed anything. Still shows variations in the same orbit...

I'd have to dig in to it a bit more, but it seems to me this is some artifact of how STK is doing the processing, and not a real plot. What exactly that is, I can't really say.

$\endgroup$
15
  • $\begingroup$ curiouser and curiouser! $\endgroup$
    – uhoh
    Dec 27, 2019 at 12:15
  • $\begingroup$ I uploaded a graph for the KMS-4 altitude (not radius vector) above the WGS 72 ellipsoid: cristianopi.altervista.org/KMS4_alt.png (calculations done with the CSpOC library). It looks very different from your graph. It seems that your graph shows the osculating perigee and apogee altitude (not radius vector), can you confirm? $\endgroup$
    – Cristiano
    Dec 27, 2019 at 16:18
  • 1
    $\begingroup$ Yeah, it came from TLEs as processed by STK. Need to dig in to what is actually causing this, I'm guessing the oblateness of Earth is somehow influencing it, but... $\endgroup$
    – PearsonArtPhoto
    Dec 30, 2019 at 22:17
  • 1
    $\begingroup$ LeoS - Converting cartesian coordinates to osculating elements is easy, and from there to perigee / apogee radius is easier still. But that will result in the so-called processing anomalies to which @Cristiano has been alluding. Converting cartesian coordinates to mean elements of any sort (e.g., TLEs are mean elements of some sort, an offshoot of Brouwer-Lyddane mean elements) is much tougher, and interpreting what those mean elements mean with regard to perigee / apogee is tougher still. $\endgroup$ Dec 31, 2019 at 5:26
  • 1
    $\begingroup$ The original plots had a time step of around 2 hours (I don't remember the exact value). The new ones I had showed 100 second time steps, so it more accurately shows the real shape. I could post the Apogee Radius, and will when I get a chance, but it is basically the same thing that is shown. $\endgroup$
    – PearsonArtPhoto
    Dec 31, 2019 at 12:33

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.