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According to all the sources cited in Why does OGO-1's trajectory and imminent reentry come from sky surveys and NEO tracking rather than normal satellite tracking? OGO-1 is in a highly elliptical orbit, and n2yo gives it a period of nearly 60 hours!

In this answer I explain that generally speaking an object in an elliptical orbit that slowly decays will first circularize before reentering, since each small delta-v at periapsis tends to lower the apoapsis.

But that's not what's happening here. n2yo gives it's TLE and this shows an eccentricity of over 0.91.

Question: Why will OGO-1 reenter from a highly elliptical orbit rather than circularize first? Is there a rule-of-thumb to gauge which orbits will circularize before reentry, and which won't?

1   879U 64054A   20233.75378097 -.00000811  00000-0  00000+0 0  9999
2   879  54.9668 307.2269 9131889 205.8880  32.8586  0.40281857 21812
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    $\begingroup$ Objects will tend to circularise. have you plotted eccentricity over time? $\endgroup$ – JCRM Aug 29 at 9:36
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    $\begingroup$ Was there some encounter with the Moon recently? Maybe this just shifted the periapsis enough for a quick reentry? $\endgroup$ – asdfex Aug 30 at 10:59
  • $\begingroup$ @asdfex I think it could be something like that. With a 352 x 141,797 orbit every eccentricity change of only 0.001 will change periapsis by 80 km. $\endgroup$ – uhoh Aug 30 at 11:11
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    $\begingroup$ I have to work with HEO-reentries regulary at work, so often the question "how does the Apo and Periapsis behave for such kind of reentries?" was part of the job interview... and even I cannot give you a satisfying answer. From my experience: Yes, the object TEND to circulise, they get more and more circular but they always reenter before reaching a near perfect circle. In the end there is not enough energy left for a last trip and the pericenter was to low for a circular orbit. BTW: forecasts of HEO reentries is the supreme discipline in the field, small errors -> one orbit more or less $\endgroup$ – CallMeTom Sep 2 at 6:35
  • $\begingroup$ @CallMeTom I guess this is an extreme case; reentering with an eccentricity of 0.91 is certainly way "before reaching a near perfect circle." :-) Feel free to post another answer if you like. $\endgroup$ – uhoh Sep 2 at 7:16
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So, I just gave it a try and have reconstructed the orbit with a numerical propagator. Using a state vector converted from the TLE-Data from the 20th Aug, 18:05:26.676 (yes, I know, not exact, but have to be sufficient*) and a BC I choose so the final re-entry epoch matches (just quick and dirty get a reentry 2h later for a 8 day propagation...). Then I plotted the eccentricity over time:

As I said intuitively in the comments below uhohs question:

Yes, the object TEND to circulise, they get more and more circular...

A day before reentry the eccentricity is still around 0.5, shortly before re-retry the eccentricity is around 0.01 ...

uhohs Question:

Why will OGO-1 reenter from a highly elliptical orbit rather than circularize first?

My Answer:

It does (nearly) circularize first!

uhohs Question:

Is there a rule-of-thumb to gauge which orbits will circularize before reentry, and which won't?

My Answer:

[All] object[s] TEND to circularise, they get more and more circular but they always reenter before reaching a near perfect circle.

If someone could re-calculate my approach and post a Ecc_over_time plot without getting in copyright issues, that would be great ;-)

*) ..::EDIT::..

DISCLAIMER: NEVER EVER CONVERT TLEs TO STATE VECTORS WITHOUT KNOWING THAT YOU ARE INDUCING A BIG ERROR! ALWAYS CONSIDER IF THE ACCURACY IS ENOUGH FOR YOUR PUROPSE!

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    $\begingroup$ I like your approach, thanks! $\endgroup$ – uhoh Sep 2 at 10:13
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Intuitively, the velocity impulse at periapsis is $$\Delta v = \int_{t_{p} - \Delta{t}}^{t_p + \Delta{t}} a_\text{drag} \mathrm{d}t$$ where $\Delta{t}$ is the duration of the periapsis pass. When the eccentricity is high enough, the periapsis pass happens very quickly and $\Delta{t}$ is small; therefore, the $\Delta{v}$ is small and is not sufficient to change the apoapsis radius by a lot. This is in part counteracted by the fact that $a_\mathrm{drag} \propto v^2$, but still not enough to generate a significant $\Delta v$.

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  • $\begingroup$ Okay this is a great start! It addresses why it may take longer (i.e. more orbits) to lower the apoapsis (circularize), but doesn't yet address why the periapsis drops. It's the competition between apoapsis loss and periapsis loss per orbit that determines if it circularizes or reenters first. $\endgroup$ – uhoh Aug 29 at 23:20
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    $\begingroup$ buuuuuut ... a_drag ~ F_d ...the Drag-Force will be F_d = 0.5 * rho * v² * C_d * A ... so the smaller dt is, the faster the object is resulting in more drag .. or in short: half the time four times more drag.... $\endgroup$ – CallMeTom Aug 30 at 7:56
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    $\begingroup$ @uhoh, in the particular case of OGO-1 the periapsis dropped because of long-periodic oscillations in eccentricity caused by lunisolar perturbations; check out Figure 3 of Shute and Chiville (1965), available at ntrs.nasa.gov/api/citations/19660007965/downloads/… $\endgroup$ – LeWavite Sep 2 at 2:20
  • $\begingroup$ @CallMeTom, that's a good point.... perhaps it's better to look at the change in angular momentum and the resulting change in eccentricity. I'll think about this and edit my answer if I find anything else. $\endgroup$ – LeWavite Sep 2 at 2:20
  • $\begingroup$ @LeWavite thanks! Yes with such a highly eccentric orbit it's possible and perhaps probable that those (and/or other) perturbations are large enough to compete with the drag-induced effects. $\endgroup$ – uhoh Sep 2 at 2:24

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