# eccentricity, decay modelling

Walter Nissen (dk058@cleveland.Freenet.Edu)
Mon, 11 Dec 1995 09:47:52 -0500

```Jay Respler writes:
> Out of about 4000 satellites, the 2 most circular are:

> 19508  88 86A  ecc=0000097 35794-35795 km
> 19671  88106B      0000010 666-666 km

> 88086  A
> 1 19508U 88086  A 95324.01492477 -.00000300  00000-0  10000-3 0   179
> 2 19508   0.0390 105.4434 0000097  91.0943   3.3959  1.00268584 26405
> 88106  B
> 1 19671U 88106  B 95334.24686766  .00000060  00000-0  97511-5 0    03
> 2 19671  56.9760 110.1146 0000010 288.0998  71.9002 14.70781741    07

> Is that 'very close' enough?

Mike McCants tells us the value for Lac 1 is arbitrary.  I guess he's
implying that he doesn't think it is really that small.  That leaves
CS-3B in the most circular orbit.  Wouldn't you know it would be the
Japanese that would perfect the engineering of a concept developed
elsewhere?

Mike McCants writes:
> In general, the eccentricity of a satellite orbit normally varies a
> little bit because of lunar and solar perturbations.

You forgot to mention the radiation pressure from the brilliant smile of
the Face on Mars.

> And when the
> eccenticity of a satellite orbit is close to zero, NORAD is likely
> to issue elements that have small variances in the mean motion
> because the mean motion is the reciprocal of the anomalistic period
> (perigee to perigee)

See next paragraph.

> and the location of perigee for a nearly circular
> orbit is not well defined.

No doubt.  Therefore the calculation, if the answer is to have any
significant digits whatever, has to be carried out in some transformed
coordinate system or using different variables or functions.

(Aside to those unfamiliar with numerical analysis:  It is standard
practice to substitute variables, use haversines instead of cosines,
divide calculations into ranges, or modify algorithms to preserve
accuracy.  One tiny example:  If b is really small and a = 1, even
quadruple precision may compute (a+b) - 1 to be 0, because when 1+b is
formed all the bits of b get shifted away to the right.  But if you
compute in a different order, (a-1) + b, you save all your significant
digits, and single precision may be more than satisfactory.  Or the
current question about the period of Periodic Comet Hale-Bopp.  All we
have is fuzzy positional data from a tiny portion of the orbit near the
Sun, so the period is very hard to pin down.  But if we had the semi-major
axis to the same precision, we'd know the period very accurately.  There
is a good Usenet newsgroup called sci.math.num-analysis).

I know there sometimes are glitches in the output from SPACECOM, but if
they were as naive as you seem(?) to suggest, they would be way out of the
ballpark.  Doesn't it seem likely that there is a more complicated
explanation for the occasional glitches?

>  I believe this explains the Mir elset variations.

But Mir's e isn't that small.

Bjorn Gimle writes:
> I think both anomalistic and nodal period are slightly esoteric
> quantities. If a stable elliptic orbit had a rigid, rotating
> shape and low inclination (increasing AOP) and a constant MM,
> its nodal period would be much shorter when AOP was close to
> 180 degrees, which I think violates the principle of conservation
> of energy. If the "nodal" period, measured at an arbitrary point
> along the orbit, is a constant, the ellipse is not rigid, but
> contorted.

If I pretended to understand what you are saying here, I'd be lying.

Dave Mullenix writes:
> I'm not a physicist either.  G3RUH's prediction of a nighttime
> reentry was made in a radio interview and he didn't give the
> reasoning behind it, but look at it this way:

> The satellite is in a highly eccentric orbit.  It will get more circular
> as energy is lost to drag, but I think it will still be quite eccentric
> at the end.  That means it only dips into the atmosphere for a small
> part of each orbit, at perigee.  The rest of the time it's coasting in
> vacuum with zero drag.

> Now picture a satellite at apogee, in vacuum, well outside the earth's
> atmosphere and with no drag to speak of.  It swoops down to perigee,
>  entering the earth's atmosphere for a short time, heating up and
> slowing down.  If it has enough speed left, it soars out of the
> atmosphere and coasts back up to apogee, except that apogee will
> be a bit lower this time due to the speed lost in the atmosphere.
> However, if it gets out of the atmosphere at all, it's in vacuum and there
> is no drag to speak of to slow it down and prevent it from climbing back
> to apogee (or most of the way, anyway.)

> On the final orbit, it will swoop down into the atmosphere, slow down
> too much to coast back into vacuum and that will be the end of it.
> James speaks of the orbit "going parabolic" at that point.

> I guess the final analysis is that if OSCAR-13 has enough energy
> to get out of the atmosphere, it will coast back to apogee.  If it doesn't
> have enough energy, it will stay in the atmosphere and burn up.
> Since all the perigees (and hence all the dips into the atmosphere)
>  will be on the night side of the planet, the satellite will re-enter in
> darkness.  I just hope it's visible from 43N, 90W.

Bjorn Gimle writes:
> I have no empirical evidence ( I would love to find decay
> dates accurate to 0.01 days or better, for my prediction
> model improvement ), but I agree with Dave, and justify it
> with an extension to your 'random' model :

> Luni-solar (and/or radiation pressure) perturbations may seem
> random, but near the end the drag effect dominates. If the
> orbit is highly elliptical, the apogee may decrease by
> hundreds of kilometers per revolution. The last-1 orbit still
> has most of its energy delta near the perigee, and the chance
> that the last orbit will be circular is quite small.
> If it still has a 'high' apogee, it should survive to
> the next perigee.
> If the 'random' delta makes the 'apogee lower than the perigee'
> it will plunge very quickly.

> So, it seems even your randomization argument supports Dave.

I grant both of you everything you mention except your final conclusion.
(Actually, Dave, I'd quibble about some of your phraseology that might be
misinterpreted to make it sound like the atmosphere has a sharp boundary,
but I'm comfortable with the fact that we are talking models here and
would gladly oversimplify if it helped model reality).  It drives me crazy
that there may be empirical data that refutes my thinking.  But, Bjoern,
are you saying that you don't have any such data?

I assume that we all agree that the least likely event is that on the
final pass the energy pulled out by the atmosphere (at whatever the new
perigee elevation is) is precisely equal to the kinetic energy of the
object, causing it to come to an abrupt halt and then fall straight down
to the surface.  That kind of perfection is all but impossible.  So the
real question is:  How much energy will remain for the final plunge?  This
is where I leave the mathematical realm where I am smart and enter the
physics realm where I am not.  It seems to me that the amount remaining
would be taken from a random distribution over a range beginning at zero
and going up to some value sensibly close to enough to get it around one
more time.  Not quite that much, because the atmosphere gets denser as you
descend, but close.  Evidently, you think otherwise.  In fact, you think
the distribution is pretty close to a point at zero.

This seems like the sort of concept that a real physicist would grasp
immediately and explain in a way he would find intuitively obvious and I
would find obscure, but highly convincing when the empirical data came in.
(I might find it convincing only then, for reasons abundantly highlighted
by the different models of the collision of S/L-9 with Jupiter).

(Mike, and others), are you going to let me twist in the wind?

If James is right, maybe the explanation lies in moving from a 1-D model,
which is basically what mine is, to a 2-D model, which is what the orbit
exists in.  (I think we are correctly ignoring the out-of-orbital-plane
effects like the angular momentum of the atmosphere).

(Incidently, I forgot to mention previously that a real value mod 1 is
exactly the function available on some calculators and in some languages
as the Frac function; or as x-int(x) in others.  Also, cryptography
actually uses integers within a huge realm, so it is a close analogy to
floating point mod 1).

Cheers.

Walter Nissen            dk058@cleveland.freenet.edu            1-216-243-4980

---

Proud to be a supernova remnant.
```