Re: decay modelling
Sun, 24 Dec 1995 03:48:36 -0500

Walter Nissen wrote:

>> > 2. Most decays I saw were southbound and shortly after perigee. 
>Is this the empirical evidence which obviates my proposal that the decays 
>should be distributed about the orbit?  


>> 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. 
>> (Mike, and others), are you going to let me twist in the wind? 
>No response yet. 

I'd like to point out what I see as an important distinction.  This thread
started out as a discussion about the decay of Oscar 13 (88 51B).  Later the
discussion of decay modelling was concerned with Cosmos 398.  I've come
to the conclusion that the decay mechanisms of these two satellites are quite different.

Cosmos 398 is an example of "natural decay."  For all practical purposes, the
only factor involved in natural decay is atmospheric drag.  Satellites
undergoing natural decay exhibit *decreasing* eccentricity as decay approaches.
For these satellites, I agree with Walter.  Decay can occur anywhere in the
orbit, seemingly randomly.  I also share Walter's interest in seeing actual
evidence of decay tendencies at perigee, if such evidence exists.

Oscar 13 is an example of lunar-solar decay.  Unlike natural decays, the
eccentricity of these satellites *increase* as decay approaches.  Here are
two TLEs that bear this out:

OSCAR 13 (AO-13)      
1 19216U 88051B   95353.49808363 -.00000295  00000-0  64619-4 0  1222
2 19216  57.4226 146.2437 7359390  26.6948 357.0343  2.09731387 26053
88051  B
1 19216U 88051  B 95334.90268617  .00000135  00000-0  12582-3 0  1176
2 19216  57.4280 149.8219 7353179  25.1344 357.3474  2.09726631 25663

I think drag at perigee is making the apogee fall, and the lunar-solar 
perturbations are increasing e.  As e goes up, perigee is forced lower.  
I think what will happen is there will be one last tug by the sun and/or 
moon out near apogee that will send Oscar 13 hurtling down to one last 
very low perigee where it will be incinerated quickly.  I see it as much
more violent than natural decay; it's really a collision with the lower
atmosphere.  Natural decay, on the other hand, is more of a grazing

This is just thinking out loud... if I'm wrong, please point it out!

Okay, I promise: no more theories from me for the rest of the year :)

Happy Holidays Everyone...

Jim Varney          |  121^ 23' 54" W,  38^ 27' 28" N   |     Sacramento, CA
Civil Engineer      |            Elev. 20 ft.           |