# Re: OSCAR-13 re-entry

bjorn@tt-tech.se
Fri, 8 Dec 1995 10:16:24 +0100

```Dave Mullenix writes:
> >>James says it will re-enter at night, so perhaps somebody will get to =20=
=20

> >>see the fireball.
and continues:
> I think it's safe to assume that the last orbit will finish with a
> perigee that turns into a fireball.
...
> the near guaranteed night time reentry

=20

>How do you jump from the first statement, which seems quite logical, to
>the second, which seems implausible?  (I should mention that I'm no
>physicist and that I see this in terms of a mathematical model,
>randomization by taking a real value mod 1, such as is used in
>encryption).  Why wouldn't you imagine that the final energy delta will
>result in a residual which is a random fraction of an orbit, originally
>eccentric or not?  If there is empirical evidence for what James says, I =20=
=20

>would readily believe it and chalk it up as a lesson learned, but it  =20
makes
>no sense to me.
=20

>Which can never amount to any excuse for not looking.

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.

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