STS-84 Rentry profile...question

Bill Krosney (bkrosney@MBnet.MB.CA)
Tue, 27 May 1997 01:41:32 -0700

I am trying to understand the details of the Shuttle re-entry to help
explain the marginal observing conditions presented for observing the
rentry  from my location: Winnipeg, Canada.  (Hey...I'm not complaining,
even marginal is better than not!)

This is all of course academic, after all:
a) the Shuttle has already landed
b) it never did land on its originally planned orbit 144
c) I was clouded out anyways


The STS-84 Web page was predicting that from Winnipeg I should of seen
the Shuttle at a max altitude above the horizon of about 12 degrees
(azimuth 209) at 06:35 CDT.  If the Shuttle did not re-enter on that
orbit it would of presented a near zenith pass at 06:34 CDT.  In fact,
when at an azimuth of 209 degrees it would be about 82 degrees above the
horizon  (this is based on a QuickSat prediction with orbital elements
from orbit 138).  So, as observed from my location, what accounts for
the difference between the altitude above the horizon for an orbital
pass and the re-entry?

Is it due strictly to the difference in the Shuttle's height above the
surface of the Earth?

I've always been under the impression that the Shuttle de-orbits much
like a decaying satellite.  That is it drops out of orbit, but stays
within the plane of its orbit until it reaches some height above the
ground (whatever that may be) where it can become a maneuverable gliding
aircraft.  When the Shuttle appears 12 degrees above my horizon (still
20 minutes from its planned touchdown) is it still in its orbital
plane?  If it is then could I essentially calculate its true height
above the ground with simple trignometry?  After all, if the orbital
pass places it 82 degrees above the horizon at an true height of about
393 kms., I can calculate a distance to the satellite's ground track. 
If I assume that the satellite (when at an altiude of 12 degrees) is
still within the orbital plane, then its ground track should essentially
be the same.  I can work backwards; knowing the distance to the ground
track, knowing the projected altitude above the horizon to calculate the
satellite's true height above the Earth's surface.  Doing so, I peg the
Shuttle at an altitude of 11.5 kms.  

Is this valid?  Or is there something wrong with my math and/or logic
skills here?

Or...has the satellite by this time maneuvered out of its original
orbital plane?

I find it hard to believe that the Shuttle, still far out from the KSC
in Florida, would be only 11.5 kms above the Earth's surface.

I then wonder if the satellite by this time has maneuvered out of its
orbital plane.  If so then where and when does it do this?

Thanks to all....clear skies....Bill