From: hathaway@stsci.edu
Subject: Re: cryptocraft photography, Re: Aurora
Organization: Space Telescope Science Institute
Distribution: sci,na
Date: Wed, 23 Dec 1992 06:45:34 GMT

My attempts to mail this reply directly to the poster kept failing 'unknown 
host', so I ask the forbearance (so great it is) of the net for posting 
this directly in the hopes Mr. McGwier will see it.  WHH 

In article <N4HY.92Dec21120246@wahoo.UUCP>, n4hy@wahoo.UUCP (Bob McGwier) writes:
>>  Identification: It had all the familiar steady motion of an 
>>                  Earth satellite, but _not_ in a common Direct orbit from 
>>                  West to East.  Motion actually more like from NE by N to 
>>                  SW by S.  If a satellite, it was in a near-polar orbit, but 
>>                  Retrograde.  
> 
> 
> 
> From NE by N down to SW by S is <NOT> retrograde.  This is the descending
> half of a normal orbit with inclination less than ninety degrees.  Remember
> that in June, the north pole is in continuous sunlight.  We could figure out
> the EXACT height from your observations since we KNOW where the satellite
> will enter the `umbra' of its eclipse given by your observations.  If I have
> time, I will do this computation if not, given this someone else should be
> able to do the simple calculation.
> 
> BMc

     
     
I thank you for your comment (this is the first chance I've had to log on
since I posted).  Several people have sent me comments directly with
several pieces of useful information and I am digesting those.
     
But I do not understand this comment (please don't take it that I am
disputing it - I want to understand, so I am asking for clarification).
     
The way I understand it, most satellites go around the earth from west to
east (direct or normal as described above).  A very few may have been
launched east to west, but it takes more energy and the advantages are
few.  Many are in near-polar orbits, (for survailance, etc.) with
inclinations near 90 degrees.  I've been told many of these are 98 degrees,
which would be 'retrograde'.  What I don't understand is this
"descending half of a normal orbit".   Suppose a direct (inclination less
than 90 degrees) satellite is ascending (has crossed the ascending node) and
heading north and east.  When it reaches the maximum north, it starts heading
south.  But it will still be drifting east as seen from an observer on
the ground under it (or above it for that matter).  Something that winds up
further west in the sky as it goes toward the equator must be 'retrograde',
with an inclination greater than 90 degrees.  (Ignoring the rotation of the
earth under it - but this was moving fast enough for that to be neglible.)
Now, maybe my visualization is wrong on this (I have no claim to
infallibility), but no one else I have discussed this with had the
interpretation you did, and I make my living in figuring where celestial
stuff goes, so it is important to me to get my own head right on this.
     
  I would appreciate further explanation of this statement - I do want
to know what it is I saw.
     
BTW, most calculations claim that for this to be illuminated by the
sun, it had to be 700 to 800 miles high.  Which makes its apparent size
even more difficult to explain.  And something that big, in full sunlight
should have been quite bright - which leads us to think it may have
been much lower - possibly in the atmosphere and illuminated by
ground lighting - which of course means it was not in a satellite
orbit at all....
     
Hoping for a clarification of your assertion,  Wm. Hathaway
     
Xref: sparky sci.space:18456 sci.physics:21924