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- Newsgroups: sci.space
- Path: sparky!uunet!gatech!asuvax!ncar!noao!stsci!stosc!hathaway
- From: hathaway@stsci.edu
- Subject: Re: cryptocraft photography, Re: Aurora
- Message-ID: <1992Dec23.014534.1@stsci.edu>
- Lines: 64
- Sender: news@stsci.edu
- Organization: Space Telescope Science Institute
- References: <BzB33J.2Av.1@cs.cmu.edu> <1992Dec16.132541.18610@mnemosyne.cs.du.edu> <N4HY.92Dec21120246@wahoo.UUCP>
- 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
-
-