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Location: UFOUpDatesList.Com > 2007 > Nov > Nov 24

Re: Skylab 3

From: Bruce Maccabee <brumac.nul>
Date: Fri, 23 Nov 2007 23:28:28 -0500
Archived: Sat, 24 Nov 2007 08:48:44 -0500
Subject: Re: Skylab 3


>From: Michael Tarbell <mtarbell.nul>
>To: ufoupdates.nul
>Date: Wed, 21 Nov 2007 09:09:09 -0700
>Subject: Re: Skylab 3

>>From: Bruce Maccabee <brumac.nul>
>>To: <ufoupdates.nul>
>>Date: Thu, 15 Nov 2007 11:26:26 -0500
>>Subject: Re: Skylab 3


>>>If you want to find an astronaut sighting of an unidentified
>>>space object - USO; this one was not submerged - then here you
>>>are.

>>>Comments welcome

>>>http://brumac.8k.com/Skylab3/SKYLAB3.html

Change to /Skylab3/SL3.html

<snip>


>Your revised analysis (SL3.html) does cover the distance/size
>issue in more depth than the original (and now deceased) link
>you provided above. While the arguments presented against the
>'Skylab debris' hypothesis are generally persuasive, the case
>could actually be made ironclad in a straightforward - if
>tedious - manner:

>While an analytic derivation would be difficult (perhaps
>intractable), a brute-force march through trial perturbations of
>the Skylab orbit would reveal whether it is possible even in
>principle for a piece of Skylab debris to satisfy the given
>constraints (shadow entry delay, angular displacement relative
>to Skylab, etc). One need only launch an array of 'test
>particles' at various points upstream along the Skylab orbit,
>the elements of which are provided in the Oberg document. These
>particles would be given an instantaneous velocity increment of
>variable magnitude and direction, and then propagated forward
>along their new orbits to the initial time of the sighting. If
>none of them are consistent with the observed subsequent object
>motion and sunlight/shadow timing, then it's case closed: Skylab
>debris is explicitly ruled out.

>This would not rule out the more general case of orbits that are
>not obtained by impulsive perturbations of the Skylab orbit,
>i.e., satellites or debris not associated with Skylab. The above
>approach would not be practical for this broader problem without
>some analytical filtering of the orbit sampling population based
>on the observed constraints.

>I have available the necessary orbital mechanics routines to
>conduct such a study, although I would need to write a driver to
>loop through the sampling variables, and also incorporate the
>earth/sun shadow geometry. I will attend to this as time permits
>and forward the results.

Sounds like a good idea. I suppose there is a record of 'stuff'
that was ejected from SL either associated with the launch and
orbit insertion or 'trash' ejected orbit. Sizes of objects
ejected are?????

A 1 meter object would have to be quite close at the time of
photo 4 to make an image of angular size .0029 radians (e.g., 1
m/.0029 = 344 m). If one places the object this close to Skylab
at 'nighttime' (entering the shadow) then the time lag would be
much less than a second.

Another scenario to contemplate that is consistent with sunset
time lag and the angular size assumes that the object was seen
and photographed while it was quite close.

In this case the minimum distance calculated depends upon the
actual size assumed for the object because the size/distance
ratio must be .0029.

Now assume that the object slowly drifted away behind the SL3 so
that it was about 30 km away when it disappeared.

This would account for the 5 sec
(approx) time lag but raises the following question: how far away
can one see a small object such as 1 m size that had been ejected
from the Skylab at some time (maybe a liong time previously)?

One could imagine a small object 'fading out' with increasing
distance, but Garriott's description implies a sudden loss of
visual contact, not a fade out that took several minutes. Also,
any satellite close enough to the Skylab to be seen for 10
minutes would be likely to be seen again.

If it had been a piece of trash that was simply given a lateral
outward velocity (a 'sideways' push) would repeatedly cross the
SL orbit.

<snip>

>>>- Garriott makes the claim that Skylab and the object were not
>>>in a region of the orbit in which sunlight was passing through
>>>earth's atmosphere (and thus reddening it). But it is
>>>unavoidable that sunlight illuminating Skylab and the object was
>>>passing through earth's atmosphere when both objects were in the
>>>vicinity of the edge of the shadow zone.

>>That is true, but the passage through the region where the
>>sunlight is reddened would have taken place in some number of
>>seconds at the end of the 10 minute period of observation, most
>>of which occurred when the Skylab and object were far from (but
>>approaching) the reddened area.

>'...some number of seconds' may understate the interval of
>significant reddening. I should think it would begin with entry
>into the penumbral shadow (when the figure of the sun first
>intersects the limb of the earth) and proceed well beyond entry
>into the the umbral shadow (where the figure of the sun is
>completely blocked). Note that the totally eclipsed moon is
>blood/copper red even though it has no line of sight to the sun.
>Note also that Skylab and the object were approaching the shadow
>cone surface obliquely, not perpendicularly. Here it would be
>useful to have some first-hand testimony from astronauts
>regarding the typical duration of this interval, although it may
>not be obvious even to them unless they had some view of the
>exterior of their vehicle.

The angular size of the sun is 1/2 degree. Traveling around a
circle (circular orbit) from the position where one edge of the sun
disappears to where the other edge of the sun disappears is
1/2 degree (.0087 radians) times the radius of the orbit, 6800 km
or about 60 km.

Traveling at about 7.6 km/sec Skylab would travel this distance
in about 8 seconds. However, Skylab was not traveling
'perpendicular' to the shadow boundary but at an angle of about
40 degrees (rotating ctrclckwise from the 'vertical' shadow; 50
degree inclination relative to 'horizontal' lines of constant
latitde) so it would travel 60/sin40 = 93 km to cross the
boundary, taking about 12 seconds.

However, this estimate ignores bending of sunlight.

I don't know what this value is, but that fact that the moon is
lit by refracted sunlight ''bent' around the earth's limb means
that at least some light is bent by about 1 degree (radius of
the earth divided by the distance to the moon is about
4000/240000 = .01666 rad; .0174 rad is a degree).

If we add 1 degree to the 1/2 used before and include the isn
factor we get the following 'crude estimate' of distance: (1.5 x
.0174 x 6800)/sin 40 = 276 km and that would be traversed in
about 36 sec. I suspect that this is too large, but maybe
someone 'out there' has a better number.



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