Re: Titan #80000 AND #80001 Obs, 15 may 96

Bjoern Gimle (b_gimle@algonet.se)
Wed, 15 May 1996 18:52:40 +0200 (MET DST)

>2.5 degrees offset to the west (Northwest?).   This is using
>local azimuth/elevation plots to measure angular differences
>and may have little to do with element set angles.

Roughly speaking, at your latitude you are moving in a 5000 km
radius circle. If you are scanning a satellite orbit at 500 km
near zenith, it moves about 10 degrees / 4 min. (less at lower
elevations - nothing if you are watching an object that 
culminates along the N-S line )

>My newbie question to the assembled experts here.  Assuming
>that one of these objects is the payload and makes an orbit
>adjust (same inclination) just after I saw it.  What can I expect
>as an arrival time/position one day later?
>
Nothing really, but I can improve on that.

The rocket is predicted at 93.9 min nodal period. The presumed
NOSS payload at 107.4 minutes. So, if the maneuvre is soon after
your observation, the pass you expect 15 orbits later could be
delayed by 15*(107.4-93.9) minutes = 202.5 minutes ( = 12.3 
minutes early) If the orbit change is delayed, *any* value
between 0 and 202.5 minutes could happen.

If you are thinking of a KH communication satellite that moves
into a molniya orbit, the spread is even larger.

However, there are constraints that could pinpoint a number of
alternatives. It will probably move from an apogee, into a more
elliptical transfer orbit, then into the NOSS circular orbit,
from the apogee of the transfer orbit (100.6 min?) 

So, if the maneuvre takes place between your May 15 and May 16
observation opportunities, there may be 14 quite well-defined
search positions - but I am not sure I have the time to
compute them tonight. I will start, though.

Keep up your good work!

Bjoern

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