# Re: Hubble Bubble

Date: Thu Mar 09 2000 - 18:19:29 PST

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```I had written:

>Since Hubble is at a 28 degree inclination, ...
>If you are too far north or south on the earth
>(above or below 38 degrees N or S latitude)
>then you won't see it at all.

More than one of you, like Jonathan T. Wojack below, responded with
something like:

>Not precisely.  In December of 1999, I sighted the Space Shuttle/HST
>complex, at about 17 degrees altitude, and I live at 39.75 degrees north
>of the equator.

And you are all correct. When I originally wrote what I did I had the word
"about" before "38 degrees," using the rule of thumb of inclination plus 10
degrees. When I modified the sentence to include southern hemisphere
observers, I dropped the "about". I knew at the time that I was sticking my
neck out for the technically astute to axe me with comments -- and
deservedly so. :)

To be exact, here's an excerpt from a post to SeeSat by Ted Molczan in Sept
99:

==========================================
Here is a simple procedure to determine the Maximum Elevation that a
satellite will reach (at apogee), when observed from a latitude greater than
its orbital inclination.

Symbols used in the procedure are defined as follows:

a = semi-major axis, km
A = apogee, km
dlat = difference between latitude of observer and inclination, deg
e = eccentricity
i = inclination, deg
latobs = latitude of observer, deg
MaxEl = maximum elevation above observer's horizon, deg
n = mean motion, rev/d
Re = observer's distance from Earth's centre, km (Earth's mean radius of
6371 km is a suitable approximation for most purposes)

Procedure:

a = (8681663.653 / n) ^ (2/3)

A = a * (1 + e)

dlat = abs(latobs) - i

MaxEl = atan((A * cos(dlat) - Re) / (A * sin(dlat)))

==========================================

Using

HST             13.3  4.3  0.0  3.0 v   71
1 20580U 90037B   00066.66540525  .00005791  00000-0  57525-3 0  3015
2 20580  28.4644 168.4790 0014658  96.6493 263.5762 14.89691643341102

results in the possibility of HST, when at apogee, culminating at about 19.4
degrees above the horizon for Jonathan's latitude. At 45 degrees latitude,
culmination at apogee would be 9.3 degrees.

Ralph McConahy
34.8829N  117.0064W  670m

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```

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