# Re: Geostationary apparent declination

From: Jim Scotti (jscotti@pirl.lpl.Arizona.EDU)
Date: Mon Oct 17 2005 - 01:33:12 EDT

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```Hi Rod,
A little trig is good for the soul, so given the radius of the Earth,
Re=6378km, the radius of the geosynchronous altitude, Rg=42164 (I found a
couple of slightly different numbers, but this one came up twice....) and the
latitude of the observer is phi, the parallax, theta is then:

TAN(-theta) = Re*sin(phi)/(Rg - Re*cos(phi))

So given your latitude of -39 degrees, the geostationary satellites will be
about 6.2 north of the equator on your meridian.  They'll be a little farther
from the equator at any other longitude.  I should work out some more trig
and figure the parallax for satellites not sitting at your longitude.... :-)

Jim.

On Mon, 17 Oct 2005, Rodney Austin wrote:

> Hi All,
>       Probably a stupid question, but is there a note somewhere that
> will tell me how to compute the apparent declination of the
> geostationary belt from my latitude (-39 degrees)? Some years ago I
> recall seeing something about it in Sky & Telescope, and of course I
> could probably work it out by simple trigonometry. Sorry I'm a bit
> lazy. I'm not after any particular satellite, just a generalisation.
>> From here it should be north of the celestial equator I think, with a
> rough guess of dec +5 degrees? Thanks for any help.
> Sorry to be a bother.
> Cheers
> Rod Austin
>
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Jim Scotti
Lunar & Planetary Laboratory
University of Arizona
Tucson, AZ 85721 USA                 http://www.lpl.arizona.edu/~jscotti/

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