Re: Geostationary apparent declination

From: Jim Scotti (
Date: Mon Oct 17 2005 - 01:33:12 EDT

  • Next message: satcom: "Shenzhou 6 safe return"

    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.... :-)
    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
    > -------------------------------------------------------------------------
    > Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive:
    Jim Scotti
    Lunar & Planetary Laboratory
    University of Arizona
    Tucson, AZ 85721 USA       
    Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive:

    This archive was generated by hypermail 2b29 : Mon Oct 17 2005 - 01:36:55 EDT