Re: Calculating declination of geosat belt?

From: Jeff Umbarger (jumbarger2000@yahoo.com)
Date: Wed Feb 23 2005 - 03:17:14 EST

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    Hey Ed,
         Here is what I came up with, "back of the
    envelope", with almost no error checking:)
    
    There are 2 constants and 1 input to the following
    equation to calculate the *declination* of the Clarke
    orbit *at the observers local meridian*. 
    
    The two constants are:
    
    Re = The Earth's Equatorial radius = 6378 km
    
    Rg = The Geosats distance from Earth center = 42,241
    km
         (Note: this is *not* the height of the geosat
    above the equator)
    
    The one input is:
    
    ThetaL = Observer's Latitude
             (Note: for Northern latitudes this is entered
    as a positive number and for Southern latitudes this
    is entered as a negative number. - Sorry Greg and
    Robert!)
    
    The one output is:
    
    ThetaD = Geosat's declination at the observer's local
    meridian
    
    The formula:
    
    ThetaD = arcsin (h/Rs)
    
        where:
    
        h = Re*sin(-1.0*ThetaL)
        Rs = sqrt(Re**2 + Rg**2 - 2*Re*Rg*cos(ThetaL))
    
        Note that "h" is the observer's distance from the
    equatorial plane and that "Rs" is the distance from
    the observer to the Geosat. The math gets way to
    complicate for proper calculation of the declination
    off the local meridian. But it will be *smaller* (that
    is, closer to the celestial equator) the farther from
    the local meridian you get.
        Hope this is right!
    
         Regards,
              Jeff Umbarger
              Plano TX
    
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