# 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:

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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