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 > > ------------------------------------------------------------------------- > Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: > http://www.satobs.org/seesat/seesatindex.html > Jim Scotti Lunar & Planetary Laboratory University of Arizona Tucson, AZ 85721 USA http://www.lpl.arizona.edu/~jscotti/ ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html
This archive was generated by hypermail 2b29 : Mon Oct 17 2005 - 01:36:55 EDT