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Geo stationary orbits

  1. Mar 28, 2012 #1
    Hi,
    I know what geo stationary orbit is, my question is what does it mean when they say

    satellite is at "83.0°E", dont you need more coordinates to find it??
     
  2. jcsd
  3. Mar 28, 2012 #2

    sophiecentaur

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    For a Geostationary orbit, the satellite must be in orbit around the Equator (so it will appear to be fixed in the sky. Its orbital distance is also fixed, of course. All that remains is to know at which angular position, around the Equator that it's stationed - hence just one number is needed.
     
  4. Mar 28, 2012 #3
    So what does 83 deg means? can you elaborate a bit more or maybe point me to a web page which offers more information, i want full details and calculations, how can i find a satellite in my sky if i have the degrees (83)

    thanks for you time
     
  5. Mar 28, 2012 #4

    sophiecentaur

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    It means Longitude 083E. Vertically above that point on the Earth's Equator.
    Google Maps have a satellite finder facility (search for it - it's there) in which you put your position (Lat / Long) and the satellite position (083). It will tell you the bearing and elevation needed to point at the satellite.
    You can check by looking at other nearby dishes if you don't feel too confident.
     
  6. Mar 28, 2012 #5
    Great answer helped me formulate my real question exactly :)

    If i have Lat/Long info and the sat position(83) whats the formula to use to get the bearing and elevation. (I am actually making a satellite finder app for the smart phones, you could simple point your phone in the sky and it will show you satellites over the camera view)

    thanks again for your valuable time !
     
  7. Mar 28, 2012 #6

    sophiecentaur

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    Could be a really handy App.
    I searched "satellite finder" but I could find only sites that would do it for you - bummer.
    I did a nav course about 40 years ago (pre-GPS) and I had some notes until recently on how to determine your position from star sightings. This is the same geometry, pretty much. It isn't simple but I'm sure there are approximations which would suit a mobile phone app accuracy.

    Stop press. I found this link at the end of a search. I think it may be just what you need but you'll have to dig around in there. Of course the formulae are much easier than for astro nav because you don't need to consider TIME haha.
     
  8. Mar 28, 2012 #7
    The link looks promising, thanks a bunch !
     
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