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Special geostationary orbit condition

  1. Oct 1, 2005 #1
    For geostationary satellites moving in circular orbits in the equatorial plane, there is no distinction between the true anomaly f, the mean anomaly M, and the eccentric anomaly E. For geostationary satellites, the angles [tex]\theta[/tex]=Greenwich sidereal time, [tex]\Omega[/tex] = longitude of ascending note, [tex]\omega[/tex] = argument of periapsis and f =true anomaly are all measured in the plane of the equator. Show that for such a satellite ro remain fixed on the equator at the sub-satellite point with east longitude [tex]\lambda_{SS}[/tex] it is necessary that

    [tex]\Omega + \omega + f = \theta + \lambda_{SS}[/tex]

    and that [tex]\frac {df}{dt}[/tex] = [tex]\frac {d \theta}{dt}[/tex]

    Include in your argument an illustration showing the equinox of all angles involved.

    Here' s what I know, i = 0. From Kepler's laws, the orbit must have e = 0. From these, where do I go? What is the first thing I could calculate?

    Any help would be greatly appreciated guys...I am lost here. :confused:

    Last edited: Oct 1, 2005
  2. jcsd
  3. Oct 5, 2005 #2
    Anyone???Just give me something guys.Please..I'll work on it.
  4. Oct 5, 2005 #3
    Here' s what I get:

    When the first identity is shown, the second one should follow from the definition of a geostationarty orbit since those three angles do not change.i.e. their rates of change will be zero.

    Am I on the right track?
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