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Orbit of Geostationary Satellites

  1. May 8, 2015 #1
    Hi all, I feel like I have the answer, or am at least in the ballpark, but I'm not confident in this area and so I also feel like there should be a more concise, or "beautiful" way to express it. Am I missing something?


    1. The problem statement, all variables and given/known data

    Explain why the orbit of a geostationary satellite is circular and in the equatorial plane.

    2. Relevant equations
    None provided.

    3. The attempt at a solution
    "For a satellite to be geostationary its position vector and the position vector of the planetary surface point above which the satellite is stationed must lie along the same line at every point in time, so since a position vector rotating with time describes a plane the two must lie in the same plane. Since a satellite's orbital plane must be focused at the centre of mass of the planet the surface point must also rotate on a plane focused at the centre of mass of the planet, and since a point fitting that condition may only be found along the equator the satellite must orbit in the equatorial plane in order to be geostationary.

    From Kepler's 2nd law (1/2 r2 dθ /dt = constant) the angular frequency of the orbiting body about the centre of its orbit must vary as the inverse of the square of the radial distance of that body from the centre of its orbit, so since a geostationary satellite must have a constant angular frequency to match the constant angular frequency of the point on the planetary surface above which it is stationed it follows that the radial distance must also be constant, in which case the trajectory must be circular."
     
  2. jcsd
  3. May 9, 2015 #2
    When I researched the answer I mostly found responses which leaned toward Kepler's 3rd Law, but I can't get my head around how the 3rd law can be used to describe an orbit with a radius which must be necessarily be fixed at the centre of a planet.

    Other answers I found talked about how an orbital plane not matching the equatorial would produce a satellite position relative to the planet's surface which could have a matching period but which would oscillate about a fixed point rather than sit at one. This is easy enough to intuit, but it is not an answer when the question is "why"?
     
  4. May 9, 2015 #3

    Simon Bridge

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    Observe: satellite must travel in a closed orbit with the same angular velocity as the an observer on the surface of the Earth in order to appear stationary to that observer.
    Does the angular velocity of the observer vary?

    Now consider each of the closed orbits possible ... do their angular velocities vary? (reason they must vary or not)
    Next, describe what the observer on the Earth would see if the orbit was not aligned with the equator.
     
  5. May 9, 2015 #4
    Thanks for the prompt reply Simon but I'm not sure I follow. Haven't I covered the angular velocities/frequencies aspect in the second paragraph of my first post? Or are you saying I would be better to use the term "observer" rather than "the point on the planetary surface" to streamline it?

    As for what the observer on the Earth would see if the orbit was not aligned with the equator, I believe this is the oscillation I mentioned in my second post (a figure 8 to be more detailed). But I had thought that the first paragraph of my first post is an attempt at the "why" behind the "what" of that oscillation.
     
  6. May 9, 2015 #5

    Simon Bridge

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    You asked for a way to give your reply a bit more finesse - that is what I'm prompting you with.
    The question can be answered fairly completely in about 3 sentences - maybe 5 if you want to invoke Kepler's laws.
    If you are happy with your answer - then don't worry.

    Note: "why" is never answered ... every possible explanation has another "why" behind it.
    Thus the problem statement needs to be interpreted in terms of the coursework.
     
  7. May 9, 2015 #6
    Ok, I'm with you now, sorry it took me two tries, mind's a little numb at the moment. Your advice is much appreciated, I'll have another go with it in mind.
     
  8. May 9, 2015 #7
    its better to have a fixed point to aim your trasmitter at.

    Anyway, heres the math for negligable mass geo - stat orbit ( 1 radius only works for a given planet mass and rotation rate)
     

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