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Homework Help: Geosynchronous orbit

  1. Apr 25, 2005 #1
    A satellite in geosynchronous orbit remains above the Earth's equator as the planet rotates on its axis.

    a) calculate the radius of its orbit.

    does this mean from the center of the earth?
    I looked online and the distance is about 7 earth radii, which would make the radius of the orbit 8 earth radii. Earth's radius is 6.37x10^6m.. makeing the radius 5.096x10^7m.

    what does it mean by calculate the orbit?

    b) the satellite relays a radio signal from a transmitter near the noth pole to a receiver, also near the north pole. Traveling at the speed of light, how long is the radio wave in trasit?

  2. jcsd
  3. Apr 25, 2005 #2


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    a) For use in formulas relating to gravitational forces and geosynchronous orbits, the orbit radius is that measured from the Earth's CENTER.
    Hint: At what orbit radius will the gravitational force between Earth and satellite provide the exact centripetal force required for the satellite's rotational velocity "ω" (in radians/sec) to equal Earth's rotational "ω" (in radians/sec)??

    b) Remember that the radio signal path is 2-way ("uplink" and "downlink").

    Last edited: Apr 25, 2005
  4. Apr 26, 2005 #3


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    You may have misunderstood what they meant by the 'distance' for a geosynchronous satellite. A geosynchronous satellite has an average radius of 42,164 km, or approximately 7 times the Earth's radius. In other words, they didn't mean for you to add that distance to the Earth's radius.

    For your problem, multiplying the Earth's radius by 7 gets you an approximate answer, provided that's the way they want you to calculate the radius.

    However, I'm not sure that's the way they wanted you to calculate the radius. Since you have thread about Kepler's Second Law, does that mean you learned Kepler's Third Law, as well? If so, in order for a satellite to stay over the same spot on the Earth, it must complete one orbit in the same amount of time the Earth completes one complete rotation (86164 seconds). The orbital period is known, so you can rearrange the equation for orbital period to solve for the radius, instead, and will come much closer to an accurate answer than the rough approximation you found online.
  5. Apr 26, 2005 #4
    thx with the help...
    [tex]\frac{GM}{r^2}=\frac{4r \pi^2}{T^2}[/tex]

    that's the equation I was trying to figure out
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