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Circular Orbits and Motion - Satellites

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A spy satellite is in circular orbit around Earth. It makes one revolution in 6.02 hours. (Radius of the Earth=6.371 times 106 m)

    (a) How high above Earth's surface is the satellite?
    (b) What is the satellite's acceleration?

    2. Relevant equations

    v = sqrt(G*M earth / r)
    Kepler's 3rd Law

    3. The attempt at a solution

    I found the angular velocity to be 2.899 rad/sec...
    I really don't know where to start after that
     
  2. jcsd
  3. Sep 25, 2007 #2

    Kurdt

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    What is Kepler's 3rd law? Have you learned Newton's form of Kepler's 3rd law for circular orbits?
     
  4. Sep 25, 2007 #3
    i just found Newton's form of Kepler's 3rd law on a different website - i will see if i can get it to work...
     
  5. Sep 25, 2007 #4
    i'm still getting a ridiculous number for the distance...i assume i use Newton's form of Kepler's 3rd law but do i keep the period in hours or seconds? argh...
     
  6. Sep 25, 2007 #5

    Kurdt

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    The SI units are seconds, so you will have to convert. Remember the distance will be the radius of the Earth plus the satellites orbital height.
     
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