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Homework Help: Time it takes for a satellite to orbit the Earth

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. The definition of "near-Earth" orbit is one which is at a height above the surface of the Earth which is small compared to the radius of the Earth, so that you may take the acceleration due to gravity as essentially the same as that on the surface. Does your result depend on the mass of the satellite?

    M(earth)= 5.98x10^24 kg
    R(earth)=6.38x10^6 m
    G= 6.67x10^-11

    2. Relevant equations

    g= GM/r^2

    FG= Gm1m2/r^2

    3. The attempt at a solution

    I am not sure where to start since I don't know what equation to use.
  2. jcsd
  3. Feb 28, 2010 #2
    One more equation might help you. The satellite in orbit will have an acceleration equal to [tex]a = \frac{v^2}{r}[/tex] which will equal earth's gravity, [tex]g = \frac{GM}{r^2}[/tex]

    Setting those equal to each other, can you determine the velocity of the satellite and use that to find the period?
  4. Feb 28, 2010 #3


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    Homework Helper

    The gravitational force between the Earth and the satellite provides the centripetal force of the satellite. Are you able to make an equation?
  5. Feb 28, 2010 #4
    Okay, I'm not sure if this is right, but here's my attempt:

    I did v^2/r = GM/r^2
    & plugged in:
    V^2/(6.38x10^6) = (6.67x10^-11)(5.98x10^24)/(6.38x10^6)

    and ended up with:
    V= 1.99x10^7 m/s

    when i plugged it in to a=v^2/r i got a= 6.25x10^7

    this doesn't seem right, isn't is suppose to be near 9.8 m/s?
  6. Feb 28, 2010 #5
    You didn't square the radius on the right side.
  7. Feb 28, 2010 #6
    Wow, dumb mistake...I redid it by squaring it and got 9.796 m/s. Thank you for your help!
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