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Calculating the height of a geostationary satellite of Earth

  1. Feb 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the height of a geo-stationary satellite of earth.

    Gravitational force of earth=6.667 x 10^-11 nm^2/kgm^2
    Mass of earth=6x10^24 kgm
    Radius of earth=6400 km
    V=86400
    2. Relevant equations
    GM/r=v^2
    r=R+h

    3. The attempt at a solution
    I plugged everything into the equation and got 53,583.6 for r. Then since I need only height of the satellite to earth, I subtracted 6400 km from 53,583.6 and I got 47,183 for h. However, my professor said the answer for r should be around 36,000 km. Here is what I did.

    (6.667x10^-11)(6x10^24)/r=86400^2

    From multiplying, I got 4x10^14/r=7464960000

    From this, I got r=53,583.6
    Then h=47,183.

    Are my calculations incorrect?
     
  2. jcsd
  3. Feb 24, 2016 #2

    haruspex

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    Looks to me like you calculated the number of seconds in a day and set that as a velocity in m/s. Have I misunderstood your working?
     
  4. Feb 24, 2016 #3
    I just realized that as well. I can't believe I've been staring at this problem for half an hour without seeing this. Thanks
     
  5. Feb 24, 2016 #4

    SteamKing

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    That's why it's important to carry the units through your calculations. If you don't know what the numbers represent, how can anyone else know?
     
  6. Feb 24, 2016 #5
    Thansk everyone
     
    Last edited: Feb 24, 2016
  7. Feb 24, 2016 #6
    I still can't figure this problem out. If I need to find the velocity to calculate r, how can I do that when I need r to calculate velocity?
     
  8. Feb 24, 2016 #7

    haruspex

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    How are the radius and velocity related for a geostationary satellite?
     
  9. Feb 24, 2016 #8
    Well the larger the radius, the smaller the velocity of the satellite will be.
     
  10. Feb 24, 2016 #9

    haruspex

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    And yet remain geostationary?
     
  11. Feb 24, 2016 #10
    Yes. So it has the same velocity as Earth. But I'm still confused as to how to find the velocity of earth. Even if I use the equation 2pir/T, I get .465 m/s. And when I plug that into the equation GM/r=v^2, I get a very large number that must be incorect.
     
  12. Feb 24, 2016 #11

    haruspex

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    No, not the same velocity. What must be the same?
     
  13. Feb 24, 2016 #12
    The time it takes for one orbit?
     
  14. Feb 24, 2016 #13

    haruspex

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    Well, not quite right. The time for one orbit of the satellite must equal what?
    From that, write an equation relating radius of orbit to velocity.
     
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