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Math Help - physics- circular motion

  1. #1
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    physics- circular motion

    calculate the orbital height about the earth of a geostationay satellite.
    the following data will be required:
    earth's radius=6400km
    earth's mass 6.0x10^24 kg
    g.s G=6.67x10^-11
    i know the time is going to be 24 hours but i have no idea what equations to use....
    thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Use the fact that the grativational pull provides the centripetal force.

    F_c = F_g

    Let m be the mass of the satellite,
    r the distance from the Earth's centre,
    omega the angular velocity of the satellite,
    M the mass of the Earth.

    mr\omega ^2 = \dfrac{GMm}{r^2}

    Simplify;

    r^3\omega ^2 = GM

    Now, fill in the values that you know.

    r^3\left(\dfrac{2\pi}{24\times3600}\right)^2 = (6.67\times10^{-11})(6\times10 ^{24})

    You should be able to get the value of r, the distance of the satellite from the centre of the Earth and then, the height above the ground.


    Of course, there are assumptions made here to make it easier.

    Post the answer that you get!

    Answer I got: 3.59 x 10^7 m
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  3. #3
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    thank you very much!
    i got 3.6x10^7 m.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Use your mouse to select my whole post and you'll see the spoiler I hid in there. That's the right answer!
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