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Centripetal Acceleration

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration equal to 9.8 m/s^2?

    ____ min

    2. Relevant equations

    Centripetal acceleration is equal to velocity * (2PI/T),
    where T is the period in seconds.

    3. The attempt at a solution

    I set centripetal acceleration equal to 9.8m/s^2 and solved for T. Then I took T and divided it by 60 so I could get the answer in minutes. I came out with 4.97 min which Webassign says is incorrect. Any ideas?
    Last edited: Feb 17, 2009
  2. jcsd
  3. Feb 17, 2009 #2


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    Homework Helper

    One way to look at it is that if the object is not to fall into the center then calculate the orbit about a mass of Earth at 1 earth radius.

    So what is the period of such an orbit?

    V2/R = GM/R2 = ω2R

    ω2 = GM/R3 = (2π /T)2

    T = 2π*(R3/GM)1/2
  4. Feb 17, 2009 #3
    What does GM stand for?
  5. Feb 17, 2009 #4


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    Homework Helper

    That's sometimes written as μ which is the standard gravitational parameter for earth.

    It is the product of earth's mass and the Universal Gravitational constant.
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