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Centripetal Acceleration at Equator

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    An individual standing at the equator:
    a)What is this individuals centripetal acceleration caused by the earths rotation?
    b)How fast would the earth need to spin in order for the centripetal acceleration to be equal to g?

    T=1 day
    radius of earth = 6.4x10^6 m


    2. Relevant equations
    a=v^2/r, v=2pir/T


    3. The attempt at a solution
    We need to convert the period from day to seconds and then plug in the information.

    a=4pi^2r/T^2 = 4pi^2(6.4x10^6)/(1x60x60x60)
     
  2. jcsd
  3. May 24, 2010 #2

    ehild

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    How do you convert one day to seconds?

    ehild
     
  4. May 24, 2010 #3
    1 day x 24hrs/day x 60min/hr x 60 sec/min .... My bad, I put 60 instead of 24
     
  5. May 24, 2010 #4

    ehild

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    Well, and do not forget to square T.

    ehild
     
  6. May 24, 2010 #5
    So I'm right?
     
  7. May 24, 2010 #6
    Yeah I'm pretty sure I'm right, Thanks for your help :D :D And I figured out the last question thank you :D
     
  8. May 24, 2010 #7

    ehild

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    You are welcome :)

    ehild
     
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