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Gravity and centrifugal force

  1. Mar 9, 2006 #1
    Hello, this is my first post here :)
    The question I'm having problem with follows: how long would one day have to be so that there would be weightlessness on the equator? (the teacher said that the correct anwser would be roughly 5 minutes)

    Solving this by myself I couldn't reach any other conclusion than:
    gravitational force = centrifugal force => mg=mvv/r => g=vv/r => vv=gr
    (v-velocity of any point on the equator, r- radius of the Earth, 6,4 million meters)
    which means that: v=7920m/s => T=2*pi*r/v=5077s=84 minutes

    What did I leave out of the account?
  2. jcsd
  3. Mar 9, 2006 #2


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    Welcome madis.

    An equation of the apparent strength of gravity is:
    [tex]g_a = g - r\omega^2 \cos^2 \phi[/tex]
    Where [itex]\phi[/itex] is latitude (which would be zero at equator) and [itex]g_a[/itex] is apparant force of gravity, which in your case would be zero. I can show the derivation if you like.
  4. Mar 9, 2006 #3
    Thanks! Using this equation I still got 84 minutes for the solution so I guess this is the real right anwser...
  5. Mar 9, 2006 #4


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    Yes, I get around 85 mintues also. I would be interested in seeing how your teacher calculated the five minutes, if you could post it here please:biggrin:
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