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Homework Help: Quick easy graviational problem

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data

    How fast would the earth(Massearth =5.98X10^24 KG, Radius Earth = 6.4X10^6 m) have to spin on it axis so a 75kg person at the equator would be weightless.

    2. Relevant equations

    T^2= kr^3
    Fg = G m1 m2 / r^2

    3. The attempt at a solution

    i know that first i need 2 find the W or the person then put the weight equal to some equation that i dont know =/ .. some 1 plz help me ;P
  2. jcsd
  3. Dec 12, 2007 #2
    Perform a force balance for a person standing on the equator and set that equal to the centripedal acceleration. Next, what does it mean to be weightless?
  4. Dec 12, 2007 #3
    weightless = u weight 0 N ;/ i still havent figured it out .. do u mean

    (G)(75kg)(5.98X10^24)/6.4X10^6 = V^2/r

    ? plz respond
  5. Dec 12, 2007 #4
    For a person standing on the earth at the equator, how about:

    GMm/r^2 - N = m*v^2/r

    Sorry for the confusing post earlier, I meant mass times the centripedal acceleration. Remember the normal force (N) is still relevant here because the earth is physically pushing the person away from the center. Ok now, which of the terms in my equation should go to zero as the earth spins faster and faster (person becomes more and more weightless)? Or, if it helps, imagine the earth is spinning just fast enough and the person's feet JUST leave the ground, then what term will be zero?
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