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Assume that the earth is perfectly round

  1. May 17, 2005 #1
    r = 6400 km

    how much less does a 100kg person weigh at the equator than at the poles because of the rotation of the earth?

    Ok I have figured out that g(equator) is .03 m/s^2 less than it is at the poles due to the rotation on the earth.

    F(pole) = (100kg)(g)


    F(equator) = 100kg(g - .03m/s^2)

    My question is what should be the value for g in this case?
    If i make g = 9.8, then it will be 9.77 at the equator

    or should it be g = 9.83 ... making it 9.80 at the equator.

    not a big difference, but I'm just wondering which one is correct.
  2. jcsd
  3. May 17, 2005 #2


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    It should be less. Your centrifugal force is counteracting the force of gravity leaving a net force less than the force of gravity alone. (A good clue is that the problem asked how much less does a person weigh.)

    Of course, the problem with this problem is that all of your numbers are rounded to 2 significant digits and you're adjusting the third significant digit. The equatorial radius is actually 6378.137 km and the force of gravity at the equator is actually 9.798 m/s^2. The percentage error due to rounding off is greater than the percentage of the adjustment due to centrifugal force.

    Of course, the choice of numbers used probably isn't up to you, so you have to go with what you've been given.
  4. May 17, 2005 #3


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    Take the difference between your two forces before substituting in a value for g. Simplify the result. Then think about your final question again.
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