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Speed at equator

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    An early major objection to the idea that Earth is spinning on its axis was that Earth
    would turn so fast at the equator that people would be thrown into space.
    Given : radius of earth = 6.37 × 106 m,
    mass of earth = 5.98 × 1024 kg ,
    radius of moon = 1.74 × 106 m, and
    g = 9.8 m/s2 .
    Show the error in this logic by calculating
    the speed of a 91.7 kg person at the equator.
    Answer in units of m/s.

    well drawing a free body diagram, we have normal and gravitational forces- which equal each other..so isnt there no unbalanced force? because doesnt there need to be one to have centrifugal acceleration?
  2. jcsd
  3. Dec 18, 2008 #2


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    Homework Helper

    Hi Maiia,

    The normal and gravitational forces will not equal each other. The difference between these two forces is what will provide the centripetal acceleration, and so you could find how much they differ.

    However, to find the real speed of a person at the equator, you just need the quantities given in the problem, and also use the fact that the earth rotates once per day. What do you get?
  4. Dec 18, 2008 #3
    hmm i did v= 2pir/T so 2pi(6.37x 10^6)/ 3600sec, adn i got 11117.74m/s- isnt that too big?
  5. Dec 18, 2008 #4


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    Homework Helper

    Yes, that's too big. You divided by a time period of 3600 seconds, so that would be the speed if the earth spun around once per hour.
  6. Dec 19, 2008 #5
    but I thought it was supposed to be in seconds? cycle/sec
  7. Dec 19, 2008 #6


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    Staff Emeritus
    Science Advisor
    Homework Helper

    How long does it take the Earth to rotate?

    Hint: it is longer than 1 hour.

    Yes. Once you answer the above question, then convert that answer into seconds.
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