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Speed at equator vs pole

  1. Oct 6, 2012 #1
    Assume there's a person standing at the pole of the earth. When the earth rotates, he/she has no tangential velocity because the person is at the pole.
    Now if the person were to take a trip by plane and land at the equator, the person would now have a fairly large velocity because the equator's tangential velocity is quite high.
    What provides the acceleration that that changes the person's velocity from zero at the pole to 1,670 kph at the equator?

    source for 1,670: http://geography.about.com/library/faq/blqzearthspin.htm
  2. jcsd
  3. Oct 6, 2012 #2

    Simon Bridge

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    Why the Earth does - and the engines in the means of transportation.

    Consider, you are in free fall above an airless spherical World composed entirely of ice and you are wearing ice-skates.
    The world is turning below you. The ground gets closer - your feet touch the ground - the world keeps turning under you.

    From your POV, you are now skating very fast across the surface.
    Turn your skates and you can slide to a "stop" - now you are going at the same speed as the surface under you ... where did the acceleration come from?
    Last edited: Oct 6, 2012
  4. Oct 6, 2012 #3


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    If we ignore variation from the local wind and weather on the surface of the earth, the air at the equator is moving at 1670 kph relative to the air at the north pole. So as the airplane flies southwards, it experiences a sidewind blowing from the west that accelerates it in an eastwards direction.

    If you were walking south from the north pole, with every single step you'd be putting your foot down on a patch of earth that is moving ever so slightly faster to the east than where your foot had been. It's not much at each step, but it's enough to add up to 1670 kph over the 20,000,000 or so steps between pole and equator.

    You might want to google for "Coriolis force".
    Last edited: Oct 6, 2012
  5. Oct 6, 2012 #4
    Thank you both for the explanations!
    I understand it now
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