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Velocity At Earth's Equator

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Earth rotates in on axis that goes through both the North and South Poles. It makes one complete revolution in 24 hours. If the distance from the axis to any location on the equator is 3960 miles, find the linear speed (in miles per hour) of a location on the equator.

    2. Relevant equations

    -None-

    3. The attempt at a solution

    I used the formula V=rw to find the linear velocity, and for 'r' I used 3960, and 'w' is used 2pi/24 for making one revolution per hour, so I simplified that down to pi/12 as angular velocity.

    V=3960 x (pi/12 radians/hour)≈ 1036.73 miles/hour

    Is this correct? It got checked in school, but I think I was off in the decimals or something. Is there something I did wrong or?

    Thanks for any help!
     
  2. jcsd
  3. Sep 10, 2012 #2

    ehild

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    Gold Member

    Your method is correct, but check the numbers. 3690*3.14/12 is way off from your result.

    ehild
     
  4. Sep 11, 2012 #3
    I re-did it, and it came out to the same? Maybe it is something you inputted into your calc, I did pi/12 first then multiplied by 3960, and it came out to this?

    I don't know what I am off at? I even did it the way you wrote it out, but it still worked?
     
  5. Sep 11, 2012 #4

    Mark44

    Staff: Mentor

    Looks OK to me.
     
  6. Sep 11, 2012 #5
    Thanks! :D
     
  7. Sep 11, 2012 #6

    ehild

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    Sorry, I misread the number 3960:shy:. Your result is correct.

    ehild
     
  8. Sep 11, 2012 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Another way to do this, though essentially the same, is to use "velocity= distance/time". You are given that the radius of the earth, at the equator, is 3960 so the circumference is [itex]2\pi(3960)= 24881[/itex] miles. Divide that by 24 hours to get 1036.7 mph.
     
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