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Homework Help: Linear speeds at points on Earth

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the linear speed of a point:
    a) on the equator,
    b) on the Arctic Circle (latitude 66.5 degrees N),
    c) at a latitude of 45 degrees N, due to the Earth's rotation?

    Radius (earth)= 6.38x10^6m

    2. Relevant equations
    w= 2pi*f
    t= 1/f

    3. The attempt at a solution
    I am not sure how go about doing this problem. I wanted to solve for w with 2pi(1/24) but am not sure how to calculate the speed at different areas on Earth.
  2. jcsd
  3. Apr 11, 2010 #2
    You've got the radius of the sphere and thus the circumference. Use the time period 24 hours. For higher latitudes, the radius decreases as the COS of the latitude as does the velocity because the distance travelled decreases.
  4. Apr 11, 2010 #3
    I don't understand how to make this change though. Is there an equation I can use?
  5. Apr 11, 2010 #4


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    Homework Helper

    Hi balletgirl! :smile:

    (have a pi: π and a degree: º and try using the X2 tag just above the Reply box :wink:)

    The point goes round a circle (of latitude) …

    what is the radius of this circle? :smile:
  6. Apr 11, 2010 #5

    The radius is 6.38 x 10^6 m. Would I add [Cos(θ)] to the equation v=rω and make it v=R*Cos(θ)*ω for parts b and c? (I tried it already for part b, but got a negative answer).

    By the way, I have v=463.83 m/s for part a.
  7. Apr 11, 2010 #6
    Tiny Tim, Though your posts are usually helpful, your signature is somewhat distracting just now.

    Balletgirl, what part don't you understand?
    Last edited: Apr 11, 2010
  8. Apr 11, 2010 #7
    The circumference of the circle is proportional to its radius, which is proportional in this case to its latitude as SIN of the Lat.
  9. Apr 11, 2010 #8
    It would appear we became asynchronous.
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