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Calculating the Speed of the Earth's Rotation.

  1. Sep 19, 2007 #1
    [​IMG]
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    I can provide a couple of better screen shots if needed. And I don't have to answer #5.

    Here are some relevant questions/attempts at answers (the ones I didn't ask about are the ones I need most help with):

    1. a) Are you basically just taking the circumference equation and changing it around to make a radius one? i.e C/2π = r

    1. c) You apply Pythagora's theorm?

    3. b) Would it have anything to do with corresponding angles? (probably not)

    4. I live in Toronto, Ontario, Canada and the answer is most likely yes. Once I've found the circumference at my latitute, would I simply divide by 24 (because that's how many hours are in a day) to get km per hr?

    Sorry if I'm not being clear in some areas. And thanks in advance to anyone who can help.
     
  2. jcsd
  3. Sep 19, 2007 #2

    mgb_phys

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    Exactly correct.
    You could also write a general formula involving just the latititude which calculates the speed relative to the speed at the equator.
     
  4. Oct 4, 2007 #3
    For question number 4, how would I find the circumference at my latitude? (which is 43 degrees) At first I thought you used a^2+b^2-2abCos43=c^2 (pythagora's theorm), and then you times by pi, but I got 14687 as an answer. Yet Ottawa is only 2 degrees more northern (45) and the book says it's circumference is 28338.

    This is due tomorrow so I'd appreciate if someone could help me.
     
  5. Oct 4, 2007 #4

    mgb_phys

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    Latitude is measured from the equator, the hypotonuse is the earth's radius so is constant (roughly).
    Draw a vertical line down from Ottawa to the equator in your picture and the distance from the intersection to the centre is is simply R cos(lat).
    ( Double check by thinking what happens at the equator and the pole )
    This is the radius at that lattiude so simply multply by 2pi to get the circumference.
     
  6. Oct 4, 2007 #5

    EnumaElish

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    You have an isosceles triangle and you can calculate the angle between the two equal sides. But you cannot simply assume that angle is 43 degrees. Look at the diagram in the book again. You should be replacing 45 with 43 degrees, but that does not mean the included angle (90 degrees for Ottawa) is 43 for Toronto.
     
  7. Oct 4, 2007 #6
    = 6378 x Cos43
    = 6348 x 0.7313537
    = 4665

    C=2πr
    C=(2)(3.14)(4665)
    C=29 308 [This seems correct, since as stated before, at 45 degrees the circumference is 28 338]

    29 308 / 24
    = 1221

    Therefore, Ottawa is traveling at 1221 km/hr.

    Pretty sure that is correct. Thanks a lot. Much appreciated.

    And could you double check some of my other answers [stars are where I need help the most]:

    1. a)

    C/2π = r
    40076/6.2831851=r
    6378=r

    This is obviously correct, but is my method a sufficient way of proving it?

    **b) The diameter or length of the chord is is 9020. [I'm very skeptical about this answer]

    c)

    a^2+b^2=c^2
    a and b are the same, and would be the radius:
    r^2+r^2=c^2
    6378^2+6378^2=c^2
    ...and if you continue solving, you do get 9020.

    **d) What's left? I don't understand.

    I'll get the others in after I eat dinner.
     
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