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Which wave reaches point B first?

  1. Feb 19, 2008 #1


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    1. The problem statement, all variables and given/known data
    Two points A and B on the surface of the earth are at the same longitude and 60 deg apart in latitude. Suppose an earthquake at point A creates a P wave that reaches B by traveling straight throught he body of the earth at constant speed of 7.80m/s. The earthquake also creates a Rayleigh wave that travels along the surface of the earth at 4.50km/s.

    a) which of these 2 waves arives at B first?
    b) What is the time difference in the arrival of these 2 waves at B? Take the earth's radius to be 6,370km.

    2. Relevant equations
    well I found this online for the distance
    D= R arc cos[sin(lat 1) x sin(lat 2) + cos (lat 1) x cos (lat 2) x cos [ lon 2- lon 1]]

    y(x,t)= f(x- vt)

    3. The attempt at a solution

    well I first calculated the distance between the 2 points with the equation I found online.

    D= R arc cos[sin(lat 1) x sin(lat 2) + cos (lat 1) x cos (lat 2) x cos [ lon 2- lon 1]]

    R of earth= 6,370km
    lat 1 = 0
    lat 2= 60 deg => [tex] \pi / 3[/tex]
    lon 1= 0
    lon 2= 0

    D= 6,370km arc cos[sin(0) x sin([tex] \pi / 3[/tex]) + cos (0) x cos ([tex] \pi / 3[/tex]) x cos [ 0]]

    D= 6,370km arc cos [0 + 0.5]

    D= 6,670.64km

    x= 6,670.64km

    But as to how to find the time has me confused a bit.

    I think I use this equation where I just solve for time then subtract the 2 times for the different waves but I'm not 100% certain. (that's where I need help)

    y(x,t)= f(x- vt)

    for the 1st wave

    y(6,670,640m,t)= f(6,670,640m - 7,800m/s (t))

    same for the Raleigh wave

    y(6,670,640m,t)= f(6,670,640m - 4,500km/s (t))

    Or do I just use V= d/t and solve for t??

    can someone help me out with which equation to use to find the time it takes for a wave to reach a certain distance?

    Well I'm not sure how to solve for t since isn't y the transverse position of the wave which I just don't have in the information?

    Thanks alot :smile:
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 20, 2008 #2

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    The same eqn as always, which you have written down.

    A and B are two points on a circle of radius R. There are two paths to be considered here; one is a straight line from A to B, and the other is the arc from A to B. If you know the angle subtended at the centre, then the dist along the paths are known, and the time is got from dist/speed.

    There is absolutely no reason for considering the wave eqn here.
  4. Feb 20, 2008 #3


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    It's interesting since I got the answer incorrect from what the book has. I think the distance is what is causing this.
    You didn't say whether the way I got the distance is correct or not... Below

    If the above way I got the distance from A=> B is incorrect then I'm not sure how I'd find it. I don't understand what you said about finding the distance.

  5. Feb 20, 2008 #4

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    Since the two points are on the same longitude, simply think of the circle you'll get get if you slice it through that longitude. You now have to deal with just a 2-d circle, whose radius is that of the earth. The difference of the latitudes being 60 deg means that the radii from the two points subtend an angle of 60 deg at the centre.

    Suppose the centre of the earth is O. Angle AOB is given. Draw a simple diagram. Can you find segment AB? Can you find arc AB? Big calculations are not necessary, I think. :smile:

    After you find the distances, use v=d/t.
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