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Angular Displacement/Homework Help. Stumped

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    To attend the 2000 Summer Olympics, a fan flew from Mosselbaai, South Africa (34S, 22E) to Sydney, Australia (34S, 151E). (a) What is the smallest angular distance the fan has to travel: (1) 34 degrees (2) 12 degrees (3) 117 degrees (4) 129 degrees? Why?
    (b) Determine the approximate shortest flight distance, in kilometers.

    2. Relevant equations


    3. The attempt at a solution
    Am I just adding the vectors of the east directions? I assumed that when it asked for the quantity in angular distance that I needed to change the Cartesian coordinates to polar coordinates, and I do realize I will need to convert to radians and use S=r(theta) to solve part b, correct?

    Any help or guidance at all would be greatly appreciated. Thanks in advance!
     
    Last edited: Feb 16, 2015
  2. jcsd
  3. Feb 16, 2015 #2

    SteamKing

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    Why are you adding?

    Strictly speaking, geographic positions (i.e., latitude and longitude) are not cartesian coordinates. They are a special means of specifying a location on the surface of a sphere.

    In order to determine the distance flown, you will need to find the circumference of the earth at the parallel of latitude on which the flight was taken, in this case 34 degrees South.
     
  4. Feb 16, 2015 #3
    This wasn't provided in the prompt, but outside resources tell me it is 33258 kilometers.

    Ahhh I meant subtraction, silly mistake.
     
  5. Feb 16, 2015 #4

    haruspex

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    Not so. That would simply produce the difference of the longitudes. The shortest distance is along a great circle. That will dip further south than 34S along the way.
    TheBestMiller, consider a slice through the Earth, perpendicular to the NS axis, at 34S. You know the angle the endpoints subtend to the centre of that circle. What is the radius of the circle? How far apart, in a straight line through the earth, are the endpoints?
     
  6. Feb 16, 2015 #5

    SteamKing

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    That's approximately correct. (How could you verify this figure, knowing only the average radius of the earth, or 6371 km?)

    What's the distance of the flight taken from SA to Oz then?
     
  7. Feb 16, 2015 #6
    Well R=√x^2 + y^2 correct? In that case the radius for point 1 would be approximately 40.5 and point 2 would be 154.8. Would I then set up a triangle with the radii and use Pythagorean theorem for the distance, in a straight line, between the two points?
     
  8. Feb 16, 2015 #7
    It would have to be an arc length because the earth is a sphere. So I would use S=r(theta)?
     
  9. Feb 16, 2015 #8

    SteamKing

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    It's not clear what you are calculating here. The geographic positions of the two cities in SA and Oz are given in latitude, longitude format, both of which parts are measured in degrees. You can't treat these geographic coordinates like they were cartesian coordinates.
     
  10. Feb 16, 2015 #9
    Ahhhh alright, that makes things much more clear.
     
  11. Feb 16, 2015 #10

    haruspex

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    I'm not sure what you're doing there, maybe you misunderstood.
    Let the endpoints be A, B and the centre of the earth be O. Poles are N and S. Let P be the point on the line NOS at the latitude of A and B, i.e. APO and BPO are right angles.
    Draw a circle containing NOPS and A. What is the distance AP? What is the angle APB? So what is the distance AB?
     
  12. Feb 17, 2015 #11

    CWatters

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    Haruspex mentioned Great Circle. Google it and make sure you understand the concept. It will make answering the problem a lot easier!
     
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