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Crude geometric estimation or am I missing something?

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm doing a report for a physics lab experiment where we are calculating the radius of the earth by measuring the time it takes to see the sunset from the base of a cliff looking out into the pacific ocean till when it sets in relation to an observer at the top of the cliff. The picture I attached shows the diagram and the initial setup on how to find the radius of the earth with the pythagorean theorem, but in yellow where it suggests how to find the value L, I'm thinking that multiplying the change in time by the circumference of the earth divided by 24 hours would really give me the distance traveled along the circumference of the earth, not L. Please let me know if I'm right about that and if so can you tell me if and how it might still be used as a crude estimate for L by my professor? If I'm wrong please let me know what geometric rule I'm missing to help me out.

    2. Relevant equations

    See attached

    3. The attempt at a solution

    Part 1 here covers that.
  2. jcsd
  3. Jan 21, 2009 #2
    Sorry, I guess the attachment function doesn't work on this site. Here's the image:[​IMG]
  4. Jan 26, 2009 #3
    Without the approximations, if the time difference between sunsets is t, then the angle subtended by the sun is [tex]\theta=\omega t[/tex].

    From the figure you get,



    You also know,
    [tex]\omega=\frac{2\pi}{24\times 60\times 60}[/tex]

    and that [tex]\theta=\omega t[/tex]

    as [tex]\theta[/tex] is very small, [tex]sin\theta=\theta[/tex] (approximately).

    This gives you (after substituting and rearranging)

    [tex]R=\frac{12\times 3600}{\pi}-h[/tex] (approximately)
  5. Jan 26, 2009 #4
    The angle theta that I've taken is 90-theta in the figure.
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