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Thread: Finding the radius of the earth

  1. #1
    Member Chokfull's Avatar
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    Finding the radius of the earth

    OK---i have a problem.

    you lie on a beach and watch a sunset. then you stand and watch it set again--you are higher and can see that supposedly.

    my book says that you then use these values:
    $\displaystyle h=1.70 m$ (your height)

    $\displaystyle t=11.1 s$ (the time between sunsets)

    then using the picture and the Pythagorean theorem, with $\displaystyle d$ as the distance between your eyes and the tangent to the earth from the second sunset, you get
    $\displaystyle d^2 + r^2 = (r + h)^2 = r^2 + 2rh + h^2$
    or
    $\displaystyle d^2 = 2rh + h^2$

    but you remove $\displaystyle h^2$ because your height is insignificant compared to the radius of the earth and the term $\displaystyle 2rh$

    then you use the number of hours in a day, the rotation of the earth in a day, and $\displaystyle t$ to find $\displaystyle \theta$:

    $\displaystyle \frac \theta {360^o}$

    which solves to $\displaystyle 0.04625^o$

    then you can get $\displaystyle d = r\tan \theta$ from the picture and, substituting for $\displaystyle d$ in the first equation, $\displaystyle r^2\tan^2\theta = 2rh$

    which when you plug in the values (like $\displaystyle \theta = 0.04625$) makes $\displaystyle 5.22 * 10^6 m$

    The problem is in the fact that we replaced the value for the $\displaystyle \theta$ from the angle between the sunsets with the $\displaystyle \theta$ from the angle from the center of the earth. i cant figure out how that works.


    Here is the picture:
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  2. #2
    Grand Panjandrum
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    see:

    Attachment 12590

    Angle $\displaystyle \text{uvA}=\theta$ so $\displaystyle \text{vuA}$ is $\displaystyle 90-\theta$ and so $\displaystyle \text{uOB}=\theta$

    CB
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  3. #3
    Member Chokfull's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    see:

    Attachment 12590

    Angle $\displaystyle \text{uvA}=\theta$ so $\displaystyle \text{vuA}$ is $\displaystyle 90-\theta$ and so $\displaystyle \text{uOB}=\theta$

    CB
    wow ok thanks. i dont know why the book didnt explain that.
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