# Thread:

1. ## 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:
$h=1.70 m$ (your height)

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

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

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

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

$\frac \theta {360^o}$

which solves to $0.04625^o$

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

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

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

Here is the picture:

2. see:

Attachment 12590

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

CB

3. Originally Posted by CaptainBlack
see:

Attachment 12590

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

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