Calculating Earth's Radius from a Beach Sunset

[The legend]

Homework Statement

while lying flat on a beach you see the sun set, and when the sun disappears you start a timer. Then you stand up, changing the position of your eyes(increasing height) by 1.7m. When the sun disappears from your view again you stop the timer which reads 11.1s. Calculate the radius of the earth.

Homework Equations

The Attempt at a Solution

I made a diagram of what i understood of the problem, here the sun is small and Earth big(see http://img2.imageshack.us/img2/9104/sum.png ).
[PLAIN]http://img2.imageshack.us/img2/9104/sum.png
Taking the angle at the sun 'x' and tangent distance = 'y'... i get tan(x) = r/y
Also by calculations i got thet
y² = 2.89+3.4r

Using pythagoras theorem i get
r/sin(x) + r² = 2.89 + 3.4r
this is quadratic in r but has another unknown x, so how can i proceed? Also i can't seem to think of a way to utilize time given..

 

[Xerxes1986]

Seems you have the right idea, but wrong picture. It says you start the timer when the bottom of the sun crosses the horizon.

Also, I think you have to assume you are on the equator (assuming the picture you drew is a view of the Earth from the north pole) because the length of the sunset depends on the Earth's angular velocity as well as the distance from the axis of rotation. Think about it, if you were very near the north pole, your linear velocity is lower than if you were at the equator. This means that the sunset takes a much longer time.

 

[ehild]

See this thread

https://www.physicsforums.com/showthread.php?t=424172

ehild

 

[The legend]

wow the exact sum

 

[The legend]

thanks ehild

 

[zgozvrm]

Seems you have the right idea, but wrong picture. It says you start the timer when the bottom of the sun crosses the horizon.

Not so. The word "bottom" isn't used in the question. It states that when the sun disappears from view, the timer is started. That means the top of the sun crosses the horizon and the diagram is correct.

Also, I'm pretty sure that the intent of the question has nothing to do with angular velocity. I'm betting that the question is making an assumption that the Earth isn't spinning (or that the speed of rotation is negligible in 11.1 seconds) and that it's path around the Sun is a perfect circle with the center of the Sun being the center of that circle.

Obviously, the distance from the Earth to the sun would be needed to make the calculation

 

[The legend]

see ehilds link...it's good, but you have to neglect rotation, revolution, refraction and all that..(according to the problem)

 

[Xerxes1986]

I don't think you can neglect the Earth's rotation, because that is why the sun sets in the first place...

 

[zgozvrm]

What I meant was that I think you can look at the problem from the perspective of being on Earth. That is, to our perspective, the Sun moves around us.