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The question I'm about to pose is from a physics book I had recently bought. Since I am very interested in physics I am quite keen in understanding how this question can be solved. Before I present this question, I would like to explain the context in which this question was posed. It is a question related to the chapter entitled - Measurement, dealing with the understanding of the basic quantities like mass, time, length etc. Now let me pose the question:

Suppose that while laying down on a beach near the equator of the Earth, you watch the sun setting below the horizon. You start a stopwatch the moment the top of the sun disappears below the horizon. You then stand to a height of 1.70 metres, and stop the watch when the top of the sun disappears again. The time interval is exactly 11.1 seconds. What is the radius of the Earth?

I must inform you that this is not a homework question or any such related assignment, I am simply asking a question whose solution I have been unable to find myself or online. I hope you'll be able to answer this question.

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Have you started making a sketch of the situation? If you share it maybe we can point you where you are making a mistake. – elcojon Jan 20 '13 at 19:04

You could use the distance to the horizon formula (

For $h=1.70$ m the horizon is $d=4.7$ km far away (roughly). When it sets again, the sun passed this distance. But we know that this must be equal to an angle of

$$ \gamma=\frac{11.1 \mbox s}{24 \mbox h}2\pi$$ but $\gamma = d/R$. So

$$ R = \frac{d}{\gamma}= \frac{24*60*60 \mbox s}{11.1 \mbox s}\frac{4.7 \mbox{km}}{2\pi}=5825 \mbox{km}$$

The answer should be 6350 km, but it's a good result. Consider that a very small error on the time leads to big uncertainties on the result. A difference of one second would lead to the correct value.

Very nice question.

EDIT: I understood that the value Wikipedia reports for the distance of horizon uses the radius of Earth! This is not valid. So we can use the original $d=\sqrt{2Rh}$, then again $\gamma = d/R$ so that $$ \sqrt{2Rh}=\gamma R$$ ignoring $R=0$ solution $$ 2h = \gamma^2 R $$ from $\gamma = 2\pi *11.1\mbox s/24\mbox h$ one obtains $$R = 5223 \mbox{ km}$$

Notice that it is less precise: now we are using the experimental error twice: to extimate the horizon distance and to find $d$.

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