Suppose that, while lying on a beach near the equator watching the Sunset over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H = 1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t = 11.1 s, what is the radius r of Earth?

The height of the person h = 1.70 m

Time elapsed t = 11.1 sec ,

Considering the projection of the earth is nothing more thana circle. A person standing in the center of the circle can be seen as a line h.Drawing the tangent to the circle from the line h it forms a rectangular triangle.Therefore, using thePythagorean Theorem, the radius of the earth can be calculated.

When the person stands up, his line-of-sight changes. Find the angle $\theta$using these two different lines of sight. Using this value of $\theta$and the time in which this change took place, it is possible to find the radius of the earth.

From Pythagorean Theorem,

role="math" localid="1651835120648" $$\begin{gathered} d^2+r^2={\left(r+h\right)}^2 \\ =r^2+2rh+h^2 \end{gathered}$$

Here,h is very small as compared to r, so neglecting higher terms of h,

$d^2\approx 2rh$ … (i)

Convert 24 into seconds.

1 hr = 3600 sec

Therefore,

Earth takes 24 hrs for one complete rotation, that is, 360 degree

So, for t = 11.1 s it takes θ degrees,

$$\begin{gathered} \frac{\theta }{360}=\frac{11.1}{86,400} \\ \theta =0.04625\text{degrees} \end{gathered}$$

Thus, the angle $\theta$ is $\text{0.04625degrees}$

From the figure, it can be written that,

$d=r\tan \left(\theta \right)$

Squaring both the sides,

$d^2=r^2{\tan }^2\left(\theta \right)$… (ii)

Now substitute the value of from equation (i) to find the equation for .

$r=\frac{2h}{{\tan }^2\left(\theta \right)}$… (iii)

Substitute the values in equation (iii) to calculate the radius of the earth.

$$\begin{gathered} r=\frac{2\times 1.7\text{m}}{{\tan }^2\left(0.04625\right)} \\ =5.2\times {10}^6\text{m} \end{gathered}$$

Thus, the radius of the earth is$5.2\times {10}^6\text{m}$