\section*{Task 1 - 031111}

The 352 m high antenna mast of the Deutschlandsender in Zehlendorf, District of Oranienburg, Potsdam Region, is currently the tallest structure in Europe.

a) How large is the area that can be overlooked from the top of the mast on a clear day? (Elevations in the terrain are neglected in the calculation.)

b) What is the percentage error that occurs if the height of the antenna mast is used instead of the correct height of the spherical cap in the formula for the spherical cap \(M=2 \pi R h\)?

Why is the error very small? (Radius of the Earth \(R=6370 \text{ km}\).)
}

a) To initially calculate the area \(M=2 \pi R H\) of the visible spherical cap, one finds, using the Pythagorean theorem for the line of sight \(a\), the expression

\[
a=\sqrt{(R+h)^{2}-R^{2}}=\sqrt{2 R h+h^{2}}
\]

This is valid because the triangle \(\triangle A C S\) is a right triangle, as the line of sight can be considered a tangent to the circle (the Earth).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1696.jpg?height=408&width=566&top_left_y=1161&top_left_x=742)

With \(E\) and \(C\) as points on the base of the spherical cap and \(D\) as the point where the antenna mast touches the Earth, let \(H\) be the length of the segment \(D E\). Then \(\cos \alpha=\frac{a}{h+R}=\frac{h+H}{a}\) (the triangles \(\triangle A C S\) and \(\triangle E C S\) are similar due to a common angle and the right angle present in both triangles) and thus also

\[
H=\frac{a^{2}}{h+R}-h=\frac{2 R h+h^{2}}{h+R}-h
\]

Thus, the result is

\[
M=2 \pi R H=2 \pi R\left(\frac{2 R h+h^{2}}{h+R}-h\right)
\]

For the given example, \(H \approx 351.98 \, \text{m}\) and thus \(M \approx 1.4088 \times 10^{10} \, \text{m}^2\). b) \(h\) is \(100.005682\%\) of \(H\), so the deviation in the area is \(0.005682\%\). This deviation is so small for \(h \ll R\) because then

\[
H=\frac{2 R h+h^{2}}{h+R}-h \approx \frac{2 R h}{R}-h=h
\]

holds.