The earth's diameter $D$ is approx. $12742$ km.
a) Under the assumption of an exact spherical shape of the earth, show that the distance $f(h)$ of an observer who is at height $h> 0$ above the surface of the earth to his Horizon is given by $f (h) = \sqrt{Dh}\sqrt{1+\frac{h}{D}}$.
b) Show that $f(h)=\sqrt{Dh}(1+r(h))$, where $0 <r (h) <\frac{h}{2D}$, and evaluate the approximation $f (h) \approx \sqrt{Dh}$ for $h = 10\ m$.
Show that in this specific case the error $f(h)-\sqrt{Dh}$ is smaller than $1\ cm$.
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I haven't really understood the description. What does it mean that the distance to his horizon?
$$f(h)=\sqrt{\left(\frac{D}{2}+h\right)^2-\frac{D^2}{4}}=\sqrt{h (D+h)}=\sqrt{D h} \sqrt{1+\frac{h}{D}}$$
$$\sqrt{1+x}\sim 1+x/2;\;\text{ as }x\to 0$$ Therefore, as $h\ll D$ we have $\frac{h}{D}\sim 0$ $$\sqrt{1+\frac{h}{D}}\sim 1+\frac{h}{2D}$$ And finally $$f(h)\approx \sqrt{Dh}\left(1+\frac{h}{2D}\right)$$
If $h=10$m$=0.010$km then
$$f(0.01)=\sqrt{12742 \times0.01} \sqrt{1+\frac{0.01}{12742}}=11.2880556\text{km}$$ the approximation gives $$f(0.01)=\sqrt{12742\times 0.01}\approx 11.2880512$$ The difference is about $8.86\times 10^{-6}$km that is $0.886$cm
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