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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?

amWhy
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Mary Star
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    The distance to the horizon is the distance from the observer to the most distant point on the spherical earth's surface that the observer can see. That is the line joining the observer's eye to a tangent to the sphere. What is the specific case you mention? – WA Don Feb 16 '21 at 13:16
  • Since the observer is at a height of $h$ and the diameter of the sphere is $D$ do we consider the tangent at distance $h+D$ from the observer? @WADon – Mary Star Feb 16 '21 at 13:21
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    Use any one of the lines from the point at $h+D$ that is tangent to the sphere. Your job is then to determine the distance from the observer to the point where the line touches the sphere. All other lines from the observer to the sphere will either be shorter or, if longer, will be obscured by the sphere itself. – WA Don Feb 16 '21 at 13:29
  • Thanks a lot!! :-) @WADon – Mary Star Feb 16 '21 at 20:59

1 Answers1

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$$f(h)=\sqrt{\left(\frac{D}{2}+h\right)^2-\frac{D^2}{4}}=\sqrt{h (D+h)}=\sqrt{D h} \sqrt{1+\frac{h}{D}}$$

$$\sqrt{1+x}\sim 1+x/2;\;\text{ as }x\to 0$$ 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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Raffaele
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