Elevation

What must be an observer's elevation so that he may see an object on the Earth 536 km away? Assume the Earth to be a smooth sphere with a radius 6378.1 km.

Correct answer:

h =  22482.4 m

Step-by-step explanation:

$$\begin{gathered} R=6378.1km \\ d=536km \\ (R+h{)}^2=d^2+R^2 \\ R+h=\sqrt{d^2+R^2} \\ h=\sqrt{d^2+R^2}-R \\ h=\sqrt{536^2+6378.1^2}-6378.1 \\ h=22482.4\text{m} \end{gathered}$$

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