What percentage

What percentage of the Earth’s surface is seen by an astronaut from a height of h = 350 km. Take the Earth as a sphere with the radius R = 6370 km

Correct result:

p =  2.6042 %

Solution:

$$\begin{gathered} h=350\text{ km } \\ R=6370\text{ km } \\ \alpha =\arcsin ({\displaystyle \frac{R}{R+h}})=\arcsin ({\displaystyle \frac{6370}{6370+350}})\doteq 1.2466\text{ rad } \\ (R+h{)}^2=a^2+R^2 \\ a=\sqrt{(R+h{)}^2-R^2}=\sqrt{(6370+350{)}^2-6370^2}=70\sqrt{935}\text{ km}\doteq 2140.4439\text{ km } \\ v=a\cdot \cos (\alpha )-h=2140.4439\cdot \cos (1.2466)-350={\displaystyle \frac{15925}{48}}\doteq 331.7708\text{ km } \\ {S}_1=2\pi \cdot R\cdot v=2\cdot 3.1416\cdot 6370\cdot 331.7708\doteq 13278759.4735{\text{km}}^2 \\ {S}_2=4\pi \cdot R^2=4\cdot 3.1416\cdot 6370^2\doteq 509904363.7818{\text{km}}^2 \\ p=100\cdot {\displaystyle \frac{S_1}{S_2}}=100\cdot {\displaystyle \frac{13278759.4735}{509904363.7818}}={\displaystyle \frac{125}{48}}=2.6042\mathrm{\%} \end{gathered}$$

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