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

Result

p =  2.604 %

Solution:

h=350 km R=6370 km  α=arcsin⁡(RR+h)=arcsin⁡(63706370+350)≐1.2466 rad  (R+h)2=a2+R2  a=(R+h)2−R2=(6370+350)2−63702≐70 935 km≐2140.4439 km v=a⋅ cos⁡(α)−h=2140.4439⋅ cos⁡(1.2466)−350≐1592548≐331.7708 km S1=2π⋅ R⋅ v=2⋅ 3.1416⋅ 6370⋅ 331.7708≐13278759.4735 km2 S2=4π⋅ R2=4⋅ 3.1416⋅ 63702≐509904363.7818 km2 p=100⋅ S1S2=100⋅ 13278759.4735509904363.7818≐12548≐2.6042≐2.604%h=350 \ \text{km} \ \\ R=6370 \ \text{km} \ \\ \ \\ α=\arcsin( \dfrac{ R }{ R+h } )=\arcsin( \dfrac{ 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 } \doteq 70 \ \sqrt{ 935 } \ \text{km} \doteq 2140.4439 \ \text{km} \ \\ v=a \cdot \ \cos(α) - h=2140.4439 \cdot \ \cos(1.2466) - 350 \doteq \dfrac{ 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 \ \dfrac{ S_{1} }{ S_{2} }=100 \cdot \ \dfrac{ 13278759.4735 }{ 509904363.7818 } \doteq \dfrac{ 125 }{ 48 } \doteq 2.6042 \doteq 2.604 \%

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