Elevation

What must be the elevation of an observer in order that he may be able to see an object on the earth 536 km away? Assume the earth to be a smooth sphere with radius 6378.1 km.

Correct result:

h =  22482.4 m

Solution:

R=6378.1 km d=536 km  (R+h)2=d2+R2 R+h=d2+R2 h=d2+R2−R h=5362+6378.12−6378.1 h=22482.4 mR = 6378.1 \ km \ \\ d = 536 \ km \ \\ \ \\ (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}

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