Assuming a circular orbit, how high is this satellite above the surface of the earth?
Assuming a circular orbit, what is the altitude of the International Space Station above the Earth's surface, given that it completes 15.65 revolutions per day around the Earth?
1 Answers
T = period or time for one revolution in sec
T = (1 / 15.65)(24)(60)(60) = 5520.77 sec
T = [ (2*pi) ] sqrt [ (r ^ 3) / (u) ]
r = R + h
R = radius of earth 6.378E+6 m
h = satellite height in m
u = (G)(M)
G = gravitational constant 6.673E-11
M = mass of earth 5.974E+24 kg
(h + R) ^ 3 = [ (G)(M) ][ [ (T)^2 / (2*pi)^2 ]
(h + 6.378E+6) ^ 3 = [ (3.986E+14) ][ [ (5520.77)^2 / (2*pi)^2 ]
h = 3.731E+5 m
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