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A

Anonymous

Feb 21, 2025

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

A
Anonymous

Jan 02, 2025

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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