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Assuming a circular orbit, how high is this satellite above the surface of the earth?

The International Space Station makes 15.65 revolutions per day in its orbit around the earth.
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Jakob Littel

Feb 20, 2025

T = period or time for one revolution in secT = (1 / 15.65)(24)(60)(60) = 5520.77 sec T = [ (2*pi) ] sqrt [ (r ^ 3) / (u) ]r = R + hR = radius of earth 6.378E+6 mh = satellite height in mu = (G)(M) G = gravitational constant 6.673E-11M = 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... Show More
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