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I know I can calculate the period of a satellite orbit by Kepler's third law, but somehow it does not work out. The sattelite is 20200km from surface of the earth.

  • $r=$orbits radius=earths radius+satellites distance from surface of earth=20,200,000+6,378,000 = 26,578,000 m
  • $G=6.67\cdot10^{-11}$
  • $M = $mass of earth $= 5.9722\cdot10^{24}$

now $T=(4\pi^2r^3/GM)^{1/2} = 43108,699\ \mathrm{s} \Rightarrow T=11.975\ \mathrm{hours}$

BUT that isn't correct, as all the calculators say it is 16,53

I have no idea what I am doing wrong.

I even followed this example and I got everything right using the numbers in the example, but as soon as I put in my 26,578,000 m I got a different solution. Even though I did not change anything else.

What am I missing?

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closed as too localized by Qmechanic, Ebenezer Sklivvze, Manishearth Dec 29 '12 at 15:58

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I don't know what you mean by "all the calculators", but the period of a satellite at that height is indeed about 12 hours, not 16.5 –  Mark Eichenlaub Oct 3 '12 at 0:22
    
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1 Answer 1

up vote 0 down vote accepted

Your formula and numbers look right to me. You can check (both your math and "all the calculators") by plugging in the numbers for a geosynchronous orbit (altitude of 35,786 km, or semi-major axis of 42,164 km): the period should be 24 hours.

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