# Satellite Orbital Period [closed]

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?

• 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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