1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Mark Eichenlaub Oct 3 '12 at 0:22
  • $\begingroup$ Welcome to Physics Stack Exchange! Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) $\endgroup$ – Manishearth Dec 29 '12 at 15:57
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.