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Given that the Russian GLONASS system has an orbital altitude of 19,140 Km and a period of revolution of 11 hours and 15 minutes, the US system has an orbital altitude of 20,180 Km and a period of revolution of 11 hours and 58 minutes and finally the Galileo system an orbital altitude of 23,222 Km and a period of revolution of 14 hours and 22 minutes, how does one calculate the period of revolution for the Chinese Compass system with an orbital altitude of 21,150 Km?

EDIT: I used Kepler's third law in my original calculation, but assumed that the orbital altitudes given are referenced to the centre of the earth. I made this assumption based on how I interpreted the math, and did not know that the orbital altitudes are referenced to mean sea level on Earth. Once I added the radius of the earth which is approximately 6,371 Km to the orbital altitudes I was able to calculate the period of revolution. So the actual orbital distance from the centre of the earth to the Chinese Compass system is 21,150 Km + 6,371 Km.

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  • $\begingroup$ I didn't downvote, but your assertion that "no question is a stupid question" is unfortunately not true on Stack Exchange at all. Questions that get answers that will be useful in general are great; questions that won't, aren't, and the site is not designed to welcome the latter. $\endgroup$
    – Nathan Tuggy
    Aug 20, 2016 at 2:04
  • $\begingroup$ Your original question didn't show research effort, and looked a bit like it might be a case of someone looking for an answer to a homework question. I was about to advise a search for 'orbital period' and link a couple of pages in a comment when the answer below came in. Your edit shows that you did make an effort and further clarifies what your confusion was. That is a big improvement. As we are all volunteers here and we are trying to assemble information useful to future visitors, we are rather picky about this. $\endgroup$
    – kim holder
    Aug 20, 2016 at 2:06
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    $\begingroup$ @Kim - At the age of 55 I doubt if I am a kid doing homework assignments. LOL. I am an embedded engineer by training and I happily answer questions that people ask me. I last studied Kepler's laws some 34 years ago and am a bit rusty. We cannot be specialists in all fields. Now that I know how this site operates I shall word my questions a tad bit more carefully. $\endgroup$
    – StonedRanger
    Aug 20, 2016 at 3:05
  • $\begingroup$ @kimholder and everyone - I left a comment about the benefits of including "what have you tried so far" etc in the question, and S'Ranger quickly updated the question accordingly (there was indeed work, just not mentioned). I deleted my comment as "cleanup", but the original "a pity" comment lingers - making it look like there is still discontent where (I assume) there isn't anymore. S'Ranger you may want to delete that comment unless you are deeply attached to it. After a certain delay, people are unable to reverse their down votes until you make an edit to the question (built-in to SE's UI). $\endgroup$
    – uhoh
    Aug 20, 2016 at 3:53
  • $\begingroup$ At least in this particular SE site, down votes are not at all personal. They are actually used to help improve questions by suggesting there's an issue. Sometimes they smart a bit, but (at least here) they are usually intended as constructive devices to flag where some fine tuning or attention is needed. $\endgroup$
    – uhoh
    Aug 20, 2016 at 3:53

1 Answer 1

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You can use an online calculator such as this.

The math is explained here.

In this case the result is a period of 12 hours 37 minutes at a velocity of 3.8 Km/s.

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    $\begingroup$ Thank you. I see where I made my error. I used the altitude as the orbital radius in my calculations, hence my confusion. $\endgroup$
    – StonedRanger
    Aug 19, 2016 at 22:12
  • $\begingroup$ It's pretty simple to see where I made my mistake. I assumed that the altitude of the satellites were referenced to the centre of the earth, as the formulae I read up on all referenced from the centre of one satellite to the other. I did not know that the altitude referred to the mean height of sea level on earth. Once I added the radius of the earth which is approximately 6,371 Km to the altitude, my calculations gave me the correct answers. $\endgroup$
    – StonedRanger
    Aug 20, 2016 at 0:39
  • $\begingroup$ I can understand using altitude for aircraft and other measurements on earth, but assumed when dealing with planetary bodies everything was referenced to the centre of each celestial body respectively. $\endgroup$
    – StonedRanger
    Aug 20, 2016 at 0:56
  • $\begingroup$ That's not a bad assumption; for pure astronomical work, references from center are often used, but for man-made orbital objects it's more natural to use sea level as the reference, since launch site altitudes are usually so measured and recorded. $\endgroup$
    – Russell Borogove
    Aug 20, 2016 at 1:42

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