Geostationary remains in one place around the Earth's equator and goes up and down on it and not round the Earth, right?
Geosynchronous rotate around the Earth?
Someone who knows this, help?
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Raithed
China7077 Posts
Geostationary remains in one place around the Earth's equator and goes up and down on it and not round the Earth, right? Geosynchronous rotate around the Earth? Someone who knows this, help? | ||
B1nary
Canada1267 Posts
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micronesia
United States24343 Posts
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jtan
Sweden5891 Posts
Then think what happens when the axis of rotation of the sattelite is other then the axis of rotation of the earth, if you look at a map of the earth it will look like its going in a kind of sine-wave, but in fact its just a circular orbit. A geostationary satellite orbits the earth in 24 hours (in the same direction as the earth spins ) If you draw such a satellite on a world map it would only be a single dot. And yeah it would have to be on the equator. | ||
Raithed
China7077 Posts
On May 05 2008 09:30 B1nary wrote: Geosynchronous means the same period as the earth's rotation, so one revolution every 24 hours. Geostationary orbits are perfectly circular geosynchronous orbits around the earth's equator. So if I look up right now, geosynchronous will be up there? For the bolded, it won't be but it rotates around the Earth's equator? | ||
Freyr
United States500 Posts
Incidentally, if you happen to read the relevant wiki article it states that geosynchronous orbits have the same period as earth's sidereal period, which as I understood it meant the period of the earth's orbit around the sun (ie ~365 days). This is not correct, and as B1nary stated, geosynchronous orbit refers to an object completing 1 revolution around the earth in ~24 hrs. | ||
Raithed
China7077 Posts
Another thing, I have a question about, not sure if you guys would know it: A band-pass filter has a Q of 50 and a bandwidth of 6MHZ. Find the center freq ... I can't find the formula for it, except something like this: http://en.wikipedia.org/wiki/Band-pass_filter Distinctly, I remember, isn't Fo = BW/Q so Fo = (6x10^3)/50 right? | ||
micronesia
United States24343 Posts
On May 05 2008 09:49 Raithed wrote: Okay. Thank you. Another thing, I have a question about, not sure if you guys would know it: A band-pass filter has a Q of 50 and a bandwidth of 6MHZ. Find the center freq ... I can't find the formula for it, except something like this: http://en.wikipedia.org/wiki/Band-pass_filter Distinctly, I remember, isn't Fo = BW/Q so Fo = (6x10^3)/50 right? I remember using those when I was playing with a sound board. Never did it quantitatively though! GL | ||
Raithed
China7077 Posts
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