What is the angular velocity, in radians per second, of a satellite in geostationary orbit?

"Geostationary" means that the satellite stays above the same point on earth all the time -- and, it must be above the equator. So the satellite angular velocity is the same as the earths angular velocity at the equator.

 A complete circle is 2pi radians, time for revolution = 24 hours   so,
angular velocity = 2* 3.14 / 24   = .26 radians / hour

For seconds  1 hour = 60*60 seconds
.26  radians/hour * 1 hour/3600  =  7.22 * 10 ^ -5  radians/second

"Geostationary" means that the satellite stays above the same point on earth all the time -- and, it must be above the equator. So the satellite angular velocity is the same as the earths angular velocity at the equator.

A complete circle is 2pi radians, time for revolution = 23 hours 56 minutes 4 seconds so,
angular velocity = 2*pi/(86164 seconds)

=
angular velocity = 7.292*10^(-5) radians per second



WHAT was different about the input of my answer, and the input of Dave's answer?

Dave used the SOLAR DAY as the period of rotation of the Earth.

What needs to be done, is to use the SIDEREAL DAY as the period of rotation of Earth.  I.e. how much time between successive "Orion rises"?