# Satellite Orbits

1. Sep 29, 2004

### Jack16

There's a satellite called Meteostat which in the geostationary orbit.Could you explain me why its periodic time is not exactly equal to 24 when it is viewed by an observer positioned at a point on the equator...

2. Sep 29, 2004

### BobG

Where are you getting the idea that it doesn't take 24 hours to reach the same same point above the Earth? If you're looking at mean motion in two line element sets (about 1.0027), it's because mean motion is measuring revolutions per sidereal day, not per solar day.

A sidereal day is measuring the amount of time for the Earth to rotate 360 degrees and a solar day is measuring the amount of time from the Sun being directly overhead to the next time the Sun is directly overhead.

3. Sep 29, 2004

A sidereal day being 23 hours, 56 mins and 4 seconds.

4. Oct 1, 2004

### edtman

It may be that Meteostat is either slightly above or below the required altitude for a 24 hour orbit; or its inclination may not be exactly 0 degrees.

5. Oct 2, 2004

### BobG

If you're getting extremely precise, it's almost never exactly at a geosynchronous altitude. Orbital perturbations from the Earth's triaxiality, the Sun, the Moon, and (to a much lesser extent) the planets pull it out of its desired orbit. The rate a geosynchronous satellite drifts depends upon the longitude it's stationed over. Normally, the only significant difference you'll notice is right after the operators maneuver the satellite to compensate for how much it has drifted during the last few months.

Inclination won't affect the period. It'll just affect how big of a 'figure-8' the satellite's ground track makes each orbit.

Last edited: Oct 2, 2004