Posted by Becky on Tuesday, March 29, 2011 at 9:27am.
set the force of gravity equal to centripetal force.
GMm/r^2=mw^2*r
where w=2pi/period (set period to seconds in a day)
solve for r
The time it takes a satellite to orbit the earth, its orbital period, can be calculated from
T = 2(Pi)sqrt[a^3/µ]
where T is the orbital period in seconds, Pi = 3.1416, a = the semi-major axis of an elliptical orbit = (rp+ra)/2 where rp = the perigee (closest) radius and ra = the apogee (farthest) radius from the center of the earth, µ = the earth's gravitational constant = 1.407974x10^16 ft.^3/sec.^2. In the case of a circular orbit, a = r, the radius of the orbit.
The geostationary orbit is one where a spacecraft or satellite appears to hover over a fixed point on the Earth's surface. There is only one geostationary orbit in contrast to there being many geosynchronous orbits. What is the difference you ask? A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, is 23hr-56min-4.09sec., not 24 hours. Thus, the required altltude providing this period is ~22,238.64 miles, or ~35,787.875 kilometers.