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February 28, 2015

February 28, 2015

Posted by **Becky** on Tuesday, March 29, 2011 at 9:27am.

- Physics -
**bobpursley**, Tuesday, March 29, 2011 at 9:31amset 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

- Physics -
**tchrwill**, Tuesday, March 29, 2011 at 11:15amThe 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.

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