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# A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on Earth.

This is known as a geosynchronous orbit. Television, communication, and weather satellites use geosynchronous orbits. At what distance would a satellite have to orbit Earth in order to have a geosynchronous orbit?

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Daniel O. | Math and Physics Tutor, with a math and physics degreeMath and Physics Tutor, with a math and ...
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First convert period T into seconds.

T = 24hrs = 24*60*60 = 86400 seconds.

Then equate gravitational force between the satellite and earth, and centripetal force on the satellite, since the forces are equal. Use v = 2*pi*r/T in place of velocity in the centripetal force formula. I won't show the derivation (unless you'd like me to, then say so) but you'll get to:

r3 = T2GME/(4*pi2

which you can solve for r. G is the gravitational constant, ME is the mass of the earth. Note that r is the distance from the center of the earth, subtract the earth's radius from that if you want altitude above the earth's surface.

BRUCE S. | Learn & Master Physics & Math with Bruce SLearn & Master Physics & Math with Bruce...
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Hello Caroline,

I agree with Alex's number using 24 hours for the earth's rotational period and Daniel presents the correct mathematics for the altitude and a 24 hour period.  If you were to use the 24 hour period for your geosynchronous satellite however you would find that the spot directly below the satelite would drift steadily westward each day because the earth's actual rotational period is slightly shorter than 24 hours.  A more precise period is 23 hr., 56 min., 4.09053 sec.

Just a note if you were actually going into the satelite business.

Bruce S.

Alex J. | Experienced Tutor in Finance, Accounting, and ECONExperienced Tutor in Finance, Accounting...
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35,786 km (22,236 mi) above sea level for a satellite in geosynchronous orbit over the equator.

This is actually a special case but it's the most frequently cited number.