Tuesday

April 28, 2015

April 28, 2015

Posted by **Jesus** on Friday, March 7, 2008 at 7:43pm.

- Physics -
**drwls**, Friday, March 7, 2008 at 8:07pmThe satellite must be in a cicular orbit with a period of one sidereal day (23 hours and 56 minutes). Use Kepler's third law to compute the distance from the center of the Earth.

- Physics -
**tchrwill**, Saturday, March 8, 2008 at 4:31pmThere is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about space travel, space stations, and communications. It was probably Arthur C. Clarke who was given the major credit for the use of this orbit for the purpose of worldwide communications.

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. An orbit with this period and altitude can exist at any inclination to the equator but clearly, a satellite in any such orbit with an inclination to the equator, cannot remain over a fixed point on the Earth's surface. On the other hand, a satellite in an orbit in the plane of the earth's equator and with the required altitude and period, does remain fixed over a point on the equator. This equatorial geosynchronous orbit is what is referred to as a geostationary orbit. The orbital velocity of satellites in this orbit is ~10,088.25 feet per second or ~6,877 MPH. The point on the orbit where the circular velocity of the launching rocket reaches 10,088.25 fps, and shuts down, is the point where the separated satellite will remain. The point on the Earth's surface immediately below the satellite is moving with a velocity of 1525.85 ft./sec.

How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation?

The force exerted by the earth on the satellite derives from

...................................................F = GMm/r^2

where G = the universal gravitational constant, M = the mass of the earth, m = the mass of the satellite and r = the radius of the satellite from the center of the earth.

GM = µ = 1.407974x10^16 = the earth's gravitational constant.

The centripetal force required to hold the satellite in orbit derives from F = mV^2/r.

Since the two forces must be equal, mV^2/r = µm/r^2 or V^2 = µ/r.

The circumference of the orbit is C = 2Pir.

A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, requires 23hr-56min-4.09sec. to rotate 360º, not 24 hours. Therefore, the time to complete one orbit is 23.93446944 hours or 86,164 seconds

Squaring both sides, 4Pi^2r^2 = 86164^2

But V^2 = µ/r

Therefore, 4Pi^2r^2/(µ/r) = 86164^2 or r^3 = 86164^2µ/4Pi^2

Thus, r^3 = 86164^2(1.407974x10^16)/4Pi^2 = 2.647808686x10^24

Therefore, r = 138,344,596 feet. = 26,201.6 miles.

Subtracting the earth's radius of 3963 miles, the altitude for a geosynchronous satellite is ~22,238 miles.