1st  Top orbits 
A geostationary orbit (or Geostationary Earth Orbit  GEO) is a geosynchronous orbit directly above the Earth's equator (0° latitude), with a period equal to the Earth's rotational period and an orbital eccentricity of approximately zero. These characteristics are required so that, from locations on the surface of the Earth, geostationary objects appear motionless in the sky, making the GEO an orbit of great interest to operators of communications and weather satellites. Due to the constant 0° latitude and circularity of geostationary orbits, satellites in GEO differ in location by longitude only.
The notion of a geosynchronous satellite for communication purposes was first published in 1928 (but not widely so) by Herman Potočnik.^{[1]} The idea of a geostationary orbit was first published on a wide scale in a paper entitled "ExtraTerrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?" by Arthur C. Clarke, published in Wireless World magazine in 1945.^{[2]} In this paper, Clarke described it as a useful orbit for communications satellites. As a result this is sometimes referred to as the Clarke Orbit.^{[3]} Similarly, the Clarke Belt is the part of space approximately 36,000 km (22,000 mi) above sea level, in the plane of the equator, where neargeostationary orbits may be implemented. The Clarke Orbit is about 265,000 km (165,000 mi) long.
Geostationary orbits are useful because they cause a satellite to appear stationary with respect to a fixed point on the rotating Earth. As a result, an antenna can point in a fixed direction and maintain a link with the satellite. The satellite orbits in the direction of the Earth's rotation, at an altitude of 35,786 km (22,236 mi) above ground. This altitude is significant because it produces an orbital period equal to the Earth's period of rotation, known as the sidereal day.
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Geostationary orbits can be achieved only very close to the ring 35,786 km (22,236 mi) high, directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) or a period of 1436 minutes, which equates to almost exactly one sidereal day or 23.934461223 hours. This makes sense considering that the satellite must be locked to the Earth's rotational period in order to have a stationary footprint on the ground. In practice this means that all geostationary satellites have to exist on this ring, which poses problems for satellites that will be decommissioned at the end of their service lives (e.g. when they run out of thruster fuel). Such satellites will either continue to be used in inclined orbits (where the orbital track appears to follow a figureofeight loop centered on the equator), or else be elevated to a "graveyard" disposal orbit.
A geostationary transfer orbit is used to move a satellite from low Earth orbit (LEO) into a geostationary orbit.
A worldwide network of operational geostationary meteorological satellites is used to provide visible and infrared images of Earth's surface and atmosphere. These satellite systems include:
Most commercial communications satellites, broadcast satellites and SBAS satellites operate in geostationary orbits. (Russian television satellites have used elliptical Molniya and Tundra orbits due to the high latitudes of the receiving audience.) The first satellite placed into a geostationary orbit was the Syncom3, launched by a DeltaD rocket in 1964.
A statite, a hypothetical satellite that uses a solar sail to modify its orbit, could theoretically hold itself in a "geostationary" orbit with different altitude and/or inclination from the "traditional" equatorial geostationary orbit.
In any circular orbit, the centripetal acceleration required to maintain the orbit is provided by the gravitational force on the satellite. To calculate the geostationary orbit altitude, one begins with this equivalence, and uses the fact that the orbital period is one sidereal day.
By Newton's second law of motion, we can replace the forces F with the mass m of the object multiplied by the acceleration felt by the object due to that force:
We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite.^{[4]} So calculating the altitude simplifies into calculating the point where the magnitudes of the centripetal acceleration required for orbital motion and the gravitational acceleration provided by Earth's gravity are equal.
The centripetal acceleration's magnitude is:
where ω is the angular speed, and r is the orbital radius as measured from the Earth's center of mass.
The magnitude of the gravitational acceleration is:
where M is the mass of Earth, 5.9736 × 10^{24} kg, and G is the gravitational constant, 6.67428 ± 0.00067 × 10^{−11} m^{3} kg^{−1} s^{−2}.
Equating the two accelerations gives:
The product GM is known with much greater accuracy than either factor; it is known as the geocentric gravitational constant μ = 398,600.4418 ± 0.0008 km^{3} s^{−2}:
The angular speed ω is found by dividing the angle travelled in one revolution (360° = 2π rad) by the orbital period (the time it takes to make one full revolution: one sidereal day, or 86,164.09054 seconds).^{[5]} This gives:
The resulting orbital radius is 42,164 kilometres (26,199 mi). Subtracting the Earth's equatorial radius, 6,378 kilometres (3,963 mi), gives the altitude of 35,786 kilometres (22,236 mi).
Orbital speed (how fast the satellite is moving through space) is calculated by multiplying the angular speed by the orbital radius:
To put the above equations for finding the required altitude of geostationary orbit above a given planet into more easily understandable terms, let us take for example
Here, the product of GM is the geocentric gravitational constant, the symbol for which is μ,
Let us use this for finding the geostationary orbit of an object in relation to mars. The geocentric gravitational constant GM (which is μ) for mars has the value of 42,828 km3s2, and the known rotational period (T) of mars is 88,642.66 seconds. Since ω = 2π/T, using the formula above, the value of ω is found to be approx .00007088218. Thus, r^3 = 8,524,300,617,153.08. It is then simply a matter of finding the cube root of r^3, which is 20,427.6255 and subtracting the equatorial radius of mars (3396.2 km) to give us our answer of 17,031.42 km.
Alternately for the purposes of finding r without the known value of μ, GM can be additionally simplified. Instead of writing the gravitational constant as
we can simply say (for our purposes)
Which equals approx. 0.0000000000000000000667428. This number multiplied by the mass in kg of a given planet will equal μ.
While a geostationary orbit should hold a satellite in fixed position above the equator, orbital perturbations, such as by the Moon and from the fact that the Earth is not an exact sphere cause slow but steady drift away from the geostationary location. Satellites correct for these effects with stationkeeping maneuvers. In the absence of servicing missions from the Earth, the consumption of thruster propellant for stationkeeping places a limitation on the lifetime of the satellite.
Satellites in geostationary orbits are far enough away from Earth that communication latency becomes very high — about a quarter of a second for a oneway trip from a ground based transmitter to a geostationary satellite and then on to a ground based receiver, and close to half a second for roundtrip communication between two earth stations.
For example, for ground stations at latitudes of φ=±45° on the same meridian as the satellite, the oneway delay can be computed by using the cosine rule, given the above derived geostationary orbital radius r, the Earth's radius R and the speed of light c, as
This presents problems for latencysensitive applications such as voice communication or online gaming.^{[6]}
Satellites in geostationary orbit must all occupy a single ring above the equator. The requirement to space these satellites apart to avoid harmful radiofrequency interference during operations means that there are a limited number of orbital "slots" available, thus only a limited number of satellites can be operated in geostationary orbit. This has led to conflict between different countries wishing access to the same orbital slots (countries at the same longitude but differing latitudes) and radio frequencies. These disputes are addressed through the International Telecommunication Union's allocation mechanism.^{[7]} Countries located at the Earth's equator have also asserted their legal claim to control the use of space above their territory.^{[8]} Since the Clarke Orbit is about 265,000 km (165,000 mi) long, countries and territories in lesspopulated parts of the world have been allocated slots and reserved portions of the radiofrequency spectrum already, even though they aren't yet using them. The problem presently lies over denselypopulated areas such as the Americas and Europe/Africa, and above the middles of the three equatorial oceans.^{[citation needed]}
This article incorporates public domain material from the General Services Administration document "Federal Standard 1037C" (in support of MILSTD188).

