What is the Orbit length for a geostationary orbit?

3 The orbit, which Clarke first described as useful for broadcast and relay communications satellites,4 is sometimes called the Clarke Orbit. 5 Similarly, the Clarke Belt is the part of space about 35,786 km (22,236 mi) above sea level, in the plane of the Equator, where near-geostationary orbits may be implemented. The Clarke Orbit is about 265,000 km (165,000 mi) long.

Most commercial communications satellites, broadcast satellites and SBAS satellites operate in geostationary orbits. A geostationary transfer orbit is used to move a satellite from low Earth orbit (LEO) into a geostationary orbit. (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 Syncom-3, launched by a Delta-D rocket in 1964.

A worldwide network of operational geostationary meteorological satellites is used to provide visible and infrared images of Earth's surface and atmosphere. 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. A geostationary orbit can only be achieved at an altitude very close to 35,786 km (22,236 mi), and directly above the Equator.

This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) or a period of 1,436 minutes, which equates to almost exactly one sidereal day or 23.934461223 hours. This ensures that the satellite is locked to the Earth's rotational period and has a stationary footprint on the ground. All geostationary satellites have to be located on this ring.

A combination of lunar gravity, solar gravity, and the flattening of the Earth at its poles causes a precession motion of the orbital plane of any geostationary object, with a period of about 53 years and an initial inclination gradient of about 0.85 degrees per year, achieving a maximum inclination of 15 degrees after 26.5 years. To correct for this orbital perturbation, regular orbital stationkeeping manoeuvres are necessary, amounting to a delta-v of approximately 50 m/s per year. A second effect to be taken into account is the longitude drift, caused by the asymmetry of the Earth – the Equator is slightly elliptical.

There are two stable (at 75.3°E, and at 104.7°W) and two unstable (at 165.3°E, and at 14.7°W) equilibrium points. Any geostationary object placed between the equilibrium points would (without any action) be slowly accelerated towards the stable equilibrium position, causing a periodic longitude variation. The correction of this effect requires orbit control manoeuvres with a maximum delta-v of about 2 m/s per year, depending on the desired longitude.

Solar wind and radiation pressure also exert small forces on satellites which, over time, cause them to slowly drift away from their prescribed orbits. In the absence of servicing missions from the Earth or a renewable propulsion method, the consumption of thruster propellant for station-keeping places a limitation on the lifetime of the satellite. Satellites in geostationary orbits are far enough away from Earth that communication latency becomes significant — about a quarter of a second for a trip from one ground-based transmitter to the satellite and back to another ground-based transmitter; close to half a second for a round-trip communication from one Earth station to another and then back to the first.

This delay presents problems for latency-sensitive applications such as voice communication or online gaming. Geostationary satellites are directly overhead at the Equator, and become lower in the sky the further north or south one travels. As the observer's latitude increases, communication becomes more difficult due to factors such as atmospheric refraction, Earth's thermal emission, line-of-sight obstructions, and signal reflections from the ground or nearby structures.

At latitudes above about 81°, geostationary satellites are below the horizon and cannot be seen at all. Satellites in geostationary orbit must all occupy a single ring above the Equator. The requirement to space these satellites apart to avoid harmful radio-frequency 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 near the same longitude but differing latitudes) and radio frequencies. These disputes are addressed through the International Telecommunication Union's allocation mechanism. 89 In the 1976 Bogotá Declaration, eight countries located on the Earth's equator claimed sovereignty over the geostationary orbits above their territory, but the claims gained no international recognition.

When they run out of thruster fuel, the satellites are at the end of their service life as they are no longer able to keep in their allocated orbital position. The transponders and other onboard systems generally outlive the thruster fuel and, by stopping N-S station keeping, some satellites can continue to be used in inclined orbits (where the orbital track appears to follow a figure-eight loop centred on the Equator),1112 or else be elevated to a "graveyard" disposal orbit. In any circular orbit, the centripetal force required to maintain the orbit (Fc) is provided by the gravitational force on the satellite (Fg).

We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite. A 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. Where?

Is the angular speed, and r is the orbital radius as measured from the Earth's center of mass. Where M is the mass of Earth, 5.9736 × 1024 kg, and G is the gravitational constant, 6.67428 ± 0.00067 × 10?11 m3 kg?1 s?2. The angular speed?

Is found by dividing the angle travelled in one revolution (360° = 2? Rad) by the orbital period (the time it takes to make one full revolution). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86,164.09054 seconds).

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). By the same formula we can find the geostationary-type orbit of an object in relation to Mars (this type of orbit above is referred to as an areostationary orbit if it is above Mars).

The geocentric gravitational constant GM (which is?) for Mars has the value of 42,828 km3s?2, and the known rotational period (T) of Mars is 88,642.66 seconds. Since? = 2?

/T, using the formula above, the value of? Is found to be approx 7.088218×10?5 s?1.