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# Geostationary

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 Title: Geostationary Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Geostationary

A geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit[1] (GEO), is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation.[2] An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day), and thus appears motionless, at a fixed position in the sky, to ground observers. Communications satellites and weather satellites are often given geostationary orbits, so that the satellite antennas that communicate with them do not have to move to track them, but can be pointed permanently at the position in the sky where they stay. Using this characteristic, ocean color satellites with visible sensors (e.g. the Geostationary Ocean Color Imager (GOCI)) can be operated in geostationary orbit in order to monitor sensitive changes of ocean environments. A geostationary orbit is a particular type of geosynchronous orbit.

The notion of a geosynchronous satellite for communication purposes was first published in 1928 (but not widely so) by Herman Potočnik.[3] The first appearance of a geostationary orbit in popular literature was in the first Venus Equilateral story by George O. Smith,[4] but Smith did not go into details. British Science Fiction author Arthur C. Clarke disseminated the idea widely, with more details on how it would work, in a 1945 paper entitled "Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?", published in Wireless World magazine. Clarke acknowledged the connection in his introduction to The Complete Venus Equilateral.[5] The orbit, which Clarke first described as useful for broadcast and relay communications satellites,[6] is sometimes called the Clarke Orbit.[7] 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.

## Practical uses

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. These satellite systems include:

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.

### Orbital stability

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.

### Communications

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.

For example, for ground stations at latitudes of φ = ±45° on the same meridian as the satellite, the time taken for a signal to pass from Earth to the satellite and back again can be computed using the cosine rule, given the geostationary orbital radius r (derived below), the Earth's radius R and the speed of light c, as

$\frac\left\{1\right\}\left\{c\right\} \sqrt\left\{R^2 + r^2 - 2 R r \cos\varphi\right\} \approx253\,\mathrm\left\{ms\right\}$

(Note that r is the orbital radius, the distance from the centre of the Earth, not the height above the Equator.)

This delay presents problems for latency-sensitive applications such as voice communication.[8]

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.[9]

### Orbit allocation

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.[10][11] 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.[12]

### Limitations to usable life of geostationary satellites

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),[13][14] or else be elevated to a "graveyard" disposal orbit.

## Derivation of geostationary altitude

In any circular orbit, the centripetal force required to maintain the orbit (Fc) is provided by the gravitational force on the satellite (Fg). To calculate the geostationary orbit altitude, one begins with this equivalence:

$\mathbf\left\{F\right\}_\text\left\{c\right\} = \mathbf\left\{F\right\}_\text\left\{g\right\}$

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:

$m \mathbf\left\{a\right\}_\text\left\{c\right\} = m \mathbf\left\{g\right\}$

We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite.[lower-alpha 3] 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:

$|\mathbf\left\{a\right\}_\text\left\{c\right\}| = \omega^2 r$

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:

$|\mathbf\left\{g\right\}| = \frac\left\{G M\right\}\left\{r^2\right\}$

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.

Equating the two accelerations gives:

$r^3 = \frac\left\{G M\right\}\left\{\omega^2\right\} \to r = \sqrt\left[3\right]\left\{\frac\left\{G M\right\}\left\{\omega^2\right\}\right\}$

The product GM is known with much greater precision than either factor alone; it is known as the geocentric gravitational constant μ = 398,600.4418 ± 0.0008 km3 s−2:

$r = \sqrt\left[3\right]\left\{\frac\mu\left\{\omega^2\right\}\right\}$

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). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86,164.09054 seconds).[15] This gives:

$\omega \approx \frac\left\{2 \mathrm\pi~\mathrm\left\{rad\right\}\right\} \left\{86\,164~\mathrm\left\{s\right\}\right\} \approx 7.2921 \times 10^\left\{-5\right\}~\mathrm\left\{rad\right\} / \mathrm\left\{s\right\}$

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:

$v = \omega r \approx 3.0746~\mathrm\left\{km\right\}/\mathrm\left\{s\right\} \approx 11\,068~\mathrm\left\{km\right\}/\mathrm\left\{h\right\} \approx 6877.8~\mathrm\left\{mph\right\}\text\left\{.\right\}$

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 μ) for Mars has the value of 42,828 km3s−2, 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 7.088218×10−5 s−1. Thus, r3 = 8.5243×1012 km3, whose cube root is 20,427 km; subtracting the equatorial radius of Mars (3396.2 km) we have 17,031 km.

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