# Geostationary orbit

Two geostationary satellites in the same orbit
A 5° × 6° view of a part of the geostationary belt, showing several geostationary satellites. Those with inclination 0° form a diagonal belt across the image; a few objects with small inclinations to the Equator are visible above this line. The satellites are pinpoint, while stars have created small trails due to Earth's rotation.

A geostationary orbit, often referred to as a geosynchronous equatorial orbit[1] (GEO), is a circular geosynchronous orbit 35,786 km (22,236 mi) above Earth's equator and following the direction of Earth's rotation. An object in such an orbit appears motionless, at a fixed position in the sky, to ground observers. Communications satellites and weather satellites are often placed in geostationary orbits, so that the satellite antennae (located on Earth) that communicate with them do not have to rotate to track them, but can be pointed permanently at the position in the sky where the satellites are located. Using this characteristic, ocean-color monitoring satellites with visible and near-infrared light sensors (e.g. GOCI) can also be operated in geostationary orbit in order to monitor sensitive changes of ocean environments.

A geostationary orbit is a particular type of geosynchronous orbit, which has an orbital period equal to Earth's rotational period, or one sidereal day (23 hours, 56 minutes, 4 seconds). Thus, the distinction is that, while an object in geosynchronous orbit returns to the same point in the sky at the same time each day, an object in geostationary orbit never leaves that position. Geosynchronous orbits move around relative to a point on Earth's surface because, while geostationary orbits have an inclination of 0° with respect to the Equator, geosynchronous orbits have varying inclinations and eccentricities.

### History

The first appearance of a geostationary orbit in popular literature was in October, 1942, in the first Venus Equilateral story by George O. Smith,[2] but Smith did not go into details. British science fiction author Arthur C. Clarke popularised and expanded the concept 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.[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 collection of artificial satellites in this orbit is known as the Clarke Belt.[6]

The first geostationary satellite was designed by Harold Rosen while he was working at Hughes Aircraft in 1959. Inspired by Sputnik, he wanted to use a geostationary satellite to globalise communications. At the time, telecommunications between the US and Europe was possible between just 136 people at a time, and reliant on HF radios and an undersea cable.[7]

Conventional wisdom at the time was that it would require too much rocket power to place a satellite in a geostationary orbit and it would not survive long enough to justify the expense, so early communication satellites were placed in a low earth orbit.[8] The first of these was the passive Echo balloon satellites in 1960, followed by Telstar 1 in 1962. Although these projects had difficulties with signal strength and tracking, the geostationary concept was seen as impractical, so Hughes often withheld funds and support.[9][7]

By 1961 Rosen and a team of 30 engineers had produced a cylindrical prototype satellite with a diameter of 76 centimetres (30 in), height of 38 centimetres (15 in), weighing 25 kilograms (55 lb), light and small enough to be placed into orbit. It was spin stabalised and produced a flattened waveform.[7][10]

They lost Syncom 1 to electronics failure, but Syncom 2 was successfully placed into a geosynchronous orbit in 1963. Although its inclined orbit still required moving antennas it was able to relay TV transmissions, and allowed for US President Kennedy to phone Nigerian PM Balewa from a ship.[9][11]

The first satellite placed in a geostationary orbit was Syncom 3, which was launched by a Delta D rocket in 1963. With its increased bandwith this satellite was able to transmit live coverage of the Summer Olympics from Japan to America. Geostationary orbits have been in common use ever since, in particular for satellite television.[9]

Today there are hundreds of geostationary satellites providing remote sensing and communications.[7][12]

Although many populated land locations on the planet now have terrestrial communications facilities (microwave, fiber-optic), even undersea, with more than sufficient capacity, telephone and Internet access is still available only via satellite in many places in Africa, Latin America, and Asia, as well as isolated locations that have no terrestrial facilities, such as Canada's Arctic islands, Antarctica, the far reaches of Alaska and Greenland, and ships at sea.

### Uses

Most commercial communications satellites, broadcast satellites and SBAS satellites operate in geostationary orbits.

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:

#### 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

${\displaystyle \Delta t={\frac {2}{c}}{\sqrt {R^{2}+r^{2}-2Rr\cos \varphi }}\approx 253~{\text{ms}}.}$

(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.[13]

Geostationary satellites are directly overhead at the equator and appear lower in the sky to an observer nearer the poles. 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.[14] Because of this, some Russian communication satellites have used elliptical Molniya and Tundra orbits, which have excellent visibility at high latitudes.

#### Statite proposal

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.[15]

### Implementation

#### Launch

A geostationary transfer orbit (GTO) is used to move a satellite from low Earth orbit (LEO) into a geostationary orbit. Most launch vehicles used for geosynchronous satellites place satellites directly into GTOs, with on board satellite propulsion used to reach GEO.[16]

#### 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, and 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.[17][18] 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.[19]

### Orbital stability

A geostationary orbit can be achieved only 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) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours). This ensures that the satellite will match 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 an orbital period of about 53 years and an initial inclination gradient of about 0.85° per year, achieving a maximal inclination of 15° after 26.5 years.[20] To correct for this orbital perturbation, regular orbital stationkeeping maneuvers are necessary, amounting to a delta-v of approximately 50 m/s per year.

A second effect to be taken into account is the longitudinal drift, caused by the asymmetry of the Earth – the equator is slightly elliptical. There are two stable (at 75.3°E and 252°E) and two unstable (at 165.3°E and 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.[20] The correction of this effect requires station-keeping maneuvers with a maximal 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; over time, these 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. Hall-effect thrusters, which are currently in use, have the potential to prolong the service life of a satellite by providing high-efficiency electric propulsion.

#### 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 stay 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),[21][22] or else be elevated to a "graveyard" disposal orbit.

### Derivation of geostationary altitude

Comparison of geostationary Earth orbit with GPS, GLONASS, Galileo and Compass (medium Earth orbit) satellite navigation system orbits with the International Space Station, Hubble Space Telescope and Iridium constellation orbits, and the nominal size of the Earth. [a] The Moon's orbit is around 9 times larger (in radius and length) than geostationary orbit. [b]

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:

${\displaystyle \mathbf {F} _{\text{c}}=\mathbf {F} _{\text{g}}.}$

By Newton's second law of motion,[23] 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:

${\displaystyle m\mathbf {a} _{\text{c}}=m\mathbf {g} .}$

We note that the mass of the satellite m appears on both sides — geostationary orbit is independent of the mass of the satellite.[c] Calculating the geostationary altitude, therefore, simplifies down to calculating the altitude 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:

${\displaystyle |\mathbf {a} _{\text{c}}|=\omega ^{2}r,}$

where ω is the angular speed, and r is the orbital geocentric radius (measured from the Earth's center of mass).

The magnitude of the gravitational acceleration is:

${\displaystyle |\mathbf {g} |={\frac {GM}{r^{2}}},}$

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:

${\displaystyle r^{3}={\frac {GM}{\omega ^{2}}}\to r={\sqrt[{3}]{\frac {GM}{\omega ^{2}}}}.}$

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. Hence

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

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). In the case of a geostationary orbit, the orbital period is one sidereal day, or 86164.09054 s).[24] This gives

${\displaystyle \omega \approx {\frac {2\pi ~{\text{rad}}}{86\,164~{\text{s}}}}\approx 7.2921\times 10^{-5}~{\text{rad/s}}.}$

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 is calculated by multiplying the angular speed by the orbital radius:

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

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 (the orbital radius); subtracting the equatorial radius of Mars (3396.2 km) gives the orbital altitude of 17,031 km.

Orbital speed of a Mars geostationary orbit can be calculated as for Earth above:

${\displaystyle v=\omega r\approx 1.4479~{\text{km/s}}\approx 5\,212~{\text{km/h}}\approx 3238~{\text{mph}}.}$