Geostationary orbit

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A geostationary orbit (abbreviated GEO) is a circular orbit directly above the Earth's equator (0º latitude). Any point on the equator plane revolves about the Earth in the same direction and with the same period as the Earth's rotation. It is a special case of the geosynchronous orbit (abbreviated GSO), and the one which is of most interest to operators of artificial satellites (including communication and television satellites). Satellite locations may differ by longitude only (remember in Geostationary orbit latitude is zero).

The idea of a geosynchronous satellite for communication purposes was first published in 1928 by Herman Potocnik. Geosynchronous and geostationary orbits were first popularised by science fiction author Arthur C. Clarke in 1945 as useful orbits for communications satellites. As a result they are sometimes referred to as Clarke orbits. Similarly, the "Clarke Belt" is the part of space approximately 35,786 km above mean sea level in the plane of the equator where near-geostationary orbits may be achieved.

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 approximately 35,786 km (22,240 statute miles) 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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Use in artificial satellites

Geostationary orbits can only be achieved very close to the ring 35,786 km directly above the equator. All other circular non-active geosynchronous orbits will cross the geostationary orbit and possibly collide with satellites there. In practice this means that all geostationary satellites have to exist on this ring, which poses problems for satellites needing to be decommissioned at the end of their service life (for example when they run out of thruster fuel). Such satelites are typically raised to a 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 satellites are used by meteorological satellites to provide visible, as well as infrared images of Earth's surface and atmosphere. These satellite systems include:

Most commercial communications satellites (and all television satellites) operate in geostationary orbits.

A statite, a hypothetical satellite that uses a solar sail to modify its orbit, can theoretically hold itself in a geostationary orbit with different altitude and/or inclination from the "traditional" equatorial geostationary orbit.

Derivation of geostationary altitude

In geostationary orbit, a satellite is neither plunging towards the earth nor flying away from it. Therefore, the inward and outward forces on the satellite must equal each another (by Newton's first law of motion). To calculate the geostationary orbit altitude, one equates the two forces:

Fcentripetal = Fcentrifugal

By Newton's second law of motion, we can replace the forces F with the mass of the object multiplied by the acceleration felt by the object due to that force:

m_{sat} \cdot a_{g} = m_{sat} \cdot a_{c}

We note that the mass of the satellite, msat, appears on both sides -- geostationary orbit is independent of the mass of the satellite! So, calculating the altitude simplifies into calculating the point where the magnitudes of the centrifugal acceleration derived from orbital motion and the centripetal acceleration provided by Earth's gravity are equal.

The centrifugal acceleration's magnitude is:

|a_c| = \omega^2 \cdot r

...where ω is the angular velocity in radians per second, and r is the orbital radius in metres as measured from the Earth's centre of mass.

The magnitude of the gravitational attraction is:

|a_g| = \frac{M_e \cdot G}{r^2}

...where Me is the mass of Earth in kilograms, and G is the gravitational constant.

Equating the two accelerations gives:

r^3 = \frac{M_e \cdot G}{\omega^2}

r = \sqrt[3]{\frac{M_e \cdot G}{\omega^2}}

We can express this in a slightly different form by replacing M_e \cdot G by μ, the geocentric gravitational constant:

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

The angular velocity ω is found by dividing the angle travelled in one revolution (360^\circ = 2 \cdot \pi\ rad) by the orbital period (the time it takes to make one full revolution: one sidereal day, or 86,164 seconds). This gives:

\omega = \frac{2 \cdot \pi}{86164} = 7.29 \cdot 10^{-5}\ \mathrm{rad} \cdot \mathrm{s}^{-1}

The resulting orbital radius is 42,164 km. Subtracting the Earth's equatorial radius, 6,378 km, gives the altitude of 35,786 km.

Orbital velocity (how fast the satellite is flying through space) is calculated by multiplying the angular velocity by the orbital radius:

v = \omega \cdot r = 3.07\ \mathrm{km} \cdot \mathrm{s}^{-1} = 11,052\ \mathrm{km/h}

References

See also

External links

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