F(gravity) = GMm/r^2,

where G is the gravitational constant, M is the Earth's mass and m is the mass of the satellite, with r being the distance between the centres of the two bodies. We can further express for a simple circular orbit, the centrifugal force (which acts in opposition to the gravitational force):

F(centrifugal) = mv^2/r,

where v is the angular velocity of the satellite. For a stable stationary orbit to exist, the two forces must be equal and opposite, and so we can write that F(gravity = F(centrifugal), and hence:

GMm/r^2 = mv^2/r. By cancelling the terms, m, and rearranging, we get:

GM = v^2 r.

Assuming a circular orbit, the mean angular velocity, v is the circumference of the orbit divided by the time (t) taken for the satellite to make that orbit, i.e. v = 2 pi r/t, and so if we substitute for v, we find:

t^2 = 4 pi^2 r^3/GM.

A special case is the geostationary orbit, with a unique property which is very useful for communications and weather satellites. This is a geosynchronous orbit directly above the Earth's equator (latitude 0°), with a period equal to the Earth's rotational period and an orbital eccentricity of approximately zero. Due to the constant 0° latitude and circular nature of geostationary orbits, satellites in them differ in location only by longitude. In essence, from the point of view of an observer on the Earth's surface the orbiting satellite stands still in the sky, because it moves through its orbital cycle at the same rate as the equatorial surface point below it moves round with the Earth's rotation. Clearly the satellite must sweep through a greater distance than the equatorial surface point below it does in the same time interval and hence it moves at a greater speed, as we shall see.

To compute the size of the orbital radius (r), taken from the centre of mass (i.e. the centre of the Earth), we can rearrange the above to solve for r:

r = (t^2GM/4 x pi^2)^1/3 =

[(24 hr x 3600 s/hr) x 6.6726 x 10^-11 m^3 kg^-1 s^-1 x 6.0 x 10^24 kg/ 4 x pi^2]^1/3

= (7.57 x 10^22)^1/3 = 4.23 x 10^7 m = 42,300 km.

If we subtract the mean earth radius of 6.4 x 10^6 m, we obtain an altitude of 3.59 x 10^7 m (35,900 km).

To obtain an orbital speed, we note that the circumference of the orbit is 2 x pi x r =

2 x pi x 4.23 x 10^7 m = 2.66 x 10^8 m.

The speed is thus: 2.66 x 10^8 m/(24 hr x 3600 s/hr) = 3,079 m/s = 3.08 km/s;

x 3600 s/hr = 11,088 km/hr = 6,930 miles per hour.

[For comparison, an equatorial point at the earth's surface rotates at (2 x pi 6 x 10^6)/(24 x 3600) = 465 m/s = 0.465 km/s; x 3600 s/hr = 1,676 km/hr = 1,047 mph].

Most satellites are launched at much lower orbits, e.g. 500 km in altitude, for use in navigation, telecommunications and other purposes, e.g., the Hubble Space Telescope has an orbital altitude of 559 km.

In this case, t^2 = 4 x pi^2 x [(6.4 + 0.5) x 10^6]^3/(6.6726 x 10^-11 x 6 x 10^24) = 3.24 x 10^7 s.

Therefore the orbital period, t = (3.24 x 10^7 s)^1/2 = 5692 s = 94.9 minutes.

Its orbital speed is 2 x pi x (6.9 x 10^6)/5692 = 7.62 km/s = 27,420 km/hr = 17,137 miles/hr.

The International Space Station has an orbital altitude of 350 km, and so its speed is nearly the same (27,725 km/hr; 17,328 mph), according to an orbital period of 91.8 mins.

Escape Speed for the Earth.

This is usually incorrectly called the "escape velocity" but is just a speed i.e. distance/time since there is no direction specified.

To get this quantity, which is the kinetic energy (1/2) mv^2, required to cancel the gravitational "pull" of the earth, we can write:

(1/2) mv^2 = GMm/r

where r is the earth's radius, and M its mass, and by cancelling the terms m from both sides (which tells us that the mass of the satellite is unimportant and only that of the earth matters), we get:

v = (2GM/r)^1/2

= (2 x 6.6726 x 10^-11 x 6 x 10^24/6.4 x 10^6)^1/2 = 11,185 m/s (11.19 km/s)

= 40,267 km/hr = 25,167 mph. Thus this is about half as fast again as the speed required to maintain a 500 km orbit above the Earth. Satellites will never therefore simply fly-off into space and with the virtual absence of air-resistance above ca. 100 km, there is no mechanism for efficient energy-loss so they cannot simply tumble back to earth either.

However, since there is no atmosphere, satellites are not shielded from radiation from space which tends to concentrate in the van Allen belts around the earth and causes damage to the materials they are made from, and solar cells, integrated circuits and sensors can be damaged by radiation. The inner Van Allen Belt extends from an altitude of 700–10,000 km (0.1 to 1.5 Earth radii) above the Earth's surface, and contains high concentrations of energetic protons with energies exceeding 100 MeV and electrons in the range of hundreds of kiloelectronvolts, trapped by the strong (relative to the outer belts) magnetic fields in the region. The large outer radiation belt extends from an altitude of about three to ten Earth radii (

*R*) above the Earth's surface, and its greatest intensity is usually at an altitude of around 3–4

_{E}*R*.

_{E}It is generally understood that the inner and outer Van Allen belts result from different processes. The inner belt, consisting mainly of energetic protons, is the product of the decay of albedo neutrons which are themselves the result of cosmic ray collisions in the upper atmosphere. The outer belt consists mainly of electrons. I am involved in a project with the Yerevan Physics Institute in Armenia, to simulate the effects of energetic electrons on satellite components in space and under other extreme conditions. At an altitude of 5.6 (

*R*) a satellite in the geostationary orbit, though away from the region of maximum intensity, will nonetheless be subject to significant radiation in the outer Van Allen belt.

_{E}
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