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Course by: Sunil Kumar Singh. E-mail the author

# Artificial satellites

Module by: Sunil Kumar Singh. E-mail the author

Summary: Artificial satellites are the backbone of modern communication systems.

The motion of a satellite or space-station is a direct consequence of Earth’s gravity. Once launched in the appropriate orbit, these man-made crafts orbit around Earth without any propulsion. In this module, we shall study basics of satellite motion without going into details of the technology. Also, we shall develop analysis framework of artificial satellite, which can as well be extended to analysis of natural satellite like our moon. For the analysis here, we shall choose a simple framework of “two – body” system, one of which is Earth.

We should be aware that gravity is not the only force of gravitation working on the satellite, particularly if satellite is far off from Earth’s surface. But, Earth being the closest massive body, its gravitational attraction is dominant to the extent of excluding effect of other bodies. For this reason, our analysis of satellite motion as “isolated two body system” is good first approximation.

Mass of artificial satellite is negligible in comparison to that of Earth. The “center of mass” of the “two body system” is about same as the center of Earth. There is possibility of different orbits, which are essentially elliptical with different eccentricity. A satellite close to the surface up to 2000 km describes nearly a circular trajectory. In this module, we shall confine ourselves to the analysis of satellites having circular trajectory only.

## Speed of the satellite

Satellites have specific orbital speed to move around Earth, depending on its distance from the center of Earth. The satellite is launched from the surface with the help of a rocket, which parks it in particular orbit with a tangential speed appropriate for that orbit. Since satellite is orbiting along a circular path, there is requirement for the provision of centripetal force, which is always directed towards the center of orbit. This requirement of centripetal force is met by the force of gravity. Hence,

G M m r 2 = m v 2 r G M m r 2 = m v 2 r

v = G M r v = G M r

where “M” is Earth’s mass and “r” is linear distance of satellite from the "center of mass" of Earth.

The important thing to realize here are : (i) orbital speed of the satellite is independent of the mass of the satellite (ii) a satellite at a greater distance moves with lesser velocity. As the product “GM” appearing in the numerator of the expression is constant, we can see that

v 1 r v 1 r

This conclusion is intuitive in the sense that force of gravitation is lesser as we move away from Earth’s surface and the corresponding centripetal force as provided by gravity is smaller. As such, orbital speed is lesser.

This fact has compounding effect on the time period of the satellite. In the first place, a satellite at a greater distance has to travel a longer distance in one revolution than the satellite closer to Earth’s surface. At the same time, orbital speed is lesser as we move away. It is, then, imperative that time period of revolution increases for satellite at greater distance.

We can write the equation of orbital speed in terms of acceleration due to gravity at the surface (g = g 0 g 0 ), which is given by :

g = G M R 2 g = G M R 2

G M = g R 2 G M = g R 2

Substituting in the equation of orbital velocity, we have :

v = G M r = g R 2 r = g R 2 R + h v = G M r = g R 2 r = g R 2 R + h

where “h” is the vertical height of the satellite above the surface. Rearranging,

v = R g R + h v = R g R + h

## Time period of revolution

Time period of revolution is equal to time taken to travel the perimeter of circular path. The time period of rotation is :

T = 2 π r v T = 2 π r v

Substituting expression of “v” as obtained earlier,

T = 2 π r 3 2 G M T = 2 π r 3 2 G M

This is the expression of time period for a satellite revolving in a circular orbit. Like orbital speed, the time period is also independent of the mass of the satellite. Now, squaring both sides, we have :

T 2 = 2 π r 3 G M T 2 = 2 π r 3 G M

Clearly, square of time period of a satellite is proportional the cube of the linear distance for the circular orbit,

T 2 r 3 T 2 r 3

## Example

Problem 1: Two satellites revolve around Earth along a coplanar circular orbit in the plane of equator. They move in the same sense of direction and their periods are 6 hrs and 24 hrs respectively. The satellite having period of 6 hrs is at a distance 10000 km from the center of Earth. When the satellites are at the minimum possible separation between each other, find the magnitude of relative velocity between two satellites.

Solution : Let us denote two satellites with subscripts “1” and “2”. Let the satellite designated with “1” is closer to the Earth. The positions of satellites, corresponding to minimum possible separation, are shown in the figure.

The distance between center of Earth and the satellite “1” is 10000 km, but this data is not available for the other satellite. However, we can evaluate other distance, using the fact that square of time period of a satellite is proportional to the cube of the linear distance for the circular orbit.

r 2 3 r 1 3 = T 2 2 T 1 2 = 24 6 2 = 16 r 2 3 r 1 3 = T 2 2 T 1 2 = 24 6 2 = 16

r 2 r 1 = 2 r 2 r 1 = 2

r 2 = 2 r 1 = 2 x 10 4 = 2 X 10 4 k m r 2 = 2 r 1 = 2 x 10 4 = 2 X 10 4 k m

We can now determine velocity of each satellite as :

v = 2 π r T v = 2 π r T

For the first satellite,

v 1 = 2 π X 10000 6 = 10000 π 3 v 1 = 2 π X 10000 6 = 10000 π 3

For the second satellite,

v 2 = 2 π X 20000 24 = 10000 π 6 v 2 = 2 π X 20000 24 = 10000 π 6

Hence, magnitude of relative velocity is :

v 1 v 2 = π 6 X 10000 = 5238 k m / h r v 1 v 2 = π 6 X 10000 = 5238 k m / h r

## Energy of “Earth-satellite” system

For consideration of energy, Earth can be treated as particle of mass “M”. Thus, potential energy of “Earth – satellite” as two particles system is given by :

U = - G M m R U = - G M m R

Since expression of orbital speed of the satellite is known, we can also determine kinetic energy of the satellite as :

K = 1 2 m v 2 K = 1 2 m v 2

Putting expression of speed, “v”, as determined before,

K = G M m 2 r K = G M m 2 r

Note that kinetic energy of the satellite is positive, which is consistent with the fact that kinetic energy can not be negative. Now, mechanical energy is algebraic sum of potential and kinetic energy. Hence, mechanical energy of “Earth – satellite” system is :

E = K + U = G M m 2 r G M m r E = K + U = G M m 2 r G M m r

E = G M m 2 r E = G M m 2 r

Here, total mechanical energy of the system is negative. We shall subsequently see that this is characteristic of a system, in which bodies are bounded together by internal force.

### Relation among energy types

The expression of mechanical energy of the “Earth – satellite” system is typical of two body system in which one body revolves around other along a circular path. Particularly note the expression of each of the energy in the equation,

E = K + U E = K + U

G M m 2 r = G M m 2 r G M m r G M m 2 r = G M m 2 r G M m r

Comparing above two equations, we see that magnitude of total mechanical energy is equal to kinetic energy, but different in sign. Hence,

E = - K E = - K

Also, we note that total mechanical energy is half of potential energy. Hence,

E = U 2 E = U 2

These relations are very significant. We shall find resemblance of forms of energies in the case of Bohr’s orbit as well. In that case, nucleus of hydrogen atom and electron form the two – body system and are held together by the electrostatic force.

Importantly, it provides an unique method to determine other energies, if we know any of them. For example, if the system has mechanical energy of - 200 X 10 6 J - 200 X 10 6 J , then :

K = - E = 200 X 10 6 = 200 X 10 6 J K = - E = 200 X 10 6 = 200 X 10 6 J

and

U = 2 E = - 400 X 10 6 J U = 2 E = - 400 X 10 6 J

### Energy plots of a satellite

An inspection of the expression of energy forms reveals that that linear distance “r” is the only parameter that can be changed for a satellite of given mass, “m”. From these expressions, it is also easy to realize that they have similar structure apart from having different signs. The product “GMm” is divided by “r” or “2r”. This indicates that nature of variation in their values with linear distance “r” should be similar.

Since kinetic energy is a positive quantity, a plot of kinetic energy .vs. linear distance, “r”, is a hyperbola in the first quadrant. The expression of mechanical energy is exactly same except for the negative sign. Its plot with linear distance, therefore, is an inverted replica of kinetic energy plot in fourth quadrant. Potential energy is also negative like mechanical energy. Its plot also falls in the fourth quadrant. However, magnitude of potential energy is greater than that of mechanical energy as such the plot is displaced further away from the origin as shown in the figure.

From plots, we can conclude one important aspect of zero potential reference at infinity. From the figure, it is clear that as the distance increases and becomes large, not only potential energy, but kinetic energy also tends to become zero. We can, therefore, conclude that an object at infinity possess zero potential and kinetic energy. In other words, mechanical energy of an object at infinity is considered zero.

## Gravitational binding energy

A system is bounded when constituents of the system are held together. The “Earth-satellite” system is a bounded system as members of the system are held together by gravitational attraction. Subsequently, we shall study such other bounded systems, which exist in other contexts as well. Bounded system of nucleons in a nucleus is one such example.

The characterizing aspect of a bounded system is that mechanical energy of the system is negative. However, we need to qualify that it is guaranteed to be negative when zero reference potential energy is at infinity.

Let us check out this requirement for the case of “Earth-satellite” system. The mechanical energy of Earth- satellite system is indeed negative :

E = G Mm 2 r E = G Mm 2 r

where “M” and “m” are the mass of Earth and satellite. Hence, "Earth - satellite" system is a bounded system.

We can infer from the discussion of a bounded system that the "binding energy" is the amount of energy required to disintegrate (dismember) a bounded system. For example, we can consider a pebble lying on Earth’s surface. What is the energy required to take this pebble far off in the interstellar space, where Earth’s gravity ceases to exist? We have seen that infinity serves as a theoretical reference, where gravitational field ceases to exits. Further, if we recall, then potential energy is defined as the amount of work done by external agency to bring a particle slowly from infinity to a position in gravitational field. The work by external force is negative as its acts opposite to the displacement. Clearly, taking pebble to the infinity is reverse action. Work by external force is in the direction of displacement. As such, work done in this case is positive. Therefore, binding energy of the pebble is a positive quantity and is equal to the magnitude of potential energy for the pebble. If its mass is “m”, then binding energy of the "Earth-pebble" system is :

E B = - U = - G Mm r = G Mm r E B = - U = - G Mm r = G Mm r

where “M” and “m” are the mass of Earth and pebble respectively and “R” is the radius of Earth.

This is, however, a specific description of dismembering process. In general, a member of the system will have kinetic energy due to its motion. Let us consider the case of “Earth-satellite” system. The satellite has certain kinetic energy. If we want to take this satellite to infinity, we would first require to bring the satellite to a dead stop and then take the same to infinity. Therefore, binding energy of the system is a positive quantity, which is equal to the magnitude of the mechanical energy of the system.

Definition 1: Binding energy
Binding energy is equal to the modulus of mechanical energy.

Going by the definition, the binding energy of the “Earth-satellite” system is :

E B = - E = - G Mm 2 r = G Mm 2 r E B = - E = - G Mm 2 r = G Mm 2 r

where “r” is the linear distance between the center of Earth and satellite.

## Satellite systems

The satellites are made to specific tasks. One of the most significant applications of artificial satellite is its use in telecast around the world. Earlier it was difficult to relay telecast signals due to spherical shape of Earth. In recent time, advancements in communication have brought about astounding change in the way we live. The backbone of this communication wonder is variety of satellite systems orbiting around Earth.

Satellite systems are classified for different aspects of satellite motion. From the point of physics, it is the orbital classification of satellite systems, which is more interesting. Few of the famous orbits are described here. Almost all orbits generally describe an elliptical orbit. We shall discuss elliptical orbits in the module dedicated to Kepler’s law. For the present, however, we can approximate them to be circular for analysis purpose.

1: Geocentric orbit : It is an orbit around Earth. This is the orbit of artificial satellite, which is launched to revolve around Earth. Geocentric orbit is further classified on the basis of distance from Earth’s surface (i) low Earth orbit up to 2000 km (ii) middle Earth orbit between 2000 and geo-synchronous orbit (36000 km) and (iii) high Earth orbit above geo-synchronous orbit (36000 km).

2: Heliocentric Orbit : It is an orbit around Sun. The orbits of planets and all other celestial bodies in the solar system describe heliocentric orbits.

3: Geosynchronous Orbit : The time period of this orbit is same as the time period of Earth.

4: Geostationary Orbit : The plane of rotation is equatorial plane. The satellite in this orbit has time period equal to that of Earth. Thus, motion of satellite is completely synchronized with the motion of Earth. The sense of rotation of the satellite is same as that of Earth. The satellite, therefore, is always above a given position on the surface. The orbit is at a distance of 36000 km from Earth’s surface and about 42400 (= 36000 + 6400) km from the center of Earth. The orbit is also known as Clarke’s orbit after the name of author, who suggested this orbit.

5: Molniya Orbit – It is an orbit having inclination of 63.4° with respect to equatorial plane and orbital period equal to half that of Earth.

6: Polar orbit : The orbit has an inclination of 90° with respect to the equatorial plane and as such, passes over Earth’s poles.

Another important classification of satellite runs along the uses of satellites. Few important satellite types under this classification are :

1: Communication satellites : They facilitate communication around the world. The geostationary satellite covers ground locations, which are close to equator. Geostationary satellites appears low from a positions away from equator. For locations at different latitudes away from equator, we need to have suitably designed orbits so that the area can be covered round the clock. Molniya orbit is one such orbit, which is designed to provide satellite coverage through a satellite system, consisting of more than one satellite.

2: Astronomical satellites : They are designed for studying celestial bodies.

3: Navigational satellites : They are used to specify location on Earth and develop services based on navigation.

4: Earth observation satellites : They are designed for studying Earth system, environment and disaster management.

5: Weather satellites : They facilitate to monitor weather and related services.

6: Space station : It is an artificial structure in space for human beings to stay and do assigned experiments/works

As a matter of fact, there is quite an elaborate classification system. We have only named few important satellite systems. In particular, there are varieties of satellite systems, including reconnaissance satellites, to meet military requirement.

## Acknowledgment

Author wishes to thank Arunabha guha, Physics dept, Georgian court university, Lakewood, New jersey, USA for pointing out a mistake in the example contained in this module.

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