1. Aircraft Communication Type 1Satellites and radars AUL-Jadra

2. Frequency band designation

3. Space and Ground Segments Space Segment: The elements of the space segment of a communications satellite system includes the satellite (or satellites) in orbit in the system, and the ground station that provides the operational control of the satellite(s) in orbit. The ground station is variously referred to as the Tracking, Telemetry, Command (TT&C) or the Tracking, Telemetry, Command and Monitoring (TTC&M) station, which provides essential spacecraft management and control functions to keep the satellite operating safely in orbit

4. Ground Segment The ground segment of the communications satellite system consists of the earth surface area based terminals that utilize the communications capabilities of the Space Segment. TTC&M ground stations are not included in the ground segment. The ground segment terminals consist of three basic types: Fixed (in-place) terminals, terminals mounted on residence buildings used to receive broadcast satellite signals Transportable terminals; Mobile terminals.

5. Transportable terminals Two satellite trucks, large and small, the larger with a Ku and C-band antenna near Soldier Field in Chicago, Illinois, USA photo by Michael MeinhardtSoldier Field ChicagoIllinoisUSA

6. Kepler’s laws Johannes Kepler (December 27, 1571 – November 15, 1630) was a German mathematician, astronomer and astrologer. He is best known for his eponymous laws of planetary motion. Kepler’s laws of planetary motion apply to any two bodies in space that interact through gravitation. The laws of motion are described through three fundamental principles.

7. Kepler’s First Law Kepler’s First Law, as it applies to artificial satellite orbits, can be simply stated as follows: ‘the path followed by a satellite around the earth will be an ellipse, with the center of mass of earth as one of the two foci of the ellipse.’ If no other forces are acting on the satellite, either intentionally by orbit control or unintentionally as in gravity forces from other bodies, the satellite will eventually settle in an elliptical orbit, with the earth as one of the two foci of the ellipse. The ‘size’ of the ellipse will depend on satellite mass and its angular velocity.

8. Orbital Parameters

9. Apogee : the point farthest from earth. Perigee : the point of closest approach to earth. Line of Apsides : the line joining the perigee and apogee through the center of the earth. Ascending Node : the point where the orbit crosses the equatorial plane, going from south to north. Descending Node : the point where the orbit crosses the equatorial plane, going from north to south. Line of Nodes : the line joining the ascending and descending nodes through the center of the earth. Argument of Perigee : the angle from ascending node to perigee, measured in the orbital plane. Right Ascension of the Ascending Node : the angle measured eastward, in the equatorial plane, from the line to the first point of Aries (Y) to the ascending node. Orbital Parameters

10. The eccentricity is a measure of the ‘circularity’ of the orbit. It is determined from e =(ra − rp)/(ra + rp) The higher the eccentricity, the ‘flatter’ the ellipse. A circular orbit is the special case of an ellipse with equal major and minor axes (zero eccentricity). That is: Elliptical Orbit 0<e<1 Circular Orbit e=0

11. Orbital Mechanics Newton’ s laws of motion can be encapsulated into three equations: S = ut + ½ at² ( distance travelled from 0 to t) v=u+ at ( V : velocity and a: acceleration) P= m*a ( force applied to a moving part) Any satellite launched in space is receiving two forces: 1- A centrifugal force due to the kinetic energy of the satellites F in. This force attempts to fling the satellite into a higher orbits. 2- A centripetal force due to the gravitational attraction of the subject planet F out. It attempts to pull the satellite down toward the planet

12. Orbital Mechanics F out = m*v²/r ( a = v²/r) where : m= masse of the satellite v = velocity of the satellite in the orbit r = distance from the center of earth. ( circular orbit) F in = G*M e *m / r² ( a = μ / r², μ is called Kepler’s constant) where : Me = masse of the satellite G= Gravitational constant r = distance from the center of earth. If F in and F out are balanced we get : v²= μ/r and T=2πr/v T=(2πr 3/2 )/(μ 1/2 )

13. exercice Satellite systemOrbital height (Km)Orbital velocityOrbital period Intelsat ( Geo)35 786, 03 New- ICO ( MEO)10 255 Irridium (LEO)780 Complete the following table :

14. Kepler’s Second Law ‘for equal time intervals, the satellite sweeps out equal areas in the orbital plane.’ Figure below demonstrates this concept.’ For example, the area swept out by the satellite in a one hour period around the point farthest from the earth (the orbit’s apogee), labeled A2 on the figure, will be equal to A1, i.e.: A1 =A2. This result also shows that the satellite orbital velocity is not constant; the satellite is moving much faster at locations near the earth, and slows down as it approaches apogee.

15. Kepler’s Third Law ‘the square of the periodic time of orbit is proportional to the cube of the mean distance between the two bodies’ : Which gives This demonstrates an important result : Orbit Radius = [Constant] × (Orbit Period) 2/3 Under this condition, a specific orbit period is determined only by proper selection of the orbit radius. This allows the satellite designer to select orbit periods that best meet particular application requirements by locating the satellite at the proper orbit altitude.

16. exercice Knowing the nominal altitude, calculate the period and the velocity of those satellites, Kepler’s Constant =3.986004×10 5 km 3 /s 2.:

17. Orbits in Common Use: geosynchronous orbit or geostationary orbit. If the orbit radius is chosen so that the period of revolution of the satellite is exactly set to the period of the earth’s rotation, one mean sidereal day, a unique satellite orbit is defined. In addition, if the orbit is circular (eccentricity=0), and the orbit is in the equatorial plane (inclination angle=0◦), the satellite will appear to hover motionless above the earth at the subsatellite point above the equator. This important special orbit is the geostationary earth orbit (GEO). From Kepler’s third law, the orbit radius for the GEO, r S, is found as: where T=1 mean sidereal day=86 164.09 s. The geostationary height (altitude above the earth’s surface), hS, is then h S = r S − r E = 42 164 − 6378 = 35 786 km

18. Geosynchronous orbit A typical GEO orbit in use today would have an inclination angle slightly greater than 0 and possibly an eccentricity that also exceeds 0. The ‘real world’ GEO orbit that results is often referred to as a geosynchronous earth orbit (GSO) to differentiate it from the ideal geostationary orbit.

19. GEO -HEO Geostationary orbit is considered as a High Earth Circular Equatorial Orbit HEO

20. LEO, MEO & HEO

21. Advantages of GEO satellites Geostationary satellites appear to be fixed over one spot above the equator. Receiving and transmitting antennas on the earth do not need to track such a satellite. These antennas can be fixed in place and are much less expensive than tracking antennas. These satellites have revolutionized global communications, television broadcasting and weather forecasting, and have a number of important defense and intelligence applications.

22. Disadvantages of GEO satellites One disadvantage of geostationary satellites is a result of their high altitude: – radio signals take approximately 0.25 s to reach and return from the satellite, resulting significant signal delay. This delay increases the difficulty of telephone conversation and reduces the performance of common network protocols such as TCP/IP. – Another disadvantage of geostationary satellites is the incomplete geographical coverage, since ground stations at higher than roughly 60 degrees latitude have difficulty reliably receiving signals at low elevations. – Finally those satellites involve more power to reach its high altitudes.

23. Locating a satellite in the orbit

24. Orbital elements An almost endless number of combinations of orbital parameters are available for satellite orbits. Orbital elements defines the set of parameters needed to uniquely specify the location of an orbiting satellite. The minimum number of parameters required is six: 1. Eccentricity; 2. Semi-Major Axis; 3. Time of Perigee; 4. Right Ascension of Ascending Node; 5. Inclination Angle; 6. Argument of Perigee.

25. Describing the orbit of a satellite a(1+e) a(1-e) ae a Y0 b r0r0 X0 c o Apogee Perigee Satellite φ0φ0 The path of the satellite in the orbital plane Circumscribed circle E a

26. Describing the orbit of a satellite r 0 = p/( 1+ e.cos(φ 0 – θ 0 ) θ 0 is a constant and e the eccentricity p is the semi-latus rectum = b²/a (ellipse) or p = r ( circle). a= p/(1 – e²) b= a/(1 – e²) To make θ 0 = 0, we choose the x 0 axis so that both the apogee and perigee lie along it and x 0 axis is therefore the major axis of the ellipse.

27. Locating the satellite in the orbit The equation of the orbit is rewritten as below: r 0 = a.(1-e²)/( 1+ e.cos(φ 0 – θ 0 ) (I) The rectangular coordinates of the satellites are given by: x 0 = r 0.cosφ 0 Y 0 = r 0.sinφ 0 (II)

28. Locating the satellite in the orbit The average orbital velocity is given by: η =(2π)/T = (μ 1/2 )/(a 3/2 )(III) r 0 =a(1 – cos E) (IV) where E is the eccentric anomalie of the satellite: (θ= φ 0 ) M= η(t-t p ) =E – e.sinE (V) M: mean anomaly (in radians),it is the arc length that the satellite would traversed since the perigee passage if it were moving on the circumscribed circle at the mean angular velocity η.

29. Locating the satellite in the orbit If we know the time of perigee t p, the eccentricity, e, the length of the semimajor axis,a, we can find the position of the object in an elliptic Kepler orbit at a given time t by following these steps: 1.Calculate η using (III) 2.Calculate M using (V) 3.Then calculating E using (V) 4.Then finding r 0 from E using (IV) 5.Finding φ 0 using (I) 6.Then calculating x 0 and Y 0 using (II)

30. taylor expansion Trigonométrie : sin x = x - x 3 /3! + x 5 /5! - x 7 /7! + … cos x = 1 - x 2 /2! + x 4 /4! - x 6 /6! +...