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Transcript
Satellite Communications Dr. Malgorzata Langer The Institute of Electronics Room 310; [email protected] Why? • Distance independent and fixed broadcast costs – the same costs, regardless of the distance between the transmitting and receiving earth stations and of the number of ground terminals receiving the transmission • High capacity – high carrier frequencies • Low error rates thanks to error correction techniques • Diverse user networks may be served simultaneously, on large areas of the earth When? • The first launch: October 1957; SPUTNIK I; Soviet Union • The first idea: Arthur C. Clarke, 1945, US Why? (How it works…) • Satellite orbit determination is based on the Laws of Motion, published in 1609 (Astronomia Nova) and in 1619 (Harmonices Mundi), developed by Johannes Kepler and later refined by Isaack Newton, in 1665, in Laws of Mechanics and Gravitation Kepler’s Laws • Kepler’s laws on planetary motion apply to any two bodies in space that interact through gravitation; • In origin – they apply to the motions of the planets around the Sun Kepler’s First Law F1 and F2 - localisation of the two foci The path followed by the planet around the Sun will be an ellipse, with the centre of mass of the Sun as one of the two foci of the ellipse Do you remember? • When F1 and F2 are the foci places of the ellipse – in any localization of the planet the sum of the distances of the both foci is constant and equals to the greater diameter (doubled major axis) The eccentricity • It is a measure of the ‘circularity’ of the orbit. The higher the eccentricity, the flatter the ellipse. • The circular orbit means that the major axis equals to the minor one and the eccentricity equals to zero 0<e<1 • Elliptical Orbit • Circular Orbit e=0 • The parabola (open orbit) e = 1; the hyperbola e > 1 e= ra − rp ra + rp where e- the eccentricity of the orbit; ra- the distance from the center of the sun to the aphelion point, and rp – to the perihelion Some Numbers: • In Solar System planet eccentricities are less than 0.1 • The exception is Mercury (~ 0.2) • The solar year (the tropical year) equals to 365 days 5 hours 48 minutes and 46 seconds • The sidereal year (the determined star is the constant point for the earth’s motion): 365 days 6 hours 9 minutes and 9 seconds Can you see the problem? • Auxiliary questions: What time do we use and what we should? What is the correct number for the day period? What is γ2000? Why has the beginning of the spring been changed lately? How did ancient (and medieval) sailors find their way across sees and oceans? Geographic coordinates? • Give the per cent difference between the Solar day and the Sidereal one. • Which coordinate (latitude or longitude, i.e. parallels of latitude, meridians) shifts against geographical places? The Earth in The Solar System • Medium distance from the Sun: 149,597,887 km (1.00000011 AU – astronomical unit) • Orbit’s perimeter: 0.940*109km • Eccentricity : 0.01671022 • Perihelion: 147,098,074km • Aphelion: 152,097,701km • Sidereal year: 365.25696 days • Mean orbital velocity: 29.783 km/s (max: 30.287 km/s ; min. 29.291 km/s) For a satellite • 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. • The size of the ellipse will depend on the satellite mass and its angular velocity Kepler’s Second Law • For equal time intervals, the tracing radius of the planet sweeps out equal areas in the orbital plane ∆s 1 K = const = 2 ∆t m So, in perihelion (near the Sun) the planet moves faster than in aphelion m – the mass of the planet; K – the angular momentum (moment of momentum) of the planet Likewise….. • For equal time intervals the satellite sweeps out equal areas in the orbital plane • The area swept out by a satellite in one hour period in the orbit’s apogee (the point farthest from the earth) equals to the area swept out during one hour in perigee. • THE SATELLITE ORBITAL VELOCITY IS NOT CONSTANT 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 2 1 3 1 2 2 3 2 T T = = const a a In Details… 4π 3 2 T = a µ 2 where T – orbital period in [s]; a – distance between the two bodies in [km]; µ - Kepler’s Constant: 3.986004x105km3/s2 µ 2/3 r = 2 T 4π 1/ 3 If the orbit is circular then a=r, and ORBIT RADIUS = [Constant] x (Orbit Period)2/3 Newton’s Laws • The gravitational force Fin and the angular velocity force, Fout can be presented as: µ Fin = m 2 r What is v for Fout v2 = m r Fin = Fout • m – satellite mass; v – satellite velocity in the plane of orbit; r – distance from the centre of the earth (orbit radius), µ - Kepler’s Constant For The Satellite 3 a T = 2π G⋅m • • • • Constant of gravitation G=6.67 * 10-11 Nm2/kg2 Earth’s mass m=5.973*1024 kg Mean radius R=6.3*106m ‘a’ is R+h (the orbit’s altitude) Results • A specific orbit period is determined only by the selection of the orbit radius Orbit altitudes for specified orbital periods Revolutions/day Nominal period (hours) Nominal altitude (km) 1 24 36000 2 12 20200 3 8 13900 4 6 10400 6 4 6400 8 3 4200 Orbital Parameters • Apogee – the point farthest from the earth • Perigee – the point of the closest approach to the earth • Line of Apsides – the line joining the perigee and the 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, ω – the angle from ascending node to perigee, measured in the orbital plane Legend Letters in the image denote: A – Minor, orbiting body B – Major body being orbited by A C – Reference plane D – Orbital plane of A E – Ascending node ☊ F – Reference direction (for orbits in or near the ecliptic plane). Ω – Argument of the Ascending Node [wiki] The Inclination Angle • The angle between the orbital plane and the earth’s equatorial plane • A satellite that is in orbit with some inclination angle is in an inclined orbit • A satellite that has an inclination angle of 900 is in polar orbit • A satellite that has an inclination angle of 00 is in equatorial orbit Wiki Prograde and Retrograde Orbits • An orbit in which the • An orbit in which the satellite moves in satellite moves in a the same direction direction counter to as the earth’s (opposite) to the rotation is called a earth’s rotation is prograde orbit (the called a retrograde inclination angle is orbit (the inclination between 00 and 900 angle is between 900 and 1800 Answer the questions: Most satellites are launched in a prograde orbit. Why? Calculate the orbit radius and the height (altitude above the earth’s surface) for the satellite with the orbit in the inclination angle =0, and when the satellite appears to be motionless above the earth.