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Satellite Communications Orbital Calculations • Orbit definition & properties • Kepler’s Laws • First Law • Second Law • Third Law • Coordinate Systems • Geocentric equatorial • Astronomical • Satellite location • Geostationary orbits • Look angle calculations • Doppler shift GALAXY-11 Satellite, Hughes Space and Communications Lect 02 © 2012 Raymond P. Jefferis III 1 Orbit Definition and Properties • An orbit is a stable path around the earth traversed periodically by a satellite above the atmosphere of the earth. • Orbits are elliptical • Orbits have an Eccentricity parameter • Certain orbital properties are described by Keppler’s laws Lect 02 © 2012 Raymond P. Jefferis III 2 Definition of Ellipse • An ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant. Lect 02 © 2012 Raymond P. Jefferis III Lect 00 - 3 Axes of Ellipse An ellipse has two axes: a major axis and a minor axis a b a b a: semimajor axis, an ellipse has two semimajor axes b: semiminor axis, an ellipse has two semiminor axes Lect 02 © 2012 Raymond P. Jefferis III 4 Ellipse Properties • The sum of the distances from any point P on an ellipse to its two foci is constant and equal to the major diameter • The eccentricity of an ellipse is the ratio of the distance between the two foci and the length of the major axis Lect 02 © 2012 Raymond P. Jefferis III 5 Kepler’s First Law • A satellite, as a secondary body, follows an elliptical path around a primary body (earth). • The center of mass of the two bodies, the barycenter, will be at one of the foci. • For semimajor axis a and semiminor axis b, the orbital eccentricity e is be expressed by, ab e ab Lect 02 a2 b2 a © 2012 Raymond P. Jefferis III 6 Kepler’s Second Law • A ray from the barycenter to an orbiting satellite will sweep out equal areas in the orbital plane in equal time intervals. Lect 02 © 2012 Raymond P. Jefferis III 7 Kepler’s Third Law • The square of the orbital time is proportional to the cube of the mean distance, a, between the two bodies (semimajor axis). For a satellite motion of n radians/sec (orbital period P = 2π/n) and the gravitational parameter of the earth, G*M = = 3.986004418E5 km3/s2, then the mean distance, a, is calculated as, P a 2 n 4 2 3 Lect 02 2 © 2012 Raymond P. Jefferis III 8 Planetary Data Planet Period of revolution around the sun [yr] Mercury 0.241 Venus 0.615 Earth 1.00 Mars 1.88 Jupiter 11.86 Saturn 29.46 Uranus 84.01 Neptune 164.79 Pluto 248.43 Lect 02 Period of rotation around own axis 58.6 days 243 days 23 h 56 m 4 s 24 h 37 m 23 s 9 h 50 m 10 h 25 m 710 h 50 m 16 h 6.4 days Semi major axis( A U ) Eccentricity 0.387 0.723 1.00 1.524 5.203 9.54 19.18 30.07 39.44 0.206 0.007 0.017 0.093 0.048 0.056 0.04 0.008 0.249 © 2012 Raymond P. Jefferis III 9 Planetary Orbits - continued Planet Radius(km) Mercury 5.791E+07 Venus 1.082E+08 Earth 1.496E+08 Mars 2.279E+08 Jupiter 7.783E+08 Saturn 1.427E+09 Uranus 2.870E+09 Neptune 4.497E+09 Period(da) 8.797E+01 2.247E+02 3.653E+02 6.870E+02 4.333E+03 1.076E+04 3.069E+04 6.019E+04 R3(km3) 1.942E+23 1.267E+24 3.348E+24 1.184E+25 4.715E+26 2.906E+27 2.363E+28 9.092E+28 T2(da2) 7.739E+03 5.049E+04 1.334E+05 4.719E+05 1.877E+07 1.158E+08 9.416E+08 3.623E+09 R3/T2 (km3) (da2) 2.510E+19 2.510E+19 2.510E+19 2.509E+19 2.512E+19 2.510E+19 2.510E+19 2.510E+19 Computations by Neal McLain, Society of Broadcast Engineers, Chap. 24. Lect 02 © 2012 Raymond P. Jefferis III 10 Tangential Velocity in A Circular Orbit • From Kepler’s Third Law, the tangential orbital velocity [km/s] at radius r [km] is calculated, for a circular orbit, from: v r where, μ= 3.986004418E5 [km3/s2] is the gravitational parameter for the earth Lect 02 © 2012 Raymond P. Jefferis III 11 Tangential Velocity Calculation • r = 42, 164.17 [km] - geostationary orbit • μ= 3.986004418E5 [km3/s2] • v = 3.0746600858 [km/s] Lect 02 © 2012 Raymond P. Jefferis III 12 Orbital Period - Low Earth Orbit • From Kepler’s Third Law, T 4 2 a 3 • Note: The satellite velocity is usually not uniform over the orbit, because the path is elliptical. Lect 02 © 2012 Raymond P. Jefferis III 13 Example - Space Station For the International Space Station with altitude h = 350 km, • • • a = re + h = 6378.14 + 350 = 6728.14 km. = 3.986004418E5 km3/s2 T = 5492.29 sec/orbit (91.538 min/orbit) 1/T =15.69 orbits/sidereal day (15.73 orbits in 24 hours) Lect 02 © 2012 Raymond P. Jefferis III 14 Elliptical Orbit Calculation • The satellite NOAA-B (1980-43A) was launched in May 1980 into an orbit with perigee height of 260 km and apogee height 1440 km. • We wish to find the orbital period and the orbital eccentricity. • Data: 2a = 2re+hp + ha = 2(6378.14)+260+1440 = 14456.28 km • Calculations: a = 7228.14 km T = 6115.77 sec/orbit e = 1 - (re+hp)/a = 0.0816254 Lect 02 © 2012 Raymond P. Jefferis III 15 Sample Orbital Calculation mu = 3.986004418 10^5; ha = 1440.0; hp = 260.0; re = 6378.14; twoa = 2*re + hp + ha; a = twoa/2; t = Sqrt[(4*pi^2*a^3)/mu]; tinv = 24*60*60/t; ecc = 1.0 - (re + hp)/a; Lect 02 © 2012 Raymond P. Jefferis III 16 Two-Line Data Element Set (TLE) • The two-line element set (TLE) is a data format that lists information pertaining to the orbital parameters of Earth-orbiting satellites. • TLE format is used by NORAD and NASA • TLE data can be used, with appropriate software, to compute satellite position at a given time • Models used: SGP4 or SGP8 Lect 02 © 2012 Raymond P. Jefferis III \Lect 00 - 17 NASA Satellite Data - TLE • Line 1 – – – – Lect 02 Col 3 - 7 Col 19-20 Col 21-32 Col 34-43 Satellite number Epoch year (last two digits) Epoch Day and fraction Mean motion derivative [rev/day 2] © 2012 Raymond P. Jefferis III 18 NASA Satellite Data - TLE • Line 2 – – – – – – – Lect 02 Col 9 –16 Col 18-25 Col 27-33 Col 35-42 Col 44-51 Col 53-63 Col 64-68 Inclination [degrees] Right ascension [degrees] Eccentricity – leading decimal assumed Argument of perigee [degrees] Mean anomaly [degrees] Mean motion [rev/day] Revolution number at epoch [rev] © 2012 Raymond P. Jefferis III 19 Spacetrack (SGP4) Reference http://www.amsat.org/amsat/ftp/docs/spacetrk.pdf SPACETRACK REPORT NO. 3 Models for Propagation of NORAD Element Sets Felix R. Hoots and Ronald L. Roehrich December 1980 Package Compiled by TS Kelso 31 December 1988 Lect 02 © 2012 Raymond P. Jefferis III Lect 00 - 20 Geosynchronous Orbits • A geosynchronous orbit is an orbit (usually equatorial) having a period of one sidereal day, 23h 56m 04.0905s (23.9344695833 hours, or 86164.090530833 seconds). • A siderial day is the rotation of the earth in relation to the (relatively fixed) position of the stars. Shorter than solar day. Lect 02 © 2012 Raymond P. Jefferis III 21 Polar Orbits • A polar orbit is an orbit that passes over (or nearly passes over) both North and South poles. – Can be sun-synchronous (heliosynchronous) – Has a low altitude (800 - 1000 km), that is slightly retrograde, and leads to high resolution images with approximately constant illumination angles – Used for weather, environmental, and spy satellites Lect 02 © 2012 Raymond P. Jefferis III 22 Radius of Geostationary Orbit • A geosynchronous orbit has a period of one sidereal day, T = 86164.090530833 seconds • The radius is given by, 2 T a 3 4 2 • So a = 42, 164.17 km. Lect 02 © 2012 Raymond P. Jefferis III 23 Orbital Coordinates The orbit is, a(1 e2 ) ro 1 ecos o With eccentricity, • Point O is the center of the earth. • Point C is the center of the elli[se. • The orbital plane may be inclined to the earth’s equator. Lect 02 a2 b2 e a © 2012 Raymond P. Jefferis III 24 Other Calculations • Apogee height (radius), ra = a(1+e) • Perigee height (radius), rp = a(1-e) • The flight path angle, is, esin o arctan 1 ecos o Lect 02 © 2012 Raymond P. Jefferis III 25 Orbital Velocity • The gravitational product G*M for the earth is G*M = μ = 3.986004418E14 [m3/s2] • The gravitational acceleration g is, g = G*M/r2 = 6.67259E-11 [N-m2/kg2] • The tangential velocity is, then, v Lect 02 GMa(1 e2 ) ro cos o © 2012 Raymond P. Jefferis III 26 Definitions Lect 02 © 2012 Raymond P. Jefferis III 27 Coordinate Reference • x-axis is directed at “First Point of Ares” – Direction to Ares at vernal equinox defines the zero point of Right Ascension to the satellite • z-axis is directed along the spin axis of the earth – Approximately toward the North Star • y-axis is orthogonal to x-axis and z-axis Lect 02 © 2012 Raymond P. Jefferis III 28 Rectangular Geocentric Coordinates Lect 02 © 2012 Raymond P. Jefferis III 29 Spherical Geocentric Coordinates α is right ascension to satellite δ is declination to satellite Lect 02 © 2012 Raymond P. Jefferis III 30 Rectangular – Spherical Relation Lect 02 © 2012 Raymond P. Jefferis III 31 Earth-Centered Coordinates • The PQW unit vector is, ur ur r r r (r cos )P (r sin )Q rp P rqQ • The orbital plane of the satellite lies at angle i with respect to the earth equator • Rotation (Right Ascension) is measured from a fixed point in space, called the first point of Aries. The latter is the direction of Aries at the vernal equinox (March 20 or 21) Lect 02 © 2012 Raymond P. Jefferis III 32 Conversion Equations Spherical => rectangular x r cos cos y r cos sin z r sin Rectangular => spherical tan y / x tan z / (x 2 y 2 )1/2 r (x 2 y 2 z 2 )1/2 Lect 02 © 2012 Raymond P. Jefferis III 33 Transformation The transformation to earth coordinates is, rI r R rP r J Q rK where, (coscos sinsin cosi) ( cossin sincos cosi) rP R (sincos cossin cosi) ( sinsin coscos cosi) rQ (sin sini) (cos sini) Lect 02 © 2012 Raymond P. Jefferis III 34 Orbital Position Description Lect 02 © 2012 Raymond P. Jefferis III 35 In-Class Example Calculate orbital position indicated in Roddy Example 2.16 Lect 02 © 2012 Raymond P. Jefferis III 36 Example (Roddy, Example 2.16) • Data Ω = 300˚, ω = 60˚, i = 65˚, rP = -6500 km, rQ = 4000 km r rP2 rQ 2 7632.2 km Lect 02 © 2012 Raymond P. Jefferis III 37 Calculation for Roddy Example W w i r R = = = = = 300.0 Degree; 60.0 Degree; 65.0 Degree; {{-6500.0}, {4000.0}} {{(Cos[W] Cos[w] - Sin[W] Sin[w] Cos[i]), (-Cos[W] Sin[w] - Sin[W] Cos[w] Cos[i])}, {(Sin[W] Cos[w] + Cos[W] Sin[w] Cos[i]), (-Sin[W] Sin[w] + Cos[W] Cos[w] Cos[i])}, {(Sin[w] Sin[i]), (Cos[w] Sin[i])}} v = R.r = {{-4685.32}, {5047.71}, {-3289.14}} vmag = Sqrt[v[[1]]^2 + v[[2]]^2 + v[[3]]^2] = {7632.17} Lect 02 © 2012 Raymond P. Jefferis III 38 Satellite Look Angles • The subsatellite point (SSP) is the intersection of the orbital radius line with the earth surface. • An earth station will lie at an angle to the zenith from earth center to satellite and at azimuth angle Az to True North. • The satellite will be seen at elevation angle El to the local horizontal at the earth station • Visibility requires positive El, otherwise it is below the horizon Lect 02 © 2012 Raymond P. Jefferis III 39 Look Angle Geometry Look angle geometry, after Pratt et al Lect 02 © 2012 Raymond P. Jefferis III 40 Look Angle Calculations By the Law of Cosines, cos cos Lat ES cos LatSat cos(LonSat LonES ) sin Lat ES sin LatSat The elevation above Earth Station vertical is, El 90 The communications path length, d, along which path losses will be calculated is calculated from: 1/2 r re e d rs 1 2 cos rs rs 2 Lect 02 © 2012 Raymond P. Jefferis III 41 Elevation Angle Calculation The Elevation Angle can then be calculated from the coordinates of the subsatellite point (SSP), the coordinates of the earth station, the satellite orbital radius, and earth radius, as follows: rs sin cos El d Note: El must be positive for visibility. Lect 02 © 2012 Raymond P. Jefferis III 42 Geostationary Orbit Case • In this case the subsatellite point is on the Equator at longitude Lons, while Lats = 0. • rs = 42,164.17 km (geosynchronous) • re = 6378.137 km • rs/re = 6.6107345 • These reduce the calculations to those on the following slide: Lect 02 © 2012 Raymond P. Jefferis III 43 Geostationary Calculations cos cos Lat ES cos(LonS LonES ) d rs [1.02288235 0.30253825 cos ] 1 2 El tan 1[(6.6107345 cos ) / sin ] tan LonS LonES tan sin Lat ES 1 Ref: Pratt, et al, §2.2 Lect 02 © 2012 Raymond P. Jefferis III 44 Visibility Conditions • The Elevation angle, El, must be positive • or, re cos r 1 s Lect 02 © 2012 Raymond P. Jefferis III 45 Calculation Example Intelsat GALAXY-11 at 91W (NORAD 26038) • 39.1 dBW on C-Band (20W, 24 ch, BW: 36 MHz) 5945 (+n*20 MHz) MHz Uplink 3720 (+n*20 MHz) MHz Downlink • 47.8 dBW on Ku-Band (75/140W, 40 ch, BW: 36 MHz) 14020 (+n*20 MHz) MHz Uplink 11720 (+n*20 MHz) MHz Downlink • Power Supply: 10 kW (Xenon ion propulsion needs) • Polarization: v (odd), h (even) - Downlink opposite Lect 02 © 2012 Raymond P. Jefferis III 46 Intelsat GALAXY-11 • • • • • SSP = 91W (on Equator) LatSat = 0 N LonSat = 91 W LatES = 39.0 N LonES = 77.0 W re = 6378.137 km rs = 42164.17 km Look angle calculation results are: = 41.0566˚ = 21.6128˚ El = 42.5447˚ Az = 201.613˚ d = 37588.8 km Mathematica® notebook follows Lect 02 © 2012 Raymond P. Jefferis III 47 Galaxy-11 Look Angle Calculations re = 6378.137; rs = 42164.17; rr = re/rs; lates = 39.0 Degree; lones = -77.0 Degree; latsat = 0.00 Degree; lonsat = -91.0 Degree; gam = ArcCos[Cos[lates]*Cos[latsat]*Cos[lonsat -lones] + Sin[lates]*Sin[latsat]]; d = rs*Sqrt[1 + rr^2 - 2.0*rr*Cos[gam]]; el = ArcCos[rs*Sin[gam]/d]; alpha = ArcTan[Tan[Abs[lonsat-lones]]/Sin[lates]]; az = 180 + alpha/Degree Lect 02 © 2012 Raymond P. Jefferis III 48 Notes on More Accurate Calculations • Alternative equations are available: – Roddy, D., Satellite Communications, McGraw-Hill, 2006, §3.2. • Equations are also available that include the earth station altitude, for greater accuracy: – Ippolito, L., Satellite Communications Systems Engineering, Wiley, 2008, §2.4. Lect 02 © 2012 Raymond P. Jefferis III 49 Class Problem – Workshop 02 • Earth Station: Washington, DC – Latitude: Late = 38.895° N (+38.895°) – Longitude: Lone = 77.0363° W (-77.0363°) • Satellite: Geosynchronous at 91W – Latitude: Lats = 0° (+0°) – Longitude: Lons = 91° W(-91°) • Find range, elevation, and azimuth angle from the earth station to the satellite Lect 02 © 2012 Raymond P. Jefferis III 50 Work on the Problem • Take 45minutes • Formulate your answers as follows: – Elevation – Azimuth – Range • Hand in next week as a brief report for Workshop credit. Lect 02 © 2012 Raymond P. Jefferis III 51 Workshop 02 Calculations re = 6378.137; rs = 42164.17; rr = re/rs; lates = 38.895 Degree; lones = -77.0363 Degree; latsat = 0.0 Degree; lonsat = -91.0 Degree; gam = ArcCos[Cos[lates]*Cos[latsat]*Cos[lonsat lones] + Sin[lates]*Sin[latsat]]; d = rs*Sqrt[(1.0 + rr*rr - 2.0*rr*Cos[gam])]; el = ArcCos[rs*Sin[gam]/d]; psi = 90 + el/Degree; alpha = ArcTan[Tan[Abs[lonsat lones]]/Sin[lates]]/Degree; az = 180 + alpha Lect 02 © 2012 Raymond P. Jefferis III 52 Look Angle Results • Look angle calculation results are: = 40.9486˚ El = 42.6651˚ d = 37580.0 km Lect 02 = 21.6043˚ Az = 201.604˚ Psi = 132.665˚ © 2012 Raymond P. Jefferis III 53 Homework 02B Problem • Earth Station: West Chester, PA – Latitude: Late = 40° N (+40°) – Longitude: Lone = 76° W (-76°) • Satellite: Geosynchronous at 91W – Latitude: Lats = 0° (+0°) – Longitude: Lons = 91° W(-91°) • Find range, elevation, and azimuth angle from the earth station to the satellite Lect 02 © 2012 Raymond P. Jefferis III 54 Look Angle Results • Look angle calculation results are: El = 41.1901˚ d = 37689.7 km Lect 02 Az = 157.371˚ © 2012 Raymond P. Jefferis III 55 Radio Propagation Time Delay • Radio waves travel at the speed of light: c = 2.99792458 * 108 [m/s] (Note: The speed of light is slightly less in air.) • Ground – Geosynchronous Satellite delay: τe-s = d/c Example: τe-s = 38580.0/c = 0.128689 [s] (about 129 msec) Lect 02 © 2012 Raymond P. Jefferis III 56 Doppler Shift • Apparent frequency change, f , at wavelength, , due to relative velocity, vr of satellite with respect to an observer. • Can be experienced with satellites of Low Earth Orbit (200 - 300 km altitude) f vr / Lect 02 © 2012 Raymond P. Jefferis III 57 Doppler Calculation Terminology • r = radial distance from center of Earth [km] wavelength of data link radiation [km] • μ = 3.986004418E5 [km3/s2] Lect 02 © 2012 Raymond P. Jefferis III 58 Doppler Calculations The satellite tangential velocity is, c 2 r 1 vs 2 3 T r/ 4 r / The observer sees, re vr vs cos vs re h The Doppler shift is, f vr / Lect 02 © 2012 Raymond P. Jefferis III 59 End Lect 02 © 2012 Raymond P. Jefferis III 60