lect 02© 2012 raymond p. jefferis iii1 satellite communications orbital calculations orbit...

Lect 02 © 2012 Raymond P. Jefferis III 1

Satellite CommunicationsOrbital 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


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

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


b a

b

a

a: semimajor axis, an ellipse has two semimajor axesb: semiminor axis, an ellipse has two semiminor axes

An ellipse has two axes: a major axis and a minor axis

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



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,

e =a−ba+b

=a2 −b2

a


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.


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 = μ= 3.986004418E5 km3/s2, then the mean distance, a, is calculated as,

a3 =μn2 =

μP2

4π 2

Planetary Data


Planet Period of revolution Period of rotation Semi major axis( A U ) Eccentricityaround the sun [yr] around own axis

Mercury 0.241 58.6 days 0.387 0.206Venus 0.615 243 days 0.723 0.007Earth 1.00 23 h 56 m 4 s 1.00 0.017Mars 1.88 24 h 37 m 23 s 1.524 0.093Jupiter 11.86 9 h 50 m 5.203 0.048Saturn 29.46 10 h 25 m 9.54 0.056Uranus 84.01 710 h 50 m 19.18 0.04Neptune 164.79 16 h 30.07 0.008Pluto 248.43 6.4 days 39.44 0.249

Planetary Orbits - continued


Planet Radius(km) Period(da) R3(km3) T2(da2) R3/T2 (km3) (da2) Mercury 5.791E+07 8.797E+01 1.942E+23 7.739E+03 2.510E+19 Venus 1.082E+08 2.247E+02 1.267E+24 5.049E+04 2.510E+19 Earth 1.496E+08 3.653E+02 3.348E+24 1.334E+05 2.510E+19 Mars 2.279E+08 6.870E+02 1.184E+25 4.719E+05 2.509E+19 Jupiter 7.783E+08 4.333E+03 4.715E+26 1.877E+07 2.512E+19 Saturn 1.427E+09 1.076E+04 2.906E+27 1.158E+08 2.510E+19 Uranus 2.870E+09 3.069E+04 2.363E+28 9.416E+08 2.510E+19 Neptune 4.497E+09 6.019E+04 9.092E+28 3.623E+09 2.510E+19

Computations by Neal McLain, Society of Broadcast Engineers, Chap. 24.

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 =μr

where, μ= 3.986004418E5 [km3/s2] is the gravitational parameter for the earth

Tangential Velocity Calculation

• r = 42, 164.17 [km] - geostationary orbit

• μ= 3.986004418E5 [km3/s2]

• v = 3.0746600858 [km/s]



Orbital Period - Low Earth Orbit

• From Kepler’s Third Law,

T =4π 2a3

μ

• Note: The satellite velocity is usually not uniform over the orbit, because the path is elliptical.


Example - Space Station

For the International Space Station with altitudeh = 350 km,

• a = re + h = 6378.14 + 350 = 6728.14 km. μ= 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)


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 kmT = 6115.77 sec/orbite = 1 - (re+hp)/a = 0.0816254

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;


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– Col 3 - 7 Satellite number

– Col 19-20 Epoch year (last two digits)

– Col 21-32 Epoch Day and fraction

– Col 34-43 Mean motion derivative [rev/day 2]


NASA Satellite Data - TLE

• Line 2– Col 9 –16 Inclination [degrees]

– Col 18-25 Right ascension [degrees]

– Col 27-33 Eccentricity – leading decimal assumed

– Col 35-42 Argument of perigee [degrees]

– Col 44-51 Mean anomaly [degrees]

– Col 53-63 Mean motion [rev/day]

– Col 64-68 Revolution number at epoch [rev]


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.


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


Radius of Geostationary Orbit• A geosynchronous orbit has a period of one

sidereal day, T = 86164.090530833 seconds

• The radius is given by,

a =

μT2

4π 23

• So a = 42, 164.17 km.


Orbital Coordinates

• 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.

ro =a(1−e2 )1+ ecosφo

The orbit is,

With eccentricity,

e =a2 −b2

a


Other Calculations• Apogee height (radius), ra = a(1+e) • Perigee height (radius), rp = a(1-e)• The flight path angle, θis,

θ =arctanesinφo

1+ ecosφo

⎛

⎝⎜⎞

⎠⎟


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 =GMa(1−e2 )ro cosφo

Definitions


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


Rectangular Geocentric Coordinates


Spherical Geocentric Coordinates


α is right ascension to satellite

δ is declinationto satellite

Rectangular – Spherical Relation



Earth-Centered Coordinates• The PQW unit vector is,

rr =(rcosφ)P

ur+ (rsinφ)Q

ur=rp

rP + rq

rQ

• The orbital plane of the satellite lies at angle 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)

Conversion Equations


x =rcosδ cosαy=rcosδ sinαz=rsinδ

Spherical => rectangular

Rectangular => spherical

tanα =y/ x

tanδ =z / (x2 + y2 )1/2

r =(x2 + y2 + z2 )1/2


Transformation

rI

rJ

rK

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=R

ur rPrQ

⎡

⎣⎢

⎤

⎦⎥

where,

Rur=

(cosΩcosω −sinΩsinω cosi) (−cosΩsinω −sinΩcosω cosi)(sinΩcosω + cosΩsinω cosi) (−sinΩsinω + cosΩcosω cosi)

(sinω sini) (cosω sini)

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

rPrQ

⎡

⎣⎢

⎤

⎦⎥

The transformation to earth coordinates is,


Orbital Position Description

In-Class Example


Calculate orbital position indicated in Roddy Example 2.16


Example (Roddy, Example 2.16)

• DataΩ = 300˚, ω = 60˚, i = 65˚, rP = -6500 km, rQ = 4000 km

r = rP

2 + rQ2 =7632.2km


Calculation for Roddy ExampleW = 300.0 Degree;r = {{-6500.0}, {4000.0}}R = {{(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}

w = 60.0 Degree;i = 65.0 Degree;


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 Υto 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


Look Angle Geometry

Look angle geometry, after Pratt et al


Look Angle Calculations

cosγ =cosLatES cosLatSat cos(LonSat −LonES) + sinLatES sinLatSat

El =ψ −90o

d =rs 1+rers

⎛

⎝⎜⎞

⎠⎟

2

−2rers

⎛

⎝⎜⎞

⎠⎟cosγ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1/2

The communications path length, d, along which path losses will be calculated is calculated from:

The elevation above Earth Station vertical is,

By the Law of Cosines,


Elevation Angle Calculation

cos El =rssinγ

d

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:

Note: El must be positive for visibility.


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:


Geostationary Calculationscosγ =cosLatES cos(LonS −LonES)

d=rs[1.02288235 −0.30253825cosγ]12

El =tan−1[(6.6107345 −cosγ) / sinγ] −γ

α =tan−1 tan LonS −LonES

sinLatES

⎡

⎣⎢

⎤

⎦⎥

Ref: Pratt, et al, §2.2


Visibility Conditions

• The Elevation angle, El, must be positive

• or,

γ ≤cos−1 re

rs

⎛

⎝⎜⎞

⎠⎟


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


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:

γ= 41.0566˚ α= 21.6128˚El = 42.5447˚ Az = 201.613˚d = 37588.8 kmMathematica® notebook follows


Galaxy-11 Look Angle Calculationsre = 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

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.


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


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.


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

Look Angle Results

• Look angle calculation results are:γ= 40.9486˚ α= 21.6043˚El = 42.6651˚ Az = 201.604˚d = 37580.0 km Psi = 132.665˚


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


Look Angle Results

• Look angle calculation results are:El = 41.1901˚ Az = 157.371˚d = 37689.7 km


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)



Doppler Shift

• Apparent frequency change, Δf , at wavelength, λ, 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)

Δf = vr / λ

Doppler Calculation Terminology

• r = radial distance from center of Earth [km] λwavelength of data link radiation [km]• μ = 3.986004418E5 [km3/s2]



Doppler Calculations

vs =cT

=2πr

4π 2r3 / μ=

1r / μ

The satellite tangential velocity is,

The observer sees,

vr =vs cos θ( ) =vs

rere +h

⎛

⎝⎜⎞

⎠⎟

The Doppler shift is,

Δf = vr / λ


End

lect 02© 2012 raymond p. jefferis iii1 satellite communications orbital calculations orbit...

Documents

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major axis lect

axes of ellipse lect

jefferis iii5 slide

minor axis slide

ellipse properties

definition of ellipse

communications slide