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,
ab
e

ab
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,
 (coscos   sinsin  cosi) ( cossin   sincos  cosi) 
 rP 


R  (sincos   cossin  cosi) ( sinsin   coscos  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