lecture02_orbital mechanics 30th aug.,2012
TRANSCRIPT
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Orbital Aspects of Satellite
Communications
Prabhakar Dubey
Associate Professor
Department of Electronics & Communication
AIMT, LUCKNOW
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Agenda
Orbital MechanicsLook Angle Determination
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Orbital Mechanics
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Kinematics & Newtons Law
s = ut + (1/2)at2
v2 = u2 + 2at
v = u + at
F = ma
s = Distance traveled in time, t
u = Initial Velocity at t= 0
v = Final Velocity at time = t
a = Acceleration
F = Force acting on the object
NewtonsSecond Law
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FORCE ON A SATELLITE : 1Force = Mass Acceleration
Unit of Force is a Newton
ANewtonis the force required toaccelerate 1 kg by 1 m/s2
Underlying units of a Newtonare
therefore (kg) (m/s2
)In Imperial Units 1 Newton= 0.2248ft lb.
Next
Slide
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ACCELERATION FORMULAa = acceleration due to gravity = / r2km/s2
r =radius from center of earth
= universal gravitational constant Gmultiplied by the mass of the earth MEis Keplers constant and
= 3.9861352 105 km3/s2G= 6.672 10-11 Nm2/kg2 or 6.672 10-20km3/kg s2 in the older units
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FORCE ON A SATELLITE : 2Inward (i.e. centripetal force)
SinceForce = Mass Acceleration
If the Force inwards due to gravity =FIN then
FIN=m ( / r2
)
=m (GME / r2)
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Orbital Velocities and Periods
Satellite Orbital Orbital Orbital
System Height (km) Velocity (km/s) Period
h min s
INTELSAT 35,786.43 3.0747 23 56 4.091
ICO-Global 10,255 4.8954 5 55 48.4
Skybridge 1,469 7.1272 1 55 17.8
Iridium 780 7.4624 1 40 27.0
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Reference Coordinate Axes 1:
Earth Centric Coordinate System
The earth is at the center
of the coordinate system
Reference planes coincide
with the equator and the
polar axis
More usual
to use this
coordinate
system
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Reference Coordinate Axes 2:
Satellite Coordinate System
The earth is at the
center of the coordinate
system and reference is
the plane of the
satellites orbit
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Balancing the Forces - 2Inward Force
r
mGME
F 3r
Equation (2.7)
F
G = Gravitational constant = 6.672 10-11 Nm2/kg2
ME= Mass of the earth (and GME= = Keplers constant)
m = mass of satellite
r= satellite orbit radius from center of earth
r= unit vector in the rdirection (positive ris away from earth)
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Balancing the Forces - 3Outward Force F
2
2
dt
dmF
r
Equation (2.8)
Equating inward and outward forces we find
2
2
3 dt
d
r
rr
Equation (2.9), or we can write
032
2
rdt
d rr Equation (2.10)
Second order differential
equation with six unknowns:
the orbital elements
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We have a second order differential equation
See text p.21 for a way to find a solutionIf we re-define our co-ordinate system intopolar coordinates (see Fig. 2.4) we can re-writeequation (2.11) as two second order differential
equations in terms of r0 and 0
THE ORBIT - 1
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THE ORBIT - 2
Solving the two differential equations
leads to six constants (the orbitalconstants) which define the orbit, andthree laws of orbits (Keplers Laws ofPlanetary Motion)
Johaness Kepler (1571 - 1630) aGerman Astronomer and Scientist
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KEPLERS THREE LAWSOrbit is an ellipse with the larger body (earth) atone focus
The satellite sweeps out equal arcs (area) inequal time (NOTE: for an ellipse, this meansthat the orbital velocity varies around the orbit)
The square of the period of revolution equals aCONSTANT the THIRD POWER of SEMI-MAJOR AXIS of the ellipse
Well look at each of these in turn
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Review: Ellipse analysis
Points (-c,0) and (c,0) are the foci.
Points (-a,0) and (a,0) are the vertices.
Line between vertices is the major axis.
a is the length of the semimajor axis.
Line between (0,b) and (0,-b) is the minor axis.
b is the length of the semiminor axis.
12
2
2
2
b
y
a
x
222cba
Standard Equation:
y
V(-a,0)
P(x,y)
F(c,0)F(-c,0) V(a,0)
(0,b)
x
(0,-b)
abA
Area of ellipse:
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KEPLER 1: Elliptical OrbitsLaw 1
The orbit is an ellipse
e = ellipses eccentricity
O = center of the earth (onefocus of the ellipse)
C = center of the ellipse
a = (Apogee + Perigee)/2
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KEPLER 1: Elliptical Orbits
(cont.)(describes a conic section,
which is an ellipse if e < 1)
)cos(*10
0e
pr
e = eccentricity
e
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KEPLER 2: Equal Arc-SweepsFigure 2.5
Law 2
If t2 - t1 = t4 - t3
then A12 = A34
Velocity of satellite is
SLOWESTatAPOGEE;FASTESTatPERIGEE
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KEPLER 3: Orbital PeriodOrbital period and the Ellipse are related by
T2 = (4 2 a3) / (Equation 2.21)
That is the
squareof the period of revolution is equal to a
constant the cube of the semi-major axis.
IMPORTANT: Period of revolution is referenced to inertial space, i.e., to
the galactic background, NOT to an observer on the surface of one of the
bodies (earth).
= Keplers Constant = GME
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Numerical Example 1
The Geostationary Orbit:
Sidereal Day = 23 hrs 56 min 4.1 sec
Calculate radius and height of GEO orbit:
T2 = (4 2 a3) / (eq. 2.21)Rearrange to a3 = T2 /(4 2)
T = 86,164.1 sec
a3 = (86,164.1) 2 x 3.986004418 x 105/(4 2)
a = 42,164.172 km = orbit radiush = orbit radius earth radius = 42,164.1726378.14
= 35,786.03 km
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Solar vs. Sidereal Day
Asidereal dayis the time between consecutive crossings ofany particular longitude on the earth by any star other thanthe sun.
Asolar dayis the time between consecutive crossings of anyparticular longitude of the earth by the sun-earth axis.
Solar day = EXACTLY 24 hrsSidereal day = 23 h 56 min. 4.091 s
Why the difference?
By the time, the Earth completes a full rotation withrespect to an external point (not the sun), it has already
moved its center position with respect to the sun. Theextra time it takes to cross the sun-earth axis, averagedover 4 full years (because every 4 years one has 366 days)is of about 3.93 minutes per day.
Calculation next page
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Solar vs. Sidereal Day
Numerical Calculation:
4 years = 1461 solar days (365*4+1)
4 years : earth moves 1440 degrees(4*360) around sun.
1 solar day: earth moves 0.98
degrees (=1440/1461) aroundsun
1 solar day : earth moves 360.98degress around itself(360 +0.98)
1sidereal day = earth moves 360
degrees around itself1 solar day = 24hrs = 1440minutes
1 sidereal day = 1436.7 minutes(1440*360/360.98)
Difference = 3.93 minutes
(Source: M.Richaria, Satellite Communication Systems, Fig.2.7)
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LOCATING THE SATELLITE IN
ORBIT: 1o is the True
Anomaly
See eq. (2.22)
Cis the
center of the
orbit ellipse
O is thecenter of the
earth
NOTE:Perigee andApogee are on opposite sides of the orbit
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LOCATING THE SATELLITE IN
ORBIT: 2Need to develop a procedure that will allowthe average angular velocity to be used
If the orbit is not circular, the procedure is to
use a Circumscribed CircleA circumscribed circle is a circle that has a
radius equal to the semi-major axis length ofthe ellipse and also has the same center
See next slide
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LOCATING THE SATELLITE IN
ORBIT: 3
= Average angular velocity
E = Eccentric Anomaly
M= Mean Anomaly
M= arc length (in radians) that thesatellite would have traversed since
perigee passage if it were moving
around the circumscribed circle
with a mean angular velocity
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ORBIT ECCENTRICITYIf a= semi-major axis,
b= semi-minor axis, ande= eccentricity of the orbit ellipse,
then
ba
ba
e
NOTE: For a circular orbit,a =b and e = 0
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Time reference:
tp Time of Perigee = Time of closest
approach to the earth, at the sametime, time the satellite is crossing the x0axis, according to the reference used.
t- tp = time elapsed since satellite lastpassed the perigee.
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ORBIT DETERMINATION 1:
Procedure:Given the time of perigee tp, the
eccentricity eand the length of thesemimajor axis a:
Average Angular Velocity (eqn. 2.25)M Mean Anomaly (eqn. 2.30)
E Eccentric Anomaly (solve eqn. 2.30)
ro
Radius from orbit center (eqn. 2.27)
o True Anomaly (solve eq. 2.22)
x0 andy0 (using eqn. 2.23 and 2.24)
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ORBIT DETERMINATION 2:Orbital Constantsallow you todetermine coordinates (ro, o)and (xo,
yo)in the orbital planeNow need to locate the orbital planewith respect to the earth
More specifically: need to locate theorbital location with respect to a pointon the surface of the earth
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LOCATING THE SATELLITE WITH
RESPECT TO THE EARTH
The orbital constants define the orbit of thesatellite with respect to the CENTERof the
earthTo know where to look for the satellite inspace, we must relate the orbital plane andtime of perigee to the earths axis
NOTE: Need a Time Reference to locate the satellite. The
time reference most often used is the Time of Perigee, tp
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GEOCENTRIC EQUATORIAL
COORDINATES - 1zi axis Earths rotational axis (N-Spoles with N as positive z)
xi axis In equatorial plane towardsFIRST POINT OF ARIES
yi
axis Orthogonal to zi
andxi
NOTE: TheFirst Point of Aries is a line from the
center of the earth through the center of the sun at
the vernal equinox (spring) in the northern
hemisphere
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GEOCENTRIC EQUATORIAL
COORDINATES - 2
ToFirst Point of Aries
RA = Right Ascension
(in the xi,yi plane)
= Declination (the
angle from the xi,yi plane
to the satellite radius)
NOTE: Direction toFirst Point of Aries does NOT rotate
with earths motion around; the direction onlytranslates
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LOCATING THE SATELLITE - 1
Find theAscending Node
Point where the satellite crossesthe equatorial plane from Southto North
Define and iDefine
Inclination
Right Ascension of the Ascending
Node (=RA from Fig. 2.6 in text)
See next slide
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DEFINING PARAMETERS
Orbit passes through
equatorial plane here
First Point
of Aries
Center of earth
Argument of Perigee
Right AscensionInclination
of orbit
Equatorial plane
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DEFINING PARAMETERS 2
(Source: M.Richaria, Satellite Communication Systems, Fig.2.9)
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LOCATING THE SATELLITE - 2
and i together locate the
Orbital plane with respect to theEquatorial plane.
locates the Orbital coordinate
system with respect to theEquatorial coordinate system.
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LOCATING THE SATELLITE - 2
Astronomers use Julian Daysor Julian Dates
Space Operations are in Universal Time
Constant(UTC) taken from Greenwich Meridian(This time is sometimes referred to as Zulu)
To find exact position of an orbiting satellite at agiven instant, we need the Orbital Elements
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ORBITAL ELEMENTS (P. 29)Right Ascension of the Ascending Nodei Inclination of the orbit
Argument of Perigee (See Figures 2.6 &2.7 in the text)
tp Time of Perigee
e Eccentricity of the elliptical orbit
a Semi-major axis of the orbit ellipse (SeeFig. 2.4 in the text)
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Numerical Example 2:
Space Shuttle Circular orbit (height = h = 250km). Use earth radius = 6378 km
a. Period = ?
b. Linear velocity = ?Solution:
a) r = (re + h) = 6378 + 250 = 6628 kmFrom equation 2.21:
T2 = (4 2 a3) / = 4 2 (6628)3 / 3.986004418 105 s2
= 2.8838287 107 s2
T = 5370.13 s = 89 mins 30.13 secs
b) The circumference of the orbit is 2a = 41,644.95 km
v = 2a / T = 41,644.95 / 5370.13 = 7.755 km/s
Alternatively:
v =(/r)
2
. =7.755 km/s.
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Numerical Example 3:Elliptical Orbit: Perigee = 1,000 km, Apogee = 4,000 km
a. Period = ?
b. Eccentricity = ?Solution:
a) 2 a = 2 re + hp + ha = 2 6378 + 1000 + 4000 = 17,756 km
a = 8878 km
T2 = (4 2 a3) / = 4 2 (8878)3 / 3.986004418 105 s2
= 6.930545 107 s2
T = 8324.99 s = 138 mins 44.99 secs = 2 hrs 18 mins 44.99secs
b. At perigee, Eccentric anomaly E = 0 and r0 = re+ hp.From Equation 2.42,:
r0 = a ( 1ecos E )
re+ hp = a( 1 e)
e = 1 - (re+ hp) / a = 1 - 7,378 / 8878 = 0.169
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Look Angle Determination
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CALCULATING THE LOOK
ANGLES 1: HISTORICAL
Needsix Orbital Elements
Calculatethe orbit from these OrbitalElements
Definethe orbital plane
Locatesatellite at time twith respect to theFirst Point of Aries
Find locationof the Greenwich Meridianrelative to the first point of Aries
Use Spherical Trigonometryto find theposition of the satellite relative to a point on
the earths surface
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CALCULATING THE LOOK
ANGLES 2: AGE OF THEPC
Need two basic look-angle parameters:
Elevation Angle
Azimuth Angle
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ANGLE DEFINITIONS - 1
C
SubZenith direction
Nadir direction
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Coordinate System 1 Latitude: Angular distance, measured in
degrees, north or south of the equator.
L from -90 to +90 (or from 90S to 90N)
Longitude: Angular distance, measured in
degrees, from a given reference longitudinal
line (Greenwich, London).l from 0 to 360E (or 180W to 180E)
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Coordinate System 2
(Source: M.Richaria, Satellite Communication Systems, Fig.2.9)
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Satellite Coordinates
SUB-SATELLITE POINTLatitude Ls
Longitude lsEARTH STATION LOCATION
Latitude LeLongitude le
Calculate , ANGLE AT EARTH CENTERBetween the line that connects the earth-center to the satellite and
the line from the earth-center to the earth station.
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LOOK ANGLES 1 Azimuth: Measured eastward (clockwise)
from geographic north to the projection of
the satellite path on a (locally) horizontalplane at the earth station.
Elevation Angle: Measured upward from
the local horizontal plane at the earth stationto the satellite path.
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LOOK ANGLES NOTE: This isTrue North
(not magnetic,
from compass)
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Geometry for Elevation Calculation
El= - 90o
= central angle
rs= radius to the satellite
re = radius of the earth
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Slant path geometry
Review of spherical trigonometryLaw of Sines
Law of Cosines for angles
Law of Cosines for sides
2,
2tan
cos2
sinsinsin
222
cbad
cdd
bdadC
Cabbac
c
C
b
B
a
A
aCBCBA
Acbcba c
C
b
B
a
A
cossinsincoscoscos
cossinsincoscoscos
sinsinsin
c
A
B
C
a
b
ab
c
AB
C
Review of plane trigonometry Law of Sines Law of Cosines
Law of Tangents
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THE CENTRAL ANGLE is defined so that it is non-negative and
cos () = cos(Le) cos(Ls
) cos(ls
le
) + sin(Le
) sin(Ls
)
The magnitude of the vectors joining the center of the
earth, the satellite and the earth station are related by
the law of cosine:
2/12
cos21
s
e
s
e
s
r
r
r
rrd
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ELEVATION CALCULATION - 1
By the sine law we have
sinsindrs
Which yields
cos (El)
2/12
cos21
sin
s
e
s
e
r
r
r
r
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AZIMUTH CALCULATION - 1More complex approach for non-geo satellites. Different formulas
and corrections apply depending on the combination of positions
of the earth station and subsatellite point with relation to each of
the four quadrants (NW, NE, SW, SE).
A simplified method for calculating azimuths in the
Geostationary case is shown in the next slides.
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GEOSTATIONARY SATELLITES
SUB-SATELLITE POINT(Equatorial plane, Latitude Ls= 0
oLongitude ls)
EARTH STATION LOCATIONLatitude LeLongitude le
We will concentrate on the GEOSTATIONARY CASE
This will allow some simplifications in the formulas
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THE CENTRAL ANGLE - GEO
The original calculation previously shown:
cos () = cos(Le) cos(Ls) cos(lsle) + sin(Le) sin(Ls)
Simplifies usingLs= 0o since the satellite is
over the equator:
cos () = cos(Le) cos(lsle)
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ELEVATION CALCULATIONGEO 1
Usingrs = 42,164 kmand re = 6,378.14 kmgives
d= 42,164 [1.0228826 - 0.3025396 cos()]1/2 km
2/1cos3025396.00228826.1
sincos
El
NOTE: These are slightly different numbers than those
given in equations (2.67) and (2.68), respectively, due to
the more precise values used for rs and re
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ELEVATION CALCULATIONGEO 2
A simpler expression forEl(after Gordon and Walter, Principles
of Communications Satellites) is :
sin
cos
tan 1s
e
r
r
El
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AZIMUTH CALCULATIONGEO 1
To find the azimuth angle, an intermediate angle, , must first be
found. The intermediate angle allows the correct quadrant (see
Figs. 2.10 & 2.13) to be found since the azimuthal direction can lie
anywhere between 0o (true North) and clockwise through 360o(back to true North again). The intermediate angle is found from
e
es
L
ll
sin
tantan 1 NOTE: Simplerexpression than
eqn. (2.73)
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AZIMUTH CALCULATIONGEO 2
Case 1: Earth station in the Northern Hemisphere with
(a) Satellite to the SE of the earth station: Az = 180o -
(b) Satellite to the SW of the earth station: Az = 180o +
Case 2: Earth station in the Southern Hemisphere with
(c) Satellite to the NE of the earth station: Az =
(d) Satellite to the NW of the earth station: Az = 360o
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EXAMPLE OF A GEO
LOOK ANGLE ALCULATION - 1
FIND the Elevation and Azimuth
Look Angles for the following case:
Earth Station Latitude 52o
NEarth Station Longitude 0o
Satellite Latitude 0o
Satellite Longitude 66o
E
London, EnglandDockland region
Geostationary
INTELSAT IOR Primary
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EXAMPLE OF A GEO
LOOK ANGLE ALCULATION - 1
Step 1. Find the central angle
cos( ) = cos(Le) cos(ls-le)
= cos(52) cos(66)
= 0.2504yielding = 75.4981o
Step 2. Find the elevation angleEl
sin
cos
tan1 s
e
r
r
El
EXAMPLE OF A GEO
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EXAMPLE OF A GEO
LOOK ANGLE ALCULATION - 1
Step 2 contd.
El= tan-1[ (0.2504(6378.14 / 42164)) / sin (75.4981) ]
= 5.85o
Step 3. Find the intermediate angle,
e
es
L
ll
sin
tantan
1
= tan-1 [ (tan (66 - 0)) / sin (52) ]
= 70.6668
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EXAMPLE OF A GEO
LOOK ANGLE ALCULATION - 1
The earth station is in the Northern hemisphere and the satellite is
to the South East of the earth station. This gives
Az = 180o
- = 18070.6668 = 109.333o (clockwise from true North)
ANSWER: The look-angles to the satellite are
Elevation Angle = 5.85o
Azimuth Angle = 109.33o
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VISIBILITY TESTA simple test, called the visibility test will quickly tell you
whether you can operate a satellite into a given location.
A positive (or zero) elevation angle requires (see Fig. 2.13)
cose
s
rr
which yields
s
e
r
r1
cos
Eqns.
(2.42)
&(2.43)
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OPERATIONAL LIMITATIONS
For Geostationary Satellites 81.3o
This would give an elevation angle = 0oNot normal to operate down to zero
usual limits are C-Band
5oKu-Band 10oKa- and V-Band 20o