Lecture02_orbital Mechanics 30th Aug.,2012

7/31/2019 Lecture02_orbital Mechanics 30th Aug.,2012

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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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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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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



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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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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



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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


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


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


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

Lecture02_orbital Mechanics 30th Aug.,2012

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