2.Orbital Aspects of Satellite Communications

8/2/2019 2.Orbital Aspects of Satellite Communications

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ORBITAL MECHANICS

Compiled by

M.LENIN BABU,M.Tech.,Lecturer,Dept. of ECEBapatla engineering college


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Topics covered according tosyllabus

Keplers laws of motion

Locating the satellite in the orbit

Locating the satellite w.r.t earth. Orbital elements

Look angle determination


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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 : 1

Force = Mass Acceleration Unit of Force is a Newton

A Newtonis the force required toaccelerate 1 kg by 1 m/s2

Underlying units of a Newtonare

therefore (kg) (m/s2) In Imperial Units 1 Newton= 0.2248 ft

lb.


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ACCELERATION FORMULA a = acceleration due to gravity = / r2km/s2

r =radius from center of earth

= universal gravitational constant Gmultiplied by the mass of the earth ME

is Keplers constant and= 3.9861352 105 km3/s2

G= 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 AccelerationIf the Force inwards due to gravity =FINthen

FIN=m ( / r2)=m (GME / r2)


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Reference Coordinate Axes 1:Earth Centric Coordinate

System

Fig. 2.2 in text

The earth is at thecenter of the coordinate

system

Reference planes

coincide with theequator and the polar

axis


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Reference Coordinate Axes2: Satellite Coordinate

System

Fig. 2.3 in text

The earth is at thecenter of the

coordinate system and

reference is the plane

of the satellites orbit


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Balancing the Forces - 2

Inward Force

r

mGME

F 3r

Equation (2.7)

F

G = Gravitational constant = 6.672 10-11 Nm2/kg2ME= Mass of the earth (and GME = = Keplersconstant)

m = mass of satellite

r = satellite orbit radius from center of earthr= unit vector in the rdirection (positive ris away from earth)


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Balancing the Forces - 3

Outward 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 solution

If we re-define our co-ordinate system intopolar coordinates (see Fig. 2.4) we can re-

write equation (2.11) as two second orderdifferential equations in terms of r0 and 0

THE ORBIT - 1


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THE ORBIT - 2

Solving the two differential equationsleads to six constants (the orbitalconstants) which define the orbit, andthree laws of orbits (Keplers Laws ofPlanetary Motion)

Johaness Kepler (1571 - 1630) a

German Astronomer and Scientist


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KEPLERS THREE LAWS

Orbit is an ellipse with the larger body(earth) at one focus

The satellite sweeps out equal arcs (area) in

equal time (NOTE: for an ellipse, this meansthat the orbital velocity varies around theorbit)

The square of the period of revolution equals

a CONSTANT the THIRD POWER of SEMI-MAJOR AXIS of the ellipse


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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 Orbits

Figure 2.6 in text

Law 1

The orbit is an

ellipse

e = ellipses eccentricity

O = center of the earth(one focus of the ellipse)

C = center of the ellipse

a = (Apogee + Perigee)/2


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KEPLER 1: Elliptical Orbits

(cont.)Equation 2.17 in text:

(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-Sweeps

Figure 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 Period

Orbital period and the Ellipse are related by

T2 = (4 2 a3) / (Equation 2.21)

That is the square of 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 1The 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 a

3

= 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 radius

h = orbit radius earth radius = 42,164.172 6378.14= 35,786.03 km


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LOCATING THE SATELLITE

IN ORBIT: 1Start with Fig. 2.6 in Text o is the True

Anomaly

See eq. (2.22)

Cis the

center of

the orbit

ellipse

O is the

center of

the earthNOTE:Perigee andApogee are on opposite sides of the orbit


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LOCATING THE SATELLITE

IN ORBIT: 2 Need 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 Circle A circumscribed circle is a circle that has a

radius equal to the semi-major axis length of

the ellipse and also has the same center


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LOCATING THE SATELLITEIN ORBIT: 3

Fig. 2.7 in the text

= Average angular velocitE = Eccentric Anomal

M= Mean Anomaly

M= arc length (in radians) thatthe satellite would have traversed

since perigee passage if it were

moving around the

circumscribed circle with a mean

angular velocity


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ORBIT CHARACTERISTICS

Semi-Axis Lengths of the Orbit

2

1 e

pa

where

2h

p

andh is the magnitude

of the angular

momentum

See eq. (2.18)

and (2.16)

2/121 eab where Che2

See eqn.(2.19)

and e is the eccentricity of the

orbit


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ORBIT ECCENTRICITY

If 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 thex0 axis, according to the referenceused.

t- tp = time elapsed since satellitelast passed the perigee.


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ORBIT DETERMINATION 1:Procedure:

Given the time of perigee tp, the eccentricity eand the length of the semimajor 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 and y0 (using eqn. 2.23 and 2.24)


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ORBIT DETERMINATION 2:

Orbital Constantsallow you to determinecoordinates (ro, o)and (xo, yo)in theorbital plane

Now need to locate the orbital plane withrespect to the earth

More specifically: need to locate the orbital

location with respect to a point on thesurface of the earth


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LOCATING THE SATELLITEWITH RESPECT TO THE EARTH

The orbital constants define the orbit of thesatellite with respect to the CENTER of the earth

To know where to look for the satellite in space,we must relate the orbital plane and time ofperigee 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 - 1 zi axis Earths rotational axis (N-S poles

with N as positive z)

xi axis In equatorial plane towards FIRSTPOINT OF ARIES

yi axis Orthogonal to zi and xi

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 EQUATORIALCOORDINATES - 2

Fig. 2.8 in text

ToFirst Point of Aries

RA = Right Ascension

(in the xi,yi plane)

= Declination (theangle 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 the Ascending Node

Point where the satellite crosses

the equatorial plane from South toNorth

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

Fig. 2.9 in text

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 Orbitalplane with respect to the

Equatorial plane. locates the Orbital coordinate

system with respect to the

Equatorial 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 Node i 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 = 250 km).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 ( 1 ecos E )

re+ hp = a( 1 e)

e = 1 - (re+ hp) / a = 1 - 7,378 / 8878 = 0.169


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Look Angle Determination

C C G OO


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CALCULATING THE LOOKANGLES 1: HISTORICAL

Needsix Orbital Elements

Calculatethe orbit from these Orbital Elements

Definethe orbital plane Locatesatellite at time twith respect to the

First Point of Aries

Find locationof the Greenwich Meridian

relative to the first point of Aries Use Spherical Trigonometryto find the

position of the satellite relative to a point on theearths surface


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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 POINT

Latitude Ls ,Longitude ls EARTH 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

Fig. 2.9 in text


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Geometry for Elevation

CalculationFig. 2.11 in text

El= - 90o = central angle

rs= radius to the satellite

re = radius of the earth


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Slant path geometry

Review of spherical trigonometry Law 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

Acbcbac

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 andcos () = cos(Le) cos(L

s

) 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

Eqn. (2.57)

Which yields

cos (El)

2/12

cos21

sin

s

e

s

e

r

r

r

r

Eqn. (2.58)


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AZIMUTH CALCULATION - 1

More 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.

GEOSTATIONARY


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GEOSTATIONARYSATELLITES

SUB-SATELLITE POINT

(Equatorial plane, Latitude Ls= 0o

Longitude ls)

EARTH STATION LOCATION

Latitude LeLongitude le

We will concentrate on the GEOSTATIONARY CASEThis 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) (eqn. 2.66)


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ELEVATION CALCULATION GEO 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 CALCULATION GEO 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 CALCULATION GEO 1

To find the azimuth angle, an intermediate angle, , must first befound. 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 CALCULATION GEO 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 GEOLOOK 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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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

O G O


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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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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 TEST

A 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 = 0o

Not normal to operate down to zero usual limits are C-Band 5o

Ku-Band 10o

Ka- and V-Band 20o

2.Orbital Aspects of Satellite Communications

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