Orbital Aspects

8/12/2019 Orbital Aspects

1/49

Sat Com : Obital Mechanics 1

Orbital Aspects of SatelliteCommunication


2/49


The Equation of the Orbit

xy

z

Satellite

Earth rotation

235 s/km104=

constantsKepler'=GM

Gravitational force on the satellite F is:

r

N ew ton ' s second L aw :

F = md r

dt

2

2

...( )2

equating:

-

r

r

d r

dt3

2

2

d r

dt+

r

r

2

2 3

0

( Equation of motion )

(3)

3r

rmGMF E


3/49


r d r

dt

2

20 4......( )

Taking of each term

Since =0 r r

also d

dt r dr

dt

dr

dt

dr

dt r d r

dt[ ]

2

2

= 0 ( from def,)=0 (from(4))

r

r drdt

h = orbital angular momentum of the satellite

=constant

therefore, orbit is on a plane


4/49


To solve (3), we use the orbital plane to define a

second rectangular coordinates (x0,y0,z0)

(3) in (x0,y0,z0) becomes

( )

( )x

d x

dty

d y

dt

x x y y

x y0

0

0

0 0 0 0 02

2

2

2

0

2

0

20

(2.10)

z

y0

x0

z0

Satellite


5/49


Change to polar coordinate:

Substitute into (2.10) and equate thecomponents:

x ry r

x r

y r

0 0 0

0 0 0

0 0 0 0 0

0 0 0 0 0

cossin cos sin cos sin

0

r0

z0 x0

y0

d r

dtr d

dt r

2

2

200

0

0

(2.12)

r0


6/49


Equate the component:

(2.13) can be written as

or ( a constant) angular momentum. (2.15)

r d

dt

dr

dt

d

dt0

20

2

0 0

2 0

0

(2.13)

10

0

02 0

r

d

dtr

d

dt

r ddt

h02 0


7/49


Squaring (2.15)

Substitute into (2.12)

to find the equation relating r0 and define

r d

d t

h

r0

0

3

02 2

(2.16)

d r

dt

h

r r

2

2

2

0

3

0

0

(2.17)

ur

1

0

p


8/49


Then

and

differentiate again,

so (2.17) becomes:

00

00

0

00

20

d

duh

r

h

d

dr

dt

d

d

dr

dt

dr

dr

d u

du

d

0

0 0

12

From (2.16)

d r

dth u

d u

d

2

2

2 22

0

02

(2.22)d u

du

h

2

0

2 2


9/49


Solving

or

u

h

C

2 0 0

cos( )

rh C

h

hC

P

e

0 2

0 0

2

2

0 0

0 0

1

1

1

/ cos( )

cos( )

cos( ) (2.24)

Semilatus rectum P h

2

(2.25)


10/49


(2.24) is the equation of

an ellipse for e < 1 a circle for e = 0 (limiting condition with

Serve to orient the ellipse w.r.t. orbitalplane. We can always choose x0,y0 axis suchthat

h C2

0

0

0

0

0

1

.

cosr p

e(2.26)


11/49


The differential area swept by r0 is

dA r

d r d

dtdt

hdt 0

2

00

2

0

2 2 2

a

p

e

b a e

1

1

2

21

2( )

(2.27)

Apogee Perigeec

b r0

ae

a (1 + e) a (1 - e)

d0 r d0 0

r0


12/49


By Keplers second law, radius vector sweeps

out equal area in equal time. Orbital period T

T = area of ellipse / area swept out in one time unit

ab

hdt

ab

h

ab

P

ab

a e

a

T a

2 2

2 2

1

2

4

0

1

2

3

2

22 3

( )

(2.31)


13/49


Keplers third law:

Radius of geo-synchronous orbit of earth:

T = 24hr = 86400 sec.

(2.31) givesa = 42242 km

T a2 3


14/49


Locating the Satellite in the Orbit The equation of the orbit is

: true anomaly

The average angular velocity is

r a e

eo

o

( )

cos

1

1

2

o

x r

y ro o o

o o o

cossin

2 1

T a a

(2.32)

(2.35)


15/49


a

OCE

y

x

A

xo axis

yo axis

Orbit

o

ro

ae a(1-e)

aFig. 1

E = eccentric anomaly

C O

E M o

Circumscribed circle


16/49


The velocity of the satellite is

It can be shown that

Also from (2.16), (2.25) and (2.27)

v dxdt

dydt

drdt

r ddt

o o o

o

o2

2 2 2

2

2

v

a

a

ro

2 2 1

d

dt

h

r

a e

r

aa

rdrdt

ar

e

dr

dt ar a e a r

o

o o

o

o

o

o

o

o

2 2

2

2

2

2

2

2 2 2

1

2 1 1

( )


17/49


Solve for dtand multiply by

From Fig. 1 can show

Sub. into the equation above

20

22

00

)()(

raea

dr

a

rdt

Eaera cos0

dEEedt )cos1( (2.43)


18/49


Integrate (2.43) get:

Lettpbe the time of perigee, the time thatthe earth is closest to the satellite, i.e. the

time that the satellite is crossing the xo

axis.Where M is the mean anomaly

(2.44) t t E e E p sin


19/49


So, given tp, e, and a, we can determine the

orbital plane coordinates (ro,o) and (xo,yo) asfollow:

calculate from (2.35)

calculate M from (2.46) calculate E from (2.46)

solve E from (2.46)

find ro from (2.43)

find o from (2.32)

find (xo,yo) from (2.33 - 2.34)


20/49


Locating the Satellite w.r.t. Earth Define geocentric equatorial coordinates

(xi, yi, zi).

RA: right ascension: declination

: right ascension of theascending node

i: inclination (angle betweenorbital and equatorial planes)

: argument of perigee

zi

yi

xi

N. Pole

RA

Equatorial plane


21/49


yi

i

Ascending nodexi

zi

PerigeeSun

Earth (onMarch 21)

xiyi

(,i) together locates orbital plane w.r.t. the equatorial plane.Their relations are:

x

y

z

i

i

i

c c s c i s

s c c c i s

s i s

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )

c s s c i c

s s s c i c

s i c

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

s s i

c s i

c i

( ) ( )( ) ( )

( )

x

y

z

o

o

o

Equatorial Orbital


22/49


Define: rotating coordinate: (xr, yr, zr)

zr=zi, (xi,yi) plane is the same as

(xr

,yr

) planexr axis intersects the primegeographic meridian.

e: angular velocity

Te: Time since xraxis coincidedwith xi axis

zi / zr

e

xr

xi

yi

yr

xr


23/49


x

y

z

T T

T T

x

y

z

r

r

r

e e e e

e e e e

i

i

i

cos sin

sin cos

0

0

0 0 1


24/49


UT: universal time = Greenwich standard

time Julian days: start at noon UT where noon

Dec. 31, 1899 = J.D. 2415020

So J.D. 2446066 = Dec. 31, 1984 Julian date: adding the time to Julian day bydecimal fraction

e.g. 2446066.5 = Jan 1, 1985, 00:00:00 hr. in

UT

Angle between xi and xraxis is eTe


25/49


define: Tc = Julian centuries = time between

0 hr. UT on Julian day (JD) and noon UT onJan 1, 1900

Tc = (JD - 2415020)/36525 Julian centuries

eTe at any time t (in minutes) after midnightUT is

e

Te

=g,o

+ 0.25068477t degrees degrees


26/49


Orbital Elements

Need 6 quantities to specify the absolute co-ordinate of the satellite at time t.

For satellite communication, they are: 1. Eccentricity, e

2. Semimajor axis, a

3. Time of perigee, tp

4. RA of ascending node 5. Inclination, i

6. Argument of perigee,


27/49


Look Angle Determination For positioning the

earth station antenna

Azimuth (Az) Elevation (El)

other specifications:

RA and declination

(for radio astronomyantenna).

Local vertical

N

E

AzEl


28/49


29/49


Lz

x y z

s

o

r r r

r

90 1

2 2 2cos

zr

yr

xr

Equatorial plane

Length = x y zr r r2 2 2

S.S.P.

Ls

is given by (2.53)

meridians: longitudial lines

parallels: lantutude lines

s


30/49


Elevation (El)

: Central angle between station and satellite

given by cos cos cos cos sin sin L L l l L Le s s e e s

reEarth station

d

rs

re

cos

El

(fig. 2)

WhereLe and le are latitude andlongitude of the earth station


31/49


By cosine rule

sine rule:

d r r

r

r

rEl

s

e

s

e

so

1 2

90

2

cos

r d

s

sin sin

cos( ) sin sin

cos

El rd r

r

r

r

r km

s

e

s

e

s

e

1 2

6370

2


32/49


Azimuth (Az)

Equation depends on the relative locationsof s.s.p. and earth station

e.g.

S.S.P.

Earth station

Az

Pole

S.S.P.Az

Pole


33/49


El andAz for Synchronous Satellites

cos cos cos

cossin

. . cos

L l l

El

e s e

102274 03616

Central angle


34/49


Az calculations: 4 cases:

(a) SSP at southwest of ES

S.S.P.

c

Ga

E Az o 180

Let s = (a+c+)/2

then from spherical trigo.identity:

tansin( )sin( )

sin sin( )2

2

s s c

s s a

Wherea l l

c L Ls e

e s

| || |


35/49


Visibility

From Fig.2, theoretically the satellite isvisible if

or the central angle between E and S mustbe smaller than

for syn. satellite,

r r r

rse e

s

coscos

1

cos

1 r

r

e

s

81 3. o


36/49


Orbital Perturbation

Causes:

asymmetry of the Earths gravitational field

gravitational field of Sun and Moon solar radiation pressure

atmosphere drag (negligible for synchronoussatellite)


37/49


Treatment:

osculating orbit: the orbit defined by Keplersequation with orbital elements (a,e,tp,,I,). Assume orbital elements vary linearly with time

given by

...etc. satellite position at t, is calculated from

elements ( )

Satellite does not return to the same point in space define anomalistic period as the elapsed time

between successive perigee passages

da

dt

de

dt

,

a da

dtt t e

de

dtt t

o o ( ), ( ),1 0 1 0


38/49


39/49


Effect of Sun and Moon

Cause the orbital inclination to drift fromto in 26.6 years.

/ year for 1970 - 80.

1467. o

0 86. o


40/49


Orbit Determination

By classical method of Laplace and Guass

3 measurements in 3 time instances

2 measurements in 2 time instances using radarsince velocity can also be determined from eachobservation

2 or more stations measure the position at the

same time


41/49


Launches and Launch Vehicles

ELV: expendable launch vehicle. (Rocket)

STS: space transportation system (Shuttle)

Mechanics of launching for ELV

no Payload Assist Module (PAM) is needed

about 30 minutes to transfer orbit


42/49


V1: velocity incrementto move satellite intotransfer orbit (byPAM).

V2: velocity incrementto move from transferorbit togeosynchronous orbit(By Apogee KickMotor (AKM)).

Apogee

Perigee

296km

STS orbit

Transfer orbit

V1

V2


43/49


Expendable Launch Vehicles

Delta

1500kg, 2000kg, 2500kg $18M (1983)

Titan 4000kg $ 40M.(1983)

Ariane

2 Satellites at the same time

2100kg for A-2, 2580kg for A-3, 4300kg for A-4

launch at French Guiana, only 5 degree inclination

price: fixed


44/49


STS

payload 4000kg

maximum price: beat estimate


45/49


Orbital Effects in System Performance -

Doppler Shift

TT

R fcvcf

fRis the receive frequency

c is the velocity of light

vT is the component of transmitter

velocity towards the receiver

fT is the fixed frequency

Pronounced for low-orbit satellites

Negligible for synchronous satellites


46/49


Range Variation

Cyclic daily variation for synchronoussatellite

Variable round-trip delay would require alarge guard time in TDMA

Continuous satellite range monitoring is

needed


47/49


Eclipse

Earths shadow

Sun

Satellite

EarthSatellite


48/49


2 periods

~21 March and ~23 September

No power from solar array Solar power fluctuation

Thermal stress ofsatellite

1 21March April

70min

35minEclipsetime

Date

Full Shadow

Half Shadow


49/49


Sun Transit Outage

The sun passes through the beam of an earthstation antenna

raises the noise level occurs ~10 minutes a day for several days a

year

Very costly because of total outage duringthe daytime.

Orbital Aspects

Documents

Transcript of Orbital Aspects