rX PREFACE

Within the United States, satellites are not used for telephone (voice) links. The de-velopment of terrestrial optical fiber links has made satellite transmission of telephonetraffic uneconomic, and the delay associated with GEO satellite link is a nuisance.Domestic satellites serving the United States now carry video signals for distribution tocable TV companies or direct to homes and serve networks of VSAT stations linked tocentral hubs in major cities. The development of direct'to home satellite broadcast tele-vision (DBS-TV) has had a major impact on the marketplace. In rhe United States, digi-tal DBS-TV transmissions are nw received in 15 million homes (2001 figure), and inEurope a similar number of homes receive satellite television programming. Video distri-bution, to cable companies and direct to home, accounts for more than half of all theworldwide earnings from satellite communication systems.

The authors would like to thank heir colieagues and students who, over the years,have made many valuable suggestions to improve this text. Their advice has been heeded,and the second edition of Satellite Communication is the better for it. Many more workedexamples have been added to the second edition to illustrate how calculations are carriedout for each topic.

CONTEN']--t

Preface ix

1. Intrl.l Bacl1.2 A BrI.3 Satel.:l Over1.5 SlrmRelerence:

2. Orbi2.1

2.2

2.3

2.42.5

2.6

OrbiDeveKepll)escLocaLocaOrbitExanExanExanLookThe IElevAzirnSpeciVisibExanOrbitLongInclirExarrOrbitLaunrExpeiPlacirOrbitrDoppExarnRangrSolaSun 1

CONTENTS'oice) links. The de-ission of telePhonelink is a nuisance's for distribution toT stations linked tollite broadcast tele-United States, digi-1001 figure), and innming. Video distri-than half of all the

vho, over the Years'ce has been heeded'ManY more worked:ulations are carried

Preface ix

1- Introduction I1.1 Background 11.2 A Brief History of Satellite Communications 31.3 Satellite Communications in 2000 6I 4. Overview of Satellite Communications 151-5 Summary l6Refercnces l6

2. Ortlital Mechanics and Launchers 17Orbital Mechanics 11Developing the Equations of the Orbit l1Kepler's Three Laws of Planetary Motion 22Describing the Orbit of a Satellite 23Locating the Sateilite in the Orbit 25Locating the Satellite with Respect to the Earth 21Orbital Elements 29Exampie 2.1.1 Geostationary Satellite Orbit Radius 29Example 2.l.2Low Earth Orbit 29Example 2.1.3 ElliPtical Orbit 30Look Angle Determination 30The Subsatellite Point 31Elevation Angle Calculation 32Azimuth Angle Calculation 34Specialization to Geostationary Satellites 35Visibil Test 36Example 2.2.1 Geostationary Satellite Look Angles 36OrbitalPerturbations 38Longitudinal Changes: Effects of the Earth's Oblateness 39Inclination Changes: Effects of the Sun and the Moon 40Example 2.3.1 Drift with a Geosttionary Satellite 42

2.4 Orbit Determination 422.5 Launches and Launch Vehicles 43

Expendable Launch Vehicles (ELVs) 44Placing Satellites into Geostationary Orbit 48

2.6 Orbital Effects in Communications Systems Performance 49Doppler Shift 49Example 2.6.1 Doppler Shift for a LEO Satellite 50R.ange Variations 51Solar EcliPse 51Sun Transit Outage 53

2.1

2.2

2.3

Ii!tIItIlt*

xt

XTI SONTENTS

2;7 Summary 54References 54Problems 55

3. Satellites 573.1 Satellite Subsystems 51

Attitude and Orbit Control Systern (AOCS) 57Telemetry, Tracking, Command and Monitoring (TTC&M)Power System 59Communications Subsysterns 59Satellite Antennas 59

3.2 Attitude and Orbit Control System (AOCS) 60Attitude Control System 60Orbit Control System 66

3.3 Telemetry, Tracking, Command, and Monitoring 68Telemetry and Monitoring System 68Tracking 68Command 10

3.4 Power Systems 713.5 Communications Subsysterns 12

Description of the Communications Systeur 12Transponders 75

3.6 Satellite Antennas 80Basic Antenna Types and Relationships 80Example 3.6.1 Global Beam Antenna 82Example 3.6.2 Regional Coverage Antenna 83Satellite Antennas in Practice 83

3.7 Equipment Reliability and Space Qualification 87Space Qualification 87Reliability 88Redundancy 90

3.8 Summary 92References 93Problems 93

4. Satellite Link Design 964.1 Introduction 964.2 Basic Transmission Theory 100

Example 4.2.1 lO4Exanryle 4.2.2 104

4.3 System Noise Temperature and G/T RatioNoise Temperature 105Calculation of System Noise TemperatureExample 4.3.1 110Example 4.3.2 110Noise Figure and Noise Temperature 111Example 4.3.3 112GIT Ratio for Earth Stations 112Example 4.3.4 ll2

59

4.4 DerLinLin

4.5 SarDirEx;

1.6 UplE,I;,

4.7 DeiExaOveupluplDorSystSatf

4.8 SrslSyslKuKu.RairS urr

SystL,

InboMotSateOuttDowOptiLinlRainPathSum

4.9 SumReferencesProblems

105

rc1

5.1

5.2

5. Mod

FreqrWaveBandBaselPre-ePre-eAnalTelelS/NExan

v4.4

4.5

.t*l*l:: ,,,f,fi

OONTENTS Xi

Design of Downlinks ll2Link Budgets I 13Link Budget Example: C-Band Downlink for Earth Coverage Beam 115Satellite Systems Using Small Earth Stations 117Direct Broadcast TV 118Example 4.5.1 123Uplink Design 124Exampie 4.6.1 121Design for Specified C/N: Combining C/N and C/I Values in Satellite Links 121Example 4.7.1 129Overall (C/N)o with Uplink and Downlink Attenuation 129Uplink and Downlink Attenuation in Rain 130Uplink Attenurtion and (C/N),, 130Downlink Attenuation and (C/N)u" 131Systern Design tbr Specihc Performance 131Satellite Communication Link Design Procedure 131System Design Examples 132System Design Example 4.8.1 133Ku Band Uplink Design 133Ku Band Downlink Design 134Rain Effects at Ku Band 135Summary of Ku Band Link Performance 131System Design Example 4.8.2 Personal Communication System Using

Low Earth Orbit Sateilites 131Inbound Link: Mobile Terminal to Gateway Station l4IMobile Terminal ro Satellite Link 142Satellite to Gateway Station Link 143Outbound Link 144Downlik C/N Budget 145Optimizing System Performance L46Link Margins with FEC 147Rain Attenuation at Ku Band 147Path Blockage at L-Band 149Summary of L-band Mobile PCS System Performance 149

46

A1+-t

4.8

4.9 Summary 150References 150Problems 151

5. Modulation and Multiplexing Techniques for Satellite Links 156Frequency Modulation l5lWaveform Equation for FM 158Bandwidth of FM Signals: Carson's Rule 159Baseband S/N Ratio for FM Signals 159Pre-emphasis and de-emphasis 16lPre-emphasis 162Analog FM Transmission by Satellitc l&l*l*ision Siguals io5S/N R.atios for FM Video Transmissior i6iExample 5.2.1 168

5.1

5.2iI

F

tL

w,ii'.::'

t:

5.3

XIV CONTEITTS

FM Threshold 168SCPC FM Links 169Example 5.2.2 L70Data Transmission Using Analog FM Channels 170Example 5.2.3 171Digital Transmission 112Baseband Digital Signals 172Ba-ehand Transrnission of Digital Data 112Band-pass Transmission of Digital Data 119Example 5.3.1 181Exarnple 5.3.2 18lTransmission of QPSK Signals through a Bandlimited ChannelExarnpie 5.3.3 185Example 5.3.3 185Digitai Modulation and Demodulation 187Terminology 187Modulation and Coding 187Bit and Symbol Error Rates 188Binary Phase Shift Keying (BPSK) 189Probability of a Symbol Error 191BPSK Bit Error Rate 194QPSK Bit Error Rate 194Example 5.4.1 195Exarnpie 5.4.2 191Generation of Quadrature Phase Shift Ke-ving (QPSK) SignalsQPSK Variants 199

5.5 Digital Transmission of Analog Signals 201Sampling and Quantizing 2AlNonuniform Quantization: Compression and Expansion 204Signal-to-Noise Ratio in Digital Voice Systems 206Digital Television 208

5.6 Time Division MulPlexing 209TDM Terminology: The U-S. T1 24-Channel System 29other TDM Systems zllChannel Synchronization in TDM 212

5.7 Summary 212References 213Problems 2146. Multiple Access 2216.1 Introduction 22I6.2 Frequency Division Multiple Access (FDMA) 223

Intermodulation 226Intermodulation ExamPle 228Calculation of C/N with Intermodulation Z3OExample 6.2.1 Power Sharing in FDMA 231Example 6.2.2 Channel Capacity with Demand Access FDMA 232Tirne Division Multiple,{"ccess (TDMA) 233Bits, Symbois, and Cbarels 234TDMA Frame Structure 235

ExReUnGuSyT.,ExExSa

6.4 OrBaSa

6.5 DeEx

6.6 Ra6.1 Pa6.8 Cc

SpD:ExExEx

6.9 SuReterencrProblems

7. Er

182

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6.4

6.5

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Example 6.3.1 TDMA in a Fixed Station Network 231Reference Burst and Preamble 238Unique Word 239Guard Times 241Synchronization in TDMA Networks 242Transmitter Power in TDMA Networks 243Example 6.3.2 TDMA in a VSAI Network 244Example 6.3.2 TDMA in a Fixed Earth Station Network 244Satellite Switched TDMA 246Onboard Processing 246Baseband Processing Transponders 241Satellite Switphed TDMA with Onboard Processing 248Demand Access Multiple Access (DAMA) 249Example 6.5.1 FDMA-SCPC-DA 252Random Access 254Packet Radio Systems and Protocols 254Code Division Multiple Access (CDMA) 251Spread Spectrum Transmission and Reception 258DS-SS CDMA CaPacitY 262Example 6.8.1 CDMA in a Fixed Earth Station Network 263Example 6.8.2 CDMA in an LEO Sateilite Netu'ork 263Example 6.8.3 GPS 264

6.9 Summary 266References 261Problems 2617. Error Control for Digital Satellite Links 2737.1 Error Detection and Correction 2137.2 Channel CapacitY 2757 -3 Error Control Coding 271

Example 7.3.1 278, Linear and CYclic Block Codes 279

Golay Codes 2807 -4 Performance of Block Error Correction Codes 2817.5 Convolutional Codes 2827.6 Implementation of Error Detection on Satellite Links 284

Example 7.6.1 2877.7 Concatenated Coding and Interleaving 2887.8 Turbo Codes 2907.9 Summary 292References 292Problems 2938. Propagation Effects and their Impact on satellite-Earth Links 2958.1 Introduction 2978.2 Quantifying Attenuation and Depolarization 298

Example 8.2.1 3018.3 Propagation Effects that Are Not Associated with Hydrometeors 306

Atmospheric AbsorPtion 307Cloud Attenuation 308

8.4

8.5

Xr, GoNTENTS

Tropospheric Scintillaon and [,ow Angle Fading 308Faraday Rotation in the Atrnosphere 310Ionospheric Scintillations 312Rain and Ice Effects 312CharacterizingRain 312Rain Climate Maps 314Rainfall Rate Exceedance Contour Maps I l5Raindrop Distributions 315Prediction of Rain Attenuation 317Example 8.5.1 319Example 8.5.2 323Calculation of Long-Term Statistics for NGSO SystemsScaiing Attenuation with Elevation Angle and FrequencyCosecant Law 325Example 8.5.3 325Squared Frequency Scaling Law 326Example 8.5.4 326ITU-R Long-Term Frequency Scaling of Rain Attenuation 326

8.6 Prediction of XPD 326Canting Angle 328Tilt Angle 328Example 8.6.1 330Example 8.6.2 331Ice Crystal Depolarization 332Rain Effects on Antenna Noise 332Example 8.6.3 333Propagation Impairment Countermeasures 333Attenuation 333Power Control 334Signal Processing 335Diversity 335Depolarizatior 337,

8.8 Summary 338References 339Problems 340

9. VSAT SYSTEMS 3439.1 Introduction 3439.2 Overview of VSAI Systems 3459.3 Network Architectures 347

One-Waylmplementation 347Spt-Two-Way (Split IP) Implementation 347Two-Way Implementation 348

9.4 Access Control Protocols 349Delay Considerations 351

9.5 Basic Techniques 354Multiple Access Selection 355Signal Forrnats 362Modulation, Coding, and Interference Issues 362

9.6 VSAntTra

9.1 Cal9.8 Sys

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TABLE t,[ Othor atellito 8yt13Sy3tem Satellites Type and lifetime Application Orbhs

Global positoningSystem (GPS),operated by U.S.Air ForceUseful web sites:www. navcen. uscg.m ilwww.laafb.af.mil/SMC/CZlhomepage/http://gps.faa.gov/http://www-spacecom.af"mil

Navstar GPS13 through 2122 through 40

43, 44, 45

Design lifetime 7.5 years

Design lifetime 10 yeors

Design lifetime 10 years

All satellites broadcastCDMA signals on twoL-band frequencies

Navigation,early warning

Six orbitalplanes withfour satellitesper plane at20,200 kmaltitude.lnclinationof orbitalplane is 55"

IItial

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low gain antennas, such as those designed for use by mobile users. Consequently. GEOsatellites look set to be the largest revenue earners in space for the foreseeable fulure.Figure 1.1 shows the estirnated growth in revenue from all satellite communicationservices, projected to 2010.

All radio systems require frequency spectrum, and the delivery of high-speed datarequires a wide bandwidth. Satellite communication systems started in C trand. with anallocation of 500 MHz, shared with terrestrial microwave links. As the GEO orbit fi1ledup with satellites operating at C band, satellites were truilt for the next available tiequencyLrand, Ku band. There is a continuing demand for ever more spectrun to allow satellitesto provide new services, with high speed access to the Internet forcing a move to Ka-bandand even higher frequencies. Access to the Internet from small transrnitting Ka-band earthstations located at the home offers a way to bypass the terrestrial telephone network andachieve much higher bit rates. SES began two-way Ka-band Internet access in Europe in1998 with the Astra-K satellite, and the next generation of Ka-band satellites in the UnitedStates will offer similar services.

1 980 1 990 2010Year

FIGUBE 1,1 Growth of worldwide revenue from satellite communications1980 through 2010. Beyond 2000, the curve is a projection.

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f .r OVERIVIEW OF SATEIIJTE OOMI/|U|!|CATIONS 15

Successive World Radio Conferences have allocated oew frequency bands for com-mercial satellite services that now include L, S, C, Ku, K Ka Y and Q bands. Mobile satel-lite systems use vhf, uhl L, and S bands with carrier frequencies from 137 to 2500 MHz,and GEO satellites use frequency bands extending from 3.2 to 50 GHz. Despite the growthof fiber-optic links with very high capacity, the demand for satellite systerns continues to in-crease. Satellites have also become itegrated into complex comrunications achitectures thatuse each element of the network to its best advantage. Examples iue VSAI/WLL 1very srnal1apefture terminalVwireless local loop) in countries where the communications infrastructureis not yet mature and GEO/LMDS (local multipoint distribution s)'stems) tbr the urban fringesof developed nations where the build-out of fiber has yet to be an econotnic proposition.

1.4 OVERVIEW OF SATELLITECOMMUNICATIONS

iequentl)'. GEO)seeable future.communication

high-speed dataI band. rvith aniEO orbit fiUedilable frequencYallow satellites

rove to Ka-bandg Ka-band earth

'ne network and:ss in Europe in:es in the United

Sateliite communication systems exist because the earth is a sphere. Radto r.vaves travelin straight lines at the microwave frequencies used for wideband colnntunications, so arcpeater is needed to convey signals over long distances. Sateilites. because they can linkplaces on the earth that are thousands of miles apart, are a good place to locate a repeater,and a GEO satellite is the best place of all. A repeater is simpll a receiver linked to atransmitter, always using different radio frequencies. that can receive a signal from oneearth station, amplify it, and retransmit it to another earth station. The repeater derit'esits name from nineteenth century telegraph links, which had a maximum length of about50 miles. Telegraph repeater stations were required ei'ery 50 miles in a long-distance linkso that the Morse code signals could be re-sent before they became too weak to read.

The majority of communication satellites are in geostationary earth orbit, at analtitude of 35,786 km. Typical path length from an earth station to a GEO satellite is38,500 km. Radio signals get weaker in proportion to the square of the distance trav-eled, so signals reaching a satellite are always very weak. Similarly, signals received onearth from a satellite 38,500 km away are also very weak, because of limits on theweight of GEO satellites and the electrical power they can generate using solar cells. It

, costs roughly $25,000 per kilogram to get a geostationary satellite in orbit. This obvi-ously places severe restrictions on the size and weight of GEO satellites, since the highcost of building and launching a satellite must be recovered over a 10 to 15 year lifetimeby selling communications capacity.

Satellite communication systems are dominated by the need to receive very weaksignals. In the early days, very large receiving antennas, with diameters up to 30 m, wereneeded to collect sufficient signal power to drive video signals or multiplexed telephonechannels. As satellites have become larger, heavier, and more powerful, smaller earthstation antennas have become feasible, and Direct Broadcast Satellite TV (DBS-TV) re-ceiving systems can use dish antennas as small as 0.5 m in diameter.

Satellite systems operate in the microwave and millimeter wave frequency bands,using frequencies between 1 and 50 GHz. Above 10 GHz, rain causes significant attenu-ation of the signal and the probability that rain will occur in the path between the satelliteand an earth station must be factored into the system design. Above 20 GHz, attenuationin heavy rain (usually associated with thunderstorms) can cause sufficient attenuation thatthe link will fail.

For the rst 20 years of satellite commurications, analog signals were *'ideiy used,with most links employirrg frequency modulation (FM). Wideband FM can operate atIow carrier-to-noise ratios (C/N), in the 5 to 15 dB range, but adds a signal-to-noise

rith signal-to-noiseCio frequencY (RF)satellite links, thatt in signal-to-noise

rne and data trans-rssion over satelliters the major analogg seems destined tollow six TV signalsdual standards Per-n TV (HDTV), will

radio broadcasting,the LEO and MEO: compression tech-stream at 4'8 kbPs.r video comPression; less then 6.2 MbPs.

righ gain fixed anten-:es the caPacitY of thehas been a trend away

using verY large earthery from more Power-:rs using much smaller:llites are used for mo-Lvigation sYstems and,Information SYstems

cations, LEO earth im-:ntial to provide strong

hnology, AerosP ace SourceYork. Vol. 154, No. 3,, Jan. 15, 2001.

.4.::.1:

ORBITAL MECHANICSAND LAUNCHERS

ORBITAL MECHANICS2.1Developing the Equations of the OrbitThis chrrpter is about horv earth orbit is achieved' the laws that cle scribe the rnotion of

an

object orbiting anorher boclv, hor'v satellites mancuver in space, ancl the deternlination ofthe look angle to a .atellite from the earth using epherneris data that describe the

orbital

trajectory ol the satellite.Tc achieve a stable orbit around the ear-th, a spacecraft must first be beyond the bulk

ol the earlh's atmosphere. t.e., in what is popularly called space There are many deflni-

tions of space. U.S. astronauts are awardec ''.heir ''space wings" if they fly at an altitudethat exceeds 50 ntiles 1*80 km), some international treaties hold that the space frontierabove a given counrr\ begins at a height of i00 miles (-160 krn)' Beiou' 100 miles' per-missio, must b. sought t o'er-fly any portion of the country in question On reentry' at-mospheric drag srarts to be felt ai a height of about 400,000 fi(.-16 miles : 122 km).Most satellites, fbr any mission of more than a few months, are placed into orbits of

at

least 250 miles (:-100 km) above the earth' Even at this height' atmospheric drag is sig-nificant. As an exampte, the initial payload elements of the International Space Station

(ISS) were injected into orbit at an aitiiude of 397 km when the shuttle mission left thosemodules on 9 June 1999. By the end of L999, the orbital height had decayed to about360 km, necessitating u

^-Luu", to raise the orbit. Without onboard thrusters and suffi-

cient orbital *ur"uu"riog fuel, the ISS would not last more than a few years at most insuch a low orbit. To apprlciate the basic laws that govern celestial mechanics' we

will be-

gin first with the funamental Newtonian equations that describe the motion of a body'we will then give some coordinate axes within which the orbit of the satellite can be set

and determine the various forces on the earth satellite'Newton,slarvsofmotioncanbeencapsulatedintofourequations:

s:ut+(l)at'1u2:u2+zatt:ulatP: ma

(2.1a)(2.1b)(2.1c)(2.td)

where s is the disrance traveled from time t : 0; u is the initial velocity of the object attime f : 0 and o the final velocity of the object at time ; a is the acceleration of the ob-

".t;ristheforceactingontheobject;andmisthemassoftheobject.Notethattheac-celeration can be positir,"e or negative, depending on the direction it is acting with respect

to tire veiocify vector. Of these four equations, It is the }as[ one thal he]ps us understandthe motion of a sarellite in a stable orbit (neglecting any ,jrag or other penurbing forces)'Putintowords,Eq.(2.1d)Statesthattheforceu"tingonabodyisequaltothemassof

17

fB cHAprER2 oRBITALMECHANICSANDLAUNcHERS

the body multiplied by the resulting acceleration of the body. Alternatively, the resultingacceleration is the ratio of the force acng on the body to the mass of the body. Thus, fora given force, the lighter the mass of the body, the higher the acceleration will be. Whenin a stable orbit, there are two main forces acting on a satellite: a bentrifugal force due tothe kinetic energy of the satellite, which attempts to fling the satellite into a higher orbit,and a centripetal force due to the gravitational attraition of the planet about which thesateilire is orbiting. which artempts to pull the salellite dorvn toward the planet. lf thesenvo forces are equal, the satellite will rernain in a stabie orbit. It will continually fail to-u,ard the planet's surface as it lnoves forwad in its orbit but, by virtue of its orbital ve-locity. it rviil have moved foru,ard just far enough to compensate fbr the "lall" toward theplanet and so it u i11 remain iit the salrie orbital height. This is why an object in a stableorbit is sorltetimes described as being in "free lall." Figure 2.1 shows the two opposingfbrces on a satellite in a stable orbitr.

Force -

mass X acceleration and the unit o[ force is a Newton, ,,vith the notationN. A Ner.,,ton is the force required to accclerate a mass of I kg with an acceleration ofI n/sr. The unclerly,ing units of a Newton are thcrefore (kg) x m/sr. ln hnperial Units,one Neu,ton - 0.1248 ft lb. The standard acceleration due to gravity at the earth's sur-face is 9.80665 x l0'r kn,/sr, which is often quoted as 98i cnr./s2. This value decreases

The satellite has a mass, mand s traveling with velocity,4 in the plane of the orbt

FTGURE 2.1 Forces acting on a satellite in a stable orbit around the eah (from Fig. 3.4 ofreference 1). Gravitational force is inversely proportional to the sguare of the distance betweenthe centers of gravity of the satellite and the planet the satellte is orbiting, in this case theearth. The gravitational force inward (EN, the centripetal force) is directed toward the center ofgravity of the earth. The kinetc energy of the satellite (Four, the centrifugal force) is directeddianretrically opposite tc the gravtationai force. Knetic energy is proportional to the square r:fthe veiocity of the satelite. When these inward and outward forces are balanced. the satellitemoves around the earth in a "free fall" trajectory: the satellite's orbit. For a description of theunits, please see the text.

wifrt

In

wh

Ifr

wiof

icl0for

her

If{-\-disorb

Gir

occ

TAIofl

Sat,

I nteNevskvlridi

Meaeartl

ely, e resulnge body. Thus, forrn will be. Whenugal force due toto a higher orbit,about which thee planet. If theserntinually fall to-of its orbital ve-"fail" toward thecbject in a stable.he two opposing

with the notationn acceleration ofr Imperial Units,.t the eanh's sur-i vaiue decreases

rom Fig. 3.4 ofdistance betweenthis case theard the center ofrce) is directedI to the square of:ed, the satellitelscription of the

2.toRBrrALMEcHANtcs 19

with height above the earth's surface. The acceleration, a, due to gravity at a distance rfrom the center of the earth isl

o : tf 12 kmls2 (2.1)

where the constant ,r is the product of the universal gravitational constant G and the massof the earth Mu.

The producf GMu.is called Kepler's constant and has the value 3.9860044i8 X105 kmr/sr. The universal gravitational constant is G : 6.612 X 10 Ir Nmr/kg) ot 6.612 Xl0 r0 krr/kg sl in lhe older units. Since force : mass X acceleration, the centripetalforce acting on the satellite, F, is given by

Fw:mx(t/r-): m x (GM6/r2)

In a similar fashion. the centrifugal acceleration is given byr

o : ut/,which will give the centrifugal force, F61, as

Four: mx(/r) (2'4)If the forces on the satellite are balanced, FrN : F61 and, using Eqs. (2.2a) and (2.4)'

mxp./r2:mxul/rhence the velocity ' of a satellite in a circular orbit is given by

, : Q"/r)rPIf the orbit is circular, the distance traveled by a satellite in one orbit around a planet is2zr, where r is the radius of the orbit from the satellite to the center of the planet. Sincedistance divided by velocity equals time to travel that distance, the period of the satellite'sorbit, Z, will be

7: {2rr)/o : (nr)/l(p./r)1a)T: (Zrr3P)/(t"rP)

Table 2.1 gives the velocity, u, and orbital perio4 7n, for four satelte systems thatoccupy typical LEO, MEO, and GEO orbits around the earth. In each case, the orbits are

TABLE 2.1 Orbital Velocity, Height, and Periodof Four Satellite Systems

Orbital height Orbital velocity Orbital periodSatellite system 1*ml lkm/sl lh min sl

Giving

(2.2a)(2.2b;

(2.31

() 5\

(2.6)

lntelsat (GEO)New-lCO (MEO)Skybridse (LEO)iridium {LEO}

35,786.0310,2551,46978

3.O7 474.89547.12721.4824

23 56555155i40

4.1

48.411.827.4

Mean earth radius is 6378.137 km and GEO radius from the center of theearth is 42,164.17 km.

20 cHAprER 2 oRBrrAL MEcHANtcs Ar{D LAIJfnERs

Satellite

FIGUBE 2.2 The inital coordinatesystem that could be used to de-scribe the rplationship between theearth and a satellite. A Cartesian

.

coordinate system with the geo-graphical axes of the earth as the

'principal axes s the simplest coor-dinate system to set up. The rota-tional axis of the earth is about theaxis c where c is the center of theearth and cz passes through thegeographic north pole. Axes cx, cand cz are mutually orthogonalaxes, with cx and cy passingthrough the earth's geographicequator. The vector r locates themoving satellte with respect to thecenter of the earth.

circular and the average radius of the earth is aken as 6378.137 kmr. A number of coor-dinate systems and reference planes can be used to describe the orbit of a satellite arounda planet. Figure 2.2 illustrates one of these usin-e a cartesian coordinate s),stem * ith theearth at the center and the reference planes coincidine with the equator and the pol;lr a-ris.This is referred to as a geocentric coordinate s),stem.

with the coordinate system set up as in Figure 2.2, and with the satellite mass ,??located at a vector distance r from the center of the earth, the gravitational force F on thesatellite is given by

(

beax

ere

EqraC

GMntiF:- IWhere M6 is the rnass of the earth and G : 6-672 xaceeleration and EC. (2-7) can be wrien as

,1-cl'rF:m- dt'

i d2l- 1l:

-r' d-

d'r r-

*;r. : odt' r'

(2.1)

lO-tr Nm?kg2. But force : mass x

(2.8)

(2.e)

(2.10)

and

Which yields

From Eqs. (2.7) and (2.8) we have

This is a second-order linear differential equation and its solution will involve sixundetermined constants called the orbital elements. The orbit descritred by these orbitalelements can be shown to lie in a plane and to have a constant angular momentum. Thesolution to Eq. (2.10) is difficult since the second derivative ofr involves the second de-rivative of e unit vector r. To remove this dependence, a different set of coordinates can zo

, nitial coordinatebe used to de-ship between theie. A Cartesianwith the geo-

.he earth as there simplest coor-;et up. The rota-)arth is about thethe center of the

ls through thepole. Axes cx, cY,ly orthogonalcy passings geographic)r r locates the,ith respect to thel.

number of coor-a satellite aroundI system with thend the polar axis.

: satellite mass ,?xral force F on the

(2.1)

t force : mass X

(2.8)

(2.e)

(2.10)

n will involve sixd by these orbital'momentum. The'es the second de-rf coordinates can

d'ro ( do\dr-"\a,):

2.1 oRBtrAL MEcHAflrcs 21

FIGURE 2.3 The orbital plane coor-dinate system. ln this coordinte sys-tem, the orbital plane of the satelliteis used as the reference plane. Theorthogonal axes xo and yo lie in the

Yo orbital plane. The third axis. zr, isperpendicular to the orbital plane.The geographical z-axis of the earth(which passes through the true i'lorthPole and the center of the earth, c)does not lie in the same direction asthe z6 axis except for satellite orbitsthat are exactly in the plane of :hegeographical equator.

be chosen to descibe the location of the satellite such that the unit vectors in the threeaxes are constant. This coordinate system uses the plane of the satellite's orbit as the ref-erence plane. This is shown in Figure 2.3.

Expressing Eq. (2.10) in terms of the new coordinate oXes -t, -y,,, and give::

r(roio + loin) : 0 ,2.1 1)(xfr + v:z:Equation (2.1 1) is easier to solve if it is expressed in a pglar coordinate system rather thana Cartesian coordinate system. The polar coordinate system is shown in Figure 2.'1.

With the polar coordinate system shown in Figure 2.4 and using the transformations(2.12a)(2.r2b)(2.12c){2-t2d)

and equating the vector components of r and @p in turn in Eq. (2.11) yields

,.(?j) ,.(i;t)-

x6 : 16 cos @sy6 : r sin @e

,i.in : io cos@e

- {e sin@69o : ocos {o f i6 sin {6

Q.r3)

FIGURE 2.4 Polar coordinate system in the planeof the satellite's orbit. The plane of the orbit coin-cides with the plane of the paper. The axis z6 isstraight out of the paper from the center of theearth, and is normal to the ptane of the satellite'sorbit. The satellite's position is described n termsof the radius from the center of the earth ro and the

x" angle this radius makes with the x6 axis, $6.

_t,r2g

22 cHAprER2 oRBTTALMEcHANTcsANDLAUNcHERs

and

,,(#).,(*)(#):, Q'4,Using standard rnathematical procedures, we can develop an equation for the radius

of the satellite's orbit. ru, namelynt'

I -ecos(f6 0o) (2. I 5)

Where 0u is a constant and is the eccentricity ol an ellipse whose semilatus rectum p isgiven by

, : qtt))/u (1.16)and /i is magnitude of the orbital angular momentum of the satellite. That the equation ofthe orbit is an ellipse is Kepler's first larv oi planetary n.lotion.

Kepler's Three Laws of Planetary MotonJohannes Kepler (1571-1630) was a German astronomer and scientist who developed histhree laws of planetary motion by careful observations of the behavior of the planets inthe solar s)'stem over many years, with help from some detailed planetary observationsbr the Hungarian astronorner Tycho Brahe. Kepier's three laws are

1. The orbit of any srnaller bocly about a larqer body is always an ellipse, with the cen-ter of mass of the larger body as one of the trvo fbci.

2. The orbit of the smaller body sweeps out equal areas in equal time (see Figure 2.5..

FIGURE 2.5 lllustration of Kepler's second faw of planetary motion. A satellite is in orbitabout the planet earth, E. The orbit is an ellipse with a relatively high eccentricity, that is,it is far from being circular. The figure shows two shaded portons of the elliptical plane inwhich the orbit moves, one is close to the earth and encloses the perigee while the otheris far from the earth and encloses the apogee. The perigee is the point of closest ap-proach to the earth while the apogee is the point in the orbit that is furthest from theearth. While close to perigee, the satellite moves in the orbit between times t, and f, andsweeps out an area denoted by A.r. While close to apogee, the satellite moves in the orbitbetween times f3 and to and sweeps out an area denoted by A.o. lf \ tz = f3 - f1 thenArt = Azr.

Kepler's lawryears later, bematical morton was one (ential calc!lltlwas able romathematical

De

De

Theaxe

anddi sc

Theof tl

andandchostherr

time

Remthe ris Ktthe peric

24 cHAprER 2 oRBtrAL MEcHANrcs AND LAUNcHERs

Apogee

r+a(1 +e) a(1-e) .ilttlttFIGURE 2.6 The orbit as it appears in the orbital plane. The point O is the center ofthe earth and the point C is the center of the ellipse. The two centers do not coincideunless the eccentricity, e, of the ellipse is zero (i.e., the ellipse becomes a circle anda : b). The dimensions a and b are the semimajor and semiminor axes of the orbitalellipse, respectively.

This equation is the mathematical expression of Kepler's third law of planetary mo-tion: the square of the period of revolution is proportional to the cube of the semimajoraxis. (Note that this is the square of Eq. (2.6) and that in Eq. (2.6) the orbit was assumedto be circular such, that semimajor axis a : semiminor axis : circular orbit radiusfrom the center of the earth r.) Kepler's third law extends the result from Eq. (2.6), whichwas derived for a circular orbit, to the more general case of an elliptical orbit. Equa-tion (2.21) is extremely important in satellite communications systems. This equationdetermines the period of the orbit of any satellite, and it is used in every GPS receiverin the calculation of the positions of GPS satellites. Equaon (2.21) is also used to findthe orbital radius of a GEO satellite, for which the period Imust be made exactly equalto the period of one revoluon of the earth for the satellite to remain stationary over apoint on the equator.

An important point to remember is that the period of revoluon, I, is referenced toinefial space, namely, to the galactic background. The orbital period is the time the or-biting body takes to retum to the same reference point in space with respect to the galac-tic background. Nearly always, the primary body will also be rotating and so the periodof revolution of the satellite may be different from that perceived by an observer who isstanding still on the surface of the primary body. This is most obvious with a geostation-ary earth orbit (GEO) satellite (see Table 2.1). The orbital period of a GEO satellite is ex-actiy equal to the period of rotation of the earth,23h56 min4.l s, but, to ax observeron the ground, the satellite appears to have an infinite orbital period: it always stays inthe same place in the sky-

(a)cortorsat(rec{not,toa_-1 ICTOS

mea

I onga[ o]the ::i der

LocConsrnay

The alv. Lfromterm ritive .rto theare gi

olutioris thus

If the the sat,cle of ta constIas th

Cthe poirintersec(A) n:al

rrgee

the center ofnot coincide

ir circle andof the orbital

v of planetary mo-: of the semimajororbit was assumedrcular orbit radiusm Eq. (2.6), whichttical orbit. Equa'ms. This equationvery GPS receivers also used to findnade exactly equalr stationary over a

Z, is referenced tois the time the or-)spect to the galac-I and so the periodrn observer who iswith a geostation-

IEO satellite is ex-but, t0 an observeit aiways stays in

2.1 oRBnAL MEcHANtcs Zs

To be perfectly geostationary, the orbit of a satellite needs to have three features:(a) it must be exactly cicular (i.e., have an eccentricity of zero); (b) it must be at thecorrect altitude (i.e., have the correct period); and (c) it must be in the plane of the equa-tor (i.e., have a zero inclination with respect to the equator)" If the inclination of thesatellite is not zero and/or if the eccentricity is not zero, but the orbital period is cor-rect. then the satellite will be in a geosynchronous orbit. The position of a geosynchro-nous satellie .ill appear to oscillate about a mean look angle in the sky with respectto a stationary bserver on the earth's surface. The orbital period of a GEO satellite,23 h 56 rnin 4.1 s, is one sidereal day. A sidereal day is the time between consecutivecrossings of any particular longitude on the earth by any star, other than the sunr' The,r,..n ,olr, day of 24 h is the time between any consecutive crossings of any particularIongitude by the sun, and is the time between successive sunrises (or sunsets) observedat one location on earth, averaged over an entire year. Because the earth movcs roundrhe sun once per 365 z days, the solar day is 1440/365.25 : 3.94 min longer than asidereal day.

Locating the Satellite in the OrbitConsider now rhe problern of locating the sateliite in its orbit. The equation of the orbitmay be rewritten b1'combining Eqs. (2. 15) and (2'18) to obtain

a(t -

d)'o-t+".oso

The angle $o (see Figure 2.6) is measured from the xs axis and is called the true anont-aly. f,Anomall, was a measure used by astronomers to mean a planet's angular distancefrm its perihelion (closest approach to the sun), measured as if viewed from the sun' Thetenn wai adopted in celestial mechanics for all orbiting bodies.l Since we dehned the pos-itive -r:s axis so that it passes through the perigee, 4| measures the angle from the perigeeto the instantaneous position of the satellite. The rectangular coordinates of the satelliteare given by

t) ))

(2.23)(2.24)

xs : 16 cos @sy6 : re sin {6

As noted earlier, the orbital period Z is the me for the satellite to complete a rev-olution in inertial space, traveling a total of 2t tadians. The average angular velocity 4is thus

, : (Zr)/T : 1tl /2)/(a3/2) (2.25)

If the orbit is an ellipse, the instantaneous angular velocity will vary with the position ofthe satellite around the orbit. If we enclose the elliptical orbit with a circumscribed cir-c/e of radius rz (see Figure 2.1), thert an object going around the circumscribed circle witha constant angular velocity 4 would complete one revolution in exactly the same periodI as the satellite requires to complete one (elliptical) orbital revolution.

Consider the geometry of the circumscribed circle as shown in Figure 2.7. Locatethe point (indicated as A) where a vertical line drarvlt throLlgh the posliion of the satellieinteisects the circumscribed circle. A line fron the center of the eilipse (C) to this point(A) makes an angle E with the x6 axis; E is called the eccentric anomaly of the satellite'

yoax|s

Crrcumscribed Circle

i;26 cHAprERz oRBITALMEcHANEsANoLAUNcHERs

xo axis

FIGURE 2.7 lhe circumscribed circle and the eccentric anomaly E. Point O is the cente ofthe earth and point C is both the center of the orbital ellipse and the center of the circum-scribed circle. The satellte location in the orbital plane coordinate system is specified bv,x,,y). A vertical line through the satellite ntersects the circumscribed circle at point A. Theeccentric anomaly E is the angle from the x0 axs to the line joining C and A.

lt is related to the radius I by

No

l-otoAtthecatwhWefroltorlzatheThibitlhefirsrequ

caliThus

We can also develop an expression that relates eccentric anomaly E to the averageangular velocity 4, which yields

Tdt:(l-rcosD)dE (2.28)Let ro be the time of perigee. This is simultaneously the time of closest approach to theearth; the time when the satellite is crossing the "16 axis; and the time when E is zero. Ifwe integrate both sides of Eq. (2.28), we obtain

rs:a(L-ecosE)a

- fo: ae cosg

(2.26)

(2.27)

(2.2e)

(2.30)

rt(t -

tp): E -

esinE

The left side of Eq. (2.29) is called the mean anomaly, M. ThusM:rtQ-lo):E-esinE

The mean anomaly M is e arc length (in radians) that the satellite would have traversedsince the perigee passage if it were moving on the circumscribed circle at the mean an-gular velocity 4.

trf we know the time of perigee, fo, the eccentricity, e, and the length of the serni-major axis, , we now have e necessary equations to deterrnine the coordinates (ro, $o)

rt O is the center ofer of the circum-r is specified bY (x6,at point A. The14.

(2.26)

(2.27)tiy E to the average

(2.28);est approach to the: when E is zero. If

(2.2e)

(2.30)ould have traversed:cie at the mean an-

lengih of the semi-coordinates (ro, o)

2.1 oRBfrAL mEcHANtcs 27

and (xs, yo) of the satellite in the orbital plane. The process is as follows

1. Calculate 4 using Eq. (2.25).2. Calculate M using Eq. (2.30).3. Solve Eq. (2.30) for E.4. Find r,, from E using Eq'. (2-21).5. Solve Eq. Q.22) for @s. '6. Use Eqs. (2.23) and (2.24) to calculate -r"6 and 'r'n'

Now we must locate the orbital plane with respect to the earth.

Locating the Satellite with Respectto the EarthAt the encl of the last section, we summarized the process for locating the satellite atthe point (-r..

-),,, .:r) in the rectangular coordinate svstem of the orbitai plane. The lo-cation was with respect to the center of the earth. In most cases, we need to knou'where the satellite is from an observation point that is not at the center of the earth.We will therefore develop the transformations thet permit the satellite to be locatedfrom a point on the rotating surface of the earth. \\ rvill begin with a geocentric eqtru'torial coordinate s)-stem as shown in Figure 2.8. The rotational axis of the earth is thez axis, which is through the geographic North Pole. The ,r axis is from the center ofthe earth roward a fixed location in space called the.first point of Aries (see Figure 2.8,).This coordinate system moves through space: it translates as the earth moves in its or-bit around the sun, but it does not rotate as the eanh rotates. The "t direction is alwaysthe same, whatever the earth's position around the sun and is in the direction of thefirst point of Aries. The (x, y,) plane contains the earth's equator and is called theequatorial plane.

Angular distance measured eastward in the equatorial plane from the x axis iscalled right ascension and given the symbol RA. The two points at which the orbit

FIGURE 2.8 The geocentricequatorial system. This geocentricsystem differs from that shown inFigure 2.1 only in that the xi axispoints to the first point of Aries.The first point of Aries s the di-rection of a line from the centerof the earth through the center ofthe sun at the vernal equinox(about March 21 in the NorthernHemisphere), the instant when

,, the subsolar Pont crosses the'' equator from south to north. ln

the above system, an object maYbe located by its right ascensionfrA and its declination .

2a CHAPTER 2 ORBITAL MECHANICS AND LAUNCHEBS

penetrates the equatorial plane are called nodes; the satellite moves upward throughthe equatorial plane at fhe ascending node and downward through the equatirrialplane at the descending node, given the conventional picture of the earth, with northat the top, which is in the direction of the positive axis for the earth centered coor-dinate set. Remetnber that in space there is no rrp or down; that is a concept we arefamiliar with because of gravity at the earth's surface. For a weightless body in space,such as an orbiting spacecraft, up and do*n have no meaning unless they are definedwith respect to a reference poinl. The right ascension oJ the ascendirtg ttode is calledf). The angle that the orbital plane makes rvith the equatorial plane (the planes inter-sect at the line joining the nodes) is called the inclincuion, i- Figure 2.9 illustrates thesequantities.

The variables l) and i together locate the orbital plane with respect to the equato-rial plane. To locate thc orbital coordinate system with respect to the equatorial coordi-nate system we need a, the urgunrent o.f'perigee v'est. This is the angle measured alongthe orbit fiom the ascending node to thc pcligee.

Standard time for space opcrations rLntl niost other scientific and engineering pur-poses is univer.ral tintc (UT, also knowr as;.ulu tine (z). This is essentially the meansolar time at the Gree nwich Observatory near London. England. Universal time is meas-ured in hours, minutes, and seconds or in fractions of a day. It is 5 h later than EasternStandard Time, so that 07:00 EST is l2:00:00 h UT. The civil or calendar day beginsat 00:00:00 hours UT, frequently written as 0 h. This is. of course, midnight (24:00:00)on the previous day. Astronomers employ a second dating system involving Julian day,tand Jttlian dates. Iulian days start at noon UT in a counting system whereby noon onDecember 3 I , I 899, was the beginning of J ulian day 24 15020, usualiy written 241 5020.These are extensively tabulated in reference 2 and additional information is in reference14. As an example, noon on December 31, 2000, the eve of the twenty-first century, isthe start of Julian day 245 1909. Julian dates can be used to indicate time by append-ing a decimal fraction; 00:00:00 h IJT on January l, 2001-zero hour, minute, and

xi

FIGURE 2.9 Locating the orbit in the geocentric equatorial system. The satellite penetratesthe equatorial plane (while moving in the positive z drection) at the ascending node" Theright ascension of the ascending node is O and the inclination i is the angle between theequatorial plane and the orbitai plane. Angle o, measured in the orbtai piane, locates theperigee with respect to the equatorial plane.

se

e)rh

0'l-"

k:ticar(orlhaityinrtin

E)Throfr

A

D--

For

Thir

EX

The250mos

perirvelo

An662t

Henr

2.1 ORBTAL MECHANCS

second for the third milleniurn 6.p--is given by Julian date 245 1909.5. To find theexact position of an orbiting satellite at a given instant in time requires knowledge ofthe orbital elements.

Orbital ElementsTo specify the absolute (i.e., the inertial) coordinates of a satellite at time I, we need toknow six quantities. (This was evident earlier when we determined that a satellite's equa-tion of motion was a second order vector linear differential equation-) These quantitiesare called the orbital elements. More than six quantities can be used to describe a uniqueorbital path and there is some arbitrariness in exactly which six quantities are used. Wehave chosen to adopt a set that is commonly used in satellite communications: eccentric-iry (r), semirnajor axis (r), time of perigee (1p). right ascension of ascending node (O),inclination (i), and argutnent of rerigee (.r). Frequently, the mean anomaly (M) at a gir enlinie is substituted fbr n.

EXAMPLE 2.1.1 Geostationary Satellite Orbit RadiusThe carth rorafes once per sidereal day of 13 h -56 min -1.09 s. Use Eq. {?.21) to show that the radiusof the GEO is 42,164.1 7 km as given in Table 2.l.

Answer Equation (2.21t gives the square of the orbital period in secondsT) : (4:t)/tL

Rearranging the equation, the orbital radius a is given by'

a1 : T2p,/(4n2)For one sidereal day, T : 86,164.09 s. Hence

dr : (86,164.1.r x 3.9a600n418 x 105/(4172):7.4920251 x 101r km3a : 42,164.17 km

This is the orbital radius for a geostationary satellite, as given in Table 2.1- f

EXAMPLE 2.1-2 Low Earth OrbitThe Space Shuttle is an example of a low earth orbit satellite. Sometimes, it orbits at an altude of250 km above the earth's surface, where there is still a finite number of molecules from the at-mosphere. The mean earth's radius is approximately 6378.14 km. Using these figures, calculate theperiod of the shuttle orbit when the altitude is 250 km and the orbit is cicular. Find also the linearveloc of the shuttle along its orbit.

Answer The radius of the 250-krn altitude Space Shuttle orbit is (r" + h) = 6379.14 + 250.0 :6628.14 km

From Eq. 2.21, the period of the orbit is I where7z : (4tr1a3)/p:4r2 x (662s.A)3/3.986004418 x 10-5 s2

: 2.88401145 x 107 s2

Flence the period of the orbit is

f : 5370.30 s : 89 min 30.3 s"

29

r upward throughh the equatorialearth, with northth centered coor-a concePt we aress bodv in sPace.; they are definedrtyi nttde is called(the planes inter-9 illustrates these

rect to the equ1to-equatorial coordi-le rneasured alons

I en-uineering Pur-ientiali)' the lllean'rsal time is I'neas-later than Easierl.trendar daY beginsidnight (2'1:00:00 ):r:ing Jultur dat su,hereby noon on'nvritten 2-1 I 5020.ion is in referencety-first centurY, isr time bY aPPend-hour. minute, and

satellite penetratesrding noie. Theile between theane, locates the

3O.. cHAprER 2 oRBfrAL MEcHANlcs AND T.AUNcHERs

This orbit period is about as small as possible. At a lower altitude, friction with the earth's atros-phere will quickly sl,ow the Shuttle down and it will retum to earth. Thus, all spacecraft in'stableearth orbit have orbital periods exceeding 89 min 30 s.

The circumference of the orbit is 2a' : 41,645.83 km.Hence the velocity of the Shuttle in orbit is

2ra/T -

41,645.83/5370.13 = 7.755 kn/s

Alternatively.youcoulclusc Eq. {1.5): r - ttL,t t)".Thetermr - 3.98600-1418 x l05kmr/srandthe ternr r : (6378.14 + 250.0) km, yielding r - 7.7-55 kn/s.Note: If ,r, and r had bce n quoter-i in units of m'/sr and m, respectively, the answer r,ould havc bcenin meters/second. Be sure to keep the units thc same during a calculation procedure.

A velocity of about 7.8 knlis is a typical velocity tbr a low ear-th orbit satelltte. As the altitude of a satellite increases, its rclocity becomcs smaller. I

EXAMPLE 2.1.3 Elliptical orbit

Frequencirregisteredtration BoGeneva. Tor contFaamade to liltion and uUnited Stprove theIFRB. The

A satellite is in an elliptical orbit rvithmean eafih radiu.s of 178. l4 km. llndthe eccentricity of the orbit.

a perigee ol' 1000 km and an aposee of 4000 km. Using athe penod of the orbit in hours. urinutes. and seconcls. rnd n

rlbG-

r1

Tti-:

t(Po

rA1i

rttt

Answer The major axis of the elliptical orbit is a straight line betw,een the apogee and perigee.as seen in Figure 2.7. Hence. fbr a semimajor axis length r, eah radius r-. perigee height /rn. andapogee height ft,,

2a : 2r" * hp t lt., : 2 x 6318 11 + 1000.0 + 4000.0 : 17.756.28 kniThus the sernimajor axis of the orbit has a length r : 8878. i4 km. Using this vaiue ol o in Eq. (2.21 Igives an orbital period f seconds where

7t : (4r1at)/*: 47 x (98i8.07)38.986004418 x 105 sr: 6.930872802 X 107 sl

f : 8325.1864s : 138min45.19s : 2h 18min45.19 sThe eccentricity of the orbit is given by e, which can be found from Eq. (.2.27 ) by consider-

ing the instant at which the satellite is at perigee. Referring to Figure 2.7, when the satellite is atperigee, the eccentric anomaly E : 0 and ro : re + o. From F4. Q.27), at perigee

rt): a(l -

e cosE) and cosE : 1Hence

r. * hr: a(l -

e)e:1-(r"+hr)la:1-'7,378.14/8878.14:0.169 I

2.2 LOOK ANGLE DETERMINATIONNavigation around the earth's oceans became more precise when the surface of the globewas divided up into a gridlike structure of orthogonal lines: latirude and longitude. Lat-itude is the angular distance, measured in degrees, north or south of the equator andlongitude is the angular distance, measured in degrees, from a given referencelongitudinal line. At the tire that this

.srid reference became popular. there were tu,orrlajor sea{'aring nations 1,ying for dcnir:;nce: England and France. Engla*d d;'ew its ref-ererlce zero longitude through Greenw,ich, a town close to f-ondon, England, and France,

ca!am

piulF]lop(

TTIrhlirge

22 LooK ANcrE DETERMNAoN ,i;i:.

h the earth's atmos-spacecraft in stable

i8 x 10s kmr/sr and

ver would have beenedure.;ateliite. As the alti-

I

Frequencies and orbital slots for new satellites areregistered with the International Frequency Regis-tration Board (IFRB), part of the ITU located inGeneva. The initial application by an orgaizationor company that wants to orbit a new satollite ismade to fhe nationil body that controls the allcca-tion and use of radio frequencies-the FCC in theUnited States, for example-which must first ap-prove the application and then forward it to theIFRB. The first organization to frle with the IFRB

for a particular service is deemed to have protectionfr

Local vertical

Projection ofpath onto iocalhorizontal plane

32 cHAprER2 oRBTTALMEcHANtcsANDLAUNcHERs

East

FIGURE 2.lO The definition of elevation lEl) a:d azimuth (Azl.fhe elevation angle ismeasured upward from the local horizontal at th? earth station and the azimuth angle ismeasured from true north in an eastward directior to the projection of the satellite pathonto the local horizontal plane.

the earth, it is usual to give their orbital locarion in terms of thei subsatellite point. Asnoted in the example given earlier, the Inte1.;t prin'rary satellite in the Atlantic OceanRegion (AOR) is at 335.5' E longitude. Operarors of international geostationary sateilitesystems that have satellites in all three ocean regions (Atlantic, Indian, and Pacihc) tendto use longitude east to describe the subsatellite points to avoid confusion betrveen usingboth east and west longitude descriptors. For L.S. geostationary satellite operators, all ofthe satellites are located west of the Greenu,ich meridian and so it has become acceptedpractice for regional systems over the United States to describe eir geostationary sateliitelocations in terms of degrees W.

To an observer of a satellite standing at the subsatellite point, the satellite will ap-pear to be directlj, overhead, in the zenth diection from the observing location. Thezenith and nadir paths are therefore in opposite directions along the same path (seeFigure 2. 1 1). Designers of satellite antennas ret'erence the pointing direction of the satel-lite's antenna beams to the nadir direction. The communications coverage region on theearth from a satellite is defined by angles measured from nadir at the satellite to theedges of the coverage. Earth station antenna designers, however, do not reference theirpointing direction to zenith. As noted earlier they use the local horizontal plane at theearth station to define elevation angle and geographical compass points to define az-imuth angle, thus giving the two look angles for the earth station antenna toward thesatellite (Az, El).

Elevation Angle GalculationFigure 2.12 shows the geometry of the elevarion angle calculation. In Figure 2"12, r,is the vector from the center of the earth to the satellite; r" is the vector from the cen-ter of the earth to the earth stafion; and d i-i the vector from the eatil station to thesatellite. These three vectors lie in rhe sanie plane and form a triangle. The centralangle 7 measured between r" and r, is the angle between the earth station and the

Ii

Fr(cerpoipoitio rbeasubtwonun

thos

%

n angle isth angle stellite path

ellite point. As,\tlantic Ocean:ionary satelliterd Pacific) tendbetween using

rperators, all ofrcome acceptedtionary satellite

rtellite will ap-1 location. Therame path (seeon of the satel-o region on thesatellite to thereference their,al plane at thes to define az-rna toward the

Figure 2.12, r,'from the cen-station to the

e. The centraltation and the

22 LOOI( ANGLE DETERMINATION 33

FIGURE 2.f 1 Zenith and nadir pointing directions. The line loining the satellite and tlecenter of the earth, C, passes through the surface of the earth at point Sub, the subsatellitepoint. The satellte is directly overhead at this pont and so an observer at the subsateltepoint would see the saellite at zenth (i.e., at an elevation angle of 90"). The pointing direc-tion from the satellte 10 the subsatellite point is the nadir direction from the satellite. lf thebeam from the satellte anfenna is to be pointed at a location on the earth that s not at thesubsatellite point, the pointing dlrection is defined by the angle away from nadir. ln general,two off-nadir angles are given: the number of degrees north (or south) from nadir; and thenumber of degrees east (or west) from nadir. East, west, north, and south directions arethose defined by the geography of the earth.

FIGURE 2.12 The geometryof elevation angle calculation.The plane of the paper is theplane defined by the center ofthe earth. the satellite, and theearth station. The central angleis y. The elevation angle E/ ismeasured upward from thelocal horzontal at the earthstation.

rir

Satellite

d

Loc

a

13

horizontal

rs

(Earth sta tion

'\v -\\

-.,,\r"-.oCenterof earth

34 cHAprER 2 oRBrrAL MEcHANtcs AND LAUNcITEBs

satellite, and ry' is fhe angle (within the triangle) measured from r" to d. Defined so thatit is nonnegative, 7 is related to the earth station north latitude I" (i.e., L" is the num-ber of degrees in latitude that the earth station is north from the equator) and west lon-gitude l" (i.e., /" is the number of degrees in longitude that the earth station is westfrom the Greenwich meridian) and the subsatellite point at north latitude l, and westlongitude 1* by

cos (y) : cos (L") cos (1,) cos (/. -

1") + sin (L.) sin (L,)The law of cosines allows us to relate the rnagnitudes of the vectors joining the cen-

ter of the earth, the satellite. and the earth station. Thus

r ) 1)r

Since the local horizontal plane lt the earth station isairgle E1 is related to the central anrle g by

El -r!-90'

Bv the law of sines we have

perpendicular to re, the elevation

( ].JJ,)

d : ..1' .' (;)' -,(r).".r,r]'"

tl.-11 i

(1.3-+)

(2.3s)

/!

sin(r/)Combining the last three equations yields

cosll) : trII(7)d

sin(7)

sin(7), . (;)'- ,(t)"",1,,r]'"

Equations (2.35) and (2.31) permit the elevation angle El to be calculated from knowl-edge of the subsatellite point and the earth station coordinates, the orbital radius r,, ande earth's radius r.. An accurate value for the average earth radius is 6378.137 kml buta common value used il approximate determinations is 6370 krn.

Azimuth Angle GalculationBecause the earth station, the center of the earth, the satellite, and the subsatellite pointall lie in the same plane, the azimuth angle Az from the earth station to the satellite is thesame as the azimuth from the earth station to the subsatellite point. This is more difficultto compute than the elevation angle because the exact geometry involved depends onwhether the subsatellite point is east or west of the earth station, and in which of the hemi-spheres the earth station and the subsatellite point are located. The problem simplifiessomewhat for geosynchronous satellites, which wiil be treated in the next section. For thegeneral case, in particular for constellations of LEO satellites, the tedium of calculatingthe individual look angles on a second-by-second basis has been considerably eased b,v arange of commercial software packages that exist for predicting a variety of orbital

A popularlaunch servalvtical Crrprogram insubseries. r,g hen that :rcal orbit lol

Suthee1e

Fordiu

me(azit(bar

Ca-s

Cas

drpr

S

Foan

Eq

Defined so that, l" is the num-r) and west lon-station is west

rde L, and west

tl.3l.ljoining the cen-

12.32)

r., the eletation

t2.33)

(2.34)

(2.3s)

arcd from knowl-ital radius rs, and;378.L37 kmr but

subsatellite Pointthe satellite is the.s is more diffrcultolved depends onvhich of the hemi-rroblem simPlihesxt section. For theium of calculatinglerably eased bY avariety of orbital

2 L)r ANGI.E DEIEEiIuNAnoN 35

A popular suite of software employed by manylaunch service contractors is that developed by An-alytical Graphics: fhe Satetlite Tool Kir1. The coreprogram in early 2001. STK 4.0. and the subsequentsubseries, was used by Hughes to rescue AsiaSat3u hen that satellite was stranded in a highly ellipti-cal orbit following the failure of an upper stage in

the launch vehicle. Hughes used two lunar flybys toprovide the necessary additional velocity to circu-larize the orbit at geostationary altitude. A numberof organizations offer web sites that provide orbitalplots in a three-dimensional graphical format withrapid updates for a variety of satellites (e.g., theNASA sitel).

dynamics and intercept solutions (see reference 13 for a brief review of l0 softwarepackages available in early 2001).

Specialization to Geostationary SatellitesFor most geostational'y satellites, the subsatellite point is on the equator at longitude /',and the latude L. is 0. The geosynchronous radius r. is 42,164-11 kmr. Since l,. is zero.Eq. (2.31) simplilles to v \

cos(7) : cos(L")cos(i, - 1l) Q-36)Substituting r,: 42,164.17 km and r, : 6,378.137 km in Eqs' (2'32) and (2'35) givesthe following expressions for the distance d from the earth station to the satellite and theelevation angle El at the earth station

(2.31)

(2.38)d : 42,164.1711.A2288235 - 0.30253825 cos(7)ir/r km

,-,, -

'in(f .=t"'/ - [ t.ozzll23s - 0.30253825 cos(7)]'

,f tanl(/, - rJlla: tan L *,(J l

cos

For a geostationary satellite with an orbital radius of 42,164.L7 km and a mean earth ra-dius of 6378.L37 km, the rutio r"f r.: 6.6107345 giving

El: tan-1[(6.6107345 -- cosfl/sinr] - z (2.39)To nd e azimuth angle, an intermediate angle a must first be found. The inter='- "-

mediate angle permits the correct 90o quadrant to be found for the azimuth since theazimuthal angle ian lie anywhere between 0o (true north) and clockwise through 360"(back to true north again). The intermediate angle is found from

(2.40)

Having found the intermediate angle a, the azimuth look angle Az can be found from:

Case 1: Earth station in the Northern Hemisphere with(a) Satellite to the SE of the earth tation: Az : 180o - a(b) Satellite to the SW of the earth station: Az : 180' * a

Case 2: Earth station in the Southem Hemisphere vith(c) Satetlite to the NE of the earth station: ?tz : n(d) Satettite to the NW of the ear"th station: ; : 360" - a

(2.41a)(2.41b)

(2.41c)(2.41d)

' ttr, r' gl'

[,,l"i 3 cHAprER 2 oRBtrAL MEc[tANtcs AND LAUNcHERs

Subsatelitepont

Earthstation

FIGURE 2.13 The geometry of thevisibility calculaton. The satellite is said tobe visible from the earth station if theelevation angle E/ is positive. This requiresthat the orbital radius r. be greater than iheratio relcos(7) where r" is the radius of theearth and 7 is the central angle.

Visibility TestFor a satellite to be visible from an earth station, its elevation angle El must be abovesome minimum value, which is at least 0o. A positive ar zero elevafion angle requires that(see Figure 2.13)

r">l.

- cos(7]This rneans that tle maximum central angular separation between tle earth station andthe subsatellite point is mited by

(2.43)

For a nominal geostationary orbit, the last equation reduces ta y =

81.3" for the satelliteto be visible.

EXAMPLE 2.2.1 Geostationary atellite Look AnglesAn eath station situated in the Docklands of [,ondon, England, needs to calculate the iook angleto a geostationary satellite in e Indian Ocean operated by Intelsat. The details ofthe earth stationsite and the satellite are as follows: !r-" o.E_

Earth station latitude and longitude are 52.0' N and O'.Satellite longitude (subsatellite point) is 66.0" E.

_Tl

re

cos 7

(2.42)

, = *.-,(f)

Sr

St

Str

Reinis)cu!satan(Sureas

whvisr

thagenb"rifrposThiopegennolcatitracthestanThocom

etry of the;atellite is said to;tation if theve. This requires3 gi-eater than the:he radius of theangle.

i'l must be abovengle requires that

(2.42)

earth station and

(2.43)

i" for the satellite

late the look angleof the earth station

2.2 LOOK ANGLE DETERMINATION 37

Step 1: Find the central angle 7cos(7) : cos(L")cos(l.

- /.)

: cos(52.0)cos(66.0) : 0.2504Yiclding Y = 75'4981'Thc centrrl angle yis lesslhan 8l.J'so the satellite is risible from the earth station.

Step 2: Find the eleralion angk- E1.El : tan r[(6.6107345 cosT)/:inl

- -

y: tan r[(6.6107345 0.3501)/srn(75 4981)] - 75 l9El= 5 847'

Step 3: Find thc intermediate angle a,[ran'1 r r-n

-liln I -I .rlr.1- -

- tan rf(tani6.() ()rtr'sin(51.t-))

- 7t).661'

Step 4: Find the azimuth angleThe earth station is in the Nolthern He misphere and the satellite is to the southe asi ol thecarth station From Eq. (2.41.a). this gives

,,1: : l3fl, -

r : 180 -

10.667 -

199.1-r-l'(cltckuisc iroir lrue norrh) I

Note that. in the example above. the eleretion engle is relatil'el1'1ou (-5.85-').Refractive efiects in rhe atmosphere wiii cause the lnean ray path to the satellite to bendin the elevation plane (making the satellite appeff to be higher in tlie sky than it actuallyis) and to cause the amplitude of the signal to flucruate with time. These aspects are dis-cussed more fully in the propagation etl-ects chapter. While it is unusual to operate to asatellite below established elevation angle minima (typically 5' af C band, 10" at Ku band.and in most cases, 20o at Ka band and above), many times it is not possible to do this.Such cases exist for high latitude regions and for satellites attempting to reach extremeeast and west coverages from their given geostationary equatorial location. To establishwhether a particular satellite location can provide service into a given region, a simplevisibility test can be carried out, as shown earlier in Eqs. (2.42) and (2.43).

A number ofgeosynchronous orbit satellites have inclinations that are much largerthan the nominal 0.05'inclination maximum for current geosynchronous satellites. (Ingeneral, a geosynchronous satellite with an inclination of

3a cHAprER 2 oRBrrAL ,lEcHANlcs ANo LAUNcHERS

for a limited part of the day. The exceptional reliability of electronic components inspace, once they have survived t}te launch and deployment sequences, has led space-craft designers to manufacture satellites with two end-of-life criteria. These are: end ofdesign life (EODL), which ret'ers to the lifetirne expectancy of the payload componentsand end of maneuvering life (EOML), which refers to the spacecraft bus capabilities,in particular the anticipated lil-etime of the spacecraft with full maneuver capabilities inlongitude and inclination.

Current spacecraft are ciesigned with fuel tanks that have a capacity that usuallysignificantly exceeds the requirement for EODL. Once the final mass of the spacecraft(without fuel) is known, a decision can be made as to how much additional fuel to loadso that the economics of the launch and the anticipated additional return on investlnentcan be balanced. Having additional. fuel on board the spacecraft can be advantageousfor many reasons, in addition to adding on-orbit lifetime. In many cases, satellites aremoved to new locations during their operational lifetime. Examples for this are open-ing up service at a new location with an older satellite or replacing a satellite that hashad catastrophic failure with a satellite from a location that has fewer customers. Eachmaneuver, however, consumes tuel. A rule of thumb is that any change in orbital loca-tion for a geostationary sateliite reduces the maneuvering lifetirne by abot"lt 1 lnonth.Moving the satellite's location by l'in longitude takes as much additional fuel as Inov-ing the location by I80": both changes require an acceleration burn, a drift phase, anda deceleration burn. The 180' location change n'ill clearly take longer. since the driftrates are the same in both cases. Another use for additional fuel is to allow for orbitalperturbations at anY location.

2.3 ORBITAL PERTURBATIONSThe orbital equations developed in Section 2.1 modeled the earth and the satellite as pointmasses influenced only by gravitational attraction. Under these ideal conditions, a "Kep-lerian" orbit results, which is an ellipse whose properties are constant with time. In prac-tice, e satellite and the earth respond to many other influences including asyrnmetry ofthe earth's gravitational field, the gravitational fields of the sun and the moon, and solarradiation prssure. For low earth orbit satellites, atrnospheric drag can also be importanlAll of these interfering forces cause the true orbit to be different from a simple Kepler-ian ellipse; if unchecked, they would cause the subsatellite point of a nominally geosyn-chronous satellite !o move with time.

Historically, much attention has been given to techniques for incorporating addi-tional perturbing forces into orbit descriptions. The approach normally adopted for com-munications satellites is fust to derive an osculating orbit far some instant in time (theKeplerian orbit the spacecraft would follow if all perturbing forces were removed at thattime) with orbital elements (a, e, t* O, j, ar). The pernrrbations aro assumed to cause theorbital elements to vary with time and the orbit and satellite location at any instant aretaken from the osculating orbit calculated with orbital elements corresponding to that time.To visualize the process, assume that the osculating orbital elements at time to arc (a6, e,fp, f}0, i0, are). Then assume that the orbital elements vary linearly wi time at constantrates given by (da/dt, de/dt, etc.). The satellite's position at any time /1 is then calculatedfrom a Keplerian orbit with elements

da. deas + \tt - r,eo + \tt - t. etc.

Ththr

bitlutelaferc1-

m:t

LoEfTh,triMCnoitorave

c0rtoriTht

riurpoiIfaAYat tlhiII.and252sliganypoitonc

(cergralag(

spervitnor

thesatrwilwa!theoffon'celt

components in, has led space-hese are: end ofoad comPonentsbus capabilities,:r capabilities in

city that usuallYrf the spacecraftcnal fuel to load'n on investmentrc advantageouses, satellites arerr this are oPen-satellite that hascustomers. Each: in orbital loca-about I month.nal fuel as mov-drift phase, andr, since the driftallow for orbital

satellite as pointnditions, a "Kep-ith time. In prac-ng asymmetry ofmoon, and solar

lso be important.a simple Kepler-rminally geosyn-

:orporating addi-dopted for com-;tant in time (the: removed at thatmed to cause thert any instant arerding to that time.ime f6 are (ao, eo,time at constant

is then calculated

..*.,

:.::,-1i1.ff..i,2.3 oRarrAL pERTURSATToNS ,tsg

This approach is particularly useful in practice because it permits the use of eithertheoretically calculated derivatives or empirical values based on satellite observations.

As the perturbed orbit is not an ellipse, some care must be taken in defining the or-bital period. Since the satellite does not retufn to the same point in space once pr revo-lution. the quantity most frequently specified is the so-called anomalistic period: theelapsed time betrveen successive perigee passages. In addition to the orbit not being a per-fect Keplerian ellipse, there will be other influences that will cause the apparent positionof a

-qeostationary satellite to change with time. These can be viewed as those causingrnainly longitudinal changes and those that principally affect the orbital inclination.

Longitudinal Changes:Effects of the Earth's OblatenessThe earth is neither a perfect sphere nor a perfect ellipse; it can be better described as atriaxial ellipsoidr . The earth is flattened at the poles; the equatorial diameter is about 20 kn'rmore than the average polar diarneter. The equatorial radius is not constant, although thenoncircularity is sm:all: the radius does not vary by more than about 100 m around the equa-torr. In addition to these nonregular features of the earth, there are regions where theaverage densit, of the earth appears to be higher. These are referred to as regions of massconcentrarion or Mctscons. Th" ryrphgJg of the earth, th. ,tglgulu.iy of the equa-torial radius, ar.rd rhe Masconslead-to a rtg-udf-o-rr-glaJllalreua!-figlg*alggtd-"t}-g-eal-th-The force on an orbiting satellite will therefore vary with position-

For a low earth orbit satellite, the rapid change in position of the satellite with re-spect to the earth's surface will lead to an averaging out of the penurbing lorces in iinewith the orbital velocity vector. The same is not true for a geostationary (or geosynchro-ncus) satellite.4. gggqfong_ry_"8U1lg.igC-ry]gylrS!.is-q!1, The smallest force onthe satellite will cause it to accelerate and then drift away from its nominal location. Thesatellite is required to maintain a constant longitudinal position over the equator, but therewill generally be an additional force toward the nearest equatorial bulge in either an east-ward or a westward direction along the orbit plane. Since this will rarely be in line withthe main gravitational force toward the earth's center, there will be a resultant componentof force acting in the same direction as the satellite's velocity vector or against it, dependingon e precise position of the satellite in the GEO orbit. This will lead to a resultant ac-celeration or deceleration component that varies with longitudinal location of the satellite.

Due to the position of the Mascons and equatorial bulges, there are four equib-rium poiqg in the geostationary orbit: tyg-g"-thea[sfahland#s-gLs-lable. The stablep[nTlre analogous to the bottom of a valley, and the unstable points to the top of a hill.If a ball is perched on top of a hill, a small push will cause it to roll down the slope intoa valley, where it will roll backwards and forwards until it gradually comes to a final stopat the lowest point. The satellite at an unstable orbital location is at the top of a gravityhill. Given a small force, it will drift down the gravity slope into the gravity well (valley)and frnally stay there, at the stable position. The stable points are at about 75o E and252' E and the unstable points are at around 162" E and 348" E1. If a satellite is perturbedslightly from one of the stable points, it will tend to drift back to the stable point withoutany thruster firings required. A satellite that is perturbed slightly from one of the unstablepoints will immediately begin to accelerate its drift toward the nearer stable point and,once it reaches this point, it will oscillate in longitudinal position about this point until(centuries later.) it stabilizes at that point. These stable points are sometimes called thegraveyard geosynchronous orbit locations (not to be confused with the graveyard orbit fora geosynchronous satellite, which is the orbit to which the satellite is raised once the

40 CHAPTER 2 ORBITAL MECHANICS AND LAUNCHERS

satellite ceases to be useful). Note at, due to the nonsphericity of the earth, etc., the sta-ble points are neither exactly l80o apart, nor are the stable and unstable points precisely90o apart.

lnclination Changes:Effects of the un and the MoonThe plane of the earth's orbit around the sun-the ecliptir:-is at an inclination of 7.3' tothe equatorial plane of the sun (Figure 2.1-1). The earth is titled about 23" away from thenormal to the ecliptic. The moon circles the earth with an inclination of around 5o to theequatorial plane of the earth. Due to the fact that the various planes-the sun's equator.the ecliptic, the earth's equator (a plane normai to the earth's rotational axis), and thcmoon's orbital plane around the earth-are all different, a satellite in orbit around the earthwill be subjected to a vuiety ol out-of-plane forces. That is, there will generally be a netacceleration force that is not in the plane of the satellite's orbit, and this will tend to tr_r,Io chiinge the inclination of the sateilite's orbit from its initial inclination. Llnder rheseconditions. the orbit rvill precess and its inclination will change.

The rnass of the sun is significantly larger than that of the moon but the moon is con-siderably closer to the earlh than the sun (see -fable 2.2). For this reason. the accelerationfbrce induced by the moon on a

-eeostationary satellite is about twice as large as that of thesun. The net effect of the acceleration forces induced by the moon and the sun on a

FIGURE 2.14 Relationship between the orbtal planes of the sun, moon, and earth. Theplane of the earth's orbit around the sun isthe ecliptic. The geostationary orbit plane (theearth's equatorial plane) is about 23" out of the ecliptic, and leads to maximum out-of-geostationary-orbit-plane forces at the solstice periods (approximately June 21 andDecember 21). The orbt of the moon is inclined about 5" to the earth's equatorial plane. Themoon revolves around the earth in 27.3 days, the earth (and the geostationary satellite)rotates once about 24 h, and the earth revolves around the sun every 365.25 days. lnaddtion, the sun-which has a greater girth at the equator than at the poles-has its equatorinclined about 7.3' to the ecliptic. All of these various angular differences and orbital periodslead to conditions where all of the out-of-plane gravitational forces are in one drection withrespect to the equatorial (geostationary orbitali plane at a given time as well as to conditionswhere the various gravitational out-of-plane forces partiaily cancel each other out. Theprecessional forces that cause the nclination of the geosiationary satellite's orbit to moveaway from the equatoral plane therefore vary with time.

Moon

T,

SrM

Irja

geol

ralilgslil(u.o1'

rvltiaclirrhi_\'e

fueinfornec

to2Thigreioft

conclinsatesotin lrectneu,

burcharreas

of arof a.abotattinthe r

to c(

2.3 OREITAL PERTURBANONS 41

TABLE 2.2 Comparative Data for the Sun, Moon, and EarthMean radius Mass Mean orbit radius Spin period

rth, etc., the sta-points precisely

Lnation of 7.3o tol" away frorn thearound 5" to thehe sun's equator,al axis), and the. around the eafihi:nerally be i nets wiil tend to tr)'ion. Uncler these

the moon is con-. the accelelationtrge as that of thend the sun on a

nd earth. Therbit plane (theum out-of-21 anditorial plane. Thelry satellite)5 days. ln;-has its equatorrd orbital periodsre direction withil as to conditionser out. lheorbit to move

Sun 696,000 kmMoon 3,476 kmEarth 6,378.14 km

333,432 units0.012 units

1.0 units

30,000 light years 25.04 earth days384,500 km 27.3 earth days

149,597,870 km 1 earth dayThe orbit radius refers to the center of the home galaxy (Milky Way) for the sun, centerof earth for the moon, and center of the sun for the eah, respectively.

geostationary satellite is to change the plane ofthe orbit at an initial averase rate ofchanseo1'0.S5"/ycar from the equatorial planer.

When both thc sun and moon are acting on the sanre sidc ol the satellitc's orbit. therate of change of the plane of the geostationary satellite's orbil will be higher than aver-age. Whe n they are on opposite sides of the orbit, the rate of change of the piane of thcstellite's orbit will be less than average. Examples of rnaxirrurn years are 1988 and 20{l10.94'lyear) and cxartples of nrinimurn ycars are 1997 and 2015 $.l5"lyear)r. These rrle-rof change are neither constant with time nor with inclination. They are at a nraxinir.imu,hen the inclination is zero and they are zero when the inclination is 1.1.67". Fon'r an ini-tial zero inclination, the plane of the geostationary orbit will change to rr maxirnulr in-clination of 14.61" over 26.6 years. The acceleration tbrces lviil then chan-qe direction atthis maxinum inclination and the orbit inclination will move back to zero in another 26.6vears and out to

- 14.67" over a further 26.6 years, and so rn.

In sorne cases, to increase the orbital maneuver lif'etirne of a satellite for a sivet'tfuel load, rnission planners deliberately place a satellite planned fclr geostationary orbitinto an initial orbit with an inclination that is substantially larger than the nominal 0.05'for a geostationary satellite. The launch is specifically timed, however, so as to set up thenecessary precessional forces that will automatically reduce the inclination "error" to closeto zero over the required period without the use of any thruster firings on the spacecraft.This will increase the maneuvering iifetime of the satellite at the expense of requiringgreater tracking by the larger earth terrninals accessing the satellite for the hrst year or soof the satellite's operational life.

Under normal operations, ground controllers command spacecraft maneuvers tocorrect for both the in-plane changes (longitudinal drifts) and out-of-plane changes (in-clination changes) of a satellite so that it remains in the correct orbit. For a geostationarysatellite, this means that the inclination, ellipticity, and longitudinal position are controlledso that the satellite appears to stay within a "box" in the slry that is bounded by 10.05"in latitude and longitude over the subsatellite point. Some maneuvers are designed to cor-rect for both inclination and longitude drifts simultaneously in the one burn of the ma-neuvering rockets on the satellite. In others, the two maneuvers are kept separate: oneburn will correct for ellipticity and longitude drift; another will correct for inclinationchanges. The latter situation of separated maneuvers is becoming more cornmon for tworeasons. The first is due to the much larger velocity increment needed to change the planeof an orbit (the so-called north-south maneuver) as compared with e longitude/ellipticityof an orbit (the so-called east-west maneuver). The difference in energy requirement isabout 10:1. By alternately correcting for inclination changes and in-plane changes, theattitude of the satellite can be held constant and different sets of thrusters exercised foru;re required laneiiver.

The second rea-scn is the increasing use of two corepleiely different types of thruster-sto control N-S maneuvers on the one hand and E-\Y naneuvers on the other. In the

lillllirliri:t,i 42 cHAprER 2 oRBlrAL MEcHANtcs AND LAUNcHERS

mid-1990s, one of the heaviest items that was carried into orbit on a large satellite wasthe fuel to raise and control the orbit. Abqut 9Q7o g-f lhiq fuel load, once on orbit, was tocontrol the inclination of the sateltite. Newer rocket motors, particularly arc jets and ionthrusters, offer increased efficiency with lighter mass. In generai, these low thrust, highefficiency rocket rnotors are used for N-S tllaneuvers leaving the liquid propellant thrusters,with their inherently higher thrust (but lower efficiency) for orbit raising ancl in-plancchanges. In order to be able to calculate the required orbit maneuver lor a given satellitc,the controliers irust have an accurate knowledge of the satellite's orbit. Orbrt delemrina-tion is a malor aspe ct of satellite control.

EXAMPLE 2.3.1 Drift with a Geostationary SatelliteA quasi-GEO sar!.llir is in a circular equatorial orbit closc to geosynchronous altitudc. Thc quasi-GEO satellite. horve','cr. dtcs not have a per-iod of one sidereal day: its orbital period is exactly2-1 h-one solar dar'. Calculate(i) thc- r'adius o1'thc r>r'bit(ii )