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Satellite Altimetry andSatellite Altimetry andGravimetryGravimetry: : Theory and ApplicationsTheory and Applications

C.K. ShumC.K. Shum1,21,2, Alexander Bruan, Alexander Bruan2,12,1

1,21,2Laboratory for Space Geodesy & Remote SensingLaboratory for Space Geodesy & Remote Sensing 2,12,1Byrd Polar Research CenterByrd Polar Research Center

The Ohio State UniversityThe Ohio State UniversityColumbus, Ohio, USAColumbus, Ohio, USA

[email protected]@osu.eduedu, , [email protected]@osuosu..edueduhttp://geodesy.eng.ohio-state.http://geodesy.eng.ohio-state.eduedu

Norwegian Univ. of Science and TechnologyTrondheimTrondheim, Norway, Norway

2121––25 June, 200425 June, 2004

Satellite Altimetry andSatellite Altimetry and Gravimetry Gravimetry::Theory and ApplicationsTheory and Applications

C.K. Shum & Alexander Braun,C.K. Shum & Alexander Braun, Ohio State UniversityOhio State UniversityContributors:Contributors:K. Cheng, Shin-K. Cheng, Shin-chan chan Han, Chung-yenHan, Chung-yen Kuo Kuo, , Yuchan Yuchan Yi, Ohio State Yi, Ohio State UnivUniv..Acknowledgements:Acknowledgements:Brian Beckley, NASA/GSFC Altimeter Pathfinder Data CenterBrian Beckley, NASA/GSFC Altimeter Pathfinder Data CenterJérôme Benveniste, Pierre Femenias, ESA/ESRINDudley Dudley CheltonChelton, Oregon State University, Oregon State UniversityDon Chambers, JohnDon Chambers, John Ries Ries, Bob , Bob SchutzSchutz, Byron , Byron TapleyTapley, , UnivUniv. of Texas. of TexasCheinway Cheinway Hwang, National Hwang, National Chiao Tung Chiao Tung University, TaiwanUniversity, TaiwanJohn John LillibridgeLillibridge, NOAA/Lab. for Satellite Altimetry, NOAA/Lab. for Satellite AltimetryLee Fu, Margaret Lee Fu, Margaret SrinivasanSrinivasan, Robert , Robert BenadaBenada, NASA/JPL, NASA/JPLPhilip Woodworth,Philip Woodworth, Proudman Proudman Oceanographic LaboratoryOceanographic Laboratory*Last Lectures:*Last Lectures: Wuhan Wuhan Altimetry Workshop, 2002; GLOSS Malaysia Workshop, 2004Altimetry Workshop, 2002; GLOSS Malaysia Workshop, 2004


•• Orbital Dynamics and Orbit Determinations I Orbital Dynamics and Orbit Determinations I (AM) By C.K. Shum(AM) By C.K. Shum–– Keplerian Keplerian motion, general perturbation, motion, general perturbation, KaulaKaula’’s s formulationsformulations–– Periodic variations and resonances due to Periodic variations and resonances due to geopotentialgeopotential

•• Introduction to Satellite Altimetry I Introduction to Satellite Altimetry I (PM) By Alexander Braun(PM) By Alexander Braun–– What is altimetry?What is altimetry?–– Basic principles of satellite altimetry and its historyBasic principles of satellite altimetry and its history–– Interdisciplinary applications of altimetryInterdisciplinary applications of altimetry

•• Tutorial on Tutorial on iGMT iGMT –– a graphics tool a graphics tool (PM) By Alexander Braun(PM) By Alexander Braun

Monday, 21 June 2004Monday, 21 June 2004


Tuesday, 22 June 2004Tuesday, 22 June 2004•• Orbital Dynamics & Orbit Determinations IIOrbital Dynamics & Orbit Determinations II (AM) By C.K. Shum(AM) By C.K. Shum

–– Nonlinear orbit determination & parameter recoveryNonlinear orbit determination & parameter recovery–– Force, measurement, and Earth orientation models for PODForce, measurement, and Earth orientation models for POD

•• Satellite Altimetry IISatellite Altimetry II (PM) By C.K. Shum(PM) By C.K. Shum–– Principles of satellite altimetry, mission design, waveformsPrinciples of satellite altimetry, mission design, waveforms–– Altimeter crossovers, geographically correlated orbit errorsAltimeter crossovers, geographically correlated orbit errors–– Precision orbit determination and accuracy evaluationsPrecision orbit determination and accuracy evaluations–– Instrument, media and geophysical correctionsInstrument, media and geophysical corrections

•• Tutorial onTutorial on iGMT iGMT (continued:) (continued:) (PM) By Alexander Braun(PM) By Alexander Braun


Wednesday, 23 June 2004Wednesday, 23 June 2004•• Space Geodesy: An Interdisciplinary ScienceSpace Geodesy: An Interdisciplinary Science (AM) (AM) C.K. ShumC.K. Shum

–– Nonlinear orbit determination & parameter recoveryNonlinear orbit determination & parameter recovery–– Example interdisciplinary applications of satellite geodesyExample interdisciplinary applications of satellite geodesy

•• Altimeter Collinear Analysis Altimeter Collinear Analysis (PM) By Alexander Braun(PM) By Alexander Braun–– Stackfile Stackfile method for oceanography and marine geophysicsmethod for oceanography and marine geophysics–– Mean sea surface, marine gravity field determinationsMean sea surface, marine gravity field determinations–– Models and accuracy evaluations and limitationsModels and accuracy evaluations and limitations

•• Radar Altimeter Data Processing Radar Altimeter Data Processing (PM) By Alexander Braun(PM) By Alexander Braun–– Demonstration of GFZ ADS systemDemonstration of GFZ ADS system–– Other data centers (access, data ordering, data products)Other data centers (access, data ordering, data products)


•• Themes:Themes:–– Orbital dynamics and orbit determinationOrbital dynamics and orbit determination–– Instrument error budget and analysisInstrument error budget and analysis–– Geophysical inverse programGeophysical inverse program–– Interdisciplinary applicationsInterdisciplinary applications

•• Basic knowledgeBasic knowledge–– Orbital mechanics, dynamics, physical and satellite geodesyOrbital mechanics, dynamics, physical and satellite geodesy–– Mathematical tools (linear algebra, statistics, numerical analysis,Mathematical tools (linear algebra, statistics, numerical analysis,

differential equations, approximate theory, adjustment)differential equations, approximate theory, adjustment)–– Physics, astronomy, engineeringPhysics, astronomy, engineering–– Instrument and their principles (radar, optical, Instrument and their principles (radar, optical, electromagneticselectromagnetics))–– Geophysics, oceanography, atmosphere, hydrology, glaciologyGeophysics, oceanography, atmosphere, hydrology, glaciology


•• A good geodesist A good geodesist shouldshould::–– Know assumptions made in theory and data processingKnow assumptions made in theory and data processing–– Know your instrument well (Know your instrument well (precision and accuracyprecision and accuracy))–– Define your problem to solve first, then the correspondingDefine your problem to solve first, then the corresponding

observation requirementsobservation requirements–– Always provide Always provide real real uncertainty of your uncertainty of your solutions solutions or or productsproducts–– Know that the knowledge and instrument accuracy are Know that the knowledge and instrument accuracy are movingmoving

targets that changes with timetargets that changes with time–– Always question your instructorsAlways question your instructors’’ lectures: they could be wrong lectures: they could be wrong–– Look for new problems not necessarily in your field to be solvedLook for new problems not necessarily in your field to be solved

using your instrumentusing your instrument

Satellite altimetry measures with an accuracy of 1 part perSatellite altimetry measures with an accuracy of 1 part perbillion: few cm accuracy at ~1500 km altitude to center of Earthbillion: few cm accuracy at ~1500 km altitude to center of Earth

Orbital DynamicsOrbital DynamicsC.K. ShumC.K. Shum1,21,2, Shin-, Shin-chanchan Han Han11

1,21,2Laboratory for Space Geodesy & Remote SensingLaboratory for Space Geodesy & Remote Sensing 2,12,1Byrd Polar Research CenterByrd Polar Research Center

The Ohio State UniversityThe Ohio State UniversityColumbus, Ohio, USAColumbus, Ohio, USA

[email protected]@osu.eduedu, , [email protected]@osuosu..edueduhttp://geodesy.eng.ohio-state.http://geodesy.eng.ohio-state.eduedu

Norwegian Univ. of Science and TechnologyTrondheimTrondheim, Norway, Norway

2121––25 June, 200425 June, 2004

Original Figure from K. Lambeck

Useful quantities/constantsSome fundamentals: Timesystem, etc

Some Useful Geodetic Constants & AccuraciesEarth Rotation and Polar Motion Accuracy1960: x, y, 30 mas (1 ppb ~ 6 mm)2000: x, y, 0.2 mas; UT1: 20 µsec; Scale, 1+0.7 ppbGravity1 gal ~ 1 m/sec2; 1 µgal (change) ~ 5 mm (surface of the Earth)1 E = 1 mgal/10 km; 1 mE = 10-12 s-2

GOCE: ∇∇TU ~ 2nd order tensor; GRACE: δ∇U ~ δ potentialGOCE Gradiometer Accuracy: 2 mE/Hz1/2 (0.005 Hz to 0.1 Hz)GRACE Inter. ranging Acc.: 0.1–0.5 µm/s; 5 cm geoid (>250 km)1 µradian (µrad) = 1 mm/kmOtherRadar: Ku-band (13.4 GHz): 2.8 cm; C-band (5.6 GHz) 5.6 cm;Ionosphere: 1 TEC = 2.2 mm@Ku band; 16 cm@L-bandSea level: 1–2 mm/yr ~ 400 Gton/yr; 1 mm/yr = 0.2 µgal/yr

Determination of Time: BasicsTime System Conversions:

Concept_1 Sidereal day = o36024 =h

In 1995, 1 mean solar day day siderealmean 9351.00273790=

timesidereal of 55537.56324 smh ×= o36050027379093.1 ×=

0 12UT

equal to how many degree

o

L

3601.0027370.5

day sidereal1.0027370.5

daysolar 0.5UT12

××=

×=

=

Courtesy: C. Hwang

Universal Time

Mean sun: an imagined sun with a uniform motion on theplane of the equatorMean solar time: hour angle of the mean sunMean solar day: time interval between two successivetransits of the mean sunUniversal time (UT): Greenwich hour angle of the mean sun h12+

€

UT12h=12h + Greenwich hour angle of the mean sun

UT0 = UT as deduced directly from observationsUT1= UT0 + Λp(polar motion), represents the true angular rotation of the earth (i.e., includes tidal variations)

Conversion from UT1 to GAST_

€

GAST =1.00273790935UT1+ϑ 0 + Δψ cos∈ (1.37) ϑ 0 = 24110•

s54841+ 8640184•s812866T + 0•

s093104T 2 - 6•s2 ×10−6T 3 (1.38)

T = (tu − 2451545) /36525tu = Juliandays at 0h UT1

Determination of Time: Basics

Courtesy: C. Hwang

1.2.3 Dynamical Time

TDB: Barycentric Dynamic Time_Time derived from orbital motionsreferred to the barycenter center (center of mass) of the solarsystem. The coordinate time.

TDT: Terrestrial Dynamical time, referred to geocenter. The proper time

1.2.4 Atomic times

TAI: International atomic time, a weighted mean of individual clocksfrom around the word, in SI(International System of Units) second.

UTC: Coordinated Universal time, is adapted to UT1 by “leap seconds”

(1.40) 0.7sQUT1UT1-UTC

(1.39) (1s)n-TAIUTC

≤=

×=

UTC is the Broadcast Time in most countries (with leap second)


Courtesy: C. Hwang

GPST: Known also as GPS Time, which is one kind of atomic time, but isused and adopted by the GPS control system for time transfer.

s

s

90

90

0

10232C Jan,1,1992

101376C Jan,1,1989

(1.41) BIPM from integer, n ; C-snUTC-GPST

−

−

×=

×−=

×=

M

Other relationships_

offsetconstant 18432-TDTTAI

offsetconstant 00019GPSTTAIs.

s.

=

+= (1.42)


Courtesy: C. Hwang

1.2.5 Calendar

Julian date (JD): number of mean solar days elapsed since the epoch 4713 B.C.,January,1.5 Modified Julian date (MJD): JD–2,400,000.5 FK5 standard epoch: 2000 January 2,451,545JD ,51. =d

GPS standard epoch: 1980 January 52,444,244.JD,06. =d

MJD for 0.2000J : 51,544.5


Courtesy: C. Hwang

Orbital Dynamics:Keplerian motion, Kepler’sequations

1. Dynamics of Satellite Orbits 1.1 Elliptic (Keplerian) Motion

Satellite's motion = Elliptic motion + Perturbed motion

(1) => and are co-linear. The double integration leads to six integration parameters spanning an orbital plane.

They are / or .

/ is called a state vector and is called the Keplerian orbital elements.

a – semi-major axis of orbital ellipse, e – eccentricity of orbital ellipse, i – inclination, ω – argument of perigee, Ω – right ascension of ascending node, ν – true anomaly.

GM , 3

=−= µµxx

r&&

x&& x

x&

x

),,,,,( νωΩiea

x&

x

),,,,,( νωΩiea

Notes from S. Han

Satellite Orbit Geometry and Orbit ElementsSatellite Orbit Geometry and Orbit ElementsX, Y, Z: non-rotating system (J2000) or theConventional Inertial System, or CIS,(An approximation to the inertial system)

a – semi-major axis of orbital ellipse e – eccentricity of orbital ellipse i – inclination ω – argument of perigee Ω – right ascension of ascending node ν – true anomaly

io

h is the angularmomentum vector

(a, e) – size and shape of the orbit(i, ω, Ω ) –orientation of orbit wrt CTS(ν ) – time dependent variable


For the parameter indicating the orbiting particle's location on the orbital ellipse, we can use otherquantities such as eccentric anomaly, E, or mean anomaly, M. Let us introduce the orbital ellipseas follows:

Notes from S. Han


From (6), we have the angular momentum conservation equation as follows:

(7) Combining this with (5), the equation of the ellipse can be derived as follows:

(8) From the above geometry, we realize the followings:

(9) (10)

Consequently, the distance can be written in terms of the eccentric anomaly as follows:

(11) From (7), (9), and (10) we have (12)

(13)

( ) .1 22 consteah =−== µνρ &

ννρ

cos1)1(

)(2

eea

+−

=

( ) Eea sin1sin 2−=νρ

aeEa −= coscosνρ

( )EeaE cos1)( −=ρ

( ) ννρ

ρ dea

ed sin

1 2

2

−=

EdEaed sin=ρNotes from S. Han


(14) After some mathematical manipulation, we will have

(15) where The integration of (15) gives Kepler’s Equation:

(16) where , which is called mean anomaly. In summary, three anomalies (true, eccentric, mean), ν(t), E(t), M(t), of the Keplerian orbit andtheir relationships are given as follows:

(17)(18)

(19)

( )dtead 22 1−= µνρ

( ) ndtdEEe =− cos1

3an

µ=

MEeE =− sin

( )0ttnM −=

( )0)( ttntM −=)(sin)()( tEetMtE +=

−

+−=

Ee

Eet

cos1

cosarccos)(ν

Notes from S. Han


In order to describe the motion of the satellite on the orbital ellipse in the inertial coordinatesystem, let us introduce the coordinate system attached to the orbital ellipse. The first coordinatepoints the perigee, the third coordinate is normal to the orbital plane, and the second coordinate isautomatically decided by the right handed rule of the coordinate system. In this coordinatesystem, we have:

(20)

Taking derivatives of each component, we have the velocity vector as follows:

(21)

−

−

=

=

0

sin1

cos

0

sin

cos2 Ee

eE

ar ν

ν

r

−

−

=

+

−

−=

0

cos1

sin

0

cos

sin

1

22

2Ee

E

r

nae

e

naν

ν

r&

Notes from S. Han


Using both vector representations, we can derive the energy conservation law and angularmomentum conservation law as follows:

(22) (23)

where h is the constant angular momentum vector orthogonal to the orbital plane The first equation indicates Energy Conservation:

The second equation indicates Angular Momentum Conservation. In the Two-Body program, theorbital plane in invariant. This means that the inclination or the orientation of the orbital planedoes not change. In the real world, it is a good approximation.

€

12ddt

˙ r ⋅ ˙ r ( ) − µr

= const.

hrr =× &

€

v22−

µr

=µ2a

Notes from S. Han


The state vector in the inertial coordinate frame can be obtained by applying appropriate rotationsto and . From the geometry shown in the previous figure, it can be done as follows:

(24) (25)

where For the conversion algorithm between Kelperian elements and a state vector, we can referMcCarthy et al. (1993) and Seeber (1993).

( ) ( ) ( ))or , ,(,,,,,,,,, νωω EMeaiMiea rRx ⋅Ω=Ω

( ) ( ) ( ))or , ,(,,,,,,,,, νωω EMeaiMiea rRx && ⋅Ω=Ω

( ) ( ) ( ) ( )ωω −−Ω−=Ω 313,, RRRR ii

Notes from S. Han

Example Problems 1.1 Elliptic (Keplerian) Motion

The maximum latitude of the groundtrack of a retrograde circular orbitis 710. The satellite crossed the equator heading north at longitude2200E on its first revolution and at longitude 1900E on the nextrevolution. What are the inclination and the altitude of this satelliteorbit? (Hint: assume Keplerian motion, use µ⊕=3.986x105 km3/s2,Re=6378 km, ω⊕=360.9856 0/day).

Solution

For retrograde orbit, i = 1800 – 710 = 1090

In one orbital period, the change in longitude is: Δλ = 1900 – 2200 = –300

The rate of rotation of the Earth, R, is: 3600/sid. Day x 366.2422/365.2422 sid.day/solar day = 360.98560/day

Now, Δλ = –RxTP

TP = Δλ/(–R) = 7180.34 s

TP = 2π a 3/2 / µ1/2

Therefore, a = 8044.3 km or Altitiude = a – 6378 km = 1666.3 km


Kepler’s Equation.

Show that the Earth has approximately 185 days in each year during which its distance isfarther than 1 Astronomical Unit (AU) from the Sun. (Note: 1 AU is the mean distancethe mean semi-major axis, a, of the Earth’s orbit around the Sun, e=0.00167).

Hint: Use the following equations:

M = E − esin E

Δt = (M2 − M1) /n

n = 3600 / 365.26 = 0.98560 / day

r = a(1 − ecos E )


Solution

Let r = distance between the Sun and the Earth a = semi-major axis of the Earth’s orbit around the Sun

Given: a = 1 AU (mean distance of the Sun from the Earth) e = 0.0167; n (mean motion) = 3600/365.25 0/day = 0.9865 0/day

Criteria for Earth to be farther from the Sun than 1 AU: r > a

Kepler’s Equation: M = E - e sin ERelationship of r to E: r = a (1- e cos E)

Therefore: r = a (1- e cos E) > aor cos E < 0 gives 900 < E < 2700

Substitute into Kepler’s Equation and solve for values of M:

M1 = 89.040 and M2 = 270.040

∆T = (M2 – M1) / n = (270.040 – 89.040) / 0.9865 = 184.6 days (The time that the Earth is farther away from the Sun than 1 AU)

Orbital Dynamics:General perturbation,Kaula’s formulation

1. Dynamics of Satellite Orbits 1.2. Perturbed Motion

Under the influence of the 'realistic' Earth (and ignoring other forces), the point mass potentialused to derive the elliptic motion have to be replaced by adding perturbing potential as follows:

(26) where and

Even though R is small compared to , it makes the orbital elliptic plane (its size, shape, andorientation) significantly change in time. Consequently, the Keplerian elements are time-varying,i.e.,

V∇=r&&

Rr

rV +=µ

λθ ),,(

( )[ ]λλφµ

λφ mSmCPr

a

rrRR lmlmlm

l

l

l

m

e sincossin ),,(2 0

+

== ∑∑

∞

= =

rµ

)(),(),(),(),( tttiiteetaa ωω =Ω=Ω===

Notes from S. Han


(27)

(28) where i = 1, 2, and 3.

Def'n: osculating ellipse– It is an instantaneous ellipse defined by the corresponding Keplerian elements at a particularepoch. On the osculating ellipse, we will have the following constraints:

(29)

(30)

dt

ds

s

x

t

x

dt

dxx k

k

iiii ∂

∂+

∂

∂==&

dt

ds

s

x

t

x

dt

xdx k

k

iiii ∂

∂+

∂

∂==

&&&&&

Msssisesas =Ω===== 654321 , , , , , ω

t

x

dt

dx ii

∂

∂=

∂∂

=∂

∂

rxt

xi µ&

Notes from S. Han


It leads to

(31)

(32) Equations (31) and (32) can be re-written as follows:

(33)

(34) where

Therefore,

(35)

0=∂

∂

dt

ds

s

x k

k

i

i

k

k

i

xR

dt

ds

s

x

∂∂

=∂

∂&

0sA =&

[ ]

RxR

i

∇=

∂∂

=×13

sA&&

[ ] [ ]6363

,××

∂

∂≡

∂

∂≡

k

i

k

i

s

x

s

x &&AA

[ ] [ ]TT , MÙùieaMÙùiea &&&&&&& == ss

[ ][ ]16

TT

×

∂

∂=⋅−

ks

RsAAAA &&&

Notes from S. Han


The 6 by 6 square matrix, , is called Lagrange bracket. It is a skew-symmetric andtime-invariant matrix. Try to prove this. Note that the time-invariance permits to evaluate theLagrange bracket at any epoch (e.g., perigee passage time) and use it for the remaining epochswithout evaluating it every time.

AAAA TT && −

Remind

and are vectors indicating the location and motion of the satellite in the orbital ellipse(osculating ellipse) and the rotation matrix, R, orients them to the inertial frame.

( ) ( ) ( )MeaiMiea ,,,,,,,,, rRx ωω Ω=Ω

( ) ( ) ( )MeaiMiea ,,,,,,,,, rRx && ωω Ω=Ω

r r&

Notes from S. Han


To compute components of the Lagrange bracket matrix, we consider the following:

(36)

(37)

Each partial can be computed using the followings:

(38)

(39)

rRRRrrr

R

rRrRx

∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

=

⋅+⋅=

ÙÙ

ddii

ddMM

dee

daa

ddd

ωω

rRRRrrr

R

rRrRx

&&&&

&&&

∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

=

⋅+⋅=

ÙÙ

ddii

ddMM

dee

daa

ddd

ωω

−−

+−

=∂∂

0

cos23

sin1

sin23

cos

2 EMra

Ee

EMra

eE

ar

( )

−−

+−

=∂∂

0

cos1

sin

sin1

2

2

eEe

Era

Era

aer

Notes from S. Han


(40)

(41)

(42)

(43)

−

−

=∂∂

0

cos1

sin2

2

Ee

E

ra

Mr

( )

−−

−+

=∂∂

0

cossin3

1

cos3

sin

2 2

22

2

2

EEMra

e

eEMra

E

rna

ar&

( )

+−

−−−

+−

−=∂∂

011

coscos11

coscossin

22

2

3

ra

eeEEe

Ear

eEEra

rna

er&

−

−

−=∂∂

0

sin1

cos2

3

4

Ee

eE

rna

Mr&

Notes from S. Han


(44)

(45)

(46)

−

Ω−Ω−Ω−

ΩΩΩ

=∂∂

iii

iii

iii

isincoscoscossin

coscossincoscossinsincos

cossinsincossinsinsinsin

ωω

ωω

ωωR

ΩΩ−Ω−Ω−Ω

ΩΩ−ΩΩ−Ω−

=Ω∂

∂

000

sinsincoscossinsincoscossinsincoscos

sincoscoscoscossinsincossincoscossin

iii

iii

ωωωω

ωωωωR

−

Ω−Ω−Ω+Ω−

Ω+Ω−Ω−Ω−

=∂

∂

0sinsinsincos

0cossincoscossincoscoscossinsin

0cossinsincoscoscoscossinsincos

ii

ii

ii

ωω

ωωωω

ωωωω

ωR

Notes from S. Han


Each element in (36) and (37) can be computed using above equations. However, we know thereare only six non-zero elements in the Lagrange bracket matrix. The complete set of nonzeroelements is

(47)

(48)

(49)

(50)

(51)

(52)

[ ] [ ] ( ) ienai i sin1,2/122

),(TT −−=−=Ω ΩAAAA &&

[ ] [ ] ( ) iena

a a cos12

,2/12

),(TT −=−=Ω ΩAAAA &&

[ ] [ ]( ) ( ) 2/12

2

,TT

1

cos,

e

ienae e

−

−=−=Ω ΩAAAA &&

[ ] [ ]( ) ( ) 2/12,

TT 12

, ena

a a −=−= ωω AAAA &&

[ ] [ ]( ) ( ) 2/12

2

,TT

1,

e

enae e

−

−=−= ωω AAAA &&

[ ] [ ]( ) 2, ,

TT naMa e

−=−= ωAAAA &&

Notes from S. Han


After substituting each partial to equation (35), the solution of the time-derivatives of six orbitalelements follows:

(53)

(54)

(55)

(56)

(57)

(58)

This set of equations is called Lagrange Planetary Equation (LPE). The "weak" singularityappearing the equations can be easily removed by introducing Delaunay elements. For detaildescription, we refer Kaula (1966), pp.29-30.

M

R

nadt

da

∂

∂=2

ω∂∂−

−∂

∂−=

R

ena

e

M

R

ena

e

dt

de2

2

2

2 11

a

R

nae

R

ena

en

dt

dM

∂

∂−

∂

∂−−=

212

2

Ω∂

∂

−−

∂

∂

−=

R

iena

R

iena

i

dt

di

sin1

1

sin1

cos2222 ω

i

R

ienadt

d

∂

∂

−=

Ω

sin1

122

e

R

ena

e

i

R

iena

i

dt

d

∂

∂−+

∂

∂

−−=

2

2

22

1

sin1

cosω

Notes from S. Han


In LPE, we still have partials need to be specified. The perturbing potential, R, given in equation(26) is expressed in terms of the Earth fixed spherical coordinates, . Here, we will try toexpress the perturbing potential, R, using the orbital elements by investigating the relationshipbetween the orbital elements and the Earth fixed spherical coordinates.

Figure 3 shows the longitude relationship:. (59)

α – right ascensionGST – Greenwich Sidereal Time

( )λθ ,,r

equinox

Greenwich

node

satellite

Ω

α

GST λ

( ) ( )GST−Ω+Ω−= αλ

Notes from S. Han


The spherical trigonometry indicates the latitudinal relationship as follows:

(60)

(61)

(62)

equinox

node

satellite

Ωνω +

Ω−αi

φ

( ) ( )φνω

αcos

coscos

+=Ω−

( ) ( )φνω

αcos

cossinsin

i+=Ω−

( ) isincossin νωφ +=

Notes from S. Han


Equations (59) to (62) show the relationship between the Earth fixed latitude and longitude to theorbital elements. Finally, we have the following equation for the radius:

(63)

(64)

where is the eccentricity function.

After elaborated mathematical manipulation (Kaula, 1966), we finally obtain the expression ofresidual potential in terms of orbital elements as follows:

(65)

where is the inclination function.

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

( )

=

−Ω++−

−Ω++−∑∞

−∞=++

lmpq

lmpq

qlpqlleG

aGSTmpl

GSTmpl

r ψ

ψ

νω

νωsin

cos12sin

2cos111

( ) ( ) ( )GSTmMqplpllmpq −Ω++−+−= 22 ωψ

( )eGlpq

( )

∑ ∑∑ ∑∞

= = = −=

−

−

+−

+

=

Ω=

2 0 0

1

1

even :

odd : sincos

sincos )()(

,,,,,

l

l

m

l

p q

ml

mllmpqlmlmpqlm

lmpqlmlmpqlmlpqlmp

l

e

CS

SCeGiF

a

a

a

MieaRR

ψψ

ψψµ

ω

)(iFlmp

Notes from S. Han


(Exercise)At this moment, it is worth to exercise computing the frequency-independent orbital perturbationinduced by the second zonal coefficient of the residual potential. These perturbations can be usedas reference elements, because the second zonal coefficient exceeds other coefficients'contribution by a factor of 1000. The frequency, , is determined by indices, l, m, p, and q. Consequently, the frequencyindependence yields and comes from the condition of

From this, we immediately identify the even zonal harmonics provide the frequency independentperturbation. For the case of (l=2, m=0, p=1, and q=0), we can find the corresponding inclinationand eccentricity functions from the appropriate equations (or just from the tables given in Kaula,1966).

lmpqψ&

0 and 0, ,02 ===− mqpl

21sin43)( 21,0,2 −= iiF

( ) 2/320,1,2 1)( eeG −=

Notes from S. Han


The corresponding residual potential follows:

By putting it to LPE, we get

We identify the constant a, e, and i and consequently, constant change of Ω, ω, and M. Theyindicate fixed size and shape of the orbital ellipse, fixed inclination, precessing node and perigee.From these equations, we can interpret the perigee will retrograde if the inclination is bounded by63.4° and 116.6°, because the second zonal coefficient is negative. Also, the node will retrogradefor the inclinations smaller than 90°. [See example problem next]

( ) ( ) 2022/32

3

2

20 21sin431 Ciea

aR

e−−=

−µ

( )

( )( )

( )( )1cos3

14

3

cos5114

3

cos12

3

0

0

0

2

22/32

220

2

222

220

222

220

−−

−=

−−

=

−=

Ω

=

=

=

iae

aCnn

dt

dM

iae

aCn

dt

d

iae

aCn

dt

d

dt

didt

dedt

da

e

e

e

ω

Notes from S. Han

Example Problem: J2 Perturbations

A scientific geodetic satellite was launched evolving an unknown planet in a circular orbit, with aninclination of 300 and a semi-major axis of 6,500 km. The mean radius and µ of the planet areassumed known at 6,000 km and 3.986x1014 m3/s2, respectively. The ascending node of theorbit is observed to be moving from west to east (progressing) and is completing a revolutionin 40 earth-days. Assuming Keplerian motion and J2 is only perturbation on the satellite,

(a) Determine J2 of the planet. Give a physical description of the shape of the planet andcontrast the property with that of the Earth.

(b) Calculate ˙ ω . Explain the meaning of this quantity.(c) If the orbit is desired to be frozen (i.e., ˙ ω = 0), what is the orbital inclination, i?(d) State the range of the inclination values of the satellite orbit for the cases:

˙ ω < 0 , ˙ ω > 0˙ Ω < 0 , ˙ Ω > 0

Example Problem Sollution: J2 Perturbations

€

1.e = 0 (Circular orbit)i = 30°,a = 6500kmR = 6000km

µ = 3.986×1014 m3

s2

Ω⋅

=2 ×π

40 × 24 × 60× 60sec=1.81805×10−6

n =µa3 =1.20475 ×10−3

Substituting for the given values and solving for J2

Ω⋅

= −32nJ2

Ra

2 cosi1− e2( )2

J2 = −1.36335×10−3

Since J2 < 0⇒ the shape of the planet is oblongJ2 > 0⇒ the shape of the planet is oblate (in the case of Earth

Example Problem Sollution: J2 Perturbations

€

2.

ω⋅

= −34nJ2

Ra

2 1− 5cos2 i( )1− e2( )2

= −34nJ2

Ra

2

1− 5cos2 i( ) (as e = 0)

⇒ω⋅

= −14.28942°day

€

3. ω⋅

= 0⇒1− 5cos2 i = 0

⇒ cosi =15

⇒ i = 63.43°or i =116.57° Critical inclination

Example Problem Solution: J2 Perturbations

€

4. For ω⋅

> 0⇒ 5cos2 i −1> 0

⇒ cosi − 15

cosi + 1

5

> 0

⇒0 ≤ i < 63.43494° or 116.56505° < i ≤180°

if ω⋅

< 0⇒ 5cos2 i −1< 0

⇒ cosi − 15

cosi + 1

5

< 0

⇒ 63.43494°<i < 116.56505°

if Ω⋅

> 0⇒ cosi < 0⇒ 90 < i <180

if Ω⋅

< 0⇒ cosi > 0⇒ 0 < i < 90

Orbital Dynamics:Periodic variations andresonances due togeopotential


Using Kaula’s expression for perturbing potential ( Ulmpq , as a f unction of Keplerian orbit

elements) and employing the Lagrangian planetary equations, derive ∆ω and ∆Ω. Steps: integratethe Lagrangian planetary equation, and express ∆Ω as a function of F(i), G(e), and S:

€

dΩdt

=1

n a2 1− e2 sini∂R∂i

;

€

Ulmpq =µae

l

al +1 Flmp (i)Glpq (e)Slmpq (ω,M,Ω,θ)

where

[ ]

[ ])()2()2(sin

)()2()2(cos

θω

θω

−Ω++−+−

+

−Ω++−+−

−=

−

−

−

−

mMqplplC

S

mMqplplS

CS

evenml

oddmllm

lm

evenml

oddmllm

lmlmpq

If we assume that the only variations with time of the elements are θω &&&& ,,, MΩ , the argument ofthe trigonometric function can be expressed as follows:

)()()(

)()2()2()(

000 tttt

mMqplplt

ψψ

θωψ

&−+=

−Ω++−+−=


ωΔ

),,,()(

)(1

)()(

sin1

cos

1

sin1

cos

1

sin1

cos

12

2

122

2

2

22

2

2

22

θωµµ

ω

Ω

∂

∂−+

∂

∂

−−=

∂

∂−+

∂

∂

−−=

∂∂−

+∂∂

−−=

++ MSe

eGiF

aa

enae

eGi

iF

aa

iena

i

e

U

enae

i

U

iena

i

eR

enae

iR

iena

idt

d

lmpqlpq

lmpl

le

lpqlmp

l

le

lmpqlmpq

lmpq

Also,

€

Slmpqt0

t∫ dt =

Clm

−Slm

l−m odd

l−m even

cos(ψ(t)t0

t∫ dt +

Slm

Clm

l−m odd

l−m even

sin(ψ(t)t0

t∫ dt

=Clm

−Slm

l−m odd

l−m even sin ψ(t)( ) − sin ψ(t0)( )˙ ψ (t0)

−Slm

Clm

l−m odd

l−m even cos ψ(t)( ) − cos ψ(t0)( )˙ ψ (t0)

=1

˙ ψ (t0)Clm

−Slm

l−m odd

l−m even

sin ψ(t)( ) −Slm

Clm

l−m odd

l−m even

cos ψ(t)( )

≡ S lmpq

1 2 4 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 4


€

+1

˙ ψ (t0)−Clm sin ψ(t0)( ) + Slm cos ψ(t0)( )Slm sin ψ(t0)( ) + Clm cos ψ(t0)( )

l−m odd

l−m even

= 0 (Qsince the total perturbation at time t 0 is zero)1 2 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4

=1

˙ ψ (t0)S lmpq

After integration,

( ) ( ) ( )[ ][ ])()2()2(

)(/1cot/)(1

)(1)(

)(1

)()(

sin1

cos

3

2/1212

012

2

122

0

θωµ

ψµµ

ωω

&&&&

&

−Ω++−+−

∂∂−−∂∂−=

∂

∂−+

∂

∂

−−=

=Δ

+

−−

++

∫

mMqplplna

SeGiFeieGiFeea

Ste

eGiF

aa

enae

eGi

iF

aa

iena

i

dtdt

d

llmpqlpqlmplpqlmpl

e

lmpqlpq

lmpl

le

lpqlmp

l

le

t

t

lmpqlmpq

∑Δ=Δnmpq

lmpqωω


ΔΩ

),,,()()(

sin1

1

sin1

1

sin1

1

122

22

22

θωµ

Ω

∂

∂

−=

∂

∂

−=

∂∂

−=

Ω

+ MSeGi

iF

aa

iena

i

U

iena

iR

ienadt

d

lmpqlpqlmp

l

le

lmpq

lmpq

Similarly,

( )[ ])()2()2(sin1

)(/

)(1

)()(

sin1

1

23

0122

0

θωµ

ψµ

&&&&

&

−Ω++−+−−

∂∂=

∂

∂

−=

Ω=ΔΩ

+

+

∫

mMqplpliena

SeGiFa

St

eGi

iF

aa

iena

dtdt

d

l

lmpqlpqlmple

lmpqlpqlmp

l

le

t

t

lmpqlmpq

∑ΔΩ=ΔΩnmpq

lmpq

Similarly for other orbit elements


The frequency of the perturbation can be rearranged as follows (Schrama, 1989):

(66)

Form this representation, we identify three major frequencies. The first one is once-per-revolution component with and l-2p+q controls how many cycles per revolution. Thesecond one is daily component with and m controls how many cycles per day. Thethird one is long period component with . Note that many combinations of such indices canproduce the same frequency. In case of l=2p and q=0, the frequency is determined solely by thespherical harmonic order, m. producing m-daily components. The perturbation induced by aspecific order and all even degrees overlap on the same frequency (m cycles per day). Putting itanother way, the satellite with m cycles per day is sensitive to the resonant effect from the orderm and all even degree geopotential coefficients.

( )( ) ( ) ωωψ &&&&&& qTSGmMqpllmpq −−Ω+++−= 2

M&& +ωTSG && −Ω

ω&

Notes from S. Han

Example Problem: Resonances for GRACE

Find the primary and secondary resonance and the respective resonance periods (in days) for thesatellite GRACE, assuming an orbital altitude of 450 km, i = 89°.

€

˙ ω = −4.035° /day; ˙ Ω = −0.141° /day; ˙ M ≅ 5000° /day; ˙ θ ≅ 360° /day˙ ψ lmpq = (l − 2p) ˙ ω + (l − 2p + q) ˙ M + m( ˙ Ω − ˙ θ )

The resonance occurs when 0)()2()2( ≈−Ω++−+−= θωψ &&&&& mMqplpllmpq .

Assume 0=q , then

( ) ( ) ( )1487.13

360141.05000035.4

2

022

≈=−−+−

−=−Ω

+−=

−

≈−Ω+−+−

θω

θω

&&

&&

&&&&

Mpl

m

mMplpl

Example Problem: Resonances for GRACE (continued:)

For the primary resonance .12 ≈− plThen, the primary resonance occurs at .14=m

( )day

daydaydaydaymMlmpq

/009.46

)/360/141.0(14/5000/035.4o

oooo&&&&&

−=

−−×++−=−Ω++= θωψ

Primary resonance period = days82.72

=ψπ&

For the secondary resonance .22 ≈− pl

Then, .28=m( )

( )day

daydaydayday

Mlmpq

/012.92

)/360/141.0(28/50002/035.42

2822

o

oooo

&&&&&

=

−−×+×+−×=

−Ω×+×+×= θωψ

Secondary resonance period = days91.32

=ψπ&

Example Problem: Orbital variations due to geopotential

Determine all the frequencies (expressed in terms of angular rates, e.g., ˙ ω , M , ˙ Ω , et c) of

€

˙ ψ lmpqassociated with C30. Justify the ranges of the indices (l, m, p, q). Categorize frequencies, ifavailable, which are associated with secular, short period, long period, m-daily, and resonances.

Example Problem: Orbital vairations due to geopotential (Continued:)

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