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6|Orbits
207
4.
5.
6.
7.
8.
Fig.6.7
Youcanseefromthesepicturesthatthereisnostableorbit.Acircularorbitisjustzerodistance
awayfromanorbitthateithercrashesintothecenteror iesaway.Theinversecubeorbitiswhatyou
shouldexpectinauniversewithfourdimensionsofspace.Foracontrast,lookattheeectivepotential
energygraphfor
F
/1
=r
andyouwillseethatnothing canescapeanythingelseandallstarswould
coalesce. Thisisthepotentialyougetintwodimensions,sothreeisthemagicnumber.Wegotlucky.
Graph
U
eective
fortheinversecubeforce.Thatwillexplainalot.
6.10HyperbolicOrbits
IntheanalysisoftheKeplerorbits,equations(6.15)and(6.16),theeccentricity
waslessthanone.
Whatifitisn’t?ThepolarequationfortheellipsefromEq.(6.17)becomestheequationforahyperbola,
thoughitneedsasignchangewhen
>
1.
r
=
a
(
2
1)
1+
cos
Nowwhen
>
1thedenominatorcan vanish. . Thatisfor
=cos
1
( 1
=
),. Theseangleshave
>=
2and
<
=
2atthesymmetricpositions. Toverifythatthisisahyperbola,putitinthe
morefamiliarrectangularform.
r
+
r
cos
=
a
(
2
1)  !
p
x
2+
y
2+
x
=
a
(
2
1)
x
2
+
y
2
=
a
(
2
1) 
x
2
=
a
2
(
2
1)
2
2
a
(
2
1)
x
+
2
x
2
x
2
(1 
2
) 2
a
(1 
2
)
x
+(1 
2
)
a
2
2
(1 
2
)
a
2
2
+
y
2
=
a
2
(
2
1)
2
(1 
2
)(
x
a
)
2
+
y
2
=
a
2
(
2
1)
2
+
a
2
(1 
2
)
2
a
2
(
2
1)
(
x
a
)
2
a
2
y
2
a
2(
1)
=1
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6|Orbits
208
Fig.6.8
The hyperbola has two branches,but only one applies tothe orbit. . The e other r branch h is the
dashedcurve,anditwillnotbeneededuntilproblem6.54. Inordertodrawthesolidcurveseasily,you
canparametrizethemby
x
=
a
a
cosh
;
y
=
a
p
2
1sinh
(6
:
57)
Forthe other(dashed)branch,changethe signinfrontofthe cosh
in this expression for
x
. The
orbits drawn herehave eccentricities1.15, , 1.3, 1.7|whichiswhich? ? The e orbits of comets can be
ellipseswithverylargeeccentricity,sothattheywillreturnevery76or10000yearsasthecasemaybe.
Cometary orbitscanalsobehyperbolassuchasthese,andsuchacometwillthenappearonlyonce,
thenleaveforever.Thesehyperbolicorbitscorrespondtotheenergy
E
3
atEq.(6.24).
6.11TimeDependence_
AllthecalculationsconcerningtheKeplerproblemresultedintheshapeoftheorbit,buttheydidn’t
givethepositionasafunctionoftime. ThereasonI’veputthatquestionoisthatthereisnoneat,
explicitsolutiontotheequationsintermsoftime. Thereishoweverasolution,justnotanexplicitone.
Thecalculationisalittleintricate,butnosinglestepistoobad.
Thisprocedurewill not produce equationsfor
r
and
asfunctionsof
t
. Insteadyouwill l get
threeequationsthatgiveyou
r
,
,and
t
asfunctionsofafourthparameter.
Thetypicalwaytosolvefor
r
(
t
)fromEq.(6.10)istogettheenergyintegralandthentouse
separationofvariables. Theprocedurehereisclosetothat,butwithatwist. . Theenergyintegralis
Eq.(6.22),andaftersolvingfor _
r
itisseparable,justasinEq.(2.23). Ifyouproceedalongthisroute
theintegraltodotond
t
(
r
)wouldbe
E
=
1
2
m
_
r
2
+
m‘
2
2
r
2
GMm
r
!
dt
=
r
m
2
dr
r
E
m‘
2
2
r
2
+
GMm
r
(6
:
58)
Todotheintegralyoucanmultiplythenumeratorand denominatorofthe
dr
integralby
r
,making
thebig square rootaquadraticpolynomial. . Youthencompletethe e square on thatpolynomial and
integrate. That t will give you
t
(
r
), not
r
(
t
), and you can’tinvertthe resulting expression to get
r
explicitly. Thecleverthingtodoistoredenetheindependentvariable
t
. Indthesimplestwayto
presentitfollowsSundman*andchangesvariablesintheenergyequation.
* Itissometimessaidthatthegeneralsolutionforthetimedependenceofthreemutuallyorbiting
masses is impossible|the three e body y problem. . Poincare e did prove e that t it is s impossible e using the
particularmethodsthatheapplied,butSundmanshowedthatthereareotherways,demonstratinghow
tond aseriessolutioninpowersof
t
1
=
3
,albeitaveryslowlyconvergingone. The
n
-bodysolution
wasn’tfounduntilthe1990’s.
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6|Orbits
209
Doingthis,itiseasiertoworkwiththeenergyequationfromthersthalfofEq.(6.58)andnot
themanipulatedandseparatedformthatisthesecondhalf.Changevariablesfrom
t
to
s
,where
ds
dt
=
r
then
dr
dt
=
dr
ds
ds
dt
=
r
dr
ds
:
Let
r
0
=
dr
ds
E
=
m
2
2
r
2
r
0
2
GMm
r
+
m‘
2
2
r
2
!
1
2
m
r
0
2
GMm
2
r
+
m‘
2
2
2
=
E
2
r
2
(6
:
59)
This is s aharmonic c oscillatorif
E <
0. The e parameter
is still arbitrary and Ican choose it for
convenience. Twoplausiblechoicesare
=1and
=
p
GM
,butIwillletthechoicehangfornow
becauseitisjustaseasytocarrythe
alonguntilyouareforcedtodecide. Isthisreallyaharmonic
oscillator?Yes,justcompletethesquare(remember,
E<
0).
1
2
m
r
0
2
E
2
r
+
GMm
2
E
2
m‘
2
2
2
G
2
M
2
m
2
4
E
2
(6
:
60)
Thisisinthesameformastheenergyconservationequationthatyou’veseenforamassona
spring:
1
2
mv
2
+
1
2
k
(
x
x
0
)
2
=
E
0
=)
x
(
t
)=
x
0
+
A
cos(
!t
+
)
; !
2
=
k=m; kA
2
=
2=
E
0
With this you can read the frequency (
p
k=m
), the center r (
x
0
) and d the e amplitude of oscillation
(
A
=
p
2
E
0
=k
).Thetranslationtableforthepresentcaseisthen
x
(
t
)!
r
(
s
)
; m
!
m; k
! 2
E=
2
; x
0
GMm=
2
E; E
0
m‘
2
2
2
G
2
M
2
m
2
4
E
2
LookbackagainatthepictureofanellipseatEq.(6.17). Thevalueof
r
variesfromamaximumof
a
+
f
downtoaminimumof
a
f
.Translatethisintothelanguageofaharmonicoscillatorandthat
says
x
0
!
a
and
A
!
f
. Theequationfor
r
(
s
)isthen
r
r
(
s
)=
a
f
cos
!
(
s
s
0
)
(6
:
61)
Translatefurther:
!
2
=
k
m
2
E
m
2
;
x
0
!
a
GMm
2
E
;
A
2
=
2
E
0
k
!
f
2
=
2
m‘
2
2
2
2
G
2
M
2
m
2
4
E
2

2
E=
2
=
m‘
2
2
E
+
(
GMm
)
2
4
E
2
(6
:
62)
Whatabout
t
?Take
s
=0when
t
=0,andforconveniencechoose
s
0
=0.
t
ds
dt
=
r
!
dt
=
rds
=(
a
f
cos
!s
)
ds
!
t
(
s
)=
1
as
f
!
sin
!s
(6
:
63)
r
2
_
=
‘;
so
d
dt
=
d
ds
ds
dt
=
d
ds
r
=
r
2
!
0
=
r
(
s
)=
Z
s
0
ds
1
a
f
cos
!s
=
2
!b
tan
1
b
a
(1 
)
tan(
!s=
2)
(6
:
64)
=
2
!b
tan
1
r
1+
tan(
!s=
2)
!
=
2
!b
tan
1
cot(
)tan(
!s=
2)
=
2
!b
sin
1
0
@
cos(
)sin(
!s=
2)
q
sin
2
(
)cos
2
(
!s=
2)+cos
2
(
)sin
2
(
!s=
2)
1
A
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6|Orbits
210
Here
b
istheusualsemi-minoraxisoftheellipse,
b
2
=
a
2
f
2
and
f
=
a
andcos2
=
. (Gradshteyn
andRyzhik2.553.3)
r
t
s
!
r
=
a
+
f
r
=
a
f
=
c
=0
=2
,
t
=0,
t
=1yr
Fig.6.9
Thesethreegraphsshow
r
,
,and
t
asfunctionsof
s
foranorbitwitheccentricity0.66asin
thepictureatEq.(6.17). Toreadandinterpretthesegraphs,takeasanexampleaspecicpartofthe
orbit: theperihelion. . Forminimum
r
=
a
f
,
s
iszero|theleftendofthegraph. Whatis
d=dt
atthatpoint?
d
dt
=
d
ds
ds
dt
=
d
ds
dt
ds
Youcanseethatattheleft(orright)endsofthegraphtheslopeofthe
t
-curveisatitssmallest
andtheslopeofthe
-curveisatitslargest.Thisgives
_
itslargestvaluethere,atperihelion,justas
Kepler’slawsays.Theplanetdoesn’tspendmuchtimeneartheperihelion. Attheotherextreme,the
aphelioninthemiddle,youcanagainlookattheslopesandseethat
_
issmallestthere.
Whyaretheresomanydierentversionsoftheresultfor
(
s
)inEq.(6.64)?Theansweristhat
whensolvingacomplicatedproblemyoudon’timmediatelyknowwhichformoftheresultwillbethe
mostuseful. Itisprudenttowriteitinseveraldierentwaysuntilyougetabetterideaoftherouteto
take. Forexample,there’saminortechnicaldicultywhenplottingthegraphof
ifyouusetherst
form.Thearctangentismultiplevalued,soyouhavetousesomecareinevaluatingit. Ifyouuseany
oftherstthreeversionsyoumayneedtowritethetangentas(sineovercosine)andthentotreatthe
arctangentasafunctionoftwovariables. Thatwillletyoukeeptrackofwhichquadrantyouarein.
Thefourthversiondoesn’thavethisdiculty. Thethirdversionhoweverwillhaveadvantageswhen
youarestarttolookatunbound(
E>
0)orbits.
Doesthiscomplicatedlookingsetofequationsreducetothecorrectresultswhentheorbitisa
circle?
=0,
b
=
a
,
f
=0,so
r
(
s
)= 
GMm
2
E
=
a;
t
(
s
)=
1
as;
(
s
)=
2
!a
tan
1
tan(
!s=
2)=
a
s
Thestatementthat
A
=0(novariationin
r
)determines
.
A
2
=
m‘
2
2
E
+
(
GMm
)
2
4
E
2
=0  !
2
G
2
M
2
m
2
E
!
=
GM
p
2
E=m
d
dt
=
(
‘= a
)
ds
(
a=
)
ds
=
a
2
andthatis(
r
2
_
=r
2
)asitshouldbe.
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6|Orbits
211
6.12PerihelionofMercury_
Thissectionismoredemandingthanmostofthistext,butitshowsthepowerofthetoolsthatyou’ve
studied. Thesolarsystemisawhole e lotmorecomplicated than anythingthatdone sofar,andthis
sectiondevelopsoneaspectofthiscomplexity.
1. Thegravitationalforcesarenotjustfromthesun.
2. Theplanetsmoveonlyapproximately inclosedorbitsalongellipses.
3. Itispossibletogureouttheeectsthattheplanetshaveoneachother,atleasttoaverygood
approximation.(Wecan’tdoeverythinghere.)
Thedevelopmentofperturbedorbitsafewsectionsbackhas anim-
portant application thatatrstmay seem impossibletodo: : Precession n of
theperihelion of Mercury. . Each h planetispulledbythe sun, butitisalso
pulledbyalltheotherplanets,andthatwillaecttheshapesoftheirorbits.
Mercury’sorbitinparticularwas foundtohavemeasurabledeviationsfrom
asimple ellipse. . Its s orbit precessed by y an n amount that was large enough
to be measured as s far back k as 1800, , and this s measurement is associated
withasignicantmilestoneinthehistoryofphysicsbecausethecalculations
of the precession and the measurements of the precession did not agree.*
Accuratemeasurementsoftheperihelion’smotionweremadeinthe1850’s,
precise methodstocalculatethe predictedmotionshadbeen developed by then,andformore than
halfacenturythedisagreementremainedavexingproblem. OnlywhenEinsteinpublishedhistheory
ofgravity(general relativity)in1916was theproblemresolved. . Thedrawingis s greatly exaggerated,
bothintheeccentricityofMercury’sorbitandinitsrateofprecession. Thedotsarenearwherethe
perihelionisontheseorbits. TheangleofprecessioninoneEarthcenturyis
532
00
(Newtonian)+43
00
(generalrelativity)=575
00
(measured)
Howdidthey computethisprecession? ? That’swhatIwanttodohere|atleastinpart. . The
person whofound the discrepancy y between n theory and observation was Urbain Le Verrier,and you
canreadsomethingabouthiminWikipedia. (Healsowasoneofthetwopeoplewhoindependently
predictednotonlytheexistence,butthepositionofNeptune.)
Theperturbationmethodsdescribedsofarinthischapterinvolvedmakingsmallchangesinthe
potential energy andinthe correspondingradialforce,
F
r
(
r
) = 
dU
(
r
)
=dr
. In n all casesthesmall
change involved addingsomething that was spherically symmetric. . All l the changes to
U
(
r
) added
anothersmall term
U
1
(
r
),andwhetheritwas acloud ofdustinthesolarsystemorabulgeatthe
Sun’s equatorit was symmetric. . The e pull on Mercury by theplanetsVenus orJupiteris not atall
symmetric;theirattractionsarenottowardtheSunandtheyareverymuchtime-dependent. Sonow
what?
Thereis aclevertrickdue toGauss. . Henoted d thatplanets orbit the Sun in periodsthatare
measuredinyears|1
=
4
yearinthecaseofMercury,165yearsforNeptune. Themeasuredprecession
rateofMercuryisonlyabout0
:
16
inoneEarthcentury,andthisisanextremelyslowmotioncompared
tothatofanyplanet,sowhynotsmearoutandaveragethepositionsoftheplanets?They’rewhizzing
aroundso(comparatively)fastthattheirdetailedpositionasafunctionoftimeshouldn’tmatter,only
theirorbit. Theaverageinthiscasemeansthatyoureplacetheorbitofagivenplanetby y aringof
uniformlinearmassdensity,thentheeectofsucharingonthemotionofMercurywillbesymmetric.
HereIwillmaketheprettygoodsimplifyingapproximationthatalltheplanetaryorbits(exceptMercury)
are circles andallofthem orbitin exactly the sameplaneas Mercury. . Renementsbeyond this are
moredicultandI’llsavementionofthemuntiltheend. Itisquiteamazinghoweverjusthowclose
thissimplemodelcomes(about4%).
* Close,butnocigar.
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6|Orbits
212
Thisisnow,atleastinprinciple,thesamesortofperturbationproblemdevelopedearlierinthis
chapter. The e only initialdicultyisinguringouttheeld ofaringofmass. . The e answerforthe
potentialenergyis
U
1
(
r
)= 
GMm
2
Z
2
0
d
1
p
R
2+
r
2
Rr
cos
;
V
1
=
U
1
m
(6
:
65)
M
isthe massoftheouterplanet;
R
is its orbital radius;
m
is Mercury’smass,
U
is the potential
energy,and
V
isthegravitationalpotential. This
r
isintheplaneofthering,andthedenominator
in theintegral comesfrom usingthe lawofcosines. . Don’t t beintimidated. . This s sortofcalculation
becomesfairlyroutineonceyou’vetakenoneE&Mcoursebeyondtherst.(
dM
=
Md=
2
)
Asacheckonthisexpressionforthepotential,whatisitat
r
=0?*
V
1
(0)= 
GM
2
Z
2
0
d
1
p
R
2
GM
2
2
R
GM
R
and since the entire mass
M
is at the constant distance
R
, this is s correct. . Ratherthan n spending
time discussinghowtoevaluatethisintegral,Iwill state the resultforitslowest orderpowerseries
representation. Attheendofthissectionyouwillndamorecompleteresult.Outtotermsin
r
2
this
potentialis(seeproblem6.70forhowtodothis.)
V
1
(
r
) 
GM
R
1+
r
2
4
R
2
;
U
1
(
r
)=
mV
1
(
r
)
; F
r
(
r
)= 
dU
1
dr
+
GMm
2
R
3
r
(6
:
66)
Thenextpicturesshowsaviewfromabovetheplaneoftheplanet’sorbitontherightandasideview
oftheorbitontheleft.Italsoshowsagraphof
V
(
r
),andyoucanseebothitscurved(
r
2
)shapenear
r
=0anditsbehaviornear
r
=
R
. IchosetolabelitastheorbitofVenus,butitcouldaseasilybe
anyoftheotherplanetsfartherout.
Fig.6.10
TopView
SideView
V
(
r
)
R
OrbitofVenus
TheradiusofMercury’sorbitisaboutone-halfofVenus’s,soyoucanseeroughlywhereinthis
graphtoputMercury. Italsomeans s thatthebehaviorof
V
near
r
=
R
will notberelevanttothe
problem.
NowfortheeectofthisextraforceontheplanetMercury,andthistakesusbacktosection
6.5. ThecoreequationisEq.(6.14),where
u
=1
=r
and
M
isthesolarmass. Also,
f
dU=dr
andleadstoEq.(6.34).
f
(
r
)= 
GM
m
r
2
+
GMm
2
R
3
r
GM
mu
2
+
GMm
2
R
3
u
d
2
u
d
2
+
u
1
m‘
2
u
2
f
1
u
=
GM
2
GM
2
2
R
3
u
3
* andwhatisitfor
r
R
?
6|Orbits
213
Followthesameprocedureasthere,lookingforoscillationsaboutacircularorbit:
u
(
)=
u
0
+
x
(
),
with
x
u
0
=1
=r
0
,andkeepingtermsonlytorstorderin
x
.
d
2
(
u
0
+
x
)
d
2
+
u
0
+
x
=
GM
2
GM
2
R
3(
u
0
+
x
)3
d
2
x
d
2
+
u
0
+
x
=
GM
2
GM
2
2
R
3
u
3
0
1 3
x
u
0
!
u
0
=
GM
2
GM
2
2
R
3
u
3
0
and
d
2
x
d
2
+
x
=+
GM
2
2
R
3
u
3
0
3
x
u
0
Thesolutionfor
x
isthen
x
(
)=
x
0
cos
!;
where
!
2
=1 
3
GM
2
2
R
3
u
4
0
Rememberthatthisisarstorderexpansion,soinputting
u
0
intothis
!
toevaluateit,youshould
keeptermstoaconsistentorder. Theequationfor
u
0
hasthe
GM
termasasmallcorrectiontothe
rstterm,sowhenyouusethistoevaluate
!
,youitwouldbeinconsistentkeepit.Use
2
=
GM
r
0
tosimplifytheequations.
1
r
0
=
u
0
GM
2
!
!
2
=1 
3
2
GM
u
0
1
2
R
3
u
3
0
1 
3
2
M
M
r
3
0
R
3
r
=
1
u
=
1
u
0
+
x
=
1
u
0
(1 
x
u
0
)=
r
0
r
2
0
x
=
r
0
r
2
0
x
0
cos
!
(6
:
67)
with
!
=1 
3
4
M
M
r
3
0
R
3
(6
:
68)
Thisiswhatwenowhavetoanalyze,puttinginthenumbersandseeingwhatitpredicts.
First:
!<
1,soinorderthat
!
=2
,theangle
mustbegreaterthan 2
. Mercurydoesn’tget
backtotheperihelionuntilithasgoneslightlymorethanoncearoundtheSun. Thisiscalledprograde
precession,theoppositeofretrograde.
Second: Howmuchmorethanoncearound?
!
.
(2
+
)=2
,ortothisorder
3
4
M
M
r
3
0
R
3
(2
+
)=2
!
=+
3
4
M
M
r
3
0
R
3
.
2
(6
:
69)
Third: What t is
r
0
? Mercury’s s pathisreally anellipse you recall. . Yousee e fromEq. (6.67)thatits
perihelionisataradius
r
0
r
2
0
x
0
anditsaphelion(at
=
)isat
r
0
+
r
2
0
x
0
.Thatmeansthat2
r
0
is
thewholemajoraxisandthat
r
0
isthenthesemi-majoraxis,
a
.
EachoftheouterplanetswillcontributetotheyearlyadvanceoftheperihelionofMercuryan
amountgivenbyEq.(6.69).
total
=
k
=Neptune
X
k
=Venus
3
4
M
k
M
a
3
R
3
k
.
2
(6
:
70)
6|Orbits
214
R=R
E
M=M
E
Mercury
0
:
38710
0
:
05527
Venus
0
:
72334
0
:
8150
1
:
76710
6
Earth
1
:
0
1
:
0=5
:
973610
24
kg
8
:
309510
7
Moon
0
:
0123
Mars
1
:
52371
0
:
10745
2
:
493310
8
Jupiter
5
:
2029
317
:
83
1
:
8524610
6
Saturn
9
:
537
95
:
159
9
:
00510
8
Uranus
19
:
189
14
:
500
1
:
68510
9
Neptune
30
:
061
17
:
204
5
:
20410
10
Pluto
39
:
482
0
:
0022
Sun
333000
:
=1
:
98910
30
kg=
M
Shouldsomeofthelargermoonsbeincluded? Ganymede,amoonofJupiter,is0
:
025themass
ofEarth,butourownmoonishalfthemassofGanymedeandit’svetimesclosertotheSun,soit
willcountformuchmore,andit’sstillasmallcorrection.
This table is expressed as ratios s to o the orbital distance and the mass s of f Earth, and because
Eq.(6.70)involvesonlyratios,thatisallyouneed.
3
2
a
3
M
k
=Neptune
X
k
=Venus
M
k
R
3
k
=
3
2
0
:
3871
3
333000
0
:
815
0
:
723343
+
1
:
0123
13
+
0
:
1074
1
:
52373
+
317
:
83
5
:
20293
+
=4
:
568610
6
radiansperorbitofMercury
=0
:
9423
00
perorbit365
:
24100
=
87
:
97
=391
00
perEarthcentury=0
:
11
perEarthcentury
Therighthandcolumninthetableaboveismadeupofthesuccessivetermsinthisseries. Youcan
seethattheorderofin uenceisJupiter,Venus,Earth,withtheotherplanetswellbehind.Ifyougoto
higherordersinthecalculationof
U
1
(
r
),keepingtermsin
r
4
=R
4
,Venuswilltakeandkeepthelead.
Thisestimateisnotabadstart,but0
:
11
ascomparedto0
:
16
isnotyetgoodenough.
Itis possibletocalculatethisprecessionwithoutmakingtheapproximationinEq.(6.66),that
thepotentialisproportionalto
1+
r
2
=
4
R
2
. Youcanusethemethodsdescribedinthenextsectionto
manipulatethepotentialofEq.(6.65)exactly,orifyouprefer,thereisamorefamiliartypeofmethod
available: UseaseriesrepresentationfromEqs.(6.72)and(6.73). Thatis
V
1
(
r
)=
U
1
(
r
)
m
GM
R
"
1+
1
2
2
r
R
2
+
1
.
3
2
.
4
2
r
R
4
+
1
.
3
.
5
2
.
4
.
6
2
r
R
6
+
#
YouuseeachsuccessiveterminthisseriestocomputeitseectonMercurybyfollowingexactlythe
sameprocedureasforthe
r
2
term. Youwillneedalotoftermstogetareallygoodresult,whichis
why,inthenextsection,Iwanttoshowyoualessfamiliarmethodthatavoidsthisseries.
6.13EllipticIntegrals_
I’mincludingthissectionhopingtopersuadeyouthatifyouencounteranintegralorafunctionthat
you’veneverheardofbefore,youshouldn’tbeintimidated.Ifyouseea\sineintegral"whenstudying
opticsthenyoushouldbereadytolookupwhatitis.Ifyoustudyelectromagnetismoranyofdozensof
othertopicsthendon’tbesurprisedbyBesselfunctions,therearebooks,tables,andcomputerprograms
todealwiththem. Gammafunctionstooareubiquitous,andtheyareeasy. Legendrepolynomialsare
easytoo,andyou’vealreadyrunintothemhereinEq.(5.36). Legendrefunctions canbeanuisance,
6|Orbits
215
buttherearetoolsreadilyavailabletohandlethem. IfhoweveryoushouldfaceaMathieufunctionin
adarkalley,youwillhavemysympathy.
Here aretwocommonlyusedfunctions,
K
and
E
: the\completeellipticintegrals s oftherst
andsecondkinds". Alongwiththeirdenitions,herearesomeusefulformulasstraightoutofstandard
references.
K
(
k
)=
Z
=
2
0
d
1
p
k
2sin
2
=
Z
1
0
dx
1
p
(1 
x
2)(1
k
2
x
2)
=
1
2
Z
0
d
1
p
1+
k
2
k
cos
E
(
k
)=
Z
=
2
0
d
p
k
2sin
2
=
Z
1
0
dx
r
k
2
x
2
x
2
(6
:
71)
=
2
1
k
0
1
K
E
dK
dk
=
E
(
k
)
k
(1 
k
2)
K
(
k
)
k
dE
dk
=
1
k
E
(
k
K
(
k
)
K
(
k
)=
2
"
1+
1
2
2
k
2
+
1
.
3
2
.
4
2
k
4
+
1
.
3
.
5
2
.
4
.
6
2
k
6
+
#
E
(
k
)=
2
"
1
2
2
k
2
1
.
3
2
.
4
2
k
4
3
1
.
3
.
5
2
.
4
.
6
2
k
6
5

#
(6
:
72)
Fromthetwoequationsfor
dK=dk
and
dE=dk
youseethathigherderivativesof
K
and
E
canalso
be expressed interms of the original
K
and
E
functions simply by dierentiating these derivatives
repeatedly|sortoflikethesineandcosine. Equation(6.65)callsfortherstofthesefunctions
K
,
buttheperturbationequationsthatfollowitwillrequiretherstandsecondderivativesof
K
.
U
1
(
r
)= 
GMm
2
Z
2
0
d
1
p
R
2
+
r
2
2
Rr
cos
GMm
2
R
Z
2
0
d
1
p
1+(
r
2
=R
2) 2(
r=R
)cos
2
GMm
R
K
r=R
(6
:
73)
Areasonnottoshyfromusingthese integralsisthattherearesomeveryfastandecientways to
evaluate them,andthose methodsare much fasterthen astraightnumerical integration or r aseries
expansionwouldbe. Also,whatweneednowistouseEq.(6.14),whichrequiresknowingonly
f
and
f
0
,andthosecomefrom
dU=dr
and
d
2
U=dr
2
,andthoseinturnrequireevaluating
K
and
E
onlyat
a=R
k
forseveralplanets. Thisisreallynottoomucheortforaveryimportantresult. . Andwhyare
thesefunctionscalled\elliptic"? Seeproblem6.71tondout.
Thegraphaboveshowsthebehaviorof
K
and
E
inthedomain0
k<
1andtheyareobviously
evenfunctions.Theleftendofthegraphat
k
=0iseasytoevaluatefromtheequations(6.71)using
the
d
integrals. Attheotherend,at
k
=1,the
dx
formmakethecalculationof
E
easy,butifyou
lookfor
K
atthesamepointyouwillseethattheintegraldiverges. It’samilddivergence,beingonly
logarithmic,andthebehaviorof
K
near
k
=1is
K
(
k
)
1
2
ln
8
=
(1 
k
)
. Thesecondderivatives
of
K
and
E
comebydierentiatingtheexpressionsatEq.(6.72)thatgivetheirrstderivatives. It’s
tediousbutstraightforward,andtheresultsare
d
2
E
dk
2
=
E
+(1 
k
2
)
K
k
2(1
k
2)
;
d
2
K
dk
2
=
( 1+3
k
2
)
E
+(1 2
k
2
)(1 
k
2
)
K
k
2(1
k
2)2
6|Orbits
216
Nowreturntotheperturbedorbitalequations.
u
=1
=r
,and
M
and
R
aretheotherplanet’s
massandorbitalradius. UsethemethodthatstartswithEq.(6.32):
d
2
u
d
2
+
u
1
m‘
2
u
2
f
1
u
with
f
(
r
)= 
d
dr
U
(
r
)= 
d
dr
GM
m
r
2
GMm
R
K
r=R
=
GM
m
r
2
+
2
GMm
R
2
K
0
r=R
Usingthesame
u
=
u
0
+
x
with
x
(
)beingtheperturbationfromacircularorbit,
d
2
u
d
2
+
u
=
GM
2
2
GM
R
2
2
u
2
K
0
1
=Ru
d
2
(
u
0
+
x
)
d
2
+
u
0
+
x
=
GM
2
2
GM
R
2
2(
u
0
+
x
)2
K
0
1
R
(
u
0
+
x
)
d
2
x
d
2
+
x
u
0
+
GM
2
2
GM
R
2
2
u
2
0
2
x
u
0
K
0
1
Ru
0
x
u
0
u
0
+
GM
2
2
GM
R
2
2
u
2
0
2
x
u
0
K
0
1
Ru
0
K
00
1
Ru
0
x
Ru
2
0
Thetermswithout\
x
"ontherightofthenalequationdenethecircularorbit
u
0
. Thetermswith
an
x
aretheonesthatcontributetotheprecession.
d
2
x
d
2
+
x
2
GM
R
2
2
u
2
0
K
0
1
Ru
0
2
u
0
+
K
00
1
Ru
0
1
Ru
2
0

=0
!
2
=1 
2
GMr
2
0
R
2
2
2
r
0
K
0
r
0
R
+
r
2
0
R
K
00
r
0
R
Simplifythisusing
r
0
=
a
and
2
=
GM
a
tothisorder,then
!
=1 
Ma
2
M
R
2
h
2
K
0
a
R
+
a
R
K
00
a
R
i
Therstcheck: Doesthis s agreewith therstordercalculation ofEq.(6.68)? ? Ithinkthatit
does,butyou’dbetterlook,usingthepowerseriesfor
K
.Asintheprevious,lowestordercalculation
theprecessioncomesfrom
!
.
(2
+
)=2
. Then
=+
2
Ma
2
M
R
2
h
2
K
0
a
R
+
a
R
K
00
a
R
i
(6
:
74)
Tocheckthesign,lookatthegraph of
K
ontheprecedingpage. Whatarethesignsof
K
0
andof
K
00
,anddoesthatmatchthesignfor
inEq.(6.69)?Toincludetheeectsfromseveralplanets,use
thesamenotationasinEq.(6.70):
=
k
=Neptune
X
k
=Venus
2
M
k
a
2
M
R
2
k
2
K
0
a
R
k
+
a
R
k
K
00
a
R
k

=553
:
4
00
perEarthcentury
(6
:
75)
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