﻿
6|Orbits
187
Dividethese.
T
H
T
E
=
a
H
a
E
3
=
2
!
a
H
a
E
=
T
H
T
E
2
=
3
Thelongestdistancethecometcouldgooccursifitsellipticalorbitisastraightline|anextremecase.
Thatwouldmakethemaximumdistancefromthesuntwiceitssemi-majoraxis.
2
a
H
=2
a
E
T
H
T
E
2
=
3
=2
a
E
.
75
:
32
2
=
3
=35
:
67
a
E
(6
:
11)
Thetruedistancewillbealittlelessthanthis,andismeasuredtobe35.1AU,whereoneastronomical
unit(AU)equals15010
9
m=150Gm,whichisalmostexactlythesemi-majoraxisofEarth’sorbit.
6.3KeplerProblem
Nowpickaspecic
f
thatvariesas1
=r
2
.Astraightforwardattackonequation(6.10)isdicult,but
therearesomesimple-lookingresults,oneofwhichisthattheorbitsareshapedasellipses. Thatyou
Itprovides noclueaboutsolving(6.10) directly. . Theclueis s thatyoumay getasimpleresultifyou
eliminatethetimevariableinfavorofanother.Thisworks,andtheotherindependentvariabletouseis
. IfIcansolveforthefunction
r
(
)perhapsI’llrecognizeitastheequationforanellipse. Nowhow
manypeopleknowenoughanalyticgeometrytowritethepolarequationforanellipse? Rectangular
coordinatesmaybe,butprobablynotpolar.
Theequationinquestionis
r
(
)=
A
B
+
C
cos
j
C
j
<
j
B
j
(6
:
12)
Amorefamiliarrectangularequationforanellipseis
x
2
a
2
+
y
2
b
2
=1 centeredattheorigin,ormoregenerally,
(
x
x
0
)
2
a
2
+
(
y
y
0
)
2
b
2
=1
(6
:
13)
WriteEq.(6.12)as
Br
+
Cr
cos
=
A
;write
r
and
r
cos
intermsof
x
and
y
,andyoucannishit
easilyinproblem6.4,showingthatthepolarformandtherectangularformdescribethesamecurve.
Anotherwaytodeneanellipseparallelsthedenitionofacircle. Acircleisthesetofpointsallat
aconstantdistancefromonexedpoint. Anellipsetaketwopointsandlooksforthesetofpointsso
thatthesumofthedistancestothetwopointsisaconstant,asinthepictureatEq.(6.17).* Thetwo
pointsarethefocioftheellipse.
Thispolarequationforanellipsedoesn’thaveanyobviouspropertiesthatImightrecognizein
=r
?That’sproportionalto
B
+
C
cos
andwespentawhole
chapteronfunctionssuchas that|harmonicoscillators. . Workingbackwards s from the experimental
resultsthensuggeststwochangesofvariables,boththeindependentandthedependentones.
t
!
and
r
!
u
=1
=r
ofdistances? They’refuntoo.
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6|Orbits
188
Nowit’salotofapplicationofthechainrule. Youcandothechangesoneatatimeineitherorder,
Recall:
=
r
2
_
;
so
‘u
2
=
_
dr
dt
=
dr
du
du
d
d
dt
1
u
2
du
d
‘u
2
du
d
d
2
r
dt
2
d
2
u
d
2
d
dt
2
u
2
d
2
u
d
2
PutthischangeintoEq.(6.10)toget
2
u
2
d
2
u
d
2
=
1
m
f
1
u
+
2
u
3
;
or
d
2
u
d
2
+
u
1
m‘
2
u
2
f
1
u
(6
:
14)
Thisisstartingtolooklikeaharmonicoscillator. Canyousolveit?Thatdependson
f
,andfor
whichfunctions
f
isiteasy?Iftherightsideisaconstantorifitisproportionalto
u
,thenyoucando
itwithoutdiculty,otherwiseit’shard. WhathappenswiththeKeplerproblem?
f
(
r
)=
GMm
r
2
=)
d
2
u
d
2
+
u
1
m‘
2
u
2
GMmu
2
=+
GM
2
(6
:
15)
Leavetheothereasycaseforlater,section6.9,becausethoughthoseresultsareamusing,they’renot
very important. . (What
f
would that be?) ) Most t of therest of this chapterwill concern theexact
solutionstotheKeplerproblemandapproximatesolutionstoalltheothers.
Equation(6.15)iseasy.
u
=
GM
2
+
C
cos(
0
)
;
or
1
u
=
r
=
2
=GM
1+
cos(
0
)
(6
:
16)
C
was anarbitrary constant so
=
C‘
2
=GM
istoo,andit’s dimensionless. . This s is theellipse as
statedinKepler’slawsaslongastheparameter
hasmagnitudelessthanone. (Moreonthatlater,
section6.10.) Thenumeratorof(6.16)isalength,and
0
isanangle. Imayaswellassumethat
0
becauseifnot,thenIcanredenetheparameter
0
toit. Thatchangesthesigninfront
ofthecosinebacktopositiveagain.Withthisconventionon
,theangle
0
isthedirectioninwhich
thedenominatorislargest. Itpointstowardthesmallest
r
intheorbit.
perihelion,andtheonefarthestawayistheaphelion. ForobjectsorbitingtheEarth,thecorresponding
namesareperigeeandapogee,andthegenerictermsareperiapsisandapoapsis.
=0
:
66
a
b
f
r
c
r
0
foci
r
=
a
(1
2
)
1+
cos
f
=
a
b
=
a
p
2
a
2
b
2
=
f
2
r
+
r
0
=2
a
A
=
ab
c
=
a
f
=
a
(1
)
(6
:
17)
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6|Orbits
189
Thisguresummarizesmanypropertiesoftheellipse. Thetwogoverningparametersare
a
,thesemi-
majoraxis,and
,theeccentricity. Othervaluesthatcanbederivedintermsofthesestartwiththe
semi-minoraxis
b
,thedistancefromthecentertoafocus
f
,andtheperiheliondistance
c
.Theorigin
ofthepolarcoordinatesystemisattherighthandfocus. That’swheretheSunis.Theotherfocusis
symmetricallyplacedontheothersideoftheellipse,andthesumofthedistancesfromthetwofoci
(
r
+
r
0
)isaconstant. Theellipseinthedrawinghasafairlylargeeccentricity,
=0
:
66,whichisfar
largerthan the eccentricityof any planet. . Earth’sis s 0.017,andevenMercury has aneccentricityof
only0.21,thelargestofalltheplanets.* Onthescaleofthisdrawing,Earth’sorbitissonon-eccentric
thattheSun’spositionwouldbealmostinthecenteroftheellipse|justontheedgeofthedarkened
partoftheverticalaxisinthepicture. OnlyVenushasalowereccentricitythanEarth’s: 0.0068. The
Despitethis,Tycho’sobservationaldata,takenbefore1600andbeforetheinventionofthetelescope,
weresoaccuratethatKeplerwasabletousethemtodiscoverhislaws. Forthederivations s ofthese
propertiesoftheellipse,seetheexercisesonpage218
Example
Whathappensto thisellipseifitisstretchedouttoinnityalongthe majoraxis? ? Thatdepends
onhowyoudoit,becauseyoucouldstretchitwhilekeeping
b
constant. Thatwillresultintwolines
b
apart.Amoreinterestinglimitappearsifyoulet
a
!1
whilekeeping
c
constant.
c
isthedistanceofclosestapproachontherightend.
c
=
a
(1
)
;
sothisrequires
!1 as
a
!1
Gouptotherstequation,theonefor
r
.
r
=
a
(1
2
)
1+
cos
=
c
(1+
)
1+
cos
!
r
=
2
c
1+cos
as
!1
Toseewhatthisrepresents,putitinrectangularcoordinates,withtheoriginatthefocus.
r
+
r
cos
=2
c
!
p
x
2+
y
2+
x
=2
c
!
p
x
2+
y
=2
c
x
!
x
2
+
y
2
=4
c
2
4
cx
+
x
2
!
y
2
=4
c
2
4
cx
(6
:
18)
Thisisaparabolaopeningtotheleft(
x
y
2
). Ithasitsvertexat
y
=0,
x
=
c
.
HowmuchtimedoesittakeaplanettoorbittheSun?Thereareenoughrelationsheretogure
thatout.Startwith
=
r
2
_
.Solveitfor
dt
=
1
r
2
d
r
rd
contour integration n you can do the integral fairly easily. . (If f youdon’t you should learn, butthat’s
anotherstory.) There’saneasiertrickhereanyway.Thefactor
r
2
d
isanarea. Itistwicetheareaof
atrianglewithvertexattheorigin.
dA
=
1
2
r
.
rd
so
dt
=
2
dA
(6
:
19)
* NowthatPluto(0.25)hasbeendethroned.
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6|Orbits
190
Noticethatthisequationis Kepler’ssecondlawasinsection6.2! Theareasweptoutisproportional
tothetimeelapsed.Integratethisoverthewholeellipse,0
<<
2
,youhavetheperiod
T
=2
A=‘
.
Combinethiswiththeknownareaofanellipse,
ab
,Eq.(6.16),andtheotherpropertiesinEq.(6.17).
T
=
2
ab
=
2
a
2
p
2
:
Nowuse
2
=
GMa
(1
2
)fromEqs.(6.16)and(6.17)
Solvethelastexpressionfor
p
2
toget
p
=
p
GMa
then
T
=
2
a
2
.
p
GMa
=
2
p
GM
a
3
=
2
(6
:
20)
thecircle,
T
/
r
3
=
2
,andyoucanderivethisspecialcasebyelementarymethods.
Energy
Angularmomentum conservationhasplayedanimportantroleinthis analysis;countthe numberof
timesthat
=
r
2
_
E
=
1
2
mv
2
+
U
(
r
)
;
where
dU
dr
=
f
(
r
)
Inthesamecylindricalcoordinates,thisis
E
=
1
2
m
_r^r+r
_
^
2
+
U
=
1
2
m
_r
2
+
r
2
_
2
+
U
(6
:
21)
Combinethisequationwithangularmomentum,
=
r
2
_
,toeliminatethe
variable.
E
=
1
2
m
_
r
2
+
2
r
2
+
U
=
1
2
m
_
r
2
+
m‘
2
2
r
2
+
U
(
r
)
(6
:
22)
This looks like aone-dimensional problem again, , with the
x
-coordinate replaced by
r
. Therole e of
potential energy is played by the nal combinationin thelastequation,not
U
alone,butcombined
withanothertermcalledthe\centrifugalpotential"energy. Thecombinationiscalledthe\eective
potentialenergy".
U
e
=
U
(
r
)+
m‘
2
2
r
2
(6
:
23)
FortheKeplerproblemEq.(6.23)is
U
e
(
r
)=
GMm
r
+
m‘
2
2
r
2
U
e
r
E
1
E
2
E
3
/1
=r
2
/ 1
=r
(6
:
24)
The same sort of analysis s done e in section 2.3 applies here too. . In n this eectively one-dimensional
problemthetotalenergydeterminesthequalitativebehaviorofthesolution. If
E
=
E
2
,theallowed
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6|Orbits
191
motionisbackandforthbetweenthetwostoppingpointswhere the line
E
2
intersects the eective
potential energy y curve. . This s is just the elliptical orbit described in n the preceding g section, and the
stoppingpointsaretheperihelion and aphelion. . Youcan’thowevertell l from justthis simple energy
analysisthatitisanellipse.Thevalueoftheenergyinthiscaseis
=0
;
0
:
33
;
0
:
66
;
0
:
99
focus
E
2
m‘
2
2
a
2(1
2)
GMm
2
a
(6
:
25)
Youcanderivethisbyndingthestoppingpointfor
E
2
andthenexpressingitintermsoftheminimum
(ormaximum)valueof
r
fromEq.(6.17). Seeproblem6.32. Theenergydependssolelyonthesemi-
majoraxisoftheellipse,
a
. Thesetofellipsesdrawn n hereall havethesame energy andacommon
focus. Thatfocusisatthecenteroftheleftmostellipse(thecircle)andthatdotislessthantheline
thicknessfromtheleftmostpointinthenarrowestellipse. It’shardtosee.
If
E
=
E
1
,thesingleallowedenergyliesatthebottomofthecurveandthatsaysthat
r
isa
constantintime|acircle. If
E
=
E
3
themotionisunbounded. Themasscomesinfromfaraway,
stopsattheintersectionof
E
3
with
U
e
andreturnstoinnity.Thatisorbitisahyperbola,discussed
insection6.10.
The boundary between the elliptic and the hyperboliccases is the line
E
= 0. . This s orbit is
Therearemorerelationshipsamongtheseparameters,andsomeofthemareevenuseful. They
arealleasytoderivefromtheotherequationsinthelastcoupleofpages,andinsection6.11thereis
alotofmanipulationthatwillrequireseveralofthem. Let
c
denotetheorbitaldistanceatperihelion,
thenwith
<
1
c
=
a
f
=
a
(1
)
E
m‘
2
2
c
2
1+
f
=
c
2
=
GMc
(1+
)
b
=
c
r
1+
r
=
c
(1+
)
1+
cos
(6
:
26)
6.4Insolation
P
=210
30
r
overasphereofarea4
r
2
P
=
4
r
2
. Howdoes
thepowerhittingtheEarthvarywiththeseason? FromtherstofEqs.(6.17),
R
2.
P
4
r
2
=
P
R
2
4
(1+
cos
)
2
a
2(1
2)2
where
R
=0)tothatataphelionis,with
=0
:
017,
(1+
)
2
(1
)2
=1
:
07
TheperihelionisinJanuaryandtheaphelionisinJuly.\"isacommonlyusedsymbolfortheSun.
* The e term intensity is sometimesusedhere,butthatwordisalsousedforpowerperareaper
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6|Orbits
192
P
R
2
Z
dt
1
4
r
2
=
P
R
2
I
d
dt
d
1
4
r
2
=
P
R
2
I
d
r
2
1
4
r
2
=
P
R
2
2
4
=
P
R
2
2
(6
:
27)
andinthisequationtheeccentricitycancels,leavingonlythedependenceonangularmomentum. As
acheck,whatshouldthisbeforacircularorbit,where
=
vr
,anddoesitagreewiththisresult?
It is s worth looking g at t the result expressed in terms of other variables. . Combine e pieces s of
Eqs.(6.26)toget
=
p
GMa
(1
2
)
;
so
P
R
2
2
=
P
R
2
2
p
GMa
(1
2
)
(6
:
28)
to be drawn from these equations. . A A detailed analysis of the interactions of the planets does lead
toimportantconsequences however. . Thesolarsystemconsists s oftheSunandJupiterplusdetritus.
onEarth’sorbit. Thattakesyoufarbeyondthistext,butIcanquotesomeresults.
Theshape ofEarth’s orbitchangesovertimesothattheeccentricity willvary slightly,aswill
thedirectionoftheperihelionwithrespecttothedistantstars,andthetimescaleforthesechangesis
theorderof100000years.Onethingthatdoesnotchangeonthistimescaleis
a
,thesemi-majoraxis.
TheimmediateconsequenceofthisisthattheannualenergyreachingtheEarthwillvaryovertimeas
varies. ThisisapparentinthelastforminEq.(6.28).Thisisasucientlycomplicatedproblemthat
\Milankovitchcycles".ThereyouwillseethatI’vebarelystartedonthesubject.
Whatis the quantitative implicationoftheprecedingparagraph? ? Thatdepends s onhowmuch
varies. Ifitvariesfrom m 0
:
01to0
:
06(notoutofline with expectedvalues),theratioofthemean
insolationswillbe[(1
:
01
2
)
=
(1
:
06
2
)]
1
=
2
=1
:
0018. Anavecalculationwouldimplythatthiswill
:
13
C.Lifeisneverthatsimpleofcourse.Thereareprobably
interactionswiththeprecessionoftheEarthitself,anditmaymatterwhetherperihelionoccursduring
6.5ApproximateSolutions
Therearefeworbitproblems thatcanbe solvedexactlyintermsof familiarfunctions. . And d ifthere
areothersthatcanbesolvedintermsofunfamiliarfunctions,thenwhichgivesmoreinsight,anexact
complicatedsolutionoranapproximatesimplesolution?Usuallythelatter.
Someexamplesofforcelawsthatcanreallyhappen:
(1)Isoursunexactlyspherical? IfitisslightlyoblatethewaythattheEarthis,thenitsgravitational
eldisn’texactly the sameasapointmassatitscenter. . Itwillhave e asmallpartthatdropso as
1
=r
4
=r
2
force.
(2)Ifthesolarsystemisembeddedinabackgroundofauniformlydensematerial(darkmatter?dust?)
howdoesthataectorbits?
(3)Therearerelativisticcorrectionstotheequations,andtheywillaecttheorbit.
(4)Thegravitationalpullbytheotherplanets,especiallyJupiter,contribute.Thisisbyfarthelargest
ofalltheseeects,butderivingitcanwaituntiltheendofthechapter,section6.12,asitissomewhat
involved.
thetimedependenceoftheobjectinorbit?Therearetwostartingpointsforthisanalysis,Eqs.(6.10)
and(6.14). Thesecondisconcernedwithonly theshapeoftheorbitandtherstcandoeither,but
notaseasily.
6|Orbits
193
Example
A reminderofthemethods: : Ifyouapply
F
=
ma
tothecase
F
x
(
x
)=
m
x
,whathappensator
near apointofequilibrium? ? You u willuseseries expansions. . Forexample,
F
x
(
x
) =
x
x
3
and
equilibriumoccurswhenthisiszero.
F
x
(
x
0
)=0
!
F
x
(
x
)=
F
x
(
x
0
)+(
x
x
0
)
F
0
x
(
x
0
)+
mx
=
F
0
(
x
0
)(
x
x
0
)
Let
z
=
x
x
0
mz
F
0
(
x
0
)
z
=0
x
0
x
3
0
=0
x
0
=(
=
)
1
=
2
F
0
x
=
3
x
2
F
0
x
(
x
0
)= 2
F
x
(
x
)=0+ 2
(
x
x
0
)
m
x
= 2
(
x
x
0
)
m
z
+2
z
=0
Andthelastequationisaharmonicoscillator,assumingthat
ispositiveofcourse.
FirstMethod
StartwithEq.(6.10). Firstwritetheequationdeningacircularorbit,
r
=
r
0
.Thisisforanarbitrary
r
=
1
m
f
(
r
)+
2
r
3
;
then
0=
1
m
f
(
r
0
)+
2
r
3
0
(6
:
29)
Foranearly circularorbit,
r
willbenearly
r
0
.ThatmeansthatIcanwrite
r
=
r
0
+
x
andassumethat
x
issmallcomparedto
r
0
. Nextdoaseriesexpansiontodeterminethepropertiesof
x
. Aswiththe
sortofexpansionsyoudidsomanyofinchapterthree,thiswillturnouttobeaharmonicoscillator.
r=x=
1
m
f
(
r
0
+
x
)+
2
(
r
0
+
x
)
3
=
1
m
f
(
r
0
)+
xf
0
(
r
0
)+
+
2
r
3
0
(1+
x=r
0
)3
=
1
m
f
(
r
0
)+
xf
0
(
r
0
)+
+
2
r
3
0
1 3
x
r
0
x
=
1
m
f
0
(
r
0
) 3
2
r
4
0
x
=
1
m
f
0
(
r
0
)+
3
r
0
f
(
r
0
)
x
(6
:
30)
x
eliminatetheconstantterm,thentorearrangethecoecientof
x
. Isthisaharmonicoscillator? That
r
:
f
(
r
)=
GMm
r
2
;
f
0
(
r
0
)+
3
r
0
f
(
r
0
)=2
GMm
r
3
0
3
r
0
GMm
r
2
0
GMm
r
3
0
r
0
f
e
(
r
)=
GMm
r2
+
2
r3
r
6|Orbits
194
Thedierentialequation(6.30)for
x
isthenaharmonicoscillatorequation.
x=
GM
r
3
0
x
=)
r
(
t
)=
r
0
+
x
=
r
0
+
x
0
cos(
!
0
t
+
)
;
where
!
2
0
=
GM
r
3
0
=
2
r
4
0
Where did that last t equation n for
!
2
0
come from? ? That t is Eq. (6.29) for the case that
f
(
r
) =
GMm=r
2
. This s equation n tells only the e way that
r
oscillates. It t doesn’t by y itself f describe the
orbitbecauseyouneedtoknowtheangularvariable
(
t
)forthat. Thisangularcoordinatewillcome
fromEq.(6.9)forangularmomentumpermass.
_
=
r
2
=
(
r
0
+
x
)2
!
r
2
0
2
x
r
0
=
!
0
2
x
r
0
If
x
=
x
0
cos
!
0
t;
then
(
t
)=
!
0
t
2
x
0
r
0
sin
!
0
t
(6
:
31)
Ifallyouwantistheshapeoftheorbit,thesecond(small)termin
(
t
)isn’timportant,soIwill
startbyignoringit. Attheend,comebackandseewhatitdoes.
(
t
)=
!
0
t;
r
(
t
)=
r
0
+
x
0
cos
!
0
t
As
t
increasesfrom0to2
=!
0
,theangle
goes from zeroto2
. Theplanetorbitsonce. . Inthat
sameperiod,
r
startsat
r
0
+
x
0
,decreasesto
r
0
x
0
,andthenreturnstoitsstartingpoint. Itisa
closedorbit.
r
0
r
0
x
0
r
0
+
x
0
Fig.6.4
ComparethistotheexactresultfromEq.(6.16),andthenlookattherstterminitsseriesexpansion.
r
=
2
=GM
1+
cos(
0
)
r
0
cos(
0
)
Is
2
=GM
equalto
r
0
0
=
andit’sthe
same.
x
0
=
r
0
.
extraterminEq.(6.31)?
=
!
0
t
(2
x
0
=r
0
)sin
!
0
t
. Attimezero,thepositionsmatch,but
_
isa
littlesmallerthanthe
!
0
speed is less,andthatisnothingmore thanconservation ofangularmomentum,
mr
2
_
. Nowwhat
v
=(
r
0
+
x
0
)
_
=(
r
0
+
x
0
)
!
0
(1 2
x
0
=r
0
)=
r
0
!
0
(1
x
0
=r
0
)
Itisalittlelessthanthe
r
0
!
0
thatyouhavefortheoriginalcircle. Thatmakessense;theplanetis
fartheraway,soitskineticenergyisalittleless. Ithasclimbedupthepotentialwell. Ontheotherside
oftheorbitthespeediscorrespondinglylargerthan
r!
0
soitcatchesup.
6|Orbits
195
To see an example of what anothershape of orbit could be, suppose that the rate of radial
oscillationisexactlyvetimestherevolutionrate.
=
!
0
t;
r
=
r
0
+
cos(5
!
0
t
)
SecondMethod
Startwith Eq.(6.14) toexamine theorbit alone,without following thetimedependence. . Theaim
againwillbetoexaminesomealmostcircularorbits.With
u
=1
=r
,
d
2
u
d
2
+
u
1
m‘
2
u
2
f
1
u
Eq.(6.14) (6
:
32)
Acircularorbitis
r
(
)=
r
0
=constant.Forthiscasetheterm
d
2
u=d
2
iszero,so
u
0
=1
=r
0
satises
u
0
1
m‘
2
u
2
0
f
1
u
0
(6
:
33)
Thismaybeahardequationtosolveoritmaybeeasy,butleaveitalonefornowbecauseitisn’tyet
clearjusthowmuchoftheequationneeds tobesolved. IfandwhenIhavetocomebackandsolveit
Iwill. Todeterminetheshapeofanorbitthatisalmostcircular,assumethat
u
isalmostconstant,so
youlookforasolutionintheform
u
(
)=
u
0
+
x
(
)
;
sothat
d
2
x
d
2
+
u
0
+
x
1
m‘
2(
u
0
+
x
)2
f
1
u
0
+
x
When
x
0thisofcoursesatisestheimmediately precedingequationforthe circle. . If
x
is small
(
u
0
)doaseriesexpansiononeverything
d
2
x
d
2
+
u
0
+
x
1
m‘
2
u
2
0
(1+
x=u
0
)2
f
1
u
0
(1+
x=u
0
)
1
m‘
2
u
2
0
(1 2
x=u
0
)
f
1
u
0
(1
x=u
0
)
1
m‘
2
u
2
0
(1 2
x=u
0
)
f
1
u
0
f
0
1
u
0
x
u
2
0
1
m‘
2
u
2
0
f
1
u
0
+
1
m‘
2
u
2
0
f
1
u
0
2
x
u
0
+
x
u
2
0
f
0
1
u
0
d
2
x
d
2
+
x
=+
1
m‘
2
u
2
0
f
1
u
0
2
u
0
+
1
u
2
0
f
0
1
u
0
x
(6
:
34)
Thelastsimplicationusedtheequationforacircularorbit,(6.33). Thisisnowasimpleequationfor
x
(
),becauseallthecoecientsareconstants. Doesthisprovideasolutionfor
r
(
)? Justonemore
step.
r
=
1
u
=
1
r
0
+
x
=
1
u
0
x
u
0
=
r
0
r
2
0
x
(6
:
35)
6|Orbits
196
=r
2
,so
nearthestartoftheproblem,soifyougetaresultthatfailstoagreewiththeexactone,gowayback.
Example
f
f
0
.Theequation(6.34)isthen
d
2
x
d
2
+
x
=
1
m‘
2
u
2
0
f
2
u
0
x
then
d
2
x
d
2
+
x
1+2
f
0
=m‘
2
u
3
0
=0
Istillneedtheequationfor
r
0
.
u
0
=+
1
m‘
2
u
2
0
f
0
;
or
u
3
0
=
f
0
=m‘
2
Theequationfortheperturbation
x
isnow
d
2
x
d
2
+
x
1+2
=0
;
x
(
)=
A
cos(
p
3
+
)
Theorbitis
r
(
)=
r
0
+
r
1
cos(
p
3
+
),where
r
1
Ar
2
0
isanotherformforthearbitraryconstant
representingthesmalldeviationfromacircularorbit.
Fig.6.5
When
by2
2
p
3,alittle less that 13
=
4
cycles. Inthe e Keplerproblem,when theplanethas
gonearoundonce,
=2
,theplanetisbackwhereitstartedandwiththesame
velocity.Theorbitisclosed.Incontrastyouhavethisexample,inwhichnomatter
howmanytimesyougoaround,you’reneverbacktoexactlythesamepositionand
velocity.
p
3isnotarationalnumber;itcan’tbeexpressedasthequotientoftwo
integers,sonointegermultipleofitwillbeaninteger. Canthishappen?
Yes,rememberthatthecalculationsfortheKeplerproblemassumedthatthesinglethingaecting
intoaccounttheproblemisofcoursefarmoredicult,butoneconsequenceisthattheellipticalorbits
areonlya(verygood)rstapproximationandtheorbitsdon’tquite close. TheperihelionofMercury
per
explanationwasworkedoutbyLaplaceasduetothepullsbytheotherplanets.
Example
Ifthesolarsystemisledwithauniformdustcloudofdensity
theplanets:
F
r
(
r
)=
Gmr=
3. Evaluatetheeectontheplanet’sorbitcausedbythisdust. Itwill
beenoughtondtheorbitwithoutndingthetimedependence.
Thetotalforceontheplanetis
GMm=r
2
Gmr=
3,andusingthesamechangeofvariables
asinEq.(6.14),
u
=1
=r
,theequationfortheshapeoftheorbitis
d
2
u
d
2
+
u
1
m‘
2
u
2
f
1
u
1
m‘
2
u
2
h
GMmu
2
Gm=
3
u
i
=
GM
2
+
G
3
2
u
3
u
=
u
0
+
x
,andthisbecomes
d
2
x
d
2
+
u
0
+
x
=
GM
2
+
G
3
2(
u
0
+
x
)3
GM
2
+
G
3
2
u
3
0
1 3
x=u
0