the radial velocity equation almostfinal

a detailed derivation of

The Radial Velocity Equation

Kelsey I. Clubb

ABSTRACT

Of the over 300 extrasolar planets discovered to date, the vast majority have been

foundusing theRADIALVELOCITYMETHOD (alsoknownasDOPPLERSPECTROSCOPYor the DOPPLER METHOD). The purpose of this paper is to derive the theoreticalequation that is associated with the variation over time of a star’s velocity along an

observer’s line‐of‐sight – a quantity that can be precisely measured with an opticaltelescope–fromfirstprinciples.Inotherwords,wewillbeginwiththeseeminglysimplesituationofaplanetorbitingastar(oftenreferredtoasthe“parentstar”or,theterm

thatwillbeused throughout thispaper, the“hoststar”)and thenstartapplyingbasicprinciplesofphysicsandutilizemathematicaltoolsinordertoanalyzeanddescribethestar‐planet system. The end result will be a complete and detailed derivation of the

RADIALVELOCITYEQUATION.

2 THERADIALVELOCITYEQUATION

PREFACE “The most incomprehensible thing about the universe is that it is comprehensible”

– Albert Einstein Thediscoveryofextrasolarplanetshasprofoundimplicationsnotonlyforthescientiststhatfindthem,butfortheentirepopulationofEarth.Forthisreason,Ihaveattemptedtomakethisderivationaccessibletothelayperson;however,someknowledgeofadvancedmathematics,physics,andastronomyisneededinordertocompletelyfolloweverystep.Iwouldliketothankmyadviser,DebraFischer,forsuggestingthisderivationasoneofmyfirstintroductionstoextrasolarplanetresearch.Theentireprocesshashelpedmeimmenselytonotonlyunderstandthedelicateintricaciesoftheradialvelocitymethod,butalsotobeabletoseethebiggerpictureofhowresearchmakesuseofallthephysics,astronomy,andmathI’velearned.Thankyou!AsmuchasIwouldlikethisderivationtobewithouterror,nothinginlifeisperfect.Withthatsaid,ifyouhappentocomeacrossanerror,typo,oranythingelsethatneedscorrectionI’dbehappytobenotified.Youcane‐[email protected],comments,orsuggestions.

Kelsey I. Clubb

Department of Physics & Astronomy San Francisco State University San Francisco, CA USA

August 2008

THERADIALVELOCITYEQUATION 3

TableofContents

TheOrbitsofaPlanetanditsHostStar ....................................................... 4

AnalyzingTheStar‐PlanetSystem ............................................................... 6

TheCenterofMassFrameofReference...................................................... 7

TheLawsofIsaacNewton........................................................................... 8

TheLawsofJohannesKepler .................................................................... 10

AngularMomentum ................................................................................. 14

TheEquationofanEllipse ......................................................................... 15

PropertiesofanEllipse ............................................................................. 18

DefinitionsofOrbitalParameters.............................................................. 20

RadialVelocity .......................................................................................... 22

RadialVelocitySemi‐Amplitude ................................................................ 25

AppendixA:Proofthatδ = 0..................................................................... 29AppendixB:TheExpressionforz.............................................................. 33ReferenceforCalculatingRVSemi‐Amplitude........................................... 34


THE ORBITS OF A PLANET AND ITS HOST STAR

Webeginourderivationwiththesimplesituationofaplanetorbitingitshoststar(showninFigure1).Inactuality,boththestarandplanetorbittheirmutualcenterofmass(henceforthabbreviatedCM),butbecausethestarismuchmoremassivethantheplanet,theCMofthestar‐planetsystemismuchclosertothestar.Infact,theCMisonlyslightlydisplacedfromthecenterofthestar.Nowweneedtosetupacoordinatesystem(showninFigure2)andreviewsomebasicsofvectors.

(a)

(b)

NOTTOSCALE

NOTTOSCALE

Figure1–Aplanetorbitingitshoststar.

Thedashedlinerepresentsthepathoftheplanet’sorbit.

(a)Topviewofthestar‐planetsystem.(b)Sideviewofthestar‐planetsystem.


SomeBasicsofVectors

• Vectorquantities(or“VECTORS”)havebothmagnitudeanddirection,unlikescalarquantities(or“SCALARS”)whichhaveonlymagnitude

• Vectorquantitiesareusuallydenotedbyplacinganarrowabovethesymbolrepresentingthevector(e.g.

€

F )and/orplacingthesymbolinboldface(e.g.F)

• TheMAGNITUDEOFAVECTORisascalarequaltothelengthofthevector(whichisalwaysapositivenumber)

• Themagnitude(orlength)ofavectorisusuallydenotedbyplacingbarsaroundthevector(e.g.

€

F )orjustthesymbolrepresentingthevectorinnormalfont(e.g.F)

• AUNITVECTORisavectorwhosemagnitudeisequalto1

• TheDIRECTIONOFAVECTORtellsyouwhichwaythevectorpoints

• Thedirectionofavectorisusuallydenotedbyplacingahatabovethesymbolrepresentingthevector(e.g.

€

ˆ F )

• TheNEGATIVEOFAVECTORisanothervectorthathasthesamemagnitudeastheoriginalvector,butpointsintheoppositedirection(e.g.if

€

F isavectorwitha

magnitudeequaltoFandadirectionpointingtotheright,then

€

− F isavectorwitha

magnitudeequaltoFandadirectionpointingtotheleft)

Figure2–Thecoordinatesystemwewillutilizewhenanalyzingthestar‐planetsystem.

(NotethatthelocationoftheCMinthisfigureisexaggeratedinordertomoreeasilyillustratethevectorsthatdependonit)

NOTTOSCALE

distance between the star and planet

r

êr

r1

r2

CM


ANALYZING THE STAR-PLANET SYSTEM

UsingFigure2andwhatweknowaboutvectors,wecandefinethefollowing:

€

ˆ e r ≡unitvectoralongthelineconnectingthestarandplanetthatpointstotheright

€

r ≡ vectorpointingfromstartoplanet(right)

€

r 1 ≡vectorpointingfromCMtostar(left)

€

r 2 ≡ vectorpointingfromCMtoplanet(right)

€

r ≡ magnitudeof

€

r =distancebetweenthestarandplanet

€

r1 ≡magnitudeof

€

r 1 =distancebetweentheCMandstar

€

r2 ≡ magnitudeof

€

r 2 =distancebetweentheCMandplanet

€

ˆ r ≡directionof

€

r =

€

+ ˆ e r

€

ˆ r 1 ≡directionof

€

r 1 =

€

− ˆ e r

€

ˆ r 2 ≡ directionof

€

r 2 =

€

+ ˆ e r Usingthedefinitionsaboveandthefactthatavectorhasbothamagnitudeandadirection,wefind:

€

r ≡ r ˆ r = r (+ ˆ e r) r 1 ≡ r1 ˆ r 1 = r1 (− ˆ e r ) r 2 ≡ r2 ˆ r 2 = r2 (+ ˆ e r)

Thus,

€

r = r ˆ e r r 1 = −r1 ˆ e r r 2 = r2 ˆ e r

(Eq1)

UsingFigure2andEq1,wecansee:

€

r = − r 1 + r 2

= − −r1 ˆ e r( ) + r2 ˆ e r( )= r1 ˆ e r + r2 ˆ e r= r1 + r2( ) ˆ e r

Therefore,themagnitude(orlength)of

€

r is

€

r = r1 + r2 and:

€

r = r 2 − r 1 (Eq2)


THE CENTER OF MASS FRAME OF REFERENCE

Thegeneraltwo‐bodyequationforthecenterofmassis:

€

R = m1

r 1 + m2 r 2

m1 + m2

wherem1≡massofthefirstbody(which,inthisderivation,isthestar)

m2≡massofthesecondbody(which,inthisderivation,istheplanet)

€

R ≡vectorpointingfromspecifiedorigintoCM

€

r 1 ≡vectorpointingfromspecifiedorigintofirstbody

€

r 2 ≡ vectorpointingfromspecifiedorigintosecondbody Inthecenterofmassreferenceframe,thelocationoftheCMisdesignatedasthelocationoftheoriginandthefollowingsubstitutionsthenfollow:

€

R = 0

€

r 1 =vectorpointingfromCMtofirstbody

€

r 2 =vectorpointingfromCMtosecondbodyThegeneralequationthenbecomes:

€

0 =m1 r 1 + m2

r 2m1 + m2

€

0 = m1 r 1 + m2

r 2 Therefore,

€

m1 r 1 = −m2

r 2 (Eq3)Usingtheexpressionsfor

€

r 1 and

€

r 2 inEq1andsubstitutingthemintoEq3,wefind:

€

m1 r 1 = −m2

r 2m1(−r1 ˆ e r ) = −m2(r2 ˆ e r )−m1r1 ˆ e r = −m2r2 ˆ e r

Therefore,

€

m1r1 = m2r2 (Eq4)


THE LAWS OF ISAAC NEWTON

Newton’sLawofUniversalGravitationtellsusthestarexertsagravitationalforceontheplanetandtheplanetexertsagravitationalforceonthestar.Bothforcesareequalinmagnitude,oppositeindirection,andactalongthelineconnectingthestarandplanet(showninFigure3).

Notethat,inFigure3:

€

F 12 ≡ Forceonobject1(thestar)exertedbyobject2(theplanet)

€

F 21 ≡Forceonobject2(theplanet)exertedbyobject1(thestar)

Themagnitudeofanygravitationalforcebetweentwoobjectsisdirectlyproportionaltotheproductofthemassesofeachobjectandinverselyproportionaltothedistancebetweenthemsquared:

€

Fg =G m1m2

r2

TheconstantofproportionalityiscalledNewton’sgravitationalconstant:

€

G = 6.67 ×10−11 m3

kg s2

€

F 12 and

€

F 21thushavethesamemagnitude

€

Fg andonlydifferindirection:

€

F 12 = G m1m2

r2 + ˆ e r( ) F 21 = G m1m2

r2 − ˆ e r( )

Therefore,

€

F 12 = G m1m2

r2ˆ e r (EQ5)

€

F 21 = −G m1m2

r2ˆ e r (EQ6)


Now,Newton’s2ndLawtellsus:

€

F = m a ∑

Foreachbody,theonlyforceexertedonitisthegravitationalforce,so:

€

forbody1(thestar) : F = F 12∑

Thus,

(Eq7)

And,

€

forbody2(theplanet) : F = F 21∑

Thus,

(Eq8)

Usingtheexpressionsfor

€

F 12 and

€

F 21inEq5andEq6andrewriting

€

a as

€

d2 r dt 2

,Eq7andEq8become:

€

G m1m2

r2ˆ e r = m1

d2 r 1dt 2

(Eq9)

and

€

−G m1m2

r2ˆ e r = m2

d2 r 2dt 2

(Eq10)

SubtractingEq9fromEq10gives:

€

−Gm1

r2ˆ e r

−

Gm2

r2ˆ e r

=

d2 r 2dt 2

−

d2 r 1dt 2

−G m1 + m2( )

r2ˆ e r =

d2

dt 2 r 2 − r 1( )

€

F 12 = m1

a 1

€

F 21 = m2

a 2

€

Gm2

r2ˆ e r =

d2 r 1dt 2

€

−Gm1

r2ˆ e r =

d2 r 2dt 2


Now,let’smakethefollowingdefinition:

(Eq11)

UsingEq11andEq2,wehave:

€

−GMr2

ˆ e r =d2

dt 2 r ( )

Using

€

d2 r dt 2 ≡

˙ ̇ r and

€

r ≡ r ˆ e r ⇒ ˆ e r = r r,wehave:

(Eq12)

(Eq13)

THE LAWS OF JOHANNES KEPLER

Asthestarandplanetorbittheirmutualcenterofmass,thedistancebetweenthem,r,changes.AccordingtoKepler’s1stLaw(andextendingthetreatmenttoextrasolarplanets),planetsrevolvearoundtheirhoststarinanellipticalorbitwiththestaratonefocusoftheellipse(showninFigure4).

€

M ≡ m1+m2

€

˙ ̇ r = − GMr2

ˆ e r

€

˙ ̇ r = − GMr3 r

r

θ

Figure4–Astheplanetorbitsitshoststar,thedistancebetweenthestarandplanetchanges.(Nottoscale!)


Itisusefultoexpressrinpolarcoordinatesand,thus,asafunctionofθ.Recallthat

€

r = r ˆ e r .Now,takethederivativeof

€

r withrespecttotime:

(Eq14)

€

ˆ e r isthedirectionofthevectorpointingfromthestartotheplanetanditshouldbeunderstoodthatthisdirectionchangeswithtime.Inordertocalculatethederivativeof

€

ˆ e r withrespecttotime,wemustfirstexpress

€

ˆ e r intermsofCartesiancoordinates(thederivationofthisexpressioncanbelookedupelsewhere):

€

ˆ e r = cosθ ˆ x + sin θ ˆ y Now,wecaneasilydeterminethederivativeof

€

ˆ e r withrespecttotime:

€

ddt

ˆ e r( ) =ddt

cos θ ˆ x + sinθ ˆ y ( )

= cosθ ddt

ˆ x ( ) + ˆ x ddt

cosθ( ) + sinθ ddt

ˆ y ( ) + ˆ y ddt

sin θ( )

= 0 + ˆ x ddt

cos θ( ) + 0 + ˆ y ddt

sinθ( )

Wewereabletodothelaststepbecause

€

ˆ x and

€

ˆ y areindependentoftime(and,thus,takingthederivativeofeitherwithrespecttotimegiveszero).However,θdoesdependontime,sowemusttakethisintoaccountwhentakingthetimederivativeofsinθandcosθ:

€

ddt

ˆ e r( ) = ˆ x ddt

cos θ( ) + ˆ y ddt

sinθ( )

= ˆ x −sin θ ˙ θ ( ) + ˆ y cosθ ˙ θ ( )= − ˙ θ sinθ ˆ x + ˙ θ cos θ ˆ y

= ˙ θ −sin θ ˆ x + cosθ ˆ y ( )

TheexpressioninparenthesesisEqualto

€

ˆ e θ expressedintermsofCartesiancoordinates(thederivationofthisexpressioncanbelookedupinthesame“elsewhere”).Inotherwords:

€

ˆ e θ = −sinθ ˆ x + cos θ ˆ y Andso,

(Eq15)

€

ddt r ( ) =

ddt

r ˆ e r( ) = r ddt

ˆ e r( ) + ˆ e rddt

r( )

€

ddt

ˆ e r( ) = ˙ θ ˆ e θ


SubstitutingEq15intoEq14gives:

€

ddt r ( ) = r ˙ θ ˆ e θ( ) + ˆ e r ˙ r

Using

€

d r dt≡ ˙ r ,wehave:

(Eq16)

Now,takethederivativeof

€

˙ r withrespecttotime:

€

ddt ˙ r ( ) =

ddt

˙ r ˆ e r + r ˙ θ ˆ e θ( )

= ˙ r ddt

ˆ e r( ) + ˆ e rddt

˙ r ( ) + r ddt

˙ θ ˆ e θ( ) + ˙ θ ˆ e θddt

r( )

= ˙ r ˙ θ ˆ e θ( ) + ˆ e r ˙ ̇ r ( ) + r ˙ θ ddt

ˆ e θ( ) + ˆ e θddt

˙ θ ( )

+ ˙ θ ˆ e θ ˙ r

(Eq17)

Now,wemusttakethederivativeof

€

ˆ e θ withrespecttotime:

€

ddt

ˆ e θ( ) =ddt

−sin θ ˆ x + cosθ ˆ y ( )

= −sinθ ddt

ˆ x ( ) + ˆ x ddt

−sin θ( ) + cos θ ddt

ˆ y ( ) + ˆ y ddt

cos θ( )

= 0 + ˆ x −cos θ ˙ θ ( ) + 0 + ˆ y −sinθ ˙ θ ( )= − ˙ θ cosθ ˆ x − ˙ θ sin θ ˆ y

= − ˙ θ cos θ ˆ x + sinθ ˆ y ( )= − ˙ θ ˆ e r( )

Andso,

(Eq18)


€

ddt ˙ r ( ) = ˙ r ˙ θ ˆ e θ + ˙ ̇ r ˆ e r + r ˙ θ − ˙ θ ˆ e r( ) + r ˙ ̇ θ ˆ e θ + ˙ r ˙ θ ˆ e θ

= 2 ˙ r ˙ θ ˆ e θ + ˙ ̇ r ˆ e r − r ˙ θ 2 ˆ e r + r ˙ ̇ θ ˆ e θ= ˙ ̇ r − r ˙ θ 2( ) ˆ e r + r ˙ ̇ θ + 2 ˙ r ˙ θ ( ) ˆ e θ

€

˙ r = ˙ r ˆ e r + r ˙ θ ˆ e θ

€

ddt ˙ r ( ) = ˙ r ˙ θ ˆ e θ + ˙ ̇ r ˆ e r + r ˙ θ

ddt

ˆ e θ( ) + r ˙ ̇ θ ˆ e θ + ˙ r ˙ θ ˆ e θ

€

ddt

ˆ e θ( ) = − ˙ θ ˆ e r


Therefore,

(Eq19)


€

˙ ̇ r − r ˙ θ 2( ) ˆ e r + r ˙ ̇ θ + 2 ˙ r ˙ θ ( ) ˆ e θ = −GMr2

ˆ e r

Theonlywayfortheaboveequationtobetrueisif:

(Eq20)

and,

(Eq21)

ToextractanimportantpieceofinformationfromEq21,recallthis:

€

ddt

r2 ˙ θ ( ) = r2 ddt

˙ θ ( ) + ˙ θ ddt

r2( )= r2 ˙ ̇ θ + ˙ θ 2r˙ r ( )= r2 ˙ ̇ θ + 2r˙ r ˙ θ

= r r ˙ ̇ θ + 2˙ r ˙ θ ( )

So,theleft‐handsideofEq21isalsoequalto

€

1rddt

r2 ˙ θ ( ) andthus,

€

1rddt

r2 ˙ θ ( ) = 0

fromwhich,follows:

(Eq22)

Thequantity

€

r2 ˙ θ doesnotchangewithtime,sothequantityissaidtobeconserved.Infact,

€

r2 ˙ θ isawell‐knownquantity,asweshallnowsee.

€

r ˙ ̇ θ + 2 ˙ r ˙ θ = 0

€

˙ ̇ r = ˙ ̇ r − r ˙ θ 2( ) ˆ e r + r ˙ ̇ θ + 2 ˙ r ˙ θ ( ) ˆ e θ

€

˙ ̇ r − r ˙ θ 2 = −GMr 2

€

ddt

r 2 ˙ θ ( ) = 0


ANGULAR MOMENTUM

Recallthegeneralequationfortotalangularmomentum:

€

L = r × p

€

where : L = total angular momentum r = displacement vector p = total linear momentum = m v

Foraplanetorbitingitshoststar,

€

r and

€

v arealwaysperpendicular(showninFigure5).

So,

€

r × p = r ×m v = m r × v ( ) = m rv sin 90°( )[ ] = m rv 1( )[ ] = mrv andthus,

€

L ≡ L = mvr

Now,

€

LmisdefinedtobetheSPECIFICANGULARMOMENTUMandisdenotedbythesymbolh.

Using

€

v =ω r and

€

ω ≡dθdt

= ˙ θ ,sothat

€

v = ˙ θ r,wehave

€

h ≡ Lm

=mvrm

= vr = ˙ θ r( )r = r2 ˙ θ ,therefore,

(Eq23)

Andso,Eq21tellsusthatthespecificangularmomentumforoursituationisaconservedquantity.

€

h = r2 ˙ θ

Figure5–Astheplanetorbitsitshoststar,itspositionvector

€

r andvelocityvector

€

v arealwaysperpendicular.(Nottoscale!)

r

v


THE EQUATION OF AN ELLIPSE

NowwearereadytoutilizeEq20forthepurposeoffindingrasafunctionofθ.First,abrieftrick.Let’sexpress

€

r ˙ θ 2 intermsofh:

€

h = r2 ˙ θ ⇒ h2 = r4 ˙ θ 2 = r3 r ˙ θ 2( ) ⇒ r ˙ θ 2 =h2

r3

SubstitutingthisresultintoEq20gives:

€

˙ ̇ r − h2

r3 = −GMr2

Or,solvingfor

€

˙ ̇ r :

(Eq24)

Next,wemustremovethetimedependenceofr.Todothis,weinvokeasubstitution:

€

u ≡ 1r

⇒ r =1u

= u−1

Takingthederivativeofrwithrespecttotimegives:

€

ddt

r( ) =ddt

u−1( ) = −u−2 dudt

Using

€

dudt

=dudθ

dθdt

=dudθ

˙ θ ,theaboveequationbecomes:

€

ddt

r( ) = − u−2( ) dudθ˙ θ = − r2( ) dudθ

˙ θ = − r2 ˙ θ ( ) dudθ = − h( ) dudθ

Therefore,

(Eq25)

Takingthederivativeof

€

˙ r withrespecttotime(andrememberingthathisindependentoftime)gives:

€

ddt

˙ r ( ) =ddt

− h dudθ

= − h d

dtdudθ

€

˙ ̇ r = h2

r 3 −GMr 2

€

˙ r = − h dudθ


Using

€

ddt

dudθ

=

ddθ

dθdt

dudθ

=d2udθ 2

dθdt

=d2udθ 2

˙ θ ,wefind:

€

ddt

˙ r ( ) = − h d2udθ 2

˙ θ

Therefore,

(Eq26)


€

− h ˙ θ d2udθ 2 =

h2

r3 −GMr2

Replacingrwith

€

1u,usingEq23,andsolvingfor

€

d2udθ 2

gives:

€

d2udθ 2 = −

1h ˙ θ

h2u3 −GMu2( )

= −h( ) u3

˙ θ +GMu2

h( ) ˙ θ

= −r2 ˙ θ ( ) u3

˙ θ +GMu2

r2 ˙ θ ( ) ˙ θ

= − r2( ) u3 +GMr2 ˙ θ 2

u2( )

= −1u2

u3 +

GMr2 ˙ θ 2

1r2

= − u +GMr4 ˙ θ 2( )

= − u +GMh2( )

Therefore,

(Eq27)

Eq27isadifferentialequationwiththefollowingsolution:

€

˙ ̇ r = − h ˙ θ d2udθ 2

€

d2udθ 2 = −u +

GMh2


(Eq28)

€

where : A ≡ constantδ ≡ phase shift

both will be determinedby initial conditions

Changingvariablesbacktor,Eq28becomes:

€

1r θ( )

= A cos θ −δ( ) +GMh2

So,

(Eq29)

Multiplyingtheright‐handsideofEq29by

€

h2

h2gives:

€

r θ( ) =h2

h2A cos θ −δ( ) +GM

Dividingtheright‐handsideoftheaboveequationbyGMgives:

(Eq30)

Thegeneralequationforanellipseis:

(Eq31)

€

where : a ≡ semi -major axis of the ellipse

e ≡ eccentricity of the ellipse

ComparingEq30toEq31,wecanimmediatelysee:

(Eq32)

€

u θ( ) = A cos θ −δ( ) +GMh2

€

r θ( ) =1

A cos θ −δ( ) +GMh2

€

r θ( ) =

h2

GMh2A cos θ −δ( )

GM+1

€

r θ( ) =a 1− e2( )1+ e cosθ

€

e =h2AGM


Or,

€

h2

GM=eAandthus,

€

a 1− e2( ) =h2

GM=eA.Sothat,

(Eq33)

Therefore,theunknownconstantAinEq28intermsofknownvariablesis:

(Eq34)

and,

€

δ = 0 (aproofofthisisprovidedinAppendixA).

PROPERTIES OF AN ELLIPSE

Inasense,wehavethusfaranalyzedthestar‐planetsysteminonedimension(whenwecomputedthegravitationalforces)andintwodimensions(whenwedeterminedanexpressionforrasafunctionofθ)andournexttaskistodevelopthetoolsneededtofullyunderstandthissysteminthreedimensions.First,let’sreviewsomeimportantpropertiesofanellipse:Themajoraxisofanellipseisthestraightlinethatrunsbetweenthefocioftheellipseandextendstotheedgesoftheellipse(showninFigure6).One‐halfofthemajoraxis(aptlynamedthe“SEMI‐MAJORAXIS”)hasalengthdefinedtobeequaltoa,therefore,thelengthofthemajoraxisisequalto2a.

€

a =e

A 1− e2( )

€

A =GMeh2

Figure6–Themajoraxisofanellipse.

Focus Focus

length=2a


Theminoraxisofanellipseisthestraightlinethatliesperpendiculartothemajoraxisandpassesthroughthemidpointofthemajoraxis(showninFigure7).One‐halfoftheminoraxis(aptlynamedthe“SEMI‐MINORAXIS”)hasalengthdefinedtobeequaltob,therefore,thelengthoftheminoraxisisequalto2b.

PERIASTRONisdefinedtobethepositioninaplanet’sorbitarounditshoststarinwhichthedistancebetweenthestarandplanetisaminimumfortheorbit.Similarly,APASTRONisdefinedasthepositioninaplanet’sorbitarounditshoststarinwhichthedistancebetweenthestarandplanetisamaximumfortheorbit(showninFigure8).

Figure7–Theminoraxisofanellipse.

Focus Focus

length=2b

Figure8–Thepositionsofperiastronandapastroninaplanet’sorbitarounditshoststar.(Nottoscale!)

apastron

periastron


TheECCENTRICITYofanellipse,orhowelongatedtheellipseis,isdenotedbythesymboleandisdefinedintermsofthesemi‐majorandsemi‐minoraxes:

(Eq35)

Inthespecialcaseinwhich

€

a = b, e =1−1= 0andtheshapeofaplanet’sorbitarounditshoststarwouldbeaperfectcircle.Finally,theareaofanellipseisdefinedtobe:

(Eq36)

Noticethatforacircle

€

a = bandtheareaisequaltoπa2orπb2asonewouldexpect!

DEFINITIONS OF ORBITAL PARAMETERS

Now,let’sdefinetwoterms(showninFigure9):LINE‐OF‐SIGHT≡thestraightlineconnectinganobserverandtheobjectbeingorbited(inourcase,

wewilltakethistobethestar)PLANE‐OF‐SKY≡thetwodimensionalplanethatliesperpendiculartotheline‐of‐sightandpasses

throughtheobjectbeingorbited

€

e2 ≡ 1− b2

a2

€

Area = πab

line‐of‐sight

plane‐of‐sky(seen“edge‐on”)

plane‐of‐sky(seen“face‐on”)

Figure9–Anobserver’sline‐of‐sightandtheplane‐of‐skyseenboth“edge‐on”and“face‐on.”(Nottoscale!)


Wearenowreadytoanalyzethestar‐planetsysteminthreedimensions(showninFigure10):

TherearetwoseparateplanesinFigure10:theplane‐of‐sky(alreadydefined)andthePLANE‐OF‐ORBIT(thetwodimensionalplaneinwhichtheplanetorbitsitshoststar).Theplane‐of‐orbitistiltedwithrespecttotheplane‐of‐skybyacertainangle.WecallthistilttheORBITALINCLINATIONanddenotetheanglebythesymboli.Therearetwopointsintheplanet’sorbitthatpassthroughtheplane‐of‐sky.OneiscalledtheASCENDINGNODE(thepointatwhichtheplanetpassesfrom“below”theplane‐of‐skyto“above”theplane‐of‐sky)andtheotheriscalledtheDESCENDINGNODE(thepointatwhichtheplanetpassesfrom“above”theplane‐of‐skyto“below”it).ThelineconnectingthesetwopointsiscalledtheLINEOFNODES.Theangleθistheanglebetweentheperiastronandthepositionoftheplanet.It’sreferredtoastheTRUEANOMALYanditchangeswithtime.ThisisthesameθthatisfoundinFigure4,Eq28,&Eq31.Theangleωistheanglebetweenthelineofnodesandperiastron.It’sreferredtoastheARGUMENTOFPERIASTRONanditdoesnotchangewithtime.Bothθandωlieintheplane‐of‐orbit.

Figure10–Illustrationoforbitalparametersforthestar‐planetsystem.


RADIAL VELOCITY

Ifwedesignatetheplane‐of‐skytobethex-yplane,thenthedirectiontotheobserveralongtheline‐of‐sightisthez‐direction.Intermsofθ,ω,andi,zcanbeexpressedas:

€

z = r sin θ +ω( ) sin i (AppendixBexplainshowthisisdetermined.)Rememberthatthisentirederivationisdevotedtofindinganexpressionfortheradialvelocity(denotedbyvr)andthattheradialvelocityisdefinedtobethevelocityalongtheobserver’slineofsight.Thus,sincezisthedirectionalongtheobserver’slineofsight,thederivativeofzwithrespecttotimeshallprovideuswiththeexpressionforradialvelocity!Recallingthatθandrchangewithtime,however,ωandidonot,wefind:

€

vr ≡ ˙ z = ddt

z( ) =ddt

r sin θ +ω( ) sin i( )

= r ddt

sin θ +ω( ) sin i( ) + sin θ +ω( ) sin i ddt

r( )

= r sin θ +ω( ) ddt

sin i( ) + sin i ddt

sin θ +ω( ){ }

+ sin θ +ω( ) sin i ˙ r

= r sin θ +ω( ) cos i didt

+ sin i cos θ +ω( ) ddt

θ +ω( )

+ ˙ r sin θ +ω( ) sin i

= r 0 + sin i cos θ +ω( ) ˙ θ [ ] + ˙ r sin θ +ω( ) sin i

Therefore,

(Eq37)

Now,wemustfindexpressionsfor

€

˙ r and

€

˙ θ inordertoridEq37ofthesetwoquantities.Tofind

€

˙ r ,wetakethederivativeofEq31withrespecttotime(aandearetime‐independent):

€

˙ r ≡ ddt

r( ) =ddt

a 1− e2( )1+ e cos θ

=1+ e cos θ[ ] d

dta 1− e2( )[ ] − a 1− e2( )[ ] d

dt1+ e cosθ[ ]

1+ e cosθ[ ]2

€

vr = r ˙ θ cos θ +ω( ) + ˙ r sin θ +ω( )[ ] sin i


€

=0 − a 1− e2( )[ ] − e ˙ θ sinθ[ ]

1+ e cos θ[ ]2

=a 1− e2( )

1+ e cosθe ˙ θ sin θ

1+ e cosθ

Thus,

(Eq38)

Tofind

€

˙ θ ,wemakeuseofKepler’s2ndLaw:

(Eq39)

whereAistheareaoftheellipse.Integratingtheleft‐handsideofEq39overoneperiod(thetimeittakestheplanettomakeonecompleterevolutionarounditshoststar)yields:

(Eq40)

wherePistheperiod.Weknowtheareaofanellipseis

€

A = πab (fromEq36),buttogettheareaintermsofaandeonly,weneedtosolveEq35intermsofb:

€

e2 ≡1− b2

a2⇒

b2

a2=1− e2 ⇒ b2 = a2 1− e2( )

Andso,

€

b = a 1− e2 andputtingthisintotheequationfortheareaofanellipsegives:

€

A = πab = πa a 1− e2( )

Therefore,

(Eq41)


€

πa2 1− e2

P=

12r2 ˙ θ

Insteadofsolvingtheaboveequationfor

€

˙ θ ,wewillsolveitfor

€

r ˙ θ sincethisquantityisfoundinEq37,whichwearetryingtosimplify:

(Eq42)

€

˙ r = r e ˙ θ sinθ1+ e cosθ

€

dAdt

=12r 2 ˙ θ

€

AP

=12r 2 ˙ θ

€

A = πa2 1− e2

€

r ˙ θ =2πa2 1− e2

r P


Now,wehavefoundexpressionsfor

€

˙ r (Eq38)and

€

r ˙ θ (Eq42),buttheexpressionfor

€

˙ r hasan

€

r ˙ θ initandtheexpressionfor

€

r ˙ θ hasanrinit.So,let’sridtheseexpressionsofanyror

€

˙ θ dependence:

€

r ˙ θ =2πa2 1− e2

r P

=1r

2πa2 1− e2

P

=1+ e cos θa 1− e2( )

2πa2 1− e2

P

Thus,

(Eq43)

And,

€

˙ r = r e ˙ θ sinθ1+ e cos θ

= r ˙ θ e sinθ

1+ e cos θ

=2πa 1+ e cosθ( )

P 1− e2

e sin θ1+ e cosθ

Thus,

(Eq44)

SubstitutingEq43andEq44intoEq37gives:

€

Vr = r ˙ θ cos θ +ω( ) + ˙ r sin θ +ω( )[ ] sin i

=2πa 1+ e cosθ( ) cos θ +ω( )

P 1− e2+

2πa e sin θ sin θ +ω( )P 1− e2

sin i

(Eq45)

Next,weneedtosimplifyEq45byutilizingthefollowingtwotrigonometricidentities:

€

cos θ +ω( ) = cos θ cosω − sin θ sinωsin θ +ω( ) = sin θ cosω + cosθ sinω

€

r ˙ θ =2πa 1+ e cosθ( )

P 1− e2

€

˙ r = 2πa e sinθP 1− e2

€

Vr =2πa sin iP 1− e2

cos θ +ω( ) + e cosθ cos θ +ω( ) + e sin θ sin θ +ω( )[ ]


TheexpressionwithinthebracketsinEq45becomes:

€

[ ] = cosθ cosω − sinθ sinω + e cos θ cos θ cosω − sin θ sinω( ) + e sin θ sin θ cosω + cosθ sinω( )= cosθ cosω − sinθ sinω + e cosω cos2 θ − e cos θ sinθ sinω + e cosω sin2 θ + e cos θ sin θ sinω

= cosθ cosω − sinθ sinω + e cosω cos2 θ + sin2 θ( )= cosθ cosω − sinθ sinω + e cosω

Weusedthetrigonometricidentity

€

cos2θ + sin2θ =1inthelaststepandwecannowrevert

€

cosθ cosω − sinθ sinω backto

€

cos θ +ω( ) foramorecompactexpression:

(Eq46)

Finally,wehavederivedtheequationforradialvelocity!Thus,theradialvelocityofthehoststar(withrespecttothecenterofmass)is:

(Eq47)

andtheradialvelocityoftheorbitingplanet(withrespecttothecenterofmass)is:

(Eq48)

RADIAL VELOCITY SEMI-AMPLITUDE

Wehaveonemorequantitytoprovideanameandsymbolfor:

(Eq49)

Theradialvelocitysemi‐amplitudeofthehoststaris:

(Eq50)

andtheradialvelocitysemi‐amplitudeoftheorbitingplanetis:

(Eq51)

€

Vr =2πa sin iP 1− e2

cos θ +ω( ) + e cosω[ ]

€

K ≡ radialvelocitysemi ‐ amplitude =2πa sin iP 1− e2

€

Vr (star ) =2πa1 sin iP 1− e2


€

Vr (planet ) =2πa2 sin iP 1− e2


€

K1 ≡ radialvelocitysemi ‐amplitude of host star =2πa1 sin iP 1− e2

€

K2 ≡ radialvelocitysemi ‐amplitude of orbiting planet =2πa2 sin iP 1− e2


Theradialvelocitymethodofdetectingextrasolarplanetsinvolvestakingprecisemeasurementsofastar’sradialvelocitywithanopticaltelescope.Eachmeasurementisassociatedwithaspecifictimeandaplotcanbecreatedshowingthestar’sradialvelocityasafunctionoftime.Ifapreviouslyundiscoveredplanetexistsinorbitaroundtheobservedstar,thedataintheplotwillshowarepeatedtrendandacurvecanbefittothedataconnectingeachoftheindividualradialvelocitydatapoints(showninFigure11).

Figure11–Plotofradialvelocityvs.timeforthehoststarindicatinghowtheperiod,P,andradialvelocitysemi‐amplitude,K,canbedeterminedfromthedata.(ImageCredit:PlanetarySystemsandtheOriginsofLife,CambridgeUniversityPress,2007)

Theradialvelocitysemi‐amplitudeofthehoststarcanthenbedeterminedfromtheplot.Itisequaltohalfofthetotalamplitudeofthefittedcurve(asseeninFigure11).Notethattheperiod,whichisthetotalamountoftimeelapsedbetweentwoconsecutivepeaksofthefittedcurve(andisthesameforboththehoststarandorbitingplanet),canbedeterminedfromtheplotaswell(asseeninFigure11).Ifpreferred,thedependenceona1inthequantityK1canberemoved.Thiswillbeournexttask.Kepler’s3rdLawrelatesthesemi‐majoraxisofaplanet’sorbittotheperiodofitsorbitasfollows:

(Eq52)

SolvingEq52fora2gives:

€

a23 =

G m1 + m2( ) P 2

4π 2

€

P 2 =4π 2

G m1 + m2( )a23


Therefore,

(Eq53)

Nowweneedtoreferallthewaybacktopage7andEq4.Withour“new”knowledgeofellipses,wecannowrecognizethatther1inEq4,whichisthemagnitudeofthevectorpointingfromtheCMtothestar,issimplythesemi‐majoraxisofthestar’sorbitaroundthemutualCM,a1.Withthesamereasoning,ther2inEq4,whichisthemagnitudeofthevectorpointingfromtheCMtotheplanet,isthesemi‐majoraxisoftheplanet’sorbitaroundthemutualCM,a2.Therefore,

€

m1a1 = m2a2 or,solvingfora1:

(Eq54)


(Eq55)


(Eq56)

TheprocessofsimplifyingEq56goessomethinglikethis:

€

K1 =m2

m1sin i1− e2

8π 3G m1 + m2( ) P 2

4π 2P 3

13

=m2

m1sin i1− e2

2πG m1 + m2( )P

13

=2πGP

13 m2

m1m1 + m2( )

13sin i1− e2

€

a1 =m2

m1a2

€

a2 =G m1+m2( ) P2

4π 2

13

€

K1 =2π sin iP 1− e2

m2

m1a2

€

K1 =2π sin iP 1− e2

m2

m1

G m1+m2( ) P2

4π 2

13


Thisisthepointatwhichweutilizethefactthat

€

m1 >> m2 andsoouranswerisnotsignificantlyaffectedifweusethefollowingapproximation:

€

m1 + m2 ≈ m1Nowwecancontinueoursimplification:

€

K1 =2πGP

13 m2

m1m1( )

13sin i1− e2

=2πGP

13 m1( )

13

m1m2 sin i1− e2

=2πGP

13m1( )−

23m2 sin i1− e2

Therefore,initsmostcommonform,theequationfortheradialvelocitysemi‐amplitudeofthestaris:

(Eq57)

€

K1 =2πGP

13 m2 sin im1

2 / 311− e2


APPENDIX A: PROOF THAT δ = 0

Toprovideproofthat

€

δ = 0 inEq28,wewillanalyzethespecialcaseinwhichtheplanetislocatedatPERIASTRON(thepositioninaplanet’sorbitarounditshoststarinwhichthedistancebetweenthestarandplanetisaminimumfortheorbit).Thepositionofperiastronisusefulfortworeasons:(1)weknowthevalueofθisequaltozeroforthispositioninaplanet’sorbitarounditshoststarand(2)weknowtheexpressionforr(θ),thedistancebetweentheplanetanditshoststar.Thefirstreasoncanbeunderstoodbyre‐examiningFigure10orre‐readingthedefinitionoftheTRUEANOMALY.Thesecondreasonfollowsfromthefactthat,foranyellipse,thedistancebetweenthecenteroftheellipseandeitherfocusoftheellipse(oneofwhichisalsothepositionofthehoststar)isequaltoae(showninFigure12).Forexample,inthespecialcaseofacircularorbite=0andthe“ellipse”(orcircle,whichisaspecialtypeofellipse)hasonlyonefocus(thelocationofwhichisthepositionofthehoststar).Thissinglefocusislocatedatthecenteroftheellipse,therefore,thedistancebetweenthefocusandthecenteroftheellipseiszerowhichagreeswithae=a(0) =0(showninFigure13).

Theexpressionforr(θ),thedistancebetweentheplanetanditshoststar,isequaltothedistancebetweentheplanetandthecenteroftheellipse(which,inthiscase,isthelengthofthesemi‐majoraxis,a)minusthedistancebetweenthestarandthecenteroftheellipse,ae(showninFigure14).Thus,theexpressionforr(θ),thedistancebetweentheplanetanditshoststar,equals:

€

r θ( ) = a − ae

or,aftersimplifying:

(EqA.1)

€

r θ( ) = a 1− e( )

Figure12–Thedistancebetweenthecenterofanellipseandeitherofitsfociisequaltotheproductofthelengthofthesemi‐majoraxisandtheeccentricityoftheellipse.

Figure13–Thebluedotrepresentsboththecenterofthecircle(aspecialtypeofellipse)andthesinglefocusoftheellipse.Inthisspecialcase,theproductaeisequaltozero.

ae ae

centerofellipse

Focus Focus


ThetaskathandistosolveEq29forδ.WecansubstituteEqA.1fortheleft‐handsideofEq29anduse

θ=0.WecanalsorearrangeEq34as

€

GMh2

=Ae.Theresultofthesesubstitutionsis:

€

r θ( ) =1

A cos θ −δ( ) +GMh2

a 1− e( ) =1

A cos 0 −δ( ) +Ae

A cos −δ( ) +Ae

=1

a 1− e( )

A cos −δ( ) =1

a 1− e( )−Ae

(EqA.2)

Figure14–Thedistancebetweentheplanetanditshoststar(r(θ),representedbytheblackarrow)isequaltothedistancebetweentheplanetandthecenteroftheellipse(a,representedbythepurplearrow,whichisalsothesemi‐

majoraxis;notethatthisarrowisnotdrawnattheexactlocationofthesemi‐majoraxisotherwiseitwouldoverlapwiththeotherarrows)minusthedistancebetweenthestarandthecenteroftheellipse(ae,representedbytheredarrow).

apastron

periastron

a

ae r(θ)

centerofellipse

€

cos −δ( ) =1

Aa 1− e( )−1e


NowrearrangingEq33as

€

A =e

a 1− e2( )andsubstitutingthisintoEqA.2,wefind:

€

cos −δ( ) =1

Aa 1− e( )−1e

=1

ea 1− e2( )

a 1− e( )[ ]−1e

=1

e 1− e( )1− e2( )

−1e

=1− e2

e 1− e( )−1e

=1e1− e2

1− e−1

=1e1− e2

1− e−1− e1− e

=1e1− e2 − (1− e)

1− e

=1e1− e2 −1+ e1− e

=1ee − e2

1− e

=1ee (1− e)1− e

=1ee( )


Thus,

€

cos − δ( ) =1Thecosineisanevenfunction,sowecanexploitthefollowingpropertyofevenfunctions:

foranyanglea:

€

cos a( ) = cos −a( ) Andso,

€

cos − δ( ) = cos δ( ) =1Thus,

€

cos δ( ) =1 ⇒ δ = cos−1 1( ) = 0 Therefore,

€

δ = 0.


APPENDIX B: THE EXPRESSION FOR z

Itmightbeabitdifficulttounderstandwheretheequationforzcomesfrom,soI’veincludedamoredetailedillustrationoftheanglesandlengthsinquestion(showninFigure15).

Partoftheplanet’sorbitarounditshoststarhasbeenmadeboldinthediagram.Wewillassumethistobeastraightlengthandrefertoitas(bold length)inthefollowingequations.Notethattheactualvalueforthelengthisnotneeded.ωistheanglebetweentheline‐of‐nodes(y‐axisinFigure15)andperiastron.θistheanglebetweenperiastronandtheplanet’spositionvector,

€

r (themagnitudeofwhichisr,thedistancebetweentheplanetanditshoststar).Theanglebetweenthey‐axisandtheplanet’spositionvectoris

€

θ +ω (I’llleaveittothereadertomakecertainhe/sheunderstandswhythisisthecase;butifyouaren’tcertain,myhintistotrymakingupnumbersforθandωormovingtheplanettoanotherpositioninitsorbit).Wenowhavethenecessaryinformationand,withtheaidoftrigonometry,wefind:

€

sin θ +ω( ) =bold length( )

r⇒ bold length( ) = r sin θ +ω( )

€

sin i =z

bold length( )⇒ z = bold length( ) sin i ∴ z = r sin θ +ω( ) sin i

Figure15–Adiagramspecificallycreatedtoshowtheimportantlengthsandanglesneededfordetermininganexpressionforz.


REFERENCE FOR CALCULATING RV SEMI-AMPLITUDE

Ifyouknowthenumericalvalueforeachofthefollowingparametersforaplanetanditshoststar,

P(theplanetorhoststar’sorbitalperiodindaysoryears)

€

Mplanet sin i (theplanet’sminimummassintermsofthemassofEarthorJupiter)

€

Mstar (themassofthehoststarintermsofthemassoftheSun)

e(theeccentricityoftheplanet’sorbitarounditshoststar–rangesfrom0‐1)thenenteringtheappropriatevalueintooneofthefollowingfourequationswillgiveyouthenumericalvalueofthestar’sradialvelocitysemi‐amplitude:

€

Kstar = 0.6395 m s−1 1 dayP

13 Mplanet sin i

MEarth

MSun

Mstar

2 / 311− e2

Kstar = 0.0895 m s−1 1 yrP

13 Mplanet sin i

MEarth

MSun

Mstar

2 / 311− e2

Kstar = 203.255 m s−1 1 dayP

13 Mplanet sin i

MJupiter

MSun

Mstar

2 / 311− e2

Kstar = 28.435 m s−1 1 yrP

13 Mplanet sin i

MJupiter

MSun

Mstar

2 / 311− e2

the radial velocity equation almostfinal

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