Solutions to Exercises

Solutions to exercises

Exercise 1.1 Firstweneed to convertthe distance to SI units: 68 light-years =68×9.461 × 1015 m =6.5×1017 m. Then we can obtain the timetaken by dividing thedistance by thespeed of travel: distance time = speed 6.5 × 1017 = m 12 × 103 ms−1 =5.4×1013 s 5.4 × 1013 = s 365.25 dayyr−1 ×(24 × 3600) sday−1 =1.7×106 yr ≈ 2 Myr. (Myr indicates 106 yr,oramegayear.) Thus thetimetaken for Pioneer 10 to reach Aldebaran is almost 2 millionyears. Exercise 1.2 In interpretingFigure 1.8, we notethatbothaxesare logarithmic. (a)The most favourablecontrastratio occurs whenthe vertical distance between Jupiter’s curveand theSun’sisminimized.The twocurves areconvergingasthey disappear offthe right-hand side of thefigure, so themostfavourablecontrast ratio occurs for wavelengths around (or greater than) 100 µm. Thevalue of the contrastratio at 100 µmisapproximately 10 000. (b) Thespectral energy distributionofJupiter peaksataround 0.5 µm. This is very different(by afactor of around 200)tothe wavelengthofthe most favourable contrastratio. Thereason for thediscrepancy is that theSun’sspectral energy distributionalsopeaksataround 0.5 µm, andthe second, thermalemission component of Jupiter’s spectral energy distributiongives amore favourable contrastratio. (c)Wavelengths around 20 µmare at the peak of Jupiter’s thermalemission. Though thecontrastratio (almost 100 000)islessfavourablethanatlonger wavelengths,the fluxfrom Jupiter is over 20 times higherat20 µmthanitisat 100 µm. Thereislimited valueinhavingafavourablecontrastratio if the flux from bothobjects is immeasurablysmall! Exercise 1.3 Kepler’s thirdlaw in theform used for planetaryorbits is a3 G(M + M ) = ∗ P . P 2 4π2 (Eqn 1.1) orb To make theestimate, we will consider thestar as asmall mass in orbit around a much largermasspositionedatthe centreofthe Galaxy. We can therefore replace M∗ + MP with Mtotal,where this is themassofthe Galaxy. We will usethe Galactocentric distance as the valuefor a,and this will allowustomakean estimate of Porb,the timetaken for acomplete orbit around theGalaxy.Thus we

284 Solutions to exercises have 5 6 4π2a3 1/2 Porb = . (S1.1) GMtotal To usethisweneed to convertall quantities into SI units (which we can accomplishusing theinformationinthe Appendix): 12 30 12 42 Mtotal =10 M% =1.99 × 10 × 10 kg ≈ 2 × 10 kg, a =8kpc =8×103 × 3.086 × 1016 m ≈ 2 × 1020 m. In each casewehaveretained onlyone significantfigure as we aremaking a rough estimate. Thus we have 5 61/2 4π2(2 × 1020 m)3 Porb ≈ 7 × 10−11 Nm2kg−2 × 2 × 1042 kg 5 6 4 × 10 × 8 × 1060 3 1/2 ≈ m 7 × 2 × 1031 kg ms−2m2kg−2 kg 5 6 3 × 1062 3 1/2 ≈ m 1 × 1032 m3 s−2 . '1/2 ≈ 3 × 1030 s2 ≈ 2 × 1015 s. It’s difficult to grasphow long 1015 sis, so we will convertthe answertoyears: 2 × 1015 s Porb ≈ 3600 sh−1 ×24 hday−1 × 365.25 dayyr−1 ≈6×107 yr. Thus we have deduced that it takes about 60 Myr for theSun to complete its orbit around theGalaxy.(Note: this is averyrough answerdue to theinitial approximationsand theaccumulated roundingerrors.) Exercise 1.4 (a)Equation1.12 hasasimple dependenceoni: V ∝sin i.

Thefunction sin i hasamaximumvalue of 1,when i =90◦;thiscorresponds to thelineofsight to thesystem beingexactly in the planeofthe orbit,asshown in Figure S1.1a. Theminimum valueofthe radial velocity corresponds to sin i =0, which occurs when i =0◦;thiscorresponds to theplaneofthe orbit coinciding with theplaneofthe skyasviewedbythe observer, as showninFigure S1.1b. Theorbital velocities of thestar andits planet arealwaysorthogonaltothe lineof sight,and zeroradial velocity variation is observed. At intermediate orientations, 0◦

285 Solutions to exercises

i =90◦

(a) v∗

i =0◦

(b) v∗

0◦ ≤ i ≤ 90◦ i

FigureS1.1 (a) An elliptical orbit viewed from an orbital inclination of i =90◦;the z-axislies in theplaneofthe orbit.(b) Thesameorbitviewedfrom i =0◦;the planeofthe orbit coincides with theplaneofthe sky, andthere is no component of theorbital velocity in thedirectiontowards or away from the observer. (c)Atintermediate orientations, 0◦

2πaMP sin i V (t)=V0,z + √ (cos(θ(t)+ω )+ecos ω ) . 2 OP OP (MP + M∗)P 1 − e Thevariable part of this is 2πaM sin i P √ cos(θ(t)+ω ), 2 OP (MP + M∗)P 1 − e and cos(θ(t)+ωOP) varies cyclicly between −1 and +1.The radial velocity 286 Solutions to exercises amplitude is therefore 2πaM sin i A = P √ . RV 2 (Eqn 1.13) (MP + M∗)P 1 − e Forthe speciﬁccaseofJupiter orbitingaround theSun, we substitute the appropriate subscripts:

2πaMJ sin i ARV = ; . (S1.2) ( + ) 1 − 2 MJ M% PJ eJ

Thedata were conveniently all giveninSIunits exceptfor PJ,which we must convertfrom yearstoseconds:

PJ =12yr =12yr × 365.25 days yr−1 × 24 hday−1 × 60 × 60 sh−1 =3.8×108 s, wherewehavegiven our converted valuetotwo signiﬁcantﬁgures,while performingthe intermediate stepsinthe conversion to ahigherprecision. Substitutingvalues into EquationS1.2, we obtain 2π × 8 × 1011 m × 2 × 1027 kg × sin i ARV = √ 2 × 1030 kg × 3.8 × 108 s × 1 − 0.0025 =13sin i ms−1.

Exercise 1.5 (a) No.Equation1.13 alsocontains afactor P in the denominator. Planetaryorbits obeyKepler’s thirdlaw,soP2∝a3.Thismeans 3/2 1−3/2 −1/2 that P ∝ a ,soARV ∝ a = a .The radial velocity amplitude decreases as the planet’sorbital semi-major axis increases. (b) Planet mass appearsinboththe denominatorand thenumerator of Equation1.13,givingadependence

MP ARV ∝ . MP + M∗

Thequantity on theright-hand side will increasewith MP if M∗ is held ﬁxed. This proportionality alsoincludesthe onlyappearance of thestellar mass, M∗,in Equation1.13.IfMPis held ﬁxed,the right-hand side will decreaseasM∗ increases. √ Theeccentricity contributes to Equation1.13 solely through thefactor 1 − e2 in the denominator, so 1 A ∝ √ . RV 1 − e2 As e increases, (1 − e2) decreases,sothe right-hand side of theproportionality increases with increasing e. We have alreadyshown that theradial velocity amplitude decreases as the planet’s orbital semi-major axis increases,sosummarizing our analysis of Equation1.13, we seethatthe radial velocity amplitude is highest for massiveplanets in close-in eccentric orbitsaround low-mass hoststars. 287 Solutions to exercises

Exercise 2.1 (a) Equation1.21 is R ≈ ∗ . geometric transitprobability a (Eqn 1.21) To usethisweneed avalue for thesemi-majoraxis, a.Kepler’s thirdlaw (Equation 1.1) can be expressedas P2 a3=G(M +M ) , ∗ P 4π2 (S2.1) so theorbital semi-major axis is 5 6 / 5 6 21/3 P 2/3 P 2 a =(G(M +M ))1/3 ≈ GM , ∗ P 2π ∗ 2π (S2.2) where theapproximation lies in neglecting the planet’smasscompared with that −3 of thestar.Since Jupiter’s mass is ∼10 M% andother giantplanets andstars have masses of thesameorderofmagnitude,tothe precisionatwhich we are working this is agood approximation,sowewill revert to usinganequals sign in thesubsequent working.InSIunits,the quantities that we need to substitute into EquationS2.2 are G =6.67 × 10−11 Nm2kg−2, 30 30 M∗ =1.12 M% =1.12 × 1.99 × 10 kg =2.23 × 10 kg, P =3.52 days =3.52 × 24 × 3600 s =3.04 × 105 s. Consequently,wehave / 5 6 21/3 3.04 × 105 2 a = 6.67 × 10−11 × 2.23 × 1030 2 −1 2 6.28 Nm kg s

. '1/3 = 3.485 × 1029 kg ms−2m2kg−1 s2 . '1/3 = 3.485 × 1029 m3 =7.04 × 109 m. (S2.3)

We aregiven the radius, R∗,and simply need to convertthistometres: 8 8 R∗ =1.146 R% =1.146 × 6.96 × 10 m =7.98 × 10 m. (S2.4) Consequently,substitutingthe values from Equations S2.3 andS2.4 into Equation1.21,wehave 8 R∗ 7.98 × 10 m geometric transitprobability ≈ ≈ ≈ 0.113. a 7.04 × 109 m Theprobabilityofaplanet in acircular orbit likeHD209458 b’stransitingfrom anyrandom lineofsight is approximately 11%, i.e. better than 1 in 10! (b) Theassumptions implicitly made by adoptingEquation1.21 are(i) that the orbit is randomly oriented,and (ii)thatthe orbit is circular.Since HD 209458 b wasdiscoveredbythe radial velocity technique,whosesensitivity to agiven planet decreases steadily as the orbital inclination decreases,the probability of HD 209458 btransitingwas actually slightly higherthansuggested by Equation1.21.For anon-circular orbit,the planet spends timeatavariety of 288 Solutions to exercises distances,which will affect the probability of transiting. Theprobabilitydepends on e and ωOP as well as a.Aﬁnal subtlety for eccentric orbitsisthatthe planet movesmore quickly when it is closer to the star (asprescribedbyKepler’s second law, or equivalently theconservationofangular momentum), so factoring in the ﬁnite observationalcoverage rendersthe transits slightly lesslikely to be caught. To quantitatively assess therelative importanceofthese threefactorsrequires more information than is giveninthe question. Exercise 2.2 (a) We have 2πaM sin i A = P √ RV 2 (Eqn 1.13) (MP + M∗)P 1 − e and / 5 6 21/3 P 2 a ≈ G(M + M ) . ∗ P 2π (Eqn S2.1)

Assuming acircular orbit, e =0,and making theapproximation MP ) M∗,these become

2πaMP sin i ARV = (S2.5) M∗P and / 5 6 21/3 P 2 a = GM . ∗ 2π (Eqn S2.2)

RearrangingEquationS2.5, we have A M P M sin i = RV ∗ , P 2πa andsubstitutinginfor a,weobtain 5 62/3 5 61/3 ARV M∗ P 2π 1 MP sin i = 2π P GM∗ 5 6 M 2 P 1/3 = A ∗ . RV 2πG (S2.6)

ARV and P areobservables,and everything elseonthe right-hand side exceptfor M∗ is aconstant, so this is theexpression that we seek. (b) Thevalues for HD 209458 are −1 ARV =84.67 ± 0.70 ms , 30 30 M∗ =1.12 M% =1.12 × 1.99 × 10 kg =2.23 × 10 kg, P =3.52 days =3.52 × 24 × 3600 s =3.04 × 105 s,

289 Solutions to exercises so we have 5 6 M 2 P 1/3 M sin i = A ∗ P RV 2πG 5 61/3 (2.23 × 1030)2 kg2 × 3.04 × 105 s =84.67 ms−1 6.28 × 6.67 × 10−11 Nm2kg−2 5 61/3 kg2 s =84.67 × 3.609 × 1075 ms−1 kg ms−2m2kg−2 . '1/3 =84.67 × 1.53 × 1025 kg3 m−3 s3 ms−1 =1.30 × 1027 kg. In theabove,weexplicitly kept theunits of all the quantities when we substituted in,which allowedustocheck that we (i)had got adimensionally correct expression in part (a), and(ii) were usinganappropriate choice of units for each quantity.Ifwehad tried to use M∗ =1.12 M% in theexpression, our ﬁnal units 2/3 1/3 wouldhavebeen M% kg ,alertingusthatthere wassomethingamiss.Of course, assuming that theexpression that we beginwith is correct, if we useSI units throughoutthenweshouldalwaysobtain an answerinSIunits.Using this fact without working theunits through wastes avaluable check of our working. (c)Weare given i =86.71◦ ± 0.05◦,sosin i =0.9984.Consequently, 1.30 × 1027 1.30 × 1027 M = kg = kg =1.30 × 1027 . P sin i 0.9984 kg Convertingthistothe othermassunits requested: 1.30 × 1027 kg M = =0.684 M P 1.90 × 1027 −1 J kg MJ and 1.30 × 1027 M = kg =6.53 × 10−4 . P 30 −1 M% 1.99 × 10 kg M%

(d) Thus we can work outthe ratio between MP and M∗ for theHD209458 system: M 6.53 × 10−4 P = M% =5.83 × 10−4. M∗ 1.12 M% Theplanet’smassisafactor of 2 lessthanathousandththatofthe star.Since we have worked to aprecision of onlythree signiﬁcantﬁgures,the approximation MP ) M∗ wasapplicable. (e)There aretwo basicapproachesthatcouldbeused. To tackle theproblem analytically,the full versions of Kepler’s thirdlaw (Equation 1.1) andEquation1.13 wouldbeused, andthe algebra wouldneed to carry through all the instances of MP.Thiswouldresultinamuch more complex expression, butitcouldstill be solved.

To tackle theproblem iteratively,wecouldreplace the expression (MP + M∗) with Mtotal in Kepler’s thirdlaw (Equation 1.1) andEquation1.13.Thenthe 290 Solutions to exercises

method in parts (a)–(d) couldbeusedtoevaluate MP usingthe approximation Mtotal ≈ M∗.Thiswouldbeour first estimate for MP,and we couldthenrefine our estimate by using Mtotal ≈ M∗ + MP andcalculatinganewvalue for MP. This technique shouldberepeated until, at theprecision required,the newvalue of MP doesnot differ from thevalue used to estimate it. Exercise 2.3 (a) Thebandpass is broad andcentred on 700 nm,sowewill use thefiducial fluxdensity valuefor theRband, which is centred on 700 nm.Since thevalue of thesky brightness is alreadygiven in pixelunits in Table2.1, we can slightly simplifythe procedure outlined in thetext. Thesky fluxdensity perpixel is givenbyapplying thestandard conversion from magnitudes:

m1 − m2 =2.5log10(F2/F1), (S2.7) where m1 =6.8(from Table2.1), m2 =0.0,the ﬁducial magnitude, F1 is −11 −2 −1 theﬂux density corresponding to m1,and F2 =1.74 × 10 Wm nm (the Rbandvalue from TableA.4 in theAppendix). Making F1 thesubject of EquationS2.7, we have

−(m1−m2)/2.5 F1 = F2 × 10 =1.74 × 10−11 Wm−2nm−1 × 10−6.8/2.5 =1.74 × 10−11 × 1.91 × 10−3 Wm−2nm−1 =3.32 × 10−14 Wm−2nm−1. We will take 700 nm to be thetypical wavelengthofthe radiation.Thus the typical photon energy is hc E = hν = ph λ 6.63 × 10−34 Js×3.00 × 108 ms−1 = =2.84 × 10−19 J. 700 × 10−9 m

This allows us to convertfrom ﬂux to photon rate, lsky: F 3.32 × 10−14 −2 −1 l = 1 = Wm nm sky −19 Eph 2.84 × 10 J =1.169 × 105 s−1 m−2 nm−1. 5 We expect, therefore, lsky =1.17 × 10 skyphotons persecond persquare metre of telescope apertureper nanometreincludedinthe bandpass to fall on each pixel of thePASS survey CCD. Using thevalues that we aregiven:the telescope apertureisofdiameter 2.5 cm, π 2 2 −4 2 so thecollectingareaisA= 4 ×0.025 m =4.91 × 10 m .The bandpass is wide (Δλ =300 nm), so thephoton rateper pixelis dn sky =1.169 × 105 −1 −2 −1 × A × Δλ dt s m nm =1.169 × 105 s−1 m−2 nm−1 × 4.91 × 10−4 m2 × 300 nm =1.72 × 104 s−1. We expect almost 20 000 skyphotons perpixel persecond for thePASS survey.

(b) Theexpected numberofsky photons, nsky,per pixelina10 sexposure is 4 −1 5 nsky =1.72 × 10 s × 10 s =1.72 × 10 , 291 Solutions to exercises or almost 200 000 skyphotons perpixel perexposure. Exercise 2.4 We have L F = , 4πr2 (Eqn 2.2) andweknowthatsourcescan be detected if F ≥ S.Thismeansthatasource will be detected if L S ≤ , 4πr2 wherewehavesubstituted S ≤ F in Equation2.2. Thelimitingdistance will be when the ﬂuxexactly equals S,sothatthe limitingdistance, dmax,corresponds to L S = . 4πd2 max Making thedistance thesubject of theequation, this becomes 5 6 L 1/2 d = , max 4πS (Eqn 2.3) as required Exercise 2.5 (a) Thearc lengthis2aθ,where θ is theangleindicated in Figure S2.1.From this ﬁgure, we seethat R sin θ = ∗ , a

R∗

θ a θ

V W

FigureS2.1 Thetwo outer vertical lines indicate the path taken by light rays on each side of thestar to adistantobserver. Theraysare parallel. Points Vand Ware theintersectionsofthese lines with theplanet’sorbit. 292 Solutions to exercises

so 5 6 R θ =sin−1 ∗ . a Hencewehave T = P × lengthofarc from VtoW dur 2πa 2asin−1(R /a) = P × ∗ 5 2πa6 P R = sin−1 ∗ . π a (S2.8)

(b) Thedeviationbetween the exact andapproximated values for Tdur is simply thedifferencebetween the right-hand sidesofEquations 2.6 andS2.8: 5 5 6 6 P R R ΔT = sin−1 ∗ − ∗ . dur π a a (S2.9) Thefractional deviation is simply ΔT dividedbyT ,i.e. 5 5 6 6dur dur P R R sin−1 ∗ − ∗ R∗ ΔT π a a dur = 5 6 =1− a5 6. T P R R dur sin−1 ∗ sin−1 ∗ π a a

(c)Aswesaw in thesolutions to Exercises 2.1 and2.2, Kepler’s thirdlaw tells us that 5 6 P 2/3 a = (GM )1/3. 2π ∗

(d) To evaluate ΔTdur/Tdur,weneed to useour expression for a for thespeciﬁc values of M∗ =1M%andorbital period P =1day. Generally,itisadvisableto substitute in expressions to obtain the answeralgebraically,before substitutingin numerical values.Inthiscase, however, thequantity that we seek depends only on theratio R∗/a,and thealgebraic expression for a is signiﬁcantly more complicated than this. So,evaluating a with P =1day =86400 s, 30 M∗ =1.99 × 10 kg, G =6.67 × 10−11 Nm2kg−2, we have 5 6 P 2/3 a = (GM )1/3 2π ∗ . '2/3 . '1/3 = 1.375 × 104 s 1.327 × 1020 Nm2kg−1 =5.740 × 102 × 5.101 × 106 s2/3 N1/3 m2/3 kg−1/3 =2.928 × 109 s2/3 (kg ms−2)1/3m2/3kg−1/3 =2.93 × 109 m.

293 Solutions to exercises

8 In SI units, 1 R% =6.96 × 10 m, so R 6.96 × 108 ∗ = =0.238 a 2.93 × 109 and 5 6 R sin−1 ∗ =sin−1(0.238) =0.240, a so ΔT 0.238 dur =1− =0.0083 ≈ 1%. Tdur 0.240 (e) Assuming that thestar’s properties remain constant, thesemi-majoraxiswill 2/3 increaseasP with theincreasing orbital period of theplanet. Theratio R∗/a will therefore decreaseasP−2/3.The small angleapproximation sin θ ≈ θ becomesincreasinglyaccurate as theangle, θ,decreases (where θ = R∗/a in this case).Sofor longerorbital periods,the approximation will be better,and the fractional deviation will be lessthan 1%. Demonstratingthisnumerically,for P =5.2days, 5 6 5.2 2/3 a(5.2 )=a(1 ) × =2.93 × 109 × 3.00 =8.79 × 109 , days day 1 m m so R 6.96 × 108 ∗ = =0.0792 a 8.79 × 109 and 5 6 R sin−1 ∗ =sin−1(0.0792) =0.0793. a Consequently, ΔT 0.0792 dur =1− ≈ 0.1%. Tdur 0.0793 Theapproximation becomesevermore exact as the planet’sorbital period is increased. (f) Theapproximation used in Equation2.6 is good to aprecision of about 1%or better for thecalculationofthe durations of transits of planets in orbits P>1day around main sequencehoststars. This precisionshouldbesufﬁcientfor an estimate. Exercise 2.6 We have PR T ≈ ∗. dur πa (Eqn 2.6) Convertingthe values that we aregiven to metres andhours,wehave P =1.3382 × 24 h =32.12 h, 8 8 R∗ =1.15 × 6.96 × 10 m =8.00 × 10 m, a =0.023 × 1.5 × 1011 m =3.45 × 109 m,

294 Solutions to exercises andsubstitutinginthese values gives 32.12 × 8.00 × 108 hm T ≈ ≈ 2.37 h. dur π×3.45 × 109 m So thetransitdurationis2.4hours (to 2 s.f.). Exercise 2.7 (a)Equation2.34 tells us that thenumberoftransitingplanet 3/2 discoveries predicted scales as L∗ .Ifstarsweremore luminous,the signal receivedwouldbestronger, andthe signal-to-noiseratio at anygiven distance wouldbebetter.This N ∝ L3/2 dependencyisexactly the same as derivedin Equation2.5, andarisesfrom thedependencyofthe survey volume on the luminosity of thesources(assuming aﬁxedﬂux thresholdfor detection).

(b) LSN is thelimitingsignal-to-noiseratio andisproportionaltothe ﬂuxfrom theplanet hoststar at thefaint limit. This limitinghoststar brightness is an exampleofthe limitingﬂux, S, in Equation2.5. Thenumberofplanets is L−3/2 proportionalto SN according to Equation2.34,and consequently this equation expressesthe same N ∝ S−3/2 relationshipasEquation2.5. Exercise2.8 Assuming, for simplicity,thatthe 16 cameras have been constantly operationalsince 2004, we can estimate the numberofmonths of operationtodate. At thetimeofwriting(March2009), this is about 5 years, or 60 months.Thus thenumberofdata pointsisofthe order of 8 numberofpoints = Nmonths × ncameras × 5 × 10 =60×16 × 5 × 108 =4.8×1011 =5×1011 (to 1 s.f.). Thus thearchive needstoorganize andstore on theorderof1012 unique photometric data points. Exercise 2.9 Theminimum valueofthe ratio is almostzero: thevalue that occurs for abarely detectable grazingtransit. Themaximum valuewill correspond to acentral transit, i.e. impact parameter of b =0.0.From theresults of Exercise2.5 we know that we can adopt the approximate expression for thetransitduration PR T (b =0.0) ≈ ∗, dur πa (Eqn 2.6) so theratio that we must calculate is T (b =0.0) R dur ≈ ∗ . P πa (S2.10) Substitutingfor a from Kepler’s thirdlaw (Equation 1.1),i.e. 5 6 P 2/3 a =(G(M +M ))1/3 ∗ P 2π into EquationS2.10, we obtain T (b =0.0) R 22/3π2/3 dur ≈ ∗ P π 2/3 1/3 1/3 P G M∗ 5 6 4 1/3 ≈ R . ∗ 2 πP GM∗ 295 Solutions to exercises

Fromthiswecan seethatfor agiven star,the ratio will be amaximum for the minimumvalue of theorbital period as T (b =0.0) dur ∝ P −2/3. P At thetimeofwriting(October2009) theshortest-period knownplanet is CoRoT-7bwith P =0.85 days.Adoptingthisperiod andthe mass andradius of theSun, we can obtain an estimate for themaximum valuefor theratio. Theinput values are P =0.85 days =0.85 × 24 × 3600 s =7.3×104 s, 8 R∗ =6.96 × 10 m, 30 M∗ =1.99 × 10 kg, so we have 5 6 T (b =0.0) 4 1/3 dur ≈ R ∗ 2 P πP GM∗ 5 6 4 1/3 ≈ 6.96 × 108 m × π × (7.3 × 104)2 × 6.67 × 10−11 × 1.99 × 1030 s2 Nm2kg−2 kg ≈ 0.085. Thetransitdurationvaries between 0%and 8.5%ofaplanet’sorbital period. Exercise 2.10 Theradial velocity semi-amplitude is givenbyEquation1.13 as 2πaM sin i A = P , RV 2 1/2 (MP + M∗)P (1 − e ) wherethe semi-major axis is givenbyKepler’s thirdlaw (Equation 1.1) as 0 > GP 2(M + M ) 1/3 a = ∗ P . 4π2 Substitutingthe second equationintothe ﬁrst,wehave 5 6 2πG 1/3 M sin i A = P , RV 2/3 2 1/2 P (M∗ + MP) (1 − e ) andrearrangingtomake MP the subject of theequation, we get 5 6 A (M + M )2/3 (1 − e2)1/2 P 1/3 M = RV ∗ P . P sin i 2πG Now, if we assume that theradial velocities measured in thetwo spectrarepresent theextremes of thereﬂex orbital motionofthe star,the minimumorbital period of theputative planet must be 12 hours,and theminimum valueofthe radial velocity amplitude is givenbythe difference between thetwo radial velocity measurements.Since thestar hasthe same spectral typeasthe Sun, clearly M∗ = M%,and themasstermonthe right-hand side hasaminimumvalue 2/3 of M∗ .Also, we notethatthe maximumvalue of sin i is 1,sosetting this at its limit will alsogiveaminimum valuefor MP.For such ashort orbital period, the orbit wouldcircularize, so we mayassume e ∼ 0,and theequationtherefore reduces to 5 6 P 1/3 M ≥ A M 2/3 . P RV ∗ 2πG 296 Solutions to exercises

So,ifthe observedradial velocity changesare duetothe reﬂexmotioncausedby an unseen orbitingbody,the minimummassofthisobject is 5 61/3 4 −1 30 2/3 12 × 3600 s MP ≥ 10 ms ×(1.99 × 10 kg) × 2π × 6.673 × 10−11 Nm2kg−2 ≥ 7.4 × 1028 ms−1kg2/3 s1/3 N−1/3 m−2/3 kg2/3 ≥ 7.4 × 1028 ms−1kg2/3 s1/3 (kg ms−2)−1/3m−2/3kg2/3 ≥ 7.4 × 1028 kg

≥ 39 MJ. Sincethe putative planet hasamass in excessofalmost 40 times themassof Jupiter (about 0.037 M%), this object maybeabrown dwarf, butcannot be a planet, andsocan be eliminated from furtherfollow-up. Exercise 3.1 (a) Thesemi-majoraxisofthe Earth’sorbitisa=1AU andthe period of theEarth’sorbitisP =1year.So, from Equation3.1, theEarth’s orbital speed is 2πa v = P 2π × 1 = AU 1 year 2π × 1.496 × 1011 = m 365.25 × 24 × 3600 s =2.98 × 104 ms−1. So theorbital speed of theEarth is about 30 km s−1. (b) Although this planet is at the same distance from its star as the Earthisfrom theSun, theorbital periods will differ because thestar hasamass differentto that of theSun. We can calculate the orbital period from Kepler’s thirdlaw (Equation 1.1) as 5 6 4π2a3 1/2 P = G(M∗ + MP) 5 6 4π2a3 1/2 ≈ GM∗ 5 6 4π2 × (1.496 × 1011 )3 1/2 ≈ m 6.673 × 10−11 Nm2kg−2 × 0.5 × 1.99 × 1030 kg ≈ 4.46 × 107 s. This orbital period is about 1.41 years. So,using Equation3.1 once again, the orbital speed of theplanet is 2πa 2π × 1 v = = AU P 4.46 × 107 s 2π × 1.496 × 1011 m = =2.11 × 104 ms−1. 4.46 × 107 s So theorbital speed of theplanet is about 21 km s−1.

297 Solutions to exercises

Exercise 3.2 (a) We startfrom Equation3.4: /+ 2 P (R + R )2 − a2 cos2 i T = sin−1 ∗ P . dur π a

This can be rewritten as %5 6 3 P R2 2R R R2 1/2 T = sin−1 ∗ + ∗ P + P − cos2 i . dur π a2 a2 a2

Since a ' R∗ ' RP,the second andthird terms inside thebrackets aremuch smaller than theﬁrst, so 5 6 P R2 1/2 T ≈ sin−1 ∗ − cos2 i . dur π a2 Sincethe terminthe argumentofthe inversesinefunctionwill be small, and sin x ≈ x for small values of x in radians, we have 5 6 P R2 1/2 T ≈ ∗ − cos2 i , dur π a2 as required. (b) To usethisequationtodetermine i,weneed to ﬁrst obtain avalue for a. Kepler’s thirdlaw tells us that a3 GM ≈ ∗ . P 2 4π2 SincewehaveaSun-like star,the right-hand side is thesamefor theplanet in the questionand theEarth, so a3 3 = a⊕ , 2 2 P P⊕ wherea⊕andP⊕arethe semi-major axis andthe orbital period of theEarth, respectively.Hence 5 62 3 3 P a = a⊕ , P⊕ thus 5 6 P 2/3 a = a⊕ P⊕ 5 62/3 6 days =1AU 365.25 days =1.496 × 1011 m × (0.0164)2/3 =9.67 × 109 m. Nowwemustrecastthe equationtosolve for i: 5 6 P R2 1/2 T ≈ ∗ − cos2 i , dur π a2

298 Solutions to exercises so 5 6 T 2 π2 R 2 dur ≈ ∗ − cos2 i, P 2 a thus 5 6 R 2 T 2 π2 cos2 i ≈ ∗ − dur , a P 2 giving %5 6 5 6 31/2 R 2 T π 2 cos i ≈ ∗ − dur a P and ﬁnally %5 6 5 6 31/2 R 2 T π 2 i ≈ cos−1 ∗ − dur . a P

8 9 Now, substitutinginR∗ =1R%=6.96 × 10 m, a =9.67 × 10 m, Tdur =2hours and P =6days =144 hours,wehave %5 6 5 6 31/2 6.96 × 108 2 2π 2 i ≈ cos−1 m − h 9.67 × 109 m 144 h %5 6 31/2 6.96 2 ! π 82 ≈ cos−1 − 96.7 72 ≈ cos−1(0.0573) ≈ 86.7◦.

(c) If Tdur =4hand all otherparametersofthe system areasabove,thenwehave %5 6 5 6 31/2 6.96 × 108 2 4π 2 i ≈ cos−1 m − h 9.67 × 109 m 144 h %5 6 31/2 6.96 2 ! π 82 ≈ cos−1 − 96.7 36 ≈ cos−1(−0.002 43)1/2.

Sincethe square root of anegative numberisnot areal number, this equationis invalid.Thatis, thetransitdurationistoo long to be possibleinsuchasystem.We wouldbeforcedtoconclude that themeasurements areerroneous or that the candidate is an astrophysical mimic: probablyitiseither ablendedeclipsing binary or agrazing eclipsebinary. Exercise 3.3 (a) ThesignalisΔF =0.0164F ,where we have used the depth as giveninthe caption of Figure 3.5. Thenoise is 1.1 × 10−4F .So 0.0164F Signal = =149. Noise 1.1 × 10−4F

299 Solutions to exercises

(b) Thecurvatureinthe transitﬂoor causesadrop from approximately 0.9870F at second contact to approximately 0.9835F at mid-transit. The‘signal’ levelis therefore (0.9870 − 0.9835)F =0.0035F . If thescatter wasmore than about 0.3%(or 3 × 10−3F ), it wouldbedifﬁcultto detect thecurvature. Exercise 3.4 As thelight emerging from closetothe limbofthe planet has γ =80◦,soµ=cos γ =0.174. Thelinear limbdarkening law(Equation3.9) gives I(µ) =1−u(1 − µ), I(1) so I(0.174) =1−0.215 × (1 − 0.174) I(1) =0.82.

Thelogarithmic limbdarkening law(Equation3.10)gives I(µ) =1−u(1 − µ) − ν µ ln µ, I(1) l l so I(0.174) =1−0.14 × (1 − 0.174) +0.12 × 0.174 × ln 0.174 I(1) =1−0.116 +0.036 =0.92.

Thequadratic limbdarkening law(Equation3.11)gives I(µ) =1−u (1 − µ) − ν (1 − µ)2, I(1) q q so I(0.174) =1−0.29 × (1 − 0.174) +0.13 × (1 − 0.174)2 I(1) =1−0.240 +0.089 =0.85.

Clearly,these threelimbdarkening prescriptions give markedly differentamounts of limbdarkening. Theform of limbdarkening lawadopted does make a difference.

Exercise 3.5 Theimpact parameter is b =3R%/4,and theinclinationangleis i=86.5◦,sothe semi-major axis of theorbitis ◦ a=b/ cos i =3R%/4cos 86.5 =12.3R% or a =12.3×6.96 × 108 m =8.56 × 109 m.

300 Solutions to exercises

Using Kepler’s thirdlaw (Equation 1.1),the period of theorbitis 5 6 4π2a3 1/2 Porb = GM∗ 5 6 4π2 × (8.56 × 109 )3 1/2 = m 6.673 × 10−11 Nm2kg−2 × 1.99 × 1030 kg =4.32 × 105 s. Theorbital period is therefore 120 hours (or 5 days). 1 hour beforeorafter mid-transit therefore corresponds to aphase angleof ±2πt ±(2π × 1 h) ±ωt = = = ±0.052 radians. Porb 120 h We therefore need to work outthe positionofthe planet in three cases: (i) ωt =2π−0.052 =6.231 radians, (ii) ωt =0radians, (iii) ωt =0.052 radians. (i)Inthiscase, the horizontal component of theplanet’sposition, with respect to the centreofthe star’s disc,is

asin ωt =12.3R%×sin(6.231 radians)=−0.64 R%. Thevertical component of theplanet’sposition, with respect to the centreofthe star’s disc,is ◦ acos i cos ωt =12.3R%×cos86.5 ×cos(6.231 radians)=0.75 R%. (ii)Inthiscase, the horizontal component of theplanet’sposition, with respect to the centreofthe star’s disc,is

asin ωt =12.3R%×sin(0 radians)=0. Thevertical component of theplanet’sposition, with respect to the centreofthe star’s disc,is ◦ acos i cos ωt =12.3R%×cos86.5 ×cos(0 radians)=0.75 R%. (iii) In this case, thehorizontal component of theplanet’sposition, with respect to the centreofthe star’s disc,is

asin ωt =12.3R%×sin(0.052 radians)=0.64 R%. Thevertical component of theplanet’sposition, with respect to the centreofthe star’s disc,is ◦ acos i cos ωt =12.3R%×cos86.5 ×cos(0.052 radians)=0.75 R%. Thecurvatureofthe locusofthe planet is too small to notice when working at this levelofprecision,and thelocus of thetransitisasshown in Figure S3.1.

301 Solutions to exercises

locusof transit

0.75 R%

0.64 R% 0.64 R% −R% R%

−R%

FigureS3.1 Thelocus of theplanetarytransit.

Exercise 3.6 (a) Sinceweare told that thestar hasnegligible limbdarkening, we mayuse therelationshipsappropriate to astellar disc of uniformbrightness. Theobservedﬂux is givenby

F(t)=Funocculted − ΔF 2 = πI0R − I0 A (t) . ∗ e ' 2 = I0 πR∗ − Ae(t) , (S3.1) where  0 if 1+p<ξ,  / + 2  4ξ2 − (1 + ξ2 − p2)2 A = R2 p2α + α − 1 − p<ξ≤1+p, e  ∗ 1 2 2 if   2 2 πp R∗ if 1 − p ≥ ξ, (Eqn 3.27) with 2 2 2 2 RP p + ξ − 1 1+ξ −p p = , cos α1 = , cos α2 = , R∗ 2ξp 2ξ a . '1/2 2π ξ= sin2 ωt +cos2 i cos2 ωt and ω = . R∗ Porb (b) For p =0.1and ξ =0.2,clearly 1 − p ≥ ξ,sothe thirdcaseofEquation3.27 is appropriate, which corresponds to theplanet fallingentirely within the A = πp2R2 = πR2 stellar disc.Inthiscase, e ∗ P,sothe change in ﬂuxis ΔF =I A =I πR2 0 e 0 P.The relative change in ﬂuxistherefore ΔF I πR2 R2 = 0 P = P , 2 2 F I0πR∗ R∗ as required.

(c)The orbital inclination, i,isknown from thetransitduration, Tdur.First contact, at time t = t1,occurs when ξ1 =1+p,and second contact, at time t = t2,occurs when ξ2 =1−p.Knowing thesetwo times,and knowing the

302 Solutions to exercises

transitduration, Tdur,and theorbital period, P ,wecan evaluate ξ for thetwo contact points: . ' a 2 2 2 1/2 a ξ1 =1+p= sin ωt1 +cos i cos ωt1 = Γ1, R∗ R∗ . ' a 2 2 2 1/2 a ξ2 =1−p= sin ωt2 +cos i cos ωt2 = Γ2, R∗ R∗ where Γ1 and Γ2 have been introduced as shorthand for theknown,but complicated,expressions involvingthe measured times andthe determinedorbital inclination. Consequently,wehavetwo expressions relating p, a and R∗.Wecan therefore treat theseassimultaneous equations,eliminate a,and make p the subject of theresultingequation: Γ1 − Γ2 p = . (S3.2) Γ1 +Γ2

We recall that p = RP/R∗,soEquationS3.2 givesusthe ratio of thestar and planet radiiinterms of themeasured contact times andthe inclination. (d) Themethodillustrated in part (b) depends on measuring thedepth of the transit, ΔF ,and relating this to the fraction of thestellar fluxocculted, which is assumedtobeproportionaltothe area of theplanet’sdisc. If thestar is limb darkened,the definitionofΔFis complicated:ifweuse thedepth at the centreof thetransit, this will probablyresultinanerroneously large planet size. At mid-transit theocculted fraction of thetotal stellar fluxislikely to be greater than 2 (RP/R∗) because thebrightcentral regionsofthe stellar disc arebeing occulted. Limb darkening seriouslyaffects the planet size deduced from this method. On theother hand, themethodillustrated in part (c) depends onlyonmeasuring thetimingatwhich the contact pointsoccur. So long as the second andthird contacts can be clearly identified in the lightcurve, this method is unaffected by thelimbdarkening. Exercise 3.7 (a) Alarge planet will give rise to aprolongedingressand egress, andthe transit will be relatively deep;alarge impact factor meansthatthe durationofthe transit is relatively short; andverylittle limbdarkening implies that thetransitfloor will be flat. Asketch of such atransitisshown in Figure S3.2a.

(a)

(b)

FigureS3.2 (a) Atransitlight curvefor alarge planet with alarge impact factor andverylittle limbdarkening. (b) Atransitlight curvefor asmall planet with asmall impact factor andsigniﬁcantlimbdarkening. 303 Solutions to exercises

(b) Asmall planet will give rise to ashort ingressand egress, andthe transitwill be relatively shallow; asmall impact factor meansthatthe durationofthe transit is relatively long; andsigniﬁcantlimbdarkening implies that thetransitﬂoor will be curved.Asketch of such atransitisshown in Figure S3.2b. Exercise 4.1 We need to useKepler’s thirdlaw,but recognize that aplanet in orbit around asolar-massstar with an orbital period of 1 year will have a semi-major axis of 1 AU.Hence we can write a3 (1 )3 = AU . 2 2 M∗P 1 M% × (1 yr)

(a) (i) In this case, theorbital period is givenby 5 6 a3 1 ×(1 )2 1/2 P = × M% yr 3 M∗ (1 AU) 5 6 (1 )3 1 × (1 )2 1/2 = AU × M% yr 3 0.7 M% (1 AU) 5 6 1 1/2 = =1.195 =437 . 0.7 yr yr days (ii)Inthiscase, the orbital period is givenby 5 6 a3 1 ×(1 )2 1/2 P = × M% yr 3 M∗ (1 AU) 5 6 (1 )3 1 × (1 )2 1/2 = AU × M% yr 3 1.5 M% (1 AU) 5 6 1 1/2 = =0.816 =298 . 1.5 yr yr days

(b) (i) In this case, thesemi-majoraxisofthe orbit is givenby 5 6 (1 )3 1/3 a= M P2× AU ∗ 2 1 M% × (1 yr) 5 6 (1 )3 1/3 = 0.7 × (500 )2 × AU M% days 2 1 M% × (1 yr) 5 6 500 2/3 =(0.7)1/3 × =1.095 . 365.25 AU AU (ii) In this case, thesemi-majoraxisofthe orbit is givenby 5 6 (1 )3 1/3 a= M P2× AU ∗ 2 1 M% × (1 yr) 5 6 (1 )3 1/3 = 1.5 × (500 )2 × AU M% days 2 1 M% × (1 yr) 5 6 500 2/3 =(1.5)1/3 × =1.411 . 365.25 AU AU

304 Solutions to exercises

(c) (i) Foraplanet in a 1 AU orbit around alate Mtype star with M∗ =0.2M%, theorbital period wouldbe 5 6 a3 1 ×(1 )2 1/2 P = × M% yr 3 M∗ (1 AU) 5 6 (1 )3 1 × (1 )2 1/2 = AU × M% yr 3 0.2 M% (1 AU) 5 6 1 1/2 = =2.236 =817 . 0.2 yr yr days

(ii)For aplanet in a 500-day orbit around alate Mtype star with M∗ =0.2M%, thesemi-majoraxiswouldbe 5 6 (1 )3 1/3 a= M P2× AU ∗ 2 1 M% × (1 yr) 5 6 (1 )3 1/3 = 0.2 × (500 )2 × AU M% days 2 1 M% × (1 yr) 5 6 500 2/3 =(0.2)1/3 × =0.721 . 365.25 AU AU

(d) Exoplanets have been discoveredaround starswith arange of masses,from late Mtype with M∗ ∼ 0.2 M% to starsofmasssigniﬁcantly greater than that of theSun. Aplanetaryorbitwith agiven semi-major axis can therefore correspond to arange of orbital periods,spanning afactor of almost 3 in theexamples above. Similarly,aplanetaryorbitwith agiven period can correspond to arange of orbital semi-major axes,spanning afactor of around 2 in theexamples above. Exercise 4.2 (a) Figure 1.4 in thebox on ‘The neareststarsand planets’ showsthatofthe roughly 330 known starswithin 10 pc,almost 240 areMstars, i.e. about 72%ofthe stars(including thewhite dwarfs)are Mstars. Of theA–M spectral typestars, about 80%are Mstars. We arenot told howmanyofthese starsare main sequencestars, andwewill assume that they all are; this is justiﬁed because starsspend most of their lifetimes on themain sequence. Reading from Figure 4.8c, the fraction of Mstarshostingone or more planets is ∼2 ± 1%. Thus we expect the numberofMstarswithin 10 pc hostingRVplanets to be

NP,M ≈ 240 × (0.02 ± 0.01) ≈ 4.8 ± 2.4. Similarly,there are 71 F, G, Kstarswithin 10 pc,and reading offthe graph, the fraction of thesehostingplanets is ∼4.1 ± 0.6%. Thus we expect the numberof F, G, KstarshostingRVplanets to be

NP,FGK ≈ 71 × (0.041 ± 0.006) ≈ 2.9 ± 0.4. Finally,there are 4 Astarswithin 10 pc in theRECONS census, andfor the highest-massFstarsand theAstars, Figure 4.8c suggeststhat ∼9 ± 3%hostgiant planets.Thus

NP,A ≈ 4 × (0.09 ± 0.03) ≈ 0.36 ± 0.12 ≈ 0.4 ± 0.1. Gatheringthistogether,the total numberofsystemshostingRV-detectable giant planets within 10 pc is expected to be about

NP,RV ≈ (4.8 ± 2.4) +(2.9±0.4) +(0.4±0.1) ≈ 8.1 ± 2.9. 305 Solutions to exercises

Thus thefraction of starswithin 10 pc hostingRV-detectable planets is expected to be numberhostingplanets 8.1 ± 2.9 fraction hostingplanets = ≈ ≈ 0.02 ± 0.009. total number 330

(b) Over the mass rangeshown in Figure 4.8c, i.e. 0.1–1.9 M%,the main −3 4 sequenceluminosity increases from 10 L% to 10 L%,i.e.byafactor of 10 . Equation2.2 tells us that theflux from asource decreases as the inversesquare of thedistance, andconsequently thedistance at which astar of luminosity L is brighter than V ∼ 10 varies as 5 6 L 1/2 dmax = , (Eqn 2.3) 4πS10 where we have used the notation S10 for theflux corresponding to magnitude V ∼ 10.Thus themostmassive Astarsinthe range arevisible at distances about 100 times thelimitingdistance for theleastmassive Mstarsinthe range.Since thevolumeofspace includedwithin alimitingdistance dmax is givenby ∝ d3 , volume max the magnitude limit includesAstarsinavolume 106 times thevolumefor Mstars. If we assume that our local volume is representative of therelative numbers of Astarsand Mstars, then Figure 1.4 suggeststhatthe ratio of Astars to Mstarswill be roughly N 4 × 106 A ≈ ≈ 104. (S4.1) NM 239 Thus themagnitude-limited sample is stronglybiased in favour of themore luminous,more massivestars. Thenumberthatweobtained in EquationS4.1 is an overestimate of theratio of thenumbers in thelowest-mass binand the highest-massbin in Figure 4.8c sinceweusedthe extremes (M∗ =0.1M%and M∗ =1.9M%)rather than the averagemasseswithin thebinsfor theestimate, but it illustrates thepoint. Sincethe magnitude-limited sample is dominated by thehighest-massbin,we expect that the percentageofstarswith RV-detectable planets will be more or less thevalue for that bin, i.e. we expect that the percentageofstarswith planets will be ∼9 ± 3%, wherewehaveread the numberfrom Figure 4.8c. (c)The twoestimates aredifferent.The answerinpart(a) is effectively the fraction for thelowest-mass bininFigure 4.8c because thevolume-limited sample is dominated by theselowest-mass stars. Theanswerinpart(b) is effectively the fraction for thehighest-massbin in Figure 4.8c because theflux-limited sample is dominated by theseluminous high-massstars. (d) Theanswerinpart(a) suggeststhatweshouldexpect roughly 8 RV-detectable planets within 10 pc.Figure 1.4 shows 18 planets,but this includesthe 8 in the Solar System,soitshows 10 exoplanets.Thisisconsistentwith our estimate, which is not surprising as bothFigure 1.4 andFigure 4.8c drawonthe known censusofexoplanets.Weshouldexpect more exoplanets to be discoveredwithin 10 pc,however,asthe RV precisionofavailable instrumentation improves, and longertimebases aresampled.These factorswill allowpreviouslyundiscovered exoplanets to be detected. 306 Solutions to exercises

Exercise 4.3 Thereflex RV amplitude is givenbyEquation1.13 as 2πaM sin i A = P √ . RV 2 (MP + M∗)P 1 − e In all the cases here, theplanetarymassisnegligible compared to that of thestar, so MP + M∗ ≈ M∗. Hencefor GJ 581 b, thereflex RV amplitude of thestar is 2π × (0.041 × 1.496 × 1011 ) × (15.64 × 5.97 × 1024 ) A = m kg RV,b (0.31 × 1.99 × 1030 kg) × (5.369 × 24 × 3600 s) =12.6ms−1. Similarly,for GJ 581 c 2π × (0.07 × 1.496 × 1011 m) × (5.36 × 5.97 × 1024 kg) ARV,c = (0.31 × 1.99 × 1030 kg) × (12.929 × 24 × 3600 s) × (1 − 0.17)1/2 =3.4ms−1, for GJ 581d 2π × (0.22 × 1.496 × 1011 ) × (7.09 × 5.97 × 1024 ) A = m kg RV,d (0.31 × 1.99 × 1030 kg) × (66.8 × 24 × 3600 s) × (1 − 0.38)1/2 =3.1ms−1, andfor GJ 581 e 2π × (0.03 × 1.496 × 1011 ) × (1.94 × 5.97 × 1024 ) A = m kg RV,e (0.31 × 1.99 × 1030 kg) × (3.149 × 24 × 3600 s) =1.9ms−1. If, at agiven time, the contributions to thereflex RV amplitude from each planet areall in phase with each other, then themaximum total amplitude that maybe observedissimply (12.6+3.4+3.1+1.9) ms−1 =21ms−1. Exercise 4.4 In Worked Example4.1 we calculated that the pressure in the core of Jupiter is approximately 1013 Pa,or108 bar. So Figure 4.12 indicates that the core temperature of Jupiter is roughly log(T/K)=4.4,orT =25000 K. The degeneracyparameter is θ =0.03.Wecan therefore rearrange Equation4.18 to give thenumberdensity of electrons as 5 6 2πmkT 3/2 n = F h2 5 6 2πmkT 3/2 = θh2 5 6 2π × (9.109 × 10−31 ) × (1.381 × 10−23 −1)×(25000 ) 3/2 = kg JK K 0.03 × (6.626 × 10−34 Js)2 =1.8×1030 (kg J−1 s−2)3/2 =1.8×1030 (kg kg−1 m−2 s2 s−2)3/2 =1.8×1030 m−3.

307 Solutions to exercises

Exercise 4.5 (a) Theenergyradiated away is Erad = L Δt,and thechange in gravitationalenergyisΔEGR =(dEGR/dt)Δt.Wehave ζ − 1 dE L = − GR , ζ dt (Eqn 4.25) so ζ − 1 dE ζ − 1 E = L Δt = − GR Δt = − ΔE , rad ζ dt ζ GR (S4.2) which is the general expression that we require. (b) In thecaseofanideal diatomic gas, we knowthat ζ =3.2,sousing EquationS4.2 we have ζ − 1 2.2 E = − ΔE = − ΔE = −0.69 ΔE . rad ζ GR 3.2 GR GR

(c)Asthe planet contracts,its gravitationalenergydecreases,and as calculated above,around 70%ofthisenergycan be radiated away.The remainder of the ‘lost’gravitationalenergygoestoincreasing theinternalenergyofthe material of which theplanet is composed.Hence conservationofenergyismaintained.

(d) Thetemperature, pressure anddensity of theH2gasduring thecollapse must all increase. As noted above,the internalenergyofthe gaswill increase, andthis will raisethe temperature. Sincethe same amount of gasiscontained within a smaller volume,anincrease in temperature implies an increaseinpressure,via the ideal gaslaw PV = nkT .Similarly,since themassisconstantbut thevolumeis decreasing,the density of thegas must alsoincrease. Exercise 5.1 (a) Using Equation5.5 we have 0 >1/4 1 (1 − A⊕) L% Teq,⊕ = 2 , 2 σπa⊕ andweare told that A⊕ =0.30.The otherconstants that we require are 26 −1 L% =3.83 × 10 Js , 11 a⊕ =1AU =1.496 × 10 m, σ =5.671 × 10−8 Jm−2K−4s−1. Substitutingin, we have 0 >1/4 1 (1 − 0.3)3.83 × 1026 Js−1 Teq,⊕ = 2 5.67 × 10−8 Jm−2K−4s−1 ×3.14 × (1.50 × 1011 m)2 0 > 1 0.7 × 3.83 × 1026 4 1/4 = K 2 5.67 × 10−8 × 3.14 × (1.50 × 1011)2 =255 K, wherewehaverestricted ourselves to three significantfigures in thecalculations as we were givenA⊕to onlytwo significantfigures. (b) Thetemperature that we have calculated for theEarth is about −19◦C, which is in reasonableagreement with theactual temperature that we experiencehereon Earth. It is,ofcourse, slightly cooler than the actual temperature, butthere area 308 Solutions to exercises numberofassumptions that arenot completely valid.For example, the Earthdoes not have auniformtemperature anditdoesnot radiate as aperfect black body. Exercise 5.2 We have 0 > 1 (1 − A)L 1/4 T = ∗ . eq 2 σπa2 (Eqn 5.5) Sinceweare just askedfor an approximate numerical value, we will not consider theerror rangethatwehavebeen givenfor L∗,and will simply adopt L∗ ≈ 1.61 L%.Substitutinginthe values that we aregiven,therefore, we have 0 > 26 −1 1/4 1 1.61 × 3.83 × 10 Js 1/4 Teq ≈ (1 − A) 2 5.67 × 10−8 Jm−2K−4s−1 ×3.14(0.0471)2(1.50 × 1011 m)2 1 9 1 13 4 1/4 1/4 ≈ 2 6.98 × 10 K (1 − A) 1 1/4 ≈ 2 [2890] (1 − A) K ≈ 1400(1 − A)1/4 K, as required. Exercise 5.3 (a) Referring to thefigure, thetransitdepth is around 15%, i.e. ΔF/F =0.15. (b) We knowthatthe transitdepth is related to theratio of thestellar and planetaryradii by ΔF R2 = P . 2 (Eqn 1.18) F R∗ Consequently,the transitdepth of HD 209458 batthe wavelengthofLyman α implies that R2 P =0.15 2 R∗ andso √ RP= 0.15R∗ =0.387 × 1.15 × 6.96 × 108 m =3.1×108 m 1 ≈ 2 R% ≈ 4 RJ.

This inferredvalue for RP is approximately three times biggerthanthe currently accepted valuefor theradius of theplanet, as it must be sincethe transitdepth in Lyman α is approximately ten times deeper than the depthinthe optical continuum.Aswesaw in Section 4.5, no known giantplanets have radiithisbig, nor do models predict planetaryradii of this size (cf. Section 4.4). (c) Theocculted area of thestar in Lyman α is farlargerthanthe area of the planet’sdiscasinferred from theoptical continuum.Thissuggeststhatthe planet hasalarge cloudofhydrogen surroundingit, so theareathatabsorbs in the spectral lines of hydrogen is farlargerthanthe area of theplanet’sdisc. Thecloud of hydrogen will be transparentexceptatthe wavelengths absorbedbyhydrogen 309 Solutions to exercises atoms. Themostobvious explanation for theexistence of this cloud surrounding theplanet is that it is beingevaporated offthe surface of theplanet by theintense irradiation of thenearby hoststar. Exercise 5.4 (a) Equation1.13 is 2πaM sin i A = P √ . RV 2 (MP + M∗)P 1 − e Comparing this with theexpression that we require, it is clear that we need to eliminate a from theequation. We can do this by usingKepler’s thirdlaw a3 G(M + M ) = ∗ P , P 2 4π2 (Eqn 1.1) which yields 5 6 P 2G(M + M ) 1/3 a = ∗ P , 4π2 andsubstitutingthisintoEquation1.13,weobtain 2π P 2/3G1/3 (M + M )1/3 M sin i(1 − e2)−1/2 A = ∗ P P . RV 2/3 2/3 P 2 π MP + M∗ Collecting terms, 5 6 2πG 1/3 M sin i A = P (1 − e2)−1/2, RV 2/3 P (MP + M∗) as required.

(b) If MP ) M∗ and e ≈ 0.0,thenthe expression for ARV simpliﬁes to 5 6 2πG 1/3 M sin i A ≈ P . RV P 2/3 M∗

We need to expresseach of thevariables P , MP and M∗ as anormalized quantity. To do this we need to multiply anddividethrough by theappropriate quantities: 5 6 5 65 6 P −1/3 M sin i M −2/3 A ≈ (2πG)1/3 P ∗ M⊕ RV 2/3 ⊕ % 1/3 yr M M yr M% 5 61/3 5 6−1/3 5 65 6−2/3 2πG P MP sin i M∗ ≈ 2 M⊕ yr M% yr M⊕ M% 5 6 2π×6.67 × 10−11 2 −2 1/3 ≈ Nm kg 365.25 × 24 × 3600 s × (1.99 × 1030)2 kg2 5 6−1/3 5 65 6−2/3 P M sin i M∗ × 5.97 × 1024 kg P yr M⊕ M% 5 6 3.353 × 10−78 −2 2 −2 1/3 ≈ kg ms m kg skg2 5 6−1/3 5 65 6−2/3 P M sin i M∗ ×5.97 × 1024 kg P yr M⊕ M% 5 6−1/3 5 65 6−2/3 P M sin i M∗ ≈0.0894 ms−1 P . yr M⊕ M% 310 Solutions to exercises

Exercise 5.5 (a) We have 5 6 R2 A = V sin i P , S S S R2 − R2 (Eqn 5.27) ∗ P andfor an exo-Earth RP ) R∗,so R2 A ≈V sin i P . S S S 2 R∗ We arerequired to expressthisinterms of normalized variables,soagain we multiply anddividethrough by theappropriate quantities: 5 62 5 65 625 6−2 R⊕ 3 −1 VSsin iS RP R∗ AS ≈ × 5 × 10 ms −1 R% 5 km s R⊕ R% 5 6 5 65 6 5 6 6.37 × 106 2 V sin i R 2 R −2 ≈ m × 5 × 103 −1 × S S P ∗ 8 ms −1 6.96 × 10 m 5 km s R⊕ R% 5 65 625 6−2 −1 VSsin iS RP R∗ ≈0.42 ms × −1 , 5 km s R⊕ R% as required.

(b) We areasked to expressthe ratio AS/ARV for an exo-Earth in terms of normalized parameters. We alreadyobtained asuitable expression for ARV in our work for Exercise5.4: 5 6−1/3 5 65 6−2/3 −1 P MP sin i M∗ ARV ≈ 0.0894 ms , (Eqn 5.17) yr M⊕ M% which we can divide into theexpression that we just obtained for AS: 5 65 6 5 6 5 6 5 6 5 6 A 0.42 −1 V sin i R 2 R −2 P 1/3 M sin i −1 M 2/3 S ≈ ms S S P ∗ P ∗ −1 −1 ARV 0.0894 ms 5 km s R⊕ R% yr M⊕ M% 5 65 625 6−1 5 61/3 5 62/3 VS sin iS RP MPsin i P ρ∗ ≈ 4.69 −1 , 5 km s R⊕ M⊕ yr ρ% wherewegathered the MR−3 terms for thestar in questionand theSun to form theratio of densities. (c)Exo-Earths will exhibittransits of depth ΔF ∼ 10−4F onceayear. Performingphotometryofthe required precisionoverseveral yearswill be challenging,sothisisnot aterriblypromisingprospect. Thereﬂex radial velocity amplitude of exo-Earths is ∼10 cm s−1,which is beyond thecapabilities of the best currently available instruments (November 2009), andmay be lessthanthe typical intrinsic radial velocity variabilityofmain sequencestars(cf.Figure 4.1). It is possiblethatexo-Earths maybedetected by their radial velocity variations, as the intrinsic stellar variabilitycan probablybecharacterized andlargely removed, butthiswill be challenging.The amplitude of theRossiter–McLaughlin effect is about 5 times greater than thereﬂex radial velocity amplitude,and at 0.42 ms−1 it is within thecapabilities of thebestcurrentinstrumentation,e.g. HARPS. Exercise 6.1 Beforewebegin anycalculations,wewill expressthe sizes and distances that we aregiven in Table6.1 in auniformset of units.Wenotethatthe sizes anddistances always appear in the relevant equations as ratiosoftwo 311 Solutions to exercises lengths:theymustdoso, as neither temperature nor secondary eclipsedepth has dimensions of length. Consequently,itdoesn’t really matter which unitwe choosetoexpressall thelengths in,but thesimplestthing to do is to convertall threelengths to SI units,asthe conversion factorsineach caseare giveninthe Appendix. Applying theconversionfactors, we obtain TableS6.1.

Ta bleS6.1 Selected parametersofHD189733 band its hoststar for use in Exercises 6.1 and6.3.

Quantity ValueUnits ValueUnits a 0.030 99 AU 4.64 × 109 m 7 RP 1.14 RJ 8.15 × 10 m 8 R∗ 0.788 R% 5.48 × 10 m Teff 5000 K Tbright(16 µm) 4315 K

Thefirststep in calculatingthe secondary eclipsedepth, ΔFSE/F ,istocalculate the daysidetemperature of theplanet, Tday.The appropriate equationis R2 T4 =(1−P)(1 − A) ∗ T 4 , day 2a2 eff (Eqn 6.4) andweare told to assume that no heat is redistributed to thenight side of the planet, so P =0,and that A =0.05.First rearrangingEquation6.4, then substitutingthese values andthe appropriate values from TableS6.1, we obtain 5 6 T 4 R2 day =(1−P)(1 − A) ∗ 2 Teff 2a so 5 6 T 4 (5.48 × 108 )2 day =(1)(0.95) m 5000 K 2 × (4.64 × 109 m)2 =0.95 × 0.006 97 =6.63 × 10−3 thus −3 1/4 Tday =5000 K × (6.63 × 10 ) =5000 K × 0.285 =1427 K. We arenot givenany information on thestar HD 189733 beyond theradius, the effective temperature, andthe brightness temperature at 16 µm. We are told to assume that thestar emits as ablack body (though we knowthatit doesn’t because thebrightnesstemperature at 16 µmdiffers from theeffective temperature) to estimate thefractional depthofthe secondary eclipse. Without further information,wecannot comment on howjustified we areinmaking this approximation for thestellar fluxat24 µm. Thesimplestform of theequations holds if we can usethe Rayleigh–Jeanslaw rather than theunapproximated Planck function.Toassess whether this is a

312 Solutions to exercises

good approximation,weneed to evaluate hc/(kTday) and compare this to the wavelengthinquestion, λc =24µm. Using our valueofTday,wehave hc 6.626 × 10−34 ×2.998 × 108 −1 = Js ms −23 −1 kTday 1.381 × 10 JK ×1427 K =1.01 × 10−5 m =10.1µm. (S6.1) Theessenceofthe Rayleigh–Jeanslaw is thereplacementof exp(hc/(λk Tday)) − 1 with hc/(λk Tday).With thevalue that we obtained in EquationS6.1, theRayleigh–Jeansapproximation uses hc 10 µ ≈ m ≈ 0.42, λk Tday 24 µm while th5efullPlan6ck function woulduse hc exp − 1=1.52 − 1=0.52. λk Tday Thus we can seethatthe approximation is good to within 20%, which is probably adequate giventhatwehavealreadyapproximated boththe planet andstar as black body emitters. We will usethe Rayleigh–Jeanslaw,but notethatthisis perhaps overestimatingthe planet fluxby20%. AdoptingEquation6.15,which is appropriate to theRayleigh–Jeansregimeand theblack body approximationsdiscussedabove,wehave 0 > ΔF (1 − P )(1 − A) 1/4 R2 SE ≈ P F 2a2 3/2 R∗ 0 > (1)(0.95) 1/4 (8.15 × 107 )2 ≈ m 2(4.64 × 109 m)2 (5.48 × 108 m)3/2 ≈ 0.0063, i.e. thesecondary eclipsedepth is predicted to be 0.6%. Exercise 6.2 (a) We have 5 6 R 2 1 9 Δf = P B (T ) − B (T ) , P,λ d λ day λ night (Eqn 6.12) which givesusthe amplitude ΔfP,λ in fluxunits, assuming that theorbit is edge-on.Ifthe orbit is not edge-on, we will always seepartofboththe dayside andthe night side hemispheres.The amplitude giveninEquation6.12 is a maximumvalue, attained onlyfor exactly edge-on orbits. Having noted this caveat, to obtain contrastunits we simply divide through by thetotal fluxofthe system:

Δf ,λ peak to peak amplitude in contrastunits = P . f∗,λ + fP,λ

Since f∗,λ ' fP,λ we can make theapproximation

Δf ,λ peak to peak amplitude in contrastunits ≈ P . f∗,λ 313 Solutions to exercises

To findthe amplitude in abandpass centred on wavelength λc,weneed to integrateoverthe bandpass (just as we didinWorked Example6.2): ( λu ΔF ΔfP,λ Qλ dλ P = (λl , F λu f∗,λ Qλ dλ λl where, as in Worked Example6.2, Qλ is theweighting as afunctionof wavelength. If we assume that we can takethe fluxesatthe central wavelengths as proportionaltothe integrated fluxoverthe bandpass, then ΔF Δf ,λ P = P c F f∗,λ 5 c6 5 6 2 1 9 2 RP d 1 = Bλ (T ) − Bλ (T ) × d c day c night R B (T ) ∗ λc bright 5 62 Bλ (T ) − Bλ (T ) R = c day c night P , B (T ) R λc bright ∗ which is theexpression that we were askedfor. (b) Foratransitinghot Jupiter,the orbit is within afew degreesofedge-on, so Equation6.18 givesthe amplitude of thephase function more or lessexactly. Sincethe system is transiting, thequantity RP/R∗ can be determinedempirically from thetransitdepth. Theflux from thestar can be measured, andconsequently theonlyunknownsinEquation6.18 arethe daysideand night side fluxes. The amplitude of thephase function in fluxunits can alsobemeasured, for usein Equation6.12.Assuming that thedistance to thestar is known,e.g. from thestar’s spectral typeand apparent brightness, we have twoequations andeverything but theday side andnight side fluxesisempirically known.Inthe caseofatransiting planet, therefore, it is possibletodeducethe temperature difference between the daysideand night side hemispheres,solong as we assume that bothhemispheres emit as black bodies. (c)Inthe caseofahot Jupiter planet that doesnot transit, theorbital orientation will be unknown,and thefullamplitude as describedinEquation6.18 wouldnot be seen.The ratio RP/R∗ wouldbeunconstrained,sowith theseunknown factors it wouldnot be possibletodeducethe temperature difference between thetwo hemispheres. Exercise 6.3 (a) In Figure 6.5 thebaseofthe secondary eclipseisone quarter of theway up from 0.994 to 0.996 on thevertical axis.Thereforethe valueofthe relative intensity during thesecondary eclipseis0.9945.The values shownare normalized so that theout-of-eclipse levelis1.0000.The deficit during the secondary eclipseistherefore

ΔFSE =(1.0000 − 0.9945)F =0.0055F. In contrastunits,therefore, ΔF SE =0.0055. F Thesecondary eclipsedepth is justoverone half of one percent. (b) In terms of thefullPlanck function,the secondary eclipsedepth is givenby 5 6 5 6 ΔF R 2 B (T ) R 2 SE P λc day P ≈ pλ + , (Eqn 6.13) F c a B (T ) R λc bright ∗ 314 Solutions to exercises andweare told that we can ignore thereflection component,sowehave 5 62 ΔF Bλ (T ) R SE = c day P , F B (T ) R λc bright ∗ wherewehavereverted to an equality,but notethatwehaveapproximated the integral overthe bandpass andneglected thereflection term. ThefullPlanck function is rather unwieldy, butwecan simplifyitbynotingthatwecan express theequationwith aratio of Planck functionsasthe subject: 5 6 B (T ) ΔF R 2 λ day = SE ∗ . Bλ(Tbright) F RP Substitutinginfor thePlanck function,thisbecomes 5 62 exp(hc/(λ kT )) − 1 ΔF R∗ c bright = SE , (Eqn 6.19) exp(hc/(λckTday)) − 1 F RP which is thefirstexpression that we were askedtoderive. Manipulatingthisto isolate Tday,wehave 5 5 6 6 5 6 5 6 hc F R 2 hc exp − 1 P =exp − 1 λckTbright ΔFSE R∗ λckTday so 5 5 6 6 5 6 5 6 hc F R 2 hc exp − 1 P +1=exp λckTbright ΔFSE R∗ λckTday thus %5 5 6 6 5 6 3 hc F R 2 hc log exp − 1 P +1 = . e λckTbright ΔFSE R∗ λckTday

Making Tday thesubject of theequation, we have % %5 5 6 6 5 6 33−1 hc hc F R 2 T = log exp − 1 P +1 . day e λck λckTbright ΔFSE R∗ (Eqn 6.20) This equation, thesecond that we were askedtoderive, givesusa(complicated) expression for Tday.

(c) Either of theexpressions in part (b) can be used.Since Equation6.20 has Tday as its subject, we will usethatone.The expression is complicated,sowewill breakdownthe evaluationusing thespeciﬁcvalues for the 16 µmsecondary eclipseofHD189733 b. First we notethatthe quantity hc/λck appearstwice, so we will evaluate that. Thewavelength is the central wavelength, λc,ofthe bandpass is 16 µm. Consequently, hc 6.626 × 10−34 ×2.998 × 108 −1 = Js ms −6 −23 −1 λck 16 × 10 m × 1.381 × 10 JK =899.0 K. Similarly,using thevalues in the questionand TableS6.1, we can evaluate 5 6 5 6 F R 2 1 8.15 × 107 2 P = m 8 ΔFSE R∗ 0.0055 5.48 × 10 m =4.0(to 2 s.f.),

315 Solutions to exercises wherewequotethe resulttoonlytwo significantfigures because our estimate of ΔFSE from theeclipsedepth is no better than this.The exponential function on theright-hand side is 5 6 5 6 hc 899.0 exp =exp K =exp(0.20835) =1.2316. λckTbright 4315 K Substitutingthese values into Equation6.20,wehave T =899.0 [log [(1.2316 − 1) × 4.0+1]]−1 day K e =899.0 [log [0.9265 +1]]−1 K e =1.525 × 899.0 K =1371 K =1400 K (to 2 s.f.), where in thefirstlineweusedthe convention that multiplicationtakes precedence to avoidyet anotherset of brackets surroundingthe terms ‘(1.2316 − 1) × 4.0’.

(d) Theessenceofthe Rayleigh–Jeanslaw is to replace exp(hc/λckT) − 1 with hc/λckT.For thestar,wehave Tbright(16 µm)=4315,which meansthat hc =0.2084, λckT while 5 6 hc exp − 1=0.2316. λckT Theseare in thesameballpark, butdifferbymore than 10%. Sincewecan read the most uncertain of theinput quantities, ΔFSE,from thegraph at abetter precisionthanthis, it wouldprobably not be justified to usethe Rayleigh–Jeans lawinstead of thefullPlanck function to evaluate the 16 µmflux from thestar HD 189733 b. Comment :InExercise6.1 we made abiggerapproximation,but in that exercise, we were making an estimate; here we arederivingresults from state-of-the-artobservations. Theplanet hasadaysidetemperature of 1400 K, which gives hc =0.6422, λckT while 5 6 hc exp − 1=0.9006. λckT In this caseitisclear that theRayleigh–Jeanslaw wouldintroduce adiscrepancy of about 30%inthe fluxfrom theplanet. It wouldnot be ajustifiable approximation.

(e)Ifthe wavelengthofthe observationwereshorter, thequantity hc/λckT wouldbecome bigger, andthe approximation,which is valid for hc/λckT ) 1, wouldbeworse. TheRayleigh–Jeanslaw is most applicable at long wavelengths.

(f)Ifthe brightness temperature of thestar were higher, thequantity hc/λckT wouldbecome smaller,and theapproximation,which is valid for hc/λckT ) 1, wouldbebetter.The Rayleigh–Jeanslaw is most applicable at high temperatures. 316 Solutions to exercises

Exercise 7.1 Themagnitude of theforce of gravity due to theSun acting on the 2 2 EarthisF%=GM%M⊕/(1 AU ),while themagnitude of theforce of gravity 2 2 due to Jupiter actingonthe EarthisFJ=GMJM⊕/((5 − 1) AU ) when theyare at their closestseparation.The ratio of theseforcesis ×16 F% = M% =1.6×104. −3 FJ 10 M% × 1 So thegravitationalforce due to theSun is about sixteen thousandtimes stronger than thegravitationalforce due to Jupiter,and this is of courseindependent of the mass of theEarth. Exercise 7.2 Using Equation7.22, 5 6 5 6 3 1/2 5 2 QP a MP a τcirc = 21 kdP GM∗ M∗ RP 5 6 2 (7.63 × 109 )3 1/2 = × 105 × m 21 (6.673 × 10−11 Nm2kg−2) × (2.39 × 1030 kg) 5 6 9.31 × 1026 7.63 × 109 5 × kg × m 2.39 × 1030 kg 1.24 × 108 m =1.73 × 1014 s, which is around 5 millionyears. Exercise 7.3 Theequilibrium temperature of aplanet is givenby 5 6 1 (1 − A)L 1/4 T = ∗ . eq 2 σπa2 (Eqn 5.5) Rearrangingthistomake a the subject, 5 6 1 (1 − A)L 1/2 a = ∗ . 2 (2Teq) σπ

(a)So, for theAtype star,the innerboundary of thehabitable zone,where Teq =373 K, is givenby 5 6 1 (1 − 0) × 10 × 3.83 × 1026 −1 1/2 a= Js (2 × 373 K)2 5.671 × 10−8 Jm−2K−4s−1 ×π =2.635 × 1011 m =1.76 AU. Similarly,the outer boundary of thehabitable zone for theAtype star,where Teq =273 K, is givenby 5 6 1 (1 − 0) × 10 × 3.83 × 1026 −1 1/2 a= Js (2 × 273 K)2 5.671 × 10−8 Jm−2K−4s−1 ×π =4.918 × 1011 m =3.29 AU. So thewidth of thehabitable zone of theAtype star is estimated as about 1.5 AU.

317 Solutions to exercises

1/2 (b) Now, since a ∝ L∗ ,and theluminosity of theKtype star is 100 times smaller than that of theAtype star,the innerand outer limits of thehabitable zone for theKtype star will be 10 times smaller than thosefor theAtype star (all else beingequal).Asaresult, theinneredge is at about 0.18 AU andthe outer edge is at about 0.33 AU.The widthofthe habitable zone of theKtype star is therefore estimated as about 0.15 AU. Exercise 7.4 An exoplanet might existinastable orbit around anyone of the sixcomponent stars(e.g. Figure S7.1aorb). Alternatively,anexoplanet could have astable orbit around anyone of thethree binary pairsthatmakeupthe system (e.g. Figure S7.1cord), or around thequadruple system that constitutes the‘visual binary’ (Figure S7.1e).Finally,one might even have an exoplanet in a stable orbit around theentiresextuplesystem (Figure S7.1f), although such a planet wouldlikely be onlyloosely bound to thesystem as it wouldbenecessarily very distantfrom thecentreofmass.

P (a) P (b)

P (c) P (d)

P (e) (f) P

FigureS7.1 SeesolutiontoExercise7.4. ‘P’ indicates the possiblelocation of aplanet.

Exercise 7.5 (a)The maximumoffsetofthe transittimeoccurs whenthe sine functionhas avalue of ±1.Inthiscase, the offset is |δt2|≈P2a1µ1/2πa2. 318 Solutions to exercises

−6 Now, thereduced mass is µ1 ≈ M1/M∗ ,which in this caseisµ1≈3×10 .So 30 days × 0.072 AU × 3 × 10−6 |δt2|≈ . 2π × 0.19 AU Themaximum offset is therefore 5.4 × 10−6 days or 0.47 seconds. (b) If themassofthe innerplanet were 10 times larger, themaximum offset in the transittimewouldalsobe10 times larger, i.e. 4.7 seconds. Exercise 7.6 Themaximum transittimingoffsetwill occurwhenthe cosine terminthe numerator of theright-hand side of Equation7.28 is ±1 andwhenthe sine terminthe denominatoris1.The timingoffsetisthen + 2 P2µ1a1 1−e2 |δt2|≈ . 2πa2(1 − e2)

(a)When e2 =0.1,thisbecomes √ 30 × 3 × 10−6 × 0.072 × 0.99 |δt |≈ days ≈ 6.0 × 10−6 =0.52 . 2 2π × 0.19 × 0.9 days s

(b) When e2 =0.5,thisbecomes √ 30 × 3 × 10−6 × 0.072 × 0.75 |δt |≈ days ≈ 9.4 × 10−6 =0.81 . 2 2π × 0.19 × 0.5 days s

Exercise 7.7 Using Equation7.29,weﬁrstnotethatthe reduced mass is −3 µ2 ≈ M2/M∗ ≈ 0.5 × 10 .The TTV is therefore 5 6 0.05 3 δt ≈ (0.5 × 10−3 × 0.5 × 100 ) × 1 days 0.42 ≈ 4.2 × 10−5 days =3.6seconds.

Exercise 7.8 (a)Using Equation7.30,the maximumTTV is 12 3 × 10−6 δt ≈ days × ≈ 2.7 × 10−3 ≈ 4 . 2 4.5 × 3 10−3 days minutes

(b) UsingEquation7.31,the librationperiod of this TTV is −4/3 −3 −2/3 Plib ≈ 0.5 × 3 × (10 ) × 12 days ≈ 139 days. So themaximum TTV wouldrecurroughlyevery 11 or 12 transits. Exercise 7.9 (a)RearrangingEquation7.32,wehave 2πa × 21/2 × M δt a M ≈ P M M M P 2 × π × 0.05 × 21/2 × 10−3 × 30 ≈ AU M% s 4 × 24 × 3600 s −8 ≈ 3.86 × 10 AU M% .

319 Solutions to exercises

RearrangingEquation7.33,wehave 21/2[M (M + M )]1/2 δT a−1/2M ≈ P P ∗ M M M 1/2 Tdur a 21/2 × [10−3 × (10−3 +1)]1/2 × 15 ≈ M% s 3 × 3600 s × 0.051/2 AU1/2 −4 −1/2 ≈ 2.78 × 10 AU M% . Now, dividing thefirstofthese by thesecond, we get a3/2 ≈ 1.39 × 10−4 3/2 , M AU so −3 8 aM ≈ 2.68 × 10 AU =4.0×10 m (which is around four hundred thousandkilometres). So, if theTTV andthe transitdurationvariation aredue to thepresenceofan exomoon, they implythatthismoon must be orbitingfairly closetothe planet. (The radius of Jupiter is about 7 × 107 m, andthe radius of ahot Jupiter exoplanet maywell be somewhat larger.) Substitutingthisorbital radius back into the first equation, 3.86 × 10−8 M ≈ ≈ 1.44 × 10−5 , M 2.68 × 10−3 M% M% which is about fivetimes themassofthe Earth. (b) Kepler’s thirdlaw applied to theorbitofthe putative exomoon around the exoplanet maybewritten as 5 6 4π2a3 1/2 M PM = G(MP + MM) 5 6 4π2 × (4.0 × 108 )3 1/2 ≈ m 6.67 × 10−11 Nm2kg−2 × (10−3 +1.4×10−5) × 1.99 × 1030 kg ≈ 1.37 × 105 s. Theorbital period of theexomoon wouldthereforebeabout 38 hours. Exercise 7.10 Using Equation7.32, a M P δt ≈ M M M 1/2 2 MP 2πa 109 × 3 × 10−6 46 × 24 × 3600 ≈ m × s 21/2 × 10−3 2π × 0.2 × 1.5 × 1011 m ≈ 45 s. Themaximum TTV is therefore lessthan 1 minute. Using Equation7.33, 5 6 a 1/2 M T δT ≈ M dur M a 1/2 21/2 M [MP(MP + M∗)] 5 6 0.2 × 1.5 × 1011 1/2 3 × 10−6 5 × 3600 ≈ m × × s 109 m [10−3 × (10−3 +0.5)]1/2 21/2 ≈ 9.3 s.

320 Solutions to exercises

Themaximum transitdurationvariation is therefore of order 10 seconds. Exercise 7.11 TheHill radius for theplanet is givenbyEquation7.24 as 5 61/3 MP RH = a 3M∗ 5 6 −3 1/3 10 M% =0.05 AU × 3 M% =0.0035 AU. 8 Hencethe Hill radius is about 5.2 × 10 morjustover 7 RJ. Exercise 7.12 FromEquation7.16,the Roche limit is 5 61/3 2MP dR = RM MM 5 6 2 × 10−3 1/3 =6.4×106 × m 3 × 10−6 =5.6×107 m. TheRoche limit for this planet andmoon is therefore about 56 thousand kilometres. Exercise 7.13 We ﬁrst notethatthe reduced mass is M 3 × 10−6 µ ≈ T ≈ M% ≈ 3 × 10−3. T −3 MP 10 M% Themaximum TTV is givenbyEquation7.34 as 3 × 10−3 × 4 × 25◦ δt ≈ days ≈ 8.3 × 10−4 =72 . 360◦ days seconds

Exercise 8.1 We need to apply R + R R = ∗ P ≈ ∗ geometric transitprobability a a (Eqn 1.21) andsubstitute in theappropriate solar andterrestrial values.Thus

R% geometric transitprobability ≈ 1 AU 6.96 × 108 ≈ m 1.50 × 1011 m ≈ 0.0047.

We aretoldtoassume that Kepler will observe 5000 such stars, so thenumberof transitingexo-Earths detected will be total exo-Earths detected = geometric transitprobability × 5000 =0.0047 × 5000 =23.

Kepler is predicted to findjustover 23 exo-Earths,with the(optimistic) assumptionthateach star hosts asingleplanet in an orbit likeEarth’s. Theactual 321 Solutions to exercises anticipated numberislessthanthis, by afactor (of perhaps roughly 10 or so)that depends on theplaneticity andthe valueofα(a, MP) overthe range within which planets wouldbeclassedasexo-Earths (cf. Subsection 4.2.2). Exercise 8.2 (a) Thedensity is simply themassdivided by thevolume, so 3 M ρ = P 4πR3 P 3 6.55 = M⊕ . 3 (S8.1) 4π (2.68 R⊕) This gives 0.75 × 6.55 × 5.97 × 1024 ρ = kg π × (2.68 × 6.37 × 106 m)3 =1876 kg m−3. In terms of thedensity of Earth, EquationS8.1 tells us that 6.55 ρ = ρ =0.340ρ . 19.25 ⊕ ⊕ Thus we seethatGJ1214 bisjustoverone-third as denseasEarth. (b) Themagnitude of theacceleration due to gravity on thesurface of aplanet is GM g = P , P R2 (Eqn 4.48) P so on thesurface of GJ 1214 bwehave 6.673 × 10−11 2 −2 × 6.55 × 5.97 × 1024 g = Nm kg kg P (2.68 × 6.37 × 106 m)2 =8.95 ms−2. This is very similar to thevalue of g on Earth, which is 9.81 ms−2. (c)Humanswouldberather comfortableonaplanet with this valueofgravity. Golf wouldbeperfectly feasible, anditwouldappear to be possibletohit aball somewhat fartherthanonEarth (assuming comparable values of air resistance) because thedownwards acceleration during flight wouldbeslightly lessthanon Earth, so theball wouldtravelfurtherbefore landing. Exercise 8.3 (a) Theappropriate equationtouse is 5 65 625 6−1 5 61/3 5 62/3 AS VS sin iS RP MPsin i P ρ∗ ≈ 4.7 −1 , ARV 5 km s R⊕ M⊕ yr ρ% (Eqn 5.31) andeverything to theright of thefirstterminparentheses is 1,because each of the terms is normalized to the Earthorthe Sun. We aretoldthatthe Sun’sspinperiod is 25 days,sowehave 2π R% VS = Prot 2π × 6.96 × 108 = m 25 × 24 × 3600 s =2.0×103 ms−1 (to 2 s.f.).

322 Solutions to exercises

Theonlything lefttoevaluate is sin iS.Weare viewingsothatthe Earthtransits with an impact parameter of b =0,soweknow i =90◦.The ecliptic is the plane ◦ of theEarth’sorbit, so iS =(90 − 7.15) .Puttingthese values into Equation5.31, we have 5 6 3 −1 ◦ AS 2.0 × 10 ms sin82.85 ≈ 4.7 −1 ARV 5 km s ≈ 1.9(to 2 s.f.). Theamplitude of theRossiter–McLaughlin effect is almosttwice the reflexradial velocity amplitude. (b) With 8 planets,the Sun’sreflex radial velocity curvewill be asuperposition of theeffects of all of them.Since theEarth is alow-masscomponent of theSolar System,itwill be difficult to unambiguously isolate the Sun’sreflex radial velocity due to theEarth. Measurements wouldneed to be made for longerthan theorbital period of Jupiter.The Rossiter–McLaughlin effect couldbemeasured in afew hours,and wouldverifythataterrestrial-planet-sized body wastransiting thehoststar.The reflexradial velocity amplitude allows themassofthe planet to be deduced;without ameasurementofthisamplitude,the mass of theplanet wouldremain unknown.

323