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◦ <i<90◦,afinite radial velocity variation with amplitude lessthanthatofthe true orbital velocity, v∗,ofthe star is observed, as showninFigure S1.1c: V = |v∗| sin I. 285 Solutions to exercises I =90◦ (a) v∗ I =0◦ (b) v∗ 0◦ ≤ I ≤ 90◦ I (c) v∗ 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◦ <i<90◦,onlythe component v∗ sin I of theorbital motionisinthe directiontowards or away from theobserver. (b) Theeccentricity appearstwice in Equation1.12.Insidethe brackets in the numerator,itmultiplies aconstant(cos ωOP,where ωOP is aconstantparameter of theorbit).The radial velocity variabilityisexclusively in the cos(θ(t)+ωOP) term, so this first occurrenceofthe eccentricity doesnot affect the radial velocity variations. Thesecond appearance of theeccentricity is in the denominatorof theterms that determine theamp√litude multiplyingthe time-variable cosine term. Here it contributes to the 1 − e2 term, which hasthe value 0 for e =1 andthe value 1 for e =0.Asthe eccentricity approaches 1,thisterminthe denominatorapproaches 0,and therefore theamplitude of theradial velocity variation approachesinfinity.Asthe eccentricity increases,the amplitude of the radial velocity variationsincreases. (c)The observedradial velocity is givenbyEquation1.12: 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 .