CHAPTER 9 THE TWO BODY PROBLEM IN TWO DIMENSIONS

9. Intoduction CHAPTER 9 THE TWO BODY PROBLEM IN TWO DIMENSIONS In this chapte we show how Keple s laws can be deived fom Newton s laws of motion and gavitation, and consevation of angula momentum, and we deive fomulas fo the enegy and angula momentum in an obit. We show also how to calculate the position of a planet in its obit as a function of time. It would be foolish to embak upon this chapte without familiaity with much of the mateial coveed in Chapte. The discussion hee is limited to two dimensions. The coesponding poblem in thee dimensions, and how to calculate an ephemeis of a planet o comet in the sky, will be teated in Chapte 0. 9. Keple s Laws Keple s law of planetay motion (the fist two announced in 609, the thid in 69) ae as follows:. Evey planet moves aound the Sun in an obit that is an ellipse with the Sun at a focus.. The adius vecto fom Sun to planet sweeps out equal aeas in equal time. 3. The squaes of the peiods of the planets ae popotional to the cubes of thei semi majo axes. The fist law is a consequence of the invese squae law of gavitation. An invese squae law of attaction will actually esult in a path that is a conic section that is, an ellipse, a paabola o a hypebola, although only an ellipse, of couse, is a closed obit. An invese squae law of epulsion (fo example, α-paticles being deflected by gold nuclei in the famous Geige-Masden expeiment) will esult in a hypebolic path. An attactive foce that is diectly popotional to the fist powe of the distance also esults in an elliptical path (a Lissajous ellipse) - fo example a mass whiled at the end of a Hooke s law elastic sping - but in that case the cente of attaction is at the cente of the ellipse, athe than at a focus. We shall deive, in section 9.5, Keple s fist and thid laws fom an assumed invese squae law of attaction. The poblem facing Newton was the opposite: Stating fom Keple s laws, what is the law of attaction govening the motions of the planets? To stat with, he had to invent the diffeential and integal calculus. This is a fa cy fom the popula notion that he discoveed gavity by seeing an apple fall fom a tee. The second law is a consequence of consevation of angula momentum, and would be valid fo any law of attaction (o epulsion) as long as the foce was entiely adial with no tansvese component. We deive it in section 9.3.

Although a full teatment of the fist and thid laws awaits section 9.5, the thid law is tivially easy to deive in the case of a cicula obit. Fo example, if we suppose that a planet of mass m is in a cicula obit of adius a aound a Sun of mass M, M being supposed to be so much lage than m that the Sun can be egaded as stationay, we can just equate the poduct of mass and centipetal acceleation of the planet, maω, to the gavitational foce between planet and Sun, GMm/a ; and, with the peiod being given by P π/ω, we immediately obtain the thid law: P 4 π a 3. 9.. GM The eade might like to show that, if the mass of the Sun is not so high that the Sun s motion can be neglected, and that planet and Sun move in cicula obits aound thei mutual cente of mass, the peiod is P 4π 3 a. 9.. G( M + m) Hee a is the distance between Sun and planet. Execise. Expess the peiod in tems of a, the adius of the planet s cicula obit aound the cente of mass. 9.3 Keple s Second Law fom Consevation of Angula Momentum δθ δ θ & δθ O FIGURE IX. In figue IX., a paticle of mass m is moving in some sot of tajectoy in which the only foce on it is diected towads o away fom the point O. At some time, its pola coodinates ae (, θ). At a time δt late these coodinates have inceased by δ and δθ.

3 Using the fomula one half base times height fo the aea of a tiangle, we see that the aea swept out by the adius vecto is appoximately δ A δθ + δθδ. 9.3. On dividing both sides by δt and taking the limit as δt 0, we see that the ate at which the adius vecto sweeps out aea is A & θ &. 9.3. But the angula momentum is m θ &, and since this is constant, the aeal speed is also constant. The aeal speed, in fact, is half the angula momentum pe unit mass. 9.4 Some Functions of the Masses In section 9.5 I am going to conside the motion of two masses, M and m aound thei mutual cente of mass unde the influence of thei gavitational attaction. I shall pobably want to make use of seveal functions of the masses, which I shall define hee, as follows: Total mass of the system: M M + m. 9.4. Mm Reduced mass m. 9.4. M + m Mass function : 3 M M 9.4.3 ( ). M + m m No paticula name: m + m +. 9.4.4 M Mass atio: q m /M. 9.4.5 Mass faction: µ m /( M + m). 9.4.6 The fist fou ae of dimension M; the last two ae dimensionless. When m << M, m m, M M and m + m.

4 (Fo those who may be inteested, the fonts I have used ae: M Aial bold m Centuy Gothic M Fench scipt MT) 9.5 Keple s Fist and Thid Laws fom Newton s Law of Gavitation M C ' m FIGURE IX. In figue IX. I illustate two masses (they needn t be point masses as long as they ae spheically symmetic, they act gavitationally as if they wee point masses) evolving about thei common cente of mass C. At some time they ae a distance apat, whee m M +,, 9.5. M + m M + m The equations of motion of m in pola coodinates (with C as pole) ae Radial: & θ & GM /. 9.5. Tansvese: & θ + & θ& 0. 9.5.3 Elimination of t between these equations will in pinciple give us the equation, in pola coodinates, of the path. A slightly easie appoach is to wite down expessions fo the angula momentum and the enegy. The angula momentum pe unit mass of m with espect to C is h &. 9.5.4 θ The speed of m is & + θ &, and the speed of M is m/m times this. Some effot will be equied of the eade to detemine that the total enegy E of the system is E m+ GM µ ( + θ& ). & 9.5.5 [It is possible that you may have found this line quite difficult. The eason fo the difficulty is that we ae not making the appoximation of a planet of negligible mass moving aound a stationay Sun, but we ae allowing both bodies to have compaable

5 masses and the move aound thei common cente of mass. You might fist like to ty the simple poblem of a planet of negligible mass moving aound a stationay Sun. In that case 0 and and m m, M M and m + m.] It is easy to eliminate the time between equations 9.5.4 and 9.5.5. Thus you can wite d d dθ d &. θ& and then use equation 9.5.4 to eliminate θ &. You should dt dθ dt dθ eventually obtain m h d + GM µ. 4 + E + 9.5.6 dθ This is the diffeential equation, in pola coodinates, fo the path of m. All that is now equied is to integate it to obtain as a function of θ. At fist, integation looks hopelessly difficult, but it poceeds by making one tentative substitution afte anothe to see if we can t make it look a little easie. Fo example, we have (if we multiply out the squae backet) in the denominato thee times in the equation. Let s at least ty the substitution w /. That will suely make it look a little d d dw dw easie. You will have to use, and afte a little algeba, you dθ dw dθ w dθ should obtain dw dθ + GM µ w m h + E m h + + 4 G M µ m h 4 +. 9.5.7 This may at fist sight not look like much of an impovement, but the ight hand side is just a lot of constants, and, since it is positive, let s call the ight hand side H. (In case you doubt that the ight hand side is positive, the left hand side cetainly is!) Also, make the obvious substitution u GM µ w, 9.5.8 m h + and the equation becomes almost tivial: du dθ + u H, 9.5.9 fom which we poceed to

6. dθ ± 9.5.0 H At this stage you can choose eithe the + o the and you can choose to make the next substitution u H sin φ o u H cos φ; you'll get the same esult in the end. I'll choose the plus sign and I ll let u H cos φ, and I get dθ dφ and hence du u θ φ + ω, 9.5. whee ω is the abitay constant of integation. Now you have to go back and emembe what φ was, what u was and what w was and what H was. Thus θ ω φ, âcos (θ ω) cos ( φ) cos φ u/h...and so on. You aim is to get it in the fom function of θ, and, if you pesist, you should eventually get m+ h /( GM µ ). 9.5. / Eh m + + + cos( θ ω) 4 G M µ You ll immediately ecognize this fom equation.3.37 o.4.6 o.5.8: l.3.37 + ecos ( θ ω) as being the pola equation to a conic section (ellipse, paabola o hypebola). Equation 9.5. is the equation of the path of the mass m about the cente of mass of the two bodies. The eccenticity is / o, if you now ecall what ae meant by µ and m +, (Check the dimensions of this!) Eh m + e +, 4 9.5.3 G M µ / 3 Eh. ( M + m) e +. 5 9.5.4 G M m The eccenticity is less than, equal to, o geate than (i.e. the path is an ellipse, a paabola o a hypebola) accoding to whethe the total enegy E is negative, zeo o positive.

7 The semi latus ectum of the path of m elative to the cente of mass is of length m+ h l, 9.5.5 GM µ o, if you now ecall what ae meant by m + and µ (see equations 9.3.4 and 9.3.6), l h. ( M + m). 9.5.6 3 G M (Check the dimensions of this!) We can also wite equations 9.4.6 o 9.4.7 as h G M. 9.5.7 l At this point it is useful to ecall what we mean by M and by h. M is the mass function, given by equation 9.4.3: 3 M M 9.4.3 ( ). M + m Let us suppose that the total enegy is negative, so that the obits ae elliptical. The two masses ae evolving in simila elliptic obits aound the cente of masses; the semi latus ectum of the obit of m is l, and the semi latus ectum of the obit of M is l, whee l l M. 9.5.8 m Relative to M the mass m is evolving in a lage but still simila ellipse with semi latus ectum l given by l M + m l. 9.5.9 M I am now going to define h as the angula momentum pe unit mass of m elative to M. In othe wods, we ae woking in a fame in which M is stationay and m is moving aound M in an elliptic obit of semi latus ectum l. Now angula momentum pe unit mass is popotional to the aeal speed, and theefoe it is popotional to the squae of the semi latus ectum. Thus we have

8 h h l l M + M m 9.5.0 Combining equations 9.5.8, 9.4.3, 9.5.9, 9.5.0 and 9.4. we obtain h GMl, 9.5. whee M is the total mass of the system. Once again: The angula momentum pe unit mass of m elative to the cente of mass is GM l, whee l is the semi latus ectum of the obit of m elative to the cente of mass, and it is M. GMl elative to M, whee l is the semi latus ectum of the obit of m elative to If you wee to stat this analysis with the assumption that m << M, and that M emains stationay, and that the cente of mass coincides with M, you would find that eithe equation 9.5.7 o 9.5. educes to h G M l. 9.5. The peiod of the elliptic obit is aea + aeal speed. The aea of an ellipse is π ab πa e, and the aeal speed is half the angula momentum pe unit mass (see section 9.3) ( h GM l GMa e ). Theefoe the peiod is π 3/ P a, o GM P 4 π a 3, 9.5.3 GM which is Keple s thid law. We might also, while we ae at it, expess the eccenticity (equation) in tems of h athe than h, using equation 9.5.0. We obtain: / Eh e +. ( ) 9.5.4 G Mm M + m

9 If we now substitute fo h fom equation 9.5., and invet equation 9.5.4, we obtain, fo the enegy of the system Gm( M + m)( e ) E, 9.5.5 l o fo the enegy o the system pe unit mass of m: GM( e ) ) E. 9.5.6 l Hee M is the mass of the system i.e. M+m. E in equation 9.5.5 is the total enegy of the system, which includes the kinetic enegy of both masses as well as the mutual potential enegy of the two, while E in equation 9.5.6 is meely E/m. That is, it is, as stated, the enegy of the system pe unit mass of m. Equations 9.5. and 9.5.6 apply to any conic section. Fo the diffeent types of conic section they can be witten: Fo an ellipse: h GM GMa( e ), E 9.5.7a,b a Fo a paabola: h GM q, E 0 9.5.8a,b Fo a hypebola: h GM GM a( e ), E + 9.5.9a,b a We see that the enegy of an elliptic obit is detemined by the semi majo axis, wheeas the angula momentum is detemined by the semi majo axis and by the eccenticity. Fo a given semi majo axis, the angula momentum is geatest when the obit is cicula. Still efeing the obit of m with espect to M, we can find the speed V of m by noting that GM E, V 9.5.30 and, by making use of the b-pats of equations 9.5.7-9, we find the following elations between speed of m in an obit vesus distance fom M:

0 Ellipse: V GM. a 9.5.3 Paabola: G M V. 9.5.3 Hypebola: V GM +. a 9.5.33 GM Cicle: V. 9.5.34 a Execise: Show that in an elliptic obit, the speeds at peihelion and aphelion ae, espectively, 9 to GM a + e e and + e aphelion speed is, theefoe,. e GM a e + e and that the atio of peihelion Chapte It might be noted at this point, fom the definition of the astonomical unit (Chapte 8, section 8.), that if distances ae expessed in astonomical units, peiods and time intevals in sideeal yeas, GM (whee M is the mass of the Sun) has the value 4π. The mass of a comet o asteoid is much smalle than the mass of the Sun, so that M M + m j M. Thus, using these units, and to this appoximation, equation 9.5.3 becomes meely 3 P a. A Delightful Constuction I am much indebted to an e-coespondent, D Bob Rimme, fo the following delightful constuction. D Rimme found it the ecent book Feynmann s Lost Lectue, The Motion of the Planets Aound the Sun, by D.L. and J.R. Goodstein, and Feynman in his tun ascibed it to a passage (Section IV, Lemma XV) in the Pincipia of Si Isaac Newton. It has no doubt changed slightly with each telling, and I pesent it hee as follows. C is a cicle of adius a (Figue IX.3). F is the cente of the cicle, and F' is a point inside the cicle such that the distance FF' ae, whee e <. Join F and F' to a point Q on the cicle. MP' is the pependicula bisecto of F'Q, meeting FQ at P.

The eade is invited to show that, as the point Q moves ound the cicle, the point P descibes an ellipse of eccenticity e, with F and F' as foci, and that MP ' is tangent to the ellipse. Q α P' P M C F F' FIGURE IX.3 Hint: It is vey easy no math equied! Daw the line F'P, and let the lengths of FP and F'P be and ' espectively. It will then become vey obvious that + ' is always equal to a, and hence P descibes an ellipse. By looking at an isosceles tiangle, it will also be clea that the angles F'PM and FPP' ae equal, thus satisfying the focus-to-focus eflection popety of an ellipse, so that MP' is tangent to the ellipse. But thee is bette to come. You ae asked to find the length QF' in tems of a, e and ', o a, e and. An easy way to do it is as follows. Let QF' p, so that QM p. Fom the ight-angled tiangle QMP we see that cos α p / '. Apply the cosine ule to tiangle QF'P to find anothe expession fo cos α, and eliminate cos α fom you two equations. You should quickly aive at And, since ' a, this becomes p a ( e ) ' a '. 9.5.35

a 3/ p a ( e ) a ( e ). a 9.5.36 Now the speed at a point P on an elliptic obit, in which a planet of negligible mass is in obit aound a sta of mass M is given by V GM. 9.5.37 a Thus we aive at the esult that the length of F'Q (o of F'M) is popotional to the speed of a planet P moving aound the Sun F in an elliptic obit, and of couse the diection MP', being tangent to the ellipse, is the diection of motion of the planet. Figue IX.4 shows the ellipse. Q α P' P M C F F' FIGURE IX.4 It is left to the eade to investigate what happens it F' is outside, o on, the cicle 9.6 Position in an Elliptic Obit The eade might like to efe back to Chapte, section.3, especially the pat that deals with the pola equation to an ellipse, to be eminded of the meanings of the angles θ, ω and v, which, in an astonomical context, ae called, espectively, the agument of

3 latitude, the agument of peihelion and the tue anomaly. In this section I shall choose the initial line of pola coodinates to coincide with the majo axis of the ellipse, so that ω is zeo and θ v. The equation to the ellipse is then l. 9.6. + ecosv FIGURE IX.5 * v I ll suppose that a planet is at peihelion at time t T, and the aim of this section will be to find v as a function of t. The semi majo axis of the ellipse is a, elated to the semi latus ectum by l a( e ) 9.6. and the peiod is given by P 4π 3 a. 9.6.3 GM Hee the planet, of mass m is supposed to be in obit aound the Sun of mass M, and the oigin, o pole, of the pola coodinates descibed by equation 9.6. is the Sun, athe than the cente of mass of the system. As usual, M M + m. The adius vecto fom Sun to planet does not move at constant speed (indeed Keple s second law states how it moves), but we can say that, ove a complete obit, it moves at π an aveage angula speed of π/p. The angle ( t T ) P is called the mean anomaly of the planet at a time t T afte peihelion passage. It is geneally denoted by the lette M, which is aleady ovewoked in this chapte fo vaious masses and functions of the masses. Fo mean anomaly, I ll ty Coppeplate Gothic Bold italic font, M. Thus

4 M π ( t T ) P. 9.6.4 The fist step in ou effot to find v as a function of t is to calculate the eccentic anomaly E fom the mean anomaly. This was defined in figue II. of Chapte, and it is epoduced below as figue IX.6. In time t T, the aea swept out by the adius vecto is the aea FBP, and, because the adius vecto sweeps out equal aeas in equal times, this aea is equal to the faction ( t T ) πab ( t T ) / P of the aea of the ellipse. In othe wods, this aea is. Now look at P the aea FBP'. Evey odinate of that aea is equal to b/a times the coesponding ( t T ) πa odinate of FBP, and theefoe the aea of FBP' is. The aea FBP' is also P equal to the secto OP'B minus the tiangle OP'F. The aea of the secto OP'B is E πa Ea, and the aea of the tiangle OP'F is ae a sin E a esin. π E ( t T ) πa P Ea a esin E. P' P O E F v B FIGURE IX.6 Multiply both sides by /a, and ecall equation 9.6.4, and we aive at the equied elation between the mean anomaly M and the eccentic anomaly E:

5 M E esin E. 9.6.5 This is Keple s equation. The fist step, then, is to calculate the mean anomaly M fom equation 9.6.4, and then calculate the eccentic anomaly E fom equation 9.6.5. This is a tanscendental equation, so I ll say a wod o two about solving it in a moment, but let s pess on fo the time being. We now have to calculate the tue anomaly v fom the eccentic anomaly. This is done fom the geomety of the ellipse, with no dynamics, and the elation is given in Chapte, equations.3.6 and.3.7c, which ae epoduced hee: cose e cosv. ecose.3.6 Fom tigonometic identities, this can also be witten e sin E sinv,.3.7a e cos E o e sin E tanv.3.7b cos E e + e o tan v tan E. e.3.7c If we can just solve equation 9.6.5 (Keple s equation), we shall have done what we want namely, find the tue anomaly as a function of the time. The solution of Keple s equation is in fact vey easy. We wite it as f ( E) E esin E M 9.6.6 fom which f '( E) ecos E, 9.6.7 and then, by the usual Newton-Raphson pocess: E ( E cos E sin E). e M 9.6.8 e cos E The computation is then extaodinaily apid (especially if you stoe cos E and don t calculate it twice!).

6 Example: Suppose e 0.95 and that M 45 o. Calculate E. Since the eccenticity is vey lage, one might expect the convegence to be slow, and also E is likely to be vey diffeent fom M, so it is not easy to make a fist guess fo E. You might as well ty 45 o fo a fist guess fo E. You should find that it conveges to ten significant figues in a mee fou iteations. Even if you make a mindlessly stupid fist guess of E 0 o, it conveges to ten significant figues in only nine iteations. Thee ae a few exceptional occasions, hadly eve encounteed in pactice, and only fo eccenticities geate than about 0.99, when the Newton-Raphson method will not convege when you make you fist guess fo E equal to M. Chales and Tatum (CelestialMechanics and Dynamical Astonomy 69, 357 (998)) have shown that the Newton-Raphson method will always convege if you make you fist guess E π. Nevetheless, the situations whee Newton-Raphson will not convege with a fist guess of E M ae unlikely to be encounteed except in almost paabolic obits, and usually a fist guess of E M is faste than a fist guess of E π. Τhe chaotic behaviou of Keple s equation on these exceptional occasions is discussed in the above pape and also by Stumpf (Cel. Mechs. and Dyn. Aston. 74, 95 (999)) and efeences theein. Execise: Show that a good fist guess fo E is E M + x( x ), 9.6.9 esin M whee x. 9.6.0 ecosm Execise: Wite a compute pogam in the language of you choice fo solving Keple s equation. The pogam should accept e and M as input, and etun E as output. The Newton-Raphson iteation should be teminated when ( Enew Eold ) / Eold is less than some small faction to be detemined by you. Execise: An asteoid is moving in an elliptic obit of semi majo axis 3 AU and eccenticity 0.6. It is at peihelion at time 0. Calculate its distance fom the Sun and its tue anomaly one sideeal yea late. You may take the mass of the asteoid and the mass of Eath to be negligible compaed with the mass of the Sun. In that case, equation 9.6.3 is meely P π G M 4 3 a, whee M is the mass of the Sun, and, if P is expessed in sideeal yeas and a in AU, this becomes just P a 3. Thus you can immediately calculate the peiod in yeas and hence, fom equation 9.5.4 you can find the mean anomaly. Fom thee, you have to

7 solve Keple s equation to get the eccentic anomaly, and the tue anomaly fom equation.3.6 o 7. Just make sue that you get the quadant ight. Execise: Wite a compute pogam that will give you the tue anomaly and heliocentic distance as a function of time since peihelion passage fo an asteoid whose elliptic obit is chaacteized by a, e. Run the pogam fo the asteoid of the pevious execise fo evey day fo a complete peiod. You ae now making some eal pogess towads ephemeis computation! 9.7 Position in a Paabolic Obit When a long-peiod comet comes in fom the Oot belt, it typically comes in on a highly eccentic obit, of which we can obseve only a vey shot ac. Consequently, it is often impossible to detemine the peiod o semi majo axis with any degee of eliability o to distinguish the obit fom a paabola. Thee is theefoe fequent occasion to have to undestand the dynamics of a paabolic obit. We have no mean o eccentic anomalies. We must ty to get v diectly as a function of t without going though these intemediaies. The angula momentum pe unit mass is given by equation 9.5.8a: h v& GM q, 9.7. whee v is the tue anomaly and q is the peihelion distance. But the equation to the paabola (see equation.4.6) is q, 9.7. + cosv o (see section 3.8 of Chapte 3), by making use of the identity u, cosv whee u tan v, + u 9.7.3a,b the equation to the paabola can be witten q sec v. 9.7.4 Thus, by substitution of equation 9.7.4 into 9.7. and integating, we obtain

8 v ( ) d G q v t q sec 4 v M dt. 9.7.5 T 0 Upon integation (dop me an email if you get stuck!) this becomes u + GM ( t ). 3/ q 3 3 u T 9.7.6 This equation, when solved fo u (which, emembe, is tan v ), gives us v as a function of t. As explained at the end of section 9.5, if q is in astonomical units and t T is in sideeal yeas, and if the mass of the comet is negligible compaed with the mass of the Sun, this becomes u ( t T ) 3 π + 3 u 9.7.7 3/ q 3 π 8( t T ) o 3u + u C 0, whee C. 9.7.8a,b 3/ q Thee is a choice of methods available fo solving equation 9.7.8, so it might be that the only difficulty is to decide which of the seveal methods you want to use! The constant C is sometimes called the paabolic mean anomaly. 3 Method : Just solve it by Newton-Raphson iteation. Thus f 3u + u 3 f ' 3( + u ), so that the Newton-Raphson u u f / f ' becomes C 0 and u 3 u + C, 9.7.9 3( + u ) which should convege quickly. Fo economy, calculate u only once pe iteation. Method : Let u x / x and C c / c. 9.7.0a,b Then equation 9.7.8a becomes x c / 3. 9.7. Thus, as soon as c is found, x, u and v can be calculated fom equations 9.7., 0a, and 3a o b, and the poblem is finished as soon as c is found! So, how do we find c? We have to solve equation 9.7.0b.

9 Method a: Equation 9.7.0b can be witten as a quadatic equation: c Cc 0. 9.7. Just be caeful that you choose the coect oot; you should end with v having the same sign as t T. Method b: Let C cot φ 9.7.3 and calculate φ. But by a tigonometic identity, cot φ cot φ / cot φ 9.7.4 so that, by compaison with equation 9.6.0b, we see that c cot φ. 9.7.5 Again, just make sue that you choose the ight quadant in calculating φ fom equation 9.7.3, so as to be sue that you end with v having the same sign as t T. Method 3. I am told that equation 9.7.8 has the exact analytic solution 3 3 u w w, 9.7.6 whee w C + 64 + 44C. 9.7.7 I haven t veified this fo myself, so you might like to have a go. Example: Solve the equation 3u + u 3 3.).6 by all fou methods. (Methods, a, b and Example: A comet is moving in an elliptic obit with peihelion distance 0.9 AU. Calculate the tue anomaly and heliocentic distance 0 days afte peihelion passage. (A sideeal yea is 365.5636 days.)

0 Execise: Wite a compute pogam that will etun the tue and anomaly as a function of time, given the peihelion distance of a paabolic obit. Test it with you answe fo the pevious example. 9.8 Position in a Hypebolic Obit If an intestella comet wee to encounte the sola system fom intestella space, it would pusue a hypebolic obit aound the Sun. To date, no such comet with an oiginal hypebolic obit has been found, although thee is no paticula eason why we might not find one some night. Howeve, a comet with a nea-paabolic obit fom the Oot belt may appoach Jupite on its way in to the inne sola system, and its obit may be petubed into a hypebolic obit. This will esult in its ultimate loss fom the sola system. Seveal examples of such cometay obits ae known. Thee is evidence, fom ada studies of meteos, of meteooidal dust encounteing Eath at speeds that ae hypebolic with espect to the Sun, although whethe these ae on obits that ae oiginally hypebolic (and ae theefoe fom intestella space) o whethe they ae of sola system oigin and have been petubed by Jupite into hypebolic obits is not known. I must admit to not having actually caied out a calculation fo a hypebolic obit, but I think we can just poceed in a manne simila to an ellipse o a paabola. Thus we can stat with the angula momentum pe unit mass: whee h v& GMl, 9.8. l 9.8. + ecosv and l a( e ). 9.8.3 If we use astonomical units fo distance and mass, we obtain v 0 dv π ( + ecosv ) a 3/ ( e ) 3/ t T dt. 9.8.4 Hee I am using astonomical units of distance and mass and have theefoe substituted 4π fo G M. I m going to wite this as

v dv ( + cosv) 0 π( t T ) 3/ a ( e ) 3/ ( e Q ) 3/, 9.8.5 π( t T ) whee Q. Now we have to integate this. 3/ a Method. Guided by the elliptical case, but beaing in mind that we ae now dealing with a hypebola, I m going to ty the substitution If you ty this, I think you ll end up with This is just the analogy of Keple s equation. e cosh E cosv. 9.8.6 ecosh E e sinh E E Q. 9.8.7 The pocedue, then, would be to calculate Q fom equation 9.8.5. Then calculate E fom equation 9.8.7. This could be done, fo example, by a Newton-Raphson iteation in quite the same way as was done fo Keple s equation in the elliptic case, the iteation now taking the fom ( E cosh E sinh E). Q + e E 9.8.8 ecosh E Then v is found fom equation 9.8.6, and the heliocentic distance is found fom the pola equation to a hypebola: a( e ). 9.8.9 + ecosv Method. Method should wok all ight, but it has the disadvantage that you may not be as familia with sinh and cosh as you ae with sin and cos, o thee may not be a sinh o cosh button you calculato. I believe thee ae SINH and COSH functions in FORTRAN, and thee may well be in othe computing languages. Ty it and see. But maybe we d like to ty to avoid hypebolic functions, so let s ty the billiant substitution

u( u e) + cosv. 9.8.0 u( eu ) + e You may have noticed, when you wee leaning calculus, that often the pofesso would make a billiant substitution, and you could see that it woked, but you could neve undestand what made the pofesso think of the substitution. I don t want to tell you what made me think of this substitution, because, when I do, you ll see that it isn t eally vey billiant at all. I emembeed that and then I let e E u, so E E ( e + ) cosh E e 9.8. cosh E ( u + / / u ), 9.8. and I just substituted this into equation 9.8.6 and I got equation 9.8.0. Now if you put the expession 9.8.0 fo cos v into equation 9.8.5, you eventually, afte a few lines, get something that you can integate. Please do wok though it. In the end, on integation of equation 9.8.5, you should get e ( u / u ) lnu. 9.8.3 Q You aleady know fom Chapte how to solve the equation f (x) 0, so thee is no difficulty in solving equation 9.8.3 fo u. Newton-Raphson iteation esults in [ e u( Q ln u) ], u u 9.8.4 u( eu ) + and this should convege in the usual apid fashion. So the pocedue in method is to calculate Q fom equation 9.8.5, then calculate u fom equation 9.8.4, and finally v fom equation 9.8.0 all vey staightfowad. Execise: Set youself a poblem to make sue that you can cay though the calculation. Then wite a compute pogam that will geneate v and as a function of t. 9.9 Obital Elements and Velocity Vecto In two dimensions, an obit can be completely specified by fou obital elements. Thee of them give the size, shape and oientation of the obit. They ae, espectively, a, e and ω. We ae familia with the semi majo axis a and the eccenticity e. The angle ω, the agument of peihelion, was illustated in figue II.9, which is epoduced hee as figue IX.7. It is the angle that the majo axis makes with the initial line of the pola

3 coodinates. Figue II.9 eminds us of the elation between the agument of peihelion ω, the agument of latitude θ and the tue anomaly v. We emind ouselves hee of the equation to a conic section l l + e cosv + e cos( θ ω), 9.9. whee the semi latus ectum l is a( e ) fo an ellipse, and a(e ) fo a hypebola. Fo a hypebola, the paamete a is usually called the semi tansvese axis. Fo a paabola, the size is geneally descibed by the peihelion distance q, and l q. The fouth element is needed to give infomation about whee the planet is in its obit at a paticula time. Usually this is T, the time of peihelion passage. In the case of a cicula obit this cannot be used. One could instead give the time when θ 0, o the value of θ at some specified time. θ v + ω v ω FIGURE IX.7 Refe now to figue IX.8. We ll suppose that at some time t we know the coodinates (x, y) o (, θ) of the planet, and also the velocity that is to say the speed and diection, o the x- and y- o the adial and tansvese components of the velocity. That is, we know fou quantities. The subsequent path of the planet is then detemined. In othe wods, given the fou quantities (two components of the position vecto and two components of the velocity vecto), we should be able to detemine the fou elements a, e, ω and T. Let us ty.