1 3 Gavitation and Keple s Laws In this chapte we will ecall the law of univesal gavitation and will then deive the esult that a spheically symmetic object acts gavitationally like a point mass at its cente if you ae outside the object. Following this we will look at obits unde gavity, deiving Keple s laws. The chapte ends with a consideation of the enegy in obital motion and the idea of an effective potential. 3.1 Newton s Law of Univesal Gavitation Fo two paticles of masses m 1 and m 2 sepaated by distance thee is a mutual foce of attaction of magnitude Gm 1 m 2 2; whee G=6:6710?11 m 3 kg?1 s?2 is the gavitational constant. If F 12 is the foce of paticle 2 on paticle 1 and vice-vesa, and if 12= 2? 1 is the vecto fom paticle 1 to paticle 2, as shown in figue 3.1, then the vecto fom of the law is: F 21=Gm 12=?F 1 m whee the hat (ˆ) denotes a unit vecto as usual. Gavity obeys the supeposition pinciple, so if paticle 1 is attacted by paticles 2 and 3, the total foce on 1 is F 12+F 13. The gavitational foce is exactly analogous to the electostatic Coulomb foce if you make the eplacements, m!q,?g!1=4πε 0 (of couse, masses ae always ˆ 12; F 21 m 2 F F m m Figue 3.1 Labelling fo gavitational foce between two masses (left) and gavitational potential and field fo a single mass (ight). 21
2 22 3 Gavitation and Keple s Laws positive, wheeas chages q can be of eithe sign). We will etun to this analogy late. Since gavity acts along the line joining the two masses, it is a cental foce and theefoe consevative (any cental foce is consevative why?). Fo a consevative foce, you can sensibly define a potential enegy diffeence between any two points accoding to, i)=?z V( f)?v( f i Fd: 0j= The definition is sensible because the answe depends only on the endpoints and not on which paticula path you used. Since only diffeences in potential enegy appea, we can abitaily choose a paticula point, say 0, as a efeence and declae its potential enegy to be zeo, V( 0)=0. If you e consideing a planet obiting the Sun, it is conventional to set V=0 at infinite sepaation fom the Sun, soj. This means that we can define a gavitational potential enegy by making the conventional choice that the potential is zeo when the two masses ae infinitely fa apat. Fo convenience, let s put the oigin of coodinates at paticle 1 and let = 2? 1 be the position of paticle 2. Then the gavitational foce on paticle 2 due to paticle 1 is F=F 21=?Gm 1 m 2 ˆ=2 and the gavitational potential enegy is, V()=?Z : Fd0=?Z 1 m 2 (?Gm Φ()=?Gm : 02)d0=?Gm 1 m 2 ˆ: (The pime(0)on the integation vaiable is simply to distinguish it fom the point whee we ae evaluating the potential enegy.) It is also useful to think of paticle 1 setting up a gavitational field which acts on paticle 2, with paticle 2 acting as a test mass fo pobing the field. Define the gavitational potential, which is the gavitational potential enegy pe unit mass, fo paticle 1 by (setting m 1=m now), Likewise, define the gavitational field g of paticle 1 as the gavitational foce pe unit mass: The use of g fo this field is delibeate: the familia g=9:81ms?2 is just the magnitude of the Eath s gavitational field at its suface. The field and potential ae elated in the usual way: g=? Φ: Gavitational Potential Enegy Nea the Eaths Suface If you ae thinking about a paticle moving unde gavity nea the Eath s suface, you might set the V=0 at the suface. Hee, the gavitational foce on a paticle of mass m is, F=?mg ˆk; g()=?gm 2 whee ˆk is an upwad vetical unit vecto, and g=9:81ms?2 is the magnitude of the gavitational acceleation. In components, F x=f y=0 and F z=?mg. Since the foce is puely vetical, the potential enegy is independent of x and y. We will
3 3.2 Gavitational Attaction of a Spheical Shell 23 measue z as the height above the suface. Applying the definition of potential enegy diffeence between height h and the Eath s suface (z=0), we find dz=?zh F z 0(?mg)dz=mgh: Choosing z=0 as ou efeence height, we set V(z=0)=0 and find the familia esult fo gavitational potential enegy, V(h)=mgh Gavitational potential enegy nea the Eath s suface V(h)?V(0)=?Zh 0 Note that since the gavitational foce acts vetically, on any path between two given points the wok done by gavity depends only on the changes in height between the endpoints. So, this foce is indeed consevative. 3.2 Gavitational Attaction of a Spheical Shell The poblem of detemining the gavitational attaction of spheically symmetic objects led Newton to invent calculus: it took him many yeas to pove the esult. The answe fo a thin unifom spheical shell of matte is that outside the shell the gavitational foce is the same as that of a point mass of the same total mass as the shell, located at the cente of the shell. Inside the shell, the foce is zeo. By consideing an abitay spheically symmetic object to be built up fom thin shells, we immediately find that outside the object the gavitational foce is the same as that of a point with the same total mass located at the cente. We will demonstate this esult in two ways: fist sinθdθ: by calculating the gavitational potential diectly, and then, making full use of the spheical symmety, using the analogy to electostatics and applying Gauss law Diect Calculation We conside a thin spheical shell of adius a, mass pe unit aea ρ and total mass m=4πρa 2. Use coodinates with oigin at the cente of the shell and calculate the gavitational potential at a point P distance fom the cente as shown in figue 3.2. We use the supeposition pinciple to sum up the individual contibutions to the potential fom all the mass elements in the shell. All the mass in the thin annulus of width adθ at angle θ is at the same distance R fom P, so we can use this as ou element of mass: dm=ρ2πasinθadθ=m 2 The contibution to the potential fom the annulus is, sinθ dθ dφ=?gdm R=?Gm 2 Now we want to sum all the contibutions by integating ove θ fom 0 to π. In fact, it is convenient to change the integation vaiable fom θ to R. They ae elated using the cosine ule: R cosθ: 2= 2+a 2?2a R: Fom this we find sinθ dθ=r=dr=(a), which makes the integation simple. If athe integation limits ae?a and +a, while if a they ae a? and a+.
4 24 3 Gavitation and Keple s Laws a θ adθ R P gavitational potential a?gm=?gm=a gavitational field a <a: <a:?gm= 2?Gm=a 2 Figue 3.2 Gavitational potential and field fo a thin unifom spheical shell of matte. We can specify the limits fo both cases asj?ajand +a, so that: j?ajdr=?gm= fo a?gm=a fo We obtain the gavitational field by diffeentiating: g()=?gm ˆ=2 fo a 0 fo As pomised, outside the shell, the potential is just that of a point mass at the cente. Inside, the potential is constant and so the foce vanishes. The immediate coollaies ae:a unifom o spheically statified sphee (so the density is a function of the adial coodinate only) attacts like a point mass of the same total mass at its cente, when you ae outside the sphee; Φ()=?Gm 2aZ+a Two non-intesecting spheically symmetic objects attact each othe like two point masses at thei centes The Easy Way Now we make use of the equivalence of the gavitational foce to the Coulomb foce using the elabelling summaised in table 3.1. We can now apply the integal fom of Gauss Law in the gavitational case to ou spheical shell. The law eads, ρ m dv ZS gds=?4πgzv
5 3.3 Obits: Peliminaies 25 Coulomb foce Gavitational foce chage q mass m coupling 1=(4πε 0)coupling?G potential V potential Φ electic field E=? V gavitational field g=? Φ chage density ρ q mass density ρ m Gauss law E=ρ q=ε 0 Gauss law g=?4πgρ m Table 3.1 Equivalence between electostatic Coulomb foce and gavitational foce. m 1 F ρ 1 1 CM R ρ 2?F 2 m 2 Figue 3.3 Coodinates fo a two-body system. which says that the suface integal of the nomal component of the gavitational field ove a given suface S is equal to(?4πg)times the mass contained within that suface, with the mass obtained by integating the mass density ρ m ove the volume V contained by S. The spheical symmety tells us that the gavitational field g must be adial, g=g ˆ. If we choose a concentic spheical suface with adius >a, the mass enclosed is just m, the mass of the shell, and Gauss Law says, which gives 4π 2 g=?4πgm g=?gm ˆ fo >a 2 immediately. Likewise, if we choose a concentic spheical suface inside the shell, the mass enclosed is zeo and g must vanish. 3.3 Obits: Peliminaies Two-body Poblem: Reduced Mass Conside a system of two paticles of masses m 1 at position 1 and m 2 at 2 inteacting with each othe by a consevative cental foce, as shown in figue 3.3. We imagine these two paticle to be isolated fom all othe influences so that thee is no extenal foce. Expess the position i of each paticle as the cente of mass location R plus a displacement ρ i elative to the cente of mass, as we did in equation (1.3) in chapte 1
6 26 3 2=?F: 2: Gavitation and Keple s Laws on page 2. 1=R+ρ 1; 2=R+ρ Now change vaiables fom 1 and 2 to R and 2=0; = 1? 2. Since the only foce acting is the intenal foce, F=F 12=?F 21, between paticles 1 and 2, the equations of motion ae: m 1 1=F;m 2 Fom these we find, setting M=m 1+m 2, M R=m 1 1+m 2 which says that the cente of mass moves with constant velocity, as we aleady know fom the geneal analysis in section (see page 2). Fo the new elative displacement, we find, which we wite as, F=µ ; µṙ2+v(): 2: (3.1) whee we have defined the educed mass 1 m 2 µm m 1+m Fo a consevative foce F thee is an associated potential enegy V()and the total enegy of the system becomes E=1 MṘ This is just an application of the geneal esult we deived fo the kinetic enegy of a system of paticles in equation (1.4) on page 3 we aleady applied it in the two-paticle case on page 3. Likewise, when F is cental, the angula momentum of the system is L=M RṘ+µṙ; µṙ2+v(); which is an application of the esult in equation (1.6) on page 7. You should make sue you can epoduce these two esults. Since the cente of mass R moves with constant velocity we can switch to an inetial fame with oigin at R, so that R=0. Then we have: E=1 2 (3.2) L=µṙ: F= The oiginal two-body poblem educes to an equivalent poblem of a single body of mass µ at position vecto elative to a fixed cente, acted on by the foce?( V= )ˆ. It s often the case that one of the masses is vey much lage than the othe, fo example: 2=1 1+m 2 = 1? m 1+1 m 2F=m m Sunm m Eathm m potonm planet; satellite; electon: m 1 m 2 F;
7 3.3 Obits: Peliminaies 27 If m 2m 1, then µ=m 1 m 2=(m 1+m 2)m 1 and the educed mass is nealy equal to the light mass. Futhemoe, 1 1+m 2 2 R=m 2 m 1+m 2 and the cente of mass is effectively at the lage mass. In such cases we teat the lage mass as fixed at 20, L= with the smalle mass obiting aound it, and set µ equal to the smalle mass. This is sometimes called the fixed Sun and moving planet appoximation. We will use this appoximation when we deive Keple s Laws. We will also ignoe inteactions between planets in compaison to the gavitational attaction of each planet towads the Sun Two-body Poblem: Conseved Quantities Recall that gavity is a cental foce: the gavitational attaction between two bodies acts along the line joining them. In the fomulation of equations 3.2 above, this means that the gavitational foce on the mass µ acts in the diection? and theefoe exets no toque about the fixed cente. Consequently, the angula momentum vecto L is a constant: its magnitude is fixed and it points in a fixed diection. Since p (whee p=µṙ), we see that L is always pependicula to the plane defined by the position and momentum of the mass µ. Altenatively stated, this means that and p must always lie in the fixed plane of all diections pependicula to L, and can theefoe be descibed using plane pola coodinates(;θ), with oigin at the fixed cente. Fo completeness we quote the adial and angula equations of motion in these plane pola coodinates. We set the educed mass equal to the planet s mass m and wite the gavitational foce as F=?kˆ=2, whee k=gmm and M is the Sun s mass. The equations become (the eade should execise to epoduce the following expessions): 2?k=:? θ 2=?k m 2 adial equation; 1 d dt( 2 θ)=0 angula equation: The angula equation simply expesses the consevation of the angula momentum L=m 2 θ. The second conseved quantity is the total enegy, kinetic plus potential. All cental foces ae consevative and in ou two-body obit poblem the only foce acting is the cental gavitational foce. We again set µ equal to the planet s mass m and wite the gavitational potential enegy as V()=?k=. Then the expession fo the constant total enegy becomes, using plane pola coodinates, E=1 mṙ m2 θ In section 3.5 on page 33 we will deduce a good deal of infomation about the obit staight fom this conseved total enegy Two-body Poblem: Examples Comet A comet appoaching the Sun in the plane of the Eath s obit (assumed cicula) cosses the obit at an angle of 60tavelling at 50kms?1. Its closest appoach to the Sun is 1=10 of the Eath s obital adius. Calculate the comet s speed at the point of closest appoach.
8 28 3 Gavitation and Keple s Laws Take a cicula obit of adius e fo the Eath. Ignoe the attaction of the comet to the Eath compaed to the attaction of the comet to the Sun and ignoe any complications due to the educed mass. The key to this poblem is that the angula momentum L=p=mv of the comet about the Sun is fixed. At the point of closest appoach the comet s velocity must be tangential only (why?), so that, jvj= min At the cossing point, jvj= e vsin30: ev; Equating these two expessions gives, min v max=0:1 e v max=1 2 leading to v Cygnus X1 Cygnus X1 is a max=5v=250kms?1: binay system of a supegiant sta of 25 sola masses and a black hole of 10 sola masses, each in a cicula obit about thei cente of mass with peiod 5:6 days. 3; 2: Detemine the distance between the supegiant and the black hole, given that a sola mass is 1: kg. Hee we apply the two-body equation of motion, equation (3.1) fom page 26. Labelling the two masses m 1 and m 2, thei sepaation and thei angula velocity ω, we have, Gm 1 2=m m 2 1 m 2 ω m 1+m 2 Reaanging and using the peiod T=2π=ω, gives 3=G(m 1+m 2)T 2 4π 2 =6:6710?11 m 3 kg?1 s?2(10+25)1: kg(5:686400s)2 4π 2 =27: m v max: leading to = m. 3.4 Keple s Laws Statement of Keple s Laws 1. The obits of the planets ae ellipses with the Sun at one focus. 2. The adius vecto fom the Sun to a planet sweeps out equal aeas in equal times. 3. The squae of the obital peiod of a planet is popotional to the cube of the semimajo axis of the planet s obit (T 2 a 3 ).
9 3.4 Keple s Laws 29 aphelion a b ae l θ x y peihelion focus (sun) l pola equation =1+ecosθ catesian equation(x+ae)2 2+y2 a b 2=1 semimajo axis a=l 2=?k 1?e 2E semimino axis b=l p1?e2=a1=2 l 1=2 eccenticity e semi latus ectum l=l2 mk constant k k=gmm total enegy E=?mk 2 2)=?k 2L2(1?e 2a 0e< Figue 3.4 Geomety of an ellipse and elations between its paametes. In the pola and catesian equations fo the ellipse, the oigin of coodinates is at the focus Summay of Deivation of Keple s Laws We will be efeing to the popeties of ellipses, so figue 3.4 shows an ellipse and its geometic paametes. The paametes ae also expessed in tems of the dynamical quantities: enegy E, angula momentum L, mass of the Sun M, mass of the planet m and the univesal constant of gavitation G. The semimajo axis a is fixed by the total enegy E and the semi latus ectum l is fixed by the total angula momentum L. In geneal the path of an object obiting unde an invese squae law foce can be any conic section. This means that the obit may be an ellipse with 1, paabola with e=1 o hypebola with e>1. With the definition that the zeo of potential enegy occus fo infinite sepaation, the total enegy of the system is negative fo an elliptical obit. When the total enegy is zeo the object can just escape to infinite distance, whee it will have zeo kinetic enegy: this is a paabolic obit. Fo positive enegy, the object can escape to infinite sepaation with finite kinetic enegy: this gives a hypebolic obit. Figue 3.5 illustates the possible obital
10 30 3 Gavitation and Keple s Laws e > 1 hypebola e = 1 paabola e < 1 ellipse e = 0 cicle focus Figue 3.5 Diffeent conic sections, showing possible obits unde an invese squae law foce. The figue is dawn so that each obit has the same angula momentum (same l) but diffeent enegy (the mass of the obiting object is held fixed). shapes. 2nd Law This is the most geneal and is a statement of angula momentum consevation unde the action of the cental gavitational foce. The angula equation of motion gives: 2 θ=l m=const: This immediately leads to, da θ=l dt=1 2 2m=const: 2: 2 The 2nd law is illustated in figue 3.6. An obiting planet moves along the ac segments AB and CD in equal times, and the two shaded aeas ae equal. Obit equation The fist and thid laws ae aived at by finding the equation fo the obit. The fact that the obits ae ellipses is specific to an invese squae law fo the foce, and hence the fist and thid laws ae also specific to an invese squae law foce. Poceed as follows, stating fom the adial equation of motion (with k=gmm),? θ 2=?k m
11 3.4 Keple s Laws 31 B 2: C D A Figue 3.6 Illustation of Keple s 2nd Law. An obiting planet moves along the ac segments AB and CD in equal times, and the two shaded aeas ae equal. (i) Eliminate θ using angula momentum consevation, θ=l=m 2, leading to a diffeential equation fo alone:?l 2 3=?k m 2 m (ii) Use the elation 2(1+ecosθ); d dt= θ dθ=l d d m dθ; 2 to obtain deivatives with espect to θ in place of time deivatives. This gives a diffeential equation fo in tems of θ. (iii) To obtain an equation which is easy to solve, make the substitution u=1=, to obtain the obit equation: d 2 u 2+u=mk dθ 1st Law The solution of the obit equation is 1 =mk L which fo 0e<1 gives an ellipse, with semi latus ectum l=l2=mk. This is the fist law. In figue 3.7 we show the obit of a hypothetical planet aound the Sun with semimajo axis 1: km (the same as Satun) and eccenticity e=0:56 (bigge than fo any eal planet Pluto has the most eccentic obit with e=0:25). The figue also shows how the planet s distance fom the Sun, speed and angula velocity vay duing its obit. 3d Law Stat with the 2nd law fo the ate at which aea is swept out, da dt=l 2m; a3: and integate ove a complete obital peiod T, to give T=2mA=L, whee A=πab is the aea of the ellipse. Substituting fo b in tems of a gives the thid law: T 2=4π2 GM L 2:
12 32 3 Gavitation and Keple s Laws distance fom Sun (10 9 km) speed (kms?1 ) a= angula velocity (10?8 ads?1 ) t=t Figue 3.7 On the left is shown the obit of a hypothetical planet aound the Sun with distance scales maked in units of 10 9 km. The planet has the same semimajo axis 1: km as Satun, and hence the same peiod, T=10760days. The eccenticity is e=0:56. The thee gaphs on the ight show the planet s distance fom the sun, speed and angula velocity espectively as functions of time measued in units of the obital peiod T. Keple s PocedueThe solution of the obit equation gives as a function of θ, but if you e an astonome, you may well be inteested in knowing θ(t), so that you can tack a planet s position in obit as a function of time. You could do this by bute foce by combining the angula equation of motion, 2 θ=l=m, with the equation giving the obit, l==1+e cosθ, and integating. This gives a disgusting integal which moeove leads to t as a function of θ: you have to invet this, by a seies expansion method, to get θ as a function of t. This is tedious, and equies you to keep many tems in the expansion to match the accuacy of astonomical obsevations. Keple himself devised an ingenious geometical way to detemine θ(t), and his constuction leads to a much neate numeical pocedue. I efe you to the textbook by Maion and Thonton 1 fo a desciption. 1 J B Maion and S T Thonton, Classical Dynamics of Paticles and Systems, 3d edition, Hacout Bace Jovanovich (1988) p261
13 0 3.5 Enegy Consideations: Effective Potential Scaling Agument fo Keple s 3d Law Suppose you have found a solution of the obit equation,? θ 2=?k=m2, giving and θ as functions of t. Now scale the adial and time vaiables by constants α and β espectively: 0=α;t0=βt: 2: 2); In tems of the new vaiables 0and t0, the left hand side of the obit equation becomes, d dt02?0dθ 2 dt02=α?α θ β β2=α 2 β2(? θ while the ight hand side becomes,?k m02=1 2?k α m Compaing the two sides, you can see that we will have a new solution in tems of 0and t0povided β 2=α3. But this says pecisely that if you have obits of simila shape, the peiod T and semimajo axis a (chaacteising the linea size of the obit) will be elated by T 2 a 3, which is Keple s thid law. To find the constant of popotionality and show that the obits ae conic sections, you eally have to solve the obit equation. Howeve, the scaling agument makes clea how the thid law depends on having an invese-squae foce law. 3.5 Enegy Consideations: Effective Potential Since the gavitational foce is consevative, the total enegy E of the obiting body is conseved. Witing V()fo the gavitational 2+V(): 2+V(): 2+V(): potential enegy fo a moment (so that we can substitute diffeent foms fo the potential enegy if necessay), we find E=1 mṙ m2 θ Since we know that angula momentum is also conseved (the foce is cental), we can eliminate θ using 2 θ=l=m, to leave, E=1 mṙ2+l2 2 2m : l= This is just the enegy equation you would get fo a paticle moving in one dimension in an effective potential U()=L2 2m The effective potential contains an additional centifugal tem, L 2=2m2, which aises because angula momentum has to be conseved. We can lean a good deal about the possible motion by studying the effective potential without having to solve the equation of motion fo. In ou case, eplacing V()by the gavitational potential enegy and using L 2=mk, the effective potential becomes (see figue 3.8) U()=kl 2?k 2
14 34 3 Gavitation and Keple s Laws U() 2 positive 1/ tem dominates at small E > 0 hypebola p c a E = 0 paabola E < 0 ellipse negative 1/ tem dominates at lage E = k/2l cicle 2=mk; : Figue 3.8 Effective potential U()=kl=2 2?k= fo motion in an invese-squae law foce. The allowed motion must have ṙ 20, so the enegy equation says EU()=kl 2?k E= 2 If we choose a value fo the total enegy E, we can then daw a hoizontal line at this value on the gaph of U(), and we know that the allowed motion occus only whee the U()cuve lies below ou chosen value of E. The minimum possible total enegy (fo a given angula momentum) is given by the minimum of the cuve of U(). In this situation is constant at c=l=l so the obit is a cicle and the total enegy is E=?k=2l=?mk 2=2L2. If?k=2l<E<0, you can see that the motion is allowed fo a finite ange of, p a. This is the case of an elliptical obit with peihelion p and aphelion a. You can find the values of p and a (α>0;κ>0): by finding the oots of the equation kl=2 2?k=. If E=0, you see that thee is a minimum value fo, but that escape to infinity is just possible. This is the case of a paabolic obit. Fo E>0, escape to infinity is possible with finite kinetic enegy at infinite sepaation. This is the case of a hypebolic obit. Obits in a Yukawa Potential We found that the obits poduced by an invesesquae law attactive foce wee ellipses, whee the planet epeatedly taced the same path though space. Now conside a foce given by the Yukawa potential, V()=?αe?κ
15 3.5 Enegy Consideations: Effective Potential 35 3 U() U() : Figue 3.9 Left: effective potentialu()=l2=2m2?αe?κ= with m=1, α=1, κ=0:24 and L=0:9. The inset shows U()at lage whee it has a local maximum (note the diffeences in scale, paticulaly fo the value of U). Right: osette obit of a paticle with this effective potential. Such a potential descibes, fo example, the foce of attaction between at(x;y)=(3;0) nucleons in an atomic nucleus. Of couse, in that situation, the poblem should be teated quantum mechanically, but fo now, let s just look at classical obits unde the influence of this potential. The effective potential is, U()=L2 2?αe?κ 2m To be specific, wok in dimensionless units, setting m=1, α=1, κ=0:24 and choosing L=0:9. The shape of the esulting effective potential as a function of is shown in the left hand pat of figue 3.9. If the total enegy E is negative but geate than the minimum of U(), then motion is allowed between a minimum and maximum value of the adius. On the ight hand side of figue 3.9 is the tajectoy of a paticle stating with(v x;v y)=(0;0:3)(so that L=0:9). Hee the paticle s (dimensionless) enegy is?0:117 and the motion is esticted to the egion 0:4863, whee 0:486 and 3 ae the two solutions of the equation U()=?0:117. Note that if κ=0, the Yukawa potential educes to the same fom as the standad gavitational potential. So, if κ emains small compaed to 1 we expect the situation to be a small petubation elative to the gavitational case. In ou example, U() fo the osette obit on the ight of figue 3.9, this is the case, and you can see that the obit looks like an ellipse whose oientation slowly changes. This is often denoted pecession of the peihelion and is typical of the effect of small petubations on planetay obits, fo example those due to the effects of othe planets. In fact, obseved iegulaities in the motion of Uanus led to the discovey of Neptune in The oientation of the majo axis of the Eath s obit difts by about 104 seconds of ac each centuy, mostly due the influence of Jupite. Fo Mecuy, the peihelion advances by about 574 seconds of ac pe centuy: 531 seconds of this can be explained by the Newtonian gavitational inteactions of the othe planets, while the emaining 43 seconds of ac ae famously explained by Einstein s geneal elativity. The effective potential shown in figue 3.9 displays anothe inteesting popety. At lage the L 2=2m2 dominates the exponentially falling Yukawa tem, so becomes positive. In ou example, U()has a local maximum nea =20. If the
16 36 3 Gavitation and Keple s Laws Figue 3.10 Obital tajectoies fo a planet aound two equal mass stas. total enegy is positive, but less than the value of U at the local maximum, thee ae two possibilities fo obital motion. Fo example, if E=0:0003, we find eithe 0:45116:31 o 36:48. Classically these obits ae distinct, and a paticle with E=0:0003 which stats out in the inne egion can neve sumount the baie in U()and so will neve be found in 36:48. In quantum mechanics, howeve, it is possible fo a paticle to tunnel though such a baie, so that an initially bound paticle has a (small) finite pobability of escaping to lage. This is the case fo a pocess like alpha decay. 3.6 Chaos in Planetay Obits We have shown that a single planet obiting the Sun follows a simple closed elliptical path. You might think that adding one moe object to the system would make the equations moe complicated, but that with patience and effot you might be able to figue out a solution fo the tajectoies. In fact, such a thee body poblem is notoiously intactable, and, even today, analytic solutions ae known only in a few special cases. In figue 3.10 is shown a numeical solution fo a esticted vesion of the thee body poblem. The two black dots ae stas of equal mass, held at fixed positions. This means that the total enegy is conseved, but that the linea and angula momentum ae not conseved since foces and toques have to be applied to hold the stas in place. The solid cuve shows the tajectoy of a planet which stats out with some given initial velocity at the point maked by the tiangle. The stas ae taken to have a finite adius and the planet is allowed to pass though them without suffeing any inteaction apat fom the gavitational foce (this avoids some numeical instability when the planet gets vey close to a point mass). The complexity of the solid cuve aleady hints at the difficulty of this poblem. In fact, the motion is chaotic in the scientific sense. One aspect of this is shown by the dashed cuve. This is a second solution fo a planet which also stats out
17 3.6 Chaos in Planetay Obits 37 at the point maked by the tiangle, but has one of its initial velocity components diffeing by 0.5% fom the coesponding component fo the fist case. You can see how the paths stay close togethe fo a little while, but then apidly divege and show qualitatively diffeent behaviou. This exteme (exponential) sensitivity to the initial conditions is one of the chaacteistics of chaotic systems. Contast it to the two body poblem, whee a small petubation to an elliptical obit would simply esult in a new slightly displaced obit. Fo an animated compute simulation of the thee body poblem descibed hee, togethe with many othe instuctive examples of chaotic systems, ty the pogam Chaos Demonstations by J C Spott and G Rowlands, available fom Physics Academic Softwae,
18 38 3 Gavitation and Keple s Laws