1. Review of Mechanics Newton s Laws

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1 . Revew of Mechancs.. Newton s Laws Moton of partcles. Let the poston of the partcle be gven by r. We can always express ths n Cartesan coordnates: r = xˆx + yŷ + zẑ, () where we wll always use ˆ (crcumflex) to represent a unt vector. Or we could use sphercal or cylndrcal coordnates. For a whle, we wll deal mostly wth moton n a fxed plane. We then can locate the partcle by the magntude r ( r = rˆr) and an angle θ. As the partcle moves, the tme dervatve of the (vector) poston s the velocty, v = d r/ = r. d r = d dr (rˆr) = ˆr + r dˆr. (2) Now t s easy to see that the dervatves of the unt vectors are gven by dˆr = dθ ˆθ and dˆθ = dθ ˆr (3) so the velocty n polar coordnates s gven by v = r = dr ˆr + r dθ ˆθ. (4) Consder a partcle n a crcular orbt about the orgn. Then dr/ = 0, but dθ/ = ω, the angular velocty, and v = r θˆθ 0, and drected perpendcular to the radus vector. From the velocty, we form the momentum, p = m v. Usually, the mass s constant, and we wll assume that unless we say otherwse. We also need to consder the change n v wth tme, the acceleraton, a: a = v = r = d r = ( dr = d2 r dr dˆr ˆr dr dθ = d2 r dr dθ ˆr + 2 ˆθ + dr [ ( ) ] d 2 2 r dθ = r ˆr + 2 ( r r θ ) 2 ˆr + ) dθ ˆr + r ˆθ ˆθ + r d2 θ ˆθ + r dθ dˆθ 2 ( ) dθ ˆθ + r d2 θ ˆθ dθ r 2ˆr 2 [ 2 dr ] dθ + θ rd2 ˆθ 2 ( 2ṙ θ + r θ) ˆθ (5)

2 Newton s second law s F = d p = m d v Ths s a vector equaton,.e., t mples three scalar equatons: = m a. (6) F x = ma x, F y = ma y, F z = ma z. Now suppose we have two partcles. Let the force on () exerted by (2) be called F 2. Newton s thrd law states that the force exerted on (2) by () s gven by F 2 = F 2. Then from Newton s second law we have m a = m 2 a 2,.e., a 2 = m m 2 a. (7) Thus f m 2 >> m, then a 2 << a ; for example, though an orbtng satellte exerts as much force on the earth as the earth exerts on the satellte, the acceleraton of the earth s much less than that of the satellte..2. Angular Momentum The angular momentum of a partcle s defned as L = r p = r m v, where denotes the vector cross product. The order matters, as p r = L. The vector L s perpendcular to the plane contanng r and v. In general, the cross product s gven by a b = a b sn(γ) ˆn = ˆx ŷ ẑ a x a y a z b z b y b z = (a y b z a z b y )ˆx (a x b z a z b x )ŷ + (a x b y a y b x )ẑ Here, γ s the angle between the two vectors and ˆn s a unt vector perpendcular to the plane contanng a and b. The expresson on the second lne should be expanded as the determnant of the array. Also, the torque N assocated wth the force F s defned as N = r F (8) 2

3 Note that both N and L depend upon the choce of the coordnate orgn. Now f we cross r nto both sdes of Newton s second law we get Next, note that N = r F = r d p. (9) d L = d ( r p) = d r p + r d p = v m v + r d p = r d p and we fnd that the tme dervatve of the angular momentum s equal to the appled torque: (0) N = d L () In the absence of torques, N = 0, the angular momentum L s constant..3. Systems of Partcles Suppose we have a (somewhat) solated system of partcles. It s useful to dstngush between nternal and external forces. Then Newton s 2nd law for the th partcle s d p = j F j + F (e) (2) Here, F (e) s the external force on the th partcle. Summng ths expresson over all the partcles, we have d m d r = F j + j F (e) (3) Now snce for every F j there s a correspondng F j = F j, the double sum must vansh, leavng us wth d 2 2 m r = F (e) (4) The total mass of our system of partcles s M = m. We defne the center of mass (CM) as 3

4 R = m r M (5) Then eqn(4) becomes M d2 R 2 = F (e) = F (e) (6) where F (e) s the total external force on the system. We see that the CM of the system moves as f t were a partcle of mass M acted on by F (e). We also defne the total lnear momentum as P = M d R so that we may wrte an equaton for the whole system that looks lke Newton s 2 nd law: (7) d P = F (e) (8) If there s no external force, then P = constant: the total lnear momentum of the system s conserved. Next, let s look at the angular momentum of the system. We cross r nto eqn(2) and sum over the partcles to obtan ( r d p ) = r F j + j r F (e) (9) Now the double sum s the sum of pars of torques: r F j + r j F j = ( r j r ) F j (20) But the vector ( r j r ) s along the lne between the two partcles, as s the force F j. Thus the cross product s zero for all j pars, and the double sum must vansh. As we saw n eqn(0), we can wrte the left hand sde of eqn(9) as ( r d p ) = d ( r p ) (2) Let us defne the total angular momentum of the system of partcles as L = ( r p ) and the total external torque as N (e) = ( r F ) (e). Then we see that eqn(9) reduces to 4

5 d L = N (e) (22) In the case where the external torque, N (e), vanshes, we see that L = constant: the total angular momentum of the system s conserved. Now the value of L depends upon the choce of the coordnate system. Let r the ponts relatve to the CM (center of mass), whose poston s R: refer to Also, let the velocty of the CM be V = d R/, so that r = R + r (23) v = V + v where v = d r (24) Then t s easy to show (!) that L = R P + ( r p ) (25) I.e., the total angular momentum s the sum of the angular momentum of the partcles referenced to the center of mass, plus a second term whch s the angular momentum of a mass M movng wth the CM. Now, f the CM of the system s at rest n the orgnal coordnate system, then P = M V = 0, and L s just the angular momentum relatve to the center of mass, regardless to the orgn. A very mportant quantty s the total knetc energy of our system of partcles: T = m v 2 = m v v (v = v ) (26) 2 2 Now v = V + v, so v v = V V v + (v ) 2. Thus T = 2 MV 2 + m (v 2 ) 2 + m V d r But the argument of the last sum s V d m r, and ths sum vanshes by defnton of the CM. ( m r = m r m R = MR MR = 0.) (27) 5

6 So the total knetc energy of the system has two parts, just lke the angular momentum: T = 2 MV 2 + m (v 2 ) 2 (28).4. Reducton of the Two-Body System Now we want to look at a system of only two masses, wth no external forces. The total mass s M = m + m 2 and the center of mass s located at R = M m r = m M r + m 2 M r 2 (29) Now r = r R, so r = r m M r m 2 M r 2 = (m + m 2 ) m r m 2 M M r 2 = m 2 M ( r r 2 ) (30) Defne r = r 2 r, whch s the the vector pontng from r towards r 2. Also, treatng r 2 n the same way, we get the followng relatons: r = m 2 M r r 2 = m M Next, let us defne the reduced mass, µ : r (3) µ = m m 2 M = m m 2 m + m 2 or µ = m + m 2 (32) then r = µ m r r 2 = µ m 2 r (33) We see that the one vector, r, s suffcent to specfy the poston of both partcles. Next, let us consder the knetc energy of the system. We wll assume that the CM s at rest n the coordnate system, so V = 0, and f there are no external forces, then by eqn(8), t wll reman so. Then from eqn(28), T = 2 m (v ) m 2(v 2) 2 6

7 = = T = { [ ( d m µ )] 2 [ d r + m 2 2 m [ ] ( ) 2 d r µ µ + m m 2 ( d r ( )] } 2 µ r m 2 ) 2 = 2 µ v2 (34) We see that the knetc energy of the system looks lke that of a sngle partcle of mass µ. Snce we wll be dealng wth moton n a plane, we would lke the knetc energy n polar coordnates. Makng use of eqn(4), we see that v 2 = v v = [ ] [ ] dr ˆr + rdθ ˆθ dr ˆr + rdθ ˆθ = ( ) 2 ( ) 2 dr dθ + r 2 (35) and usng the dot notaton, the total knetc energy of the 2-body system s just T = ( ) 2 µ ṙ 2 + r 2 θ2 (36) Note that the reduced mass s closer to (but smaller than) the less massve of the two partcles: m 2 = 00 m > µ = 0.99 m m 2 = 0 m > µ = 0.9 m m 2 = m > µ = 0.5 m Next, we can evaluate the angular momentum of the system. From eqn(25) we have L = ( r d m ) r + ( r 2 d m ) r 2 2 (37) Usng eqn(33) to elmnate r and r 2 we have [( L = µ ) ( r m µ ) ] [( ) ( ) ] d r µ µ d r + r m 2 m m m 2 m 2 [( = µ 2 + ) r d r ] m m 2 L = µ r d r = r µ v (38) 7

8 Once agan, the result looks just lke a sngle partcle of mass µ at poston r. Now snce N (e) = 0, L s constant and perpendcular to the plane ncludng r and v. Ths means that the moton must reman n the same plane. We can then choose the drecton of L as the z-axs of a cylndrcal coordnate system, and descrbe the moton of r by the coordnates r and θ. Then, usng eqn(4), the total angular momentum s Snce ˆr ˆr = 0 and ˆr ˆθ = ẑ, we have L = (r ˆr) µ(ṙ ˆr + r θ ˆθ) (39) L = µ r 2 θ ẑ = constant (40) If we defne the angular momentum per unt mass as h = L /µ, then we have that h = r 2 θ = constant (4) Eqn(4) s an ntegral of the moton of the 2-body system and just expresses the conservaton of angular momentum. Note that we have as yet not specfed the force between the two partcles. If we consder the r vector as t moves, the tp of the vector s dsplaced by a dstance r dθ n the ˆθ drecton. We see that the area between the old and new postons of r amounts to ds = (/2)r(rdθ). (The change n area caused by the dsplacement n the ˆr drecton wll be proportonal to drdθ and thus vanshes n the lmt.) We thus see that the rate at whch the radus vector sweeps out area s ds/, whch by eqn(4) s constant: ds = 2 r2 dθ = 2 h = constant (42) Fg.. The radus vector sweeps out area ds. 8

9 and the area swept out n any fnte nterval of tme t must be S = h t (43) 2 Ths s just Kepler s 2 nd law. Note that ths result does not depend upon the nverse square nature of the gravtatonal force. It s just an expresson of the conservaton of angular momentum and apples to any central force moton..5. Newton s Law of Gravtaton Newton s law of unversal gravtaton states that between any two masses, there s a force exerted on the frst, m, due to the second, m 2, gven by F 2 = G m m 2 r 2 ˆr where r = r 2 r, ˆr = r/r (44) and by Newton s thrd law, the force on the second mass, m 2, due to m, s just F 2. Insertng ths force nto Newton s 2 nd law yelds d 2 r m = d p 2 = F 2 = Gm m 2 r 2 ˆr (45) where we have placed the orgn at the CM so that r = r. Usng eqn(33) to replace r by r, we have Thus [ m m ] 2 d 2 r = G m m 2 ˆr (46) M 2 r 2 r = G M r 2 ˆr or we can wrte µ r = G Mµ r 2 ˆr (47) where the second expresson shows explctly that the equaton of moton of ether of the two bodes s equvalent to the equaton of moton of a mass µ orbtng a fxed mass M at dstance r. Usng eqn(5) to wrte r n polar coordnates, we have the vector equaton ( r r θ 2 )ˆr + (2ṙ θ + r θ)ˆθ = G M r 2 ˆr (48) The ˆθ component of ths equaton s just 2ṙ θ + r θ = 0. But 2ṙ θ + r θ = d ( ) r 2 θ = 0 (49) 9

10 whch ntegrates mmedately to r 2 θ = h,.e., just eqn(4), whch we already obtaned from angular momentum conservaton. Turnng to the ˆr component of eqn(48), we fnd r = r θ 2 G M r 2 = h2 r 3 G M r 2 (50) where we have used eqn(4) to elmnate θ. Ths s the basc dfferental equaton n r that we must solve to determne the orbt..6. The Gravtatonal Potental and the Lagrangan of the Problem Before we ntegrate eqn(50) for the shape of the orbt, t s useful to take a more general approach whch wll gve us some nsght all central force problems. We frst note that the gravtatonal potental energy of a mass M s gven by V (r) = G M r Now the force s gven by the gradent of the potental, (5) f(r) = V (52) The gradent actng on a scalar functon generates a vector. In the cylndrcal coordnates we are usng for ths problem, the gradent s = ˆr r + ˆθ r θ + ẑ z so n ths case, where the potental depends only upon r we have (53) { f(r) = V (r) = G M } r = G M r 2 ˆr (54) whch s just the Newtonan gravtatonal force (per unt mass, snce we dd not nclude µ n the equaton). The Lagrangan, L, for a system s the knetc mnus the potental energy. For our central force problem t s thus L = T V (r) = 2 (ṙ2 + r 2 θ2 ) + G M r A more profound approach to mechancal problems s by means of Hamlton s Prncple, whch states that the moton of a system from tme t to tme t 2 s such that the acton, defned as (55) 0

11 A = t2 t L (56) s a mnmum. It can then be shown that ths prncple leads drectly to the Euler-Lagrange equatons of moton: d ( ) L q L q = 0 =, 2,...,n (57) where the q are the (generalzed) coordnates and the q the correspondng veloctes. In our problem, q = r and q 2 = θ. Operatng on eqn(55) wth eqn(57), we mmedately obtan ( d (ṙ) r θ 2 G M ) r 2 = 0 and d ( ) r 2 θ = 0 (58) whch are precsely eqn(49) and eqn(50) that we obtaned from the Newtonan equatons. Whle the Lagrangan s the dfference between the knetc and potental energes, t s useful to look at ther sum, the total energy. E, whch must be constant: where we have used θ = h/r 2 from eqn(4). E = T + V (r) = ) (ṙ 2 + h2 + V (r) (59) 2 r 2.7. The Equvalent One-Dmensonal Problem The expresson for the total energy, whch s conserved, can gve us a useful way of lookng at the central force problem. Suppose the central force f(r) can be derved from a potental V (r) such that f(r) = V/ r. Now let us ntroduce a fcttous potental, V (r) : V (r) = V (r) + h2 so that f (r) = V 2r 2 r = f(r) + h2 r 3 (60) n the case of Newtonan gravty, f(r) = GM/r 2, so that V (r) = GM r But ths means that eqn(50) s just + h2 and thus f (r) = h2 2r 2 r 3 G M r 2 (6)

12 r = f (r) (62) Ths s the equaton for a partcle movng n one dmenson under a force f (r). The total energy of ths partcle, n terms of ts radal velocty v = ṙ, s E = 2 v2 + V (r) or 2 v2 = E V (r) (63) If we than plot the curve V (r) vs r, and draw the horzontal lne whch represents the constant value E, we see that the dstance of E above V (r) gves the knetc energy and hence v at that r. Where the curves ntersect, v must go to zero; ths s a turnng pont of the moton. The partcle cannot move to any value of r where the V (r) curve s above E, for that would requre negatve knetc energy, and hence magnary velocty. Analyss of the fcttous potental eqn(6) shows that we can have ether bound orbts or orbts that escape to nfnty. However, as long as h > 0, a partcle cannot reach the orgn. Now consder the force as gven by Ensten s general theory of relatvty. Then for a mass M there s a crtcal radus, the Schwarzschld radus, gven by r S = 2GM/c 2. In terms of ths radus, the fcttous potental s ( ) c V 2 (r) = ( r S /r) 2 + h2 2r 2 whch corresponds to a force = c2 2 GM r + h2 2r 2 GMh2 c 2 r 3 (64) f (r) = GM r 2 + h2 r 3 3GMh2 c 2 r 4 (65) The c 2 /2 term n V (r) s of no mportance as the zero pont of a potental s arbtrary. Notce the appearance of a new force term: t s attractve and t scales as /r 4. As a result, t wll domnate the h 2 /r 3 angular momentum term for suffcently small r. We fnd, therefore, that n general relatvty, there s a whole new class of orbts, capture orbts, where a partcle wth non-zero angular momentum h can reach r = 0. Ths s what allows black holes to accrete materal. 2

13 Fg. 2. Left: A partcle wth energy E 3 s bounded by rad r and r 2. Rght: Whle the orbt wll be a closed ellpse for an nverse square force, the orbts for dfferent force laws are not necessarly closed, but are stll bounded by r and r 2. 3

14 .8. Integraton of the Orbt Equaton Now let us look at the soluton of eqn(50). Drect ntegraton of ths equaton to fnd r as a functon of tme results n an ellptc ntegral whch s not easy to nterpret. It s better to obtan r as a functon of θ, as ths wll gve us the shape of the orbt. To make ths transformaton, we wrte d = dθ d dθ but dθ Thus we can wrte the second dervatve of r as = θ = h r 2 so d = h r 2 d dθ (66) r = d dr = h ( ) d h dr r 2 dθ r 2 dθ (67) As a result, eqn(50) s now ( ) h d h dr r 2 dθ r 2 dθ We now make a further transformaton = h2 r 3 G M r 2 (68) y = r so that dy dθ = r 2 dr dθ (69) Thus eqn(68) becomes ( h d h dy ) r 2 dθ dθ = h2 d 2 y r 2 dθ 2 = h2 r 3 G M r 2 (70) Multplyng through by r 2 /h 2, we thus obtan d 2 y dθ 2 + y = G M h 2 (7) (We note n passng that the general relatvstc potental gven n eqn(64) would gve rse to the followng equaton: d 2 y dθ 2 + y = G M h 2 + 3GM c 2 y 2 (72) The soluton to eqn(72) s not a closed orbt, but rather one whch looks lke a flower f the last term s small, t s lke an ellpse whose axs slowly rotates.) It s easy to see that a soluton of the homogeneous equaton d 2 y/dθ 2 + y = 0 s y = cos(θ). A more general soluton s y = B cos(θ θ 0 ). Thus we see that f we add the 4

15 (constant) rght hand sde of eqn(7), GM/h 2, we obtan the general soluton of equaton (7): Recall that r = /y, so the equaton for the orbt s y = B cos(θ θ 0 ) + GM h 2 (73) r = B cos(θ θ 0 ) + GM h 2 (74) Or r = (h 2 /GM) + ( Bh 2 GM ) cos(θ θ0 ) (75) Now the equaton of a conc secton, whch ncludes the crcle, ellpse, parabola and hyperbola, s gven by the expresson r = a( e 2 ) + e cos(θ) (76) So t s clear that our orbt s a conc secton, and that we have the relatons a( e 2 ) = h2 GM and e = Bh2 GM (77) Notce that n eqn(76), as θ vares between 0 and π, r vares from ts mnmum value of r mn = a( e) to ts maxmum value of r max = a( + e). It s clear that θ 0 n eqn(75) s just the phase of the θ varable. We can drop θ 0 f we agree to measure θ from r mn. When r s at r mn or r max (call these values r m ), then we are at a turnng pont n the orbt, where ṙ = 0. At ths pont, the total velocty s just the transverse velocty: v = v θ = r θ. So at these ponts, the knetc energy per unt mass s T(r m ) = 2 v2 = 2 r2 m θ 2 = h2 2r 2 m (78) Where the last step follows because θ = h/r 2 for any r. Now the total energy per unt mass at r m s E = T(r m ) + V (r m ) = h2 2r 2 m GM r m (79) whch we can wrte as 5

16 ( ) 2 2GM r m h 2 ( r m ) 2E h 2 = 0 (80) Ths s a quadratc equaton for /r m. The soluton s just r m = GM h 2 ± [ G 2 M 2 h 4 + 2E ] /2 (8) h 2 where the (+) sgn corresponds to r m = r mn and the ( ) sgn to r m = r max. Go back and compare ths to eqn(74) for the turnng ponts where (θ θ 0 ) = 0 or π so the cosne s ± : r m = GM h 2 ± B (82) and we see that B = [ G 2 M 2 h 4 + 2E ] /2 (83) h 2 puttng ths nto eqn(77) e 2 = h 4 G 2 M 2 B2 = + 2Eh2 G 2 M 2 (84) and usng ths for ( e 2 ) n the frst part of eqn(77) gves us ) a ( 2Eh2 G 2 M 2 = h2 GM (85) E = GM 2a Now we also have another expresson for the total energy E, (86) E = 2 v2 GM r (87) Here, v s the total velocty, whch n general has both radal and transverse components. Combnng the above two equatons, we obtan the so-called vs vva equaton: v 2 = GM ( 2 r ) a (88) 6

17 In the specal case of a crcular orbt, the sem-major axs a s equal to the (constant) radus r, and we obtan the formula for the velocty n a crcular orbt v = GM r (89).9. Kepler s Thrd Law Let us now take eqn(42), whch gves the rate, ds/, that the radus vector sweeps out area, and ntegrate t over the whole orbt (for the case 0 e<, ellptcal or crcular orbts): θ=2π θ=0 ds = 2 h P 0 = hp (90) 2 where P s the orbtal perod. But the left hand sde s just the total area of the ellpse: θ=2π θ=0 where b s the sem-mnor axs of the ellpse. Snce b = a e 2, ds = πab (9) P = 2π h ab = 2π h a2 ( e 2 ) /2 = 2π h a3/2 a /2 ( e 2 ) /2 (92) But from eqn(77), a /2 ( e 2 ) /2 = h/ GM, so we obtan P = 2π GM a 3/2 (93) whch s Kepler s thrd law. Perhaps a more usual form s: a 3 = G(m + m 2 ) 4π 2 P 2 (94) Kepler, of course, dd not know the physcal sgnfcance of the constant, but only that P 2 a 3. We should probably note that the gravtatonal constant G s not known to hgh accuracy: G = ± erg cm 2 g 2. Gravty s a very weak force and hard to measure n the laboratory. As a result, the mass of the sun s also not that accurately known. On the other hand, the perod of the Earth n orbt about the sun (the year) s known accurately, as s the Earth s sem-major axs. Thus from Kepler s 3rd law we can 7

18 fnd the product GM sun to hgh accuracy. As a result, astronomers never use G for accurate work, but rather the Gaussan gravtatonal constant, k, whch s defned as k 2 = GM sun. Then Kepler s thrd law for a planet s wrtten a 3 = k2 ( + m) 4π 2 P 2 where m = M planet M sun (95) The Gaussan constant has a value of k = rad. Yes, the unts are radans. In ths equaton, the unt of length for a s the astronomcal unt, the unt of tme for P s the mean solar day, and the mass s n solar masses. Now Gauss used P = days for the perod of the Earth s orbt. Subsequently, astronomers decded to keep the Gaussan value of k, even after they had a better value for the Earth-Sun dstance and for the length of the year. So nstead of changng k, they adjusted the astronomcal unt (AU) whch s supposed to be the sem-major axs of the Earth s orbt. As a result, the Earth s sem-major axs s not exactly t s AU! Just one of those lttle thngs astronomers do to keep the physcsts out The Shape of Orbts and the Element of Tme: Ellptcal Orbts An ellpse can be regarded as the projecton of a crcle. Imagne a crcle centered on the ellpse that just touches t at the ends of the major axs. Ths s the auxlary crcle. Imagne tltng ths crcle about the major axs: ts projecton s an ellpse. If we draw any lne perpendcular to the major axs (see the fgure on the next page), then the rato of the dstance between the major axs and the ellpse, RP, to the dstance between the axs and the crcle, RQ, s s a constant. Ths s K, the projecton factor: K = b a = e 2 (96) The angle of the lne to pont Q as measured at the center of the ellpse, s called the eccentrc anomaly, E. The x-component of ths lne, CR s just a cose. Thus the dstance FR s just a cose ae, negatve snce we are measurng from the focus F. But we can also express FR n terms of the angle θ, called the true anomaly. (The symbol f s often used for θ.) : r cosθ = a cose ae (97) RQ, the y-component of the auxlary crcle, s a sn E. Thus RP, the y-component of the ellpse, s just RQ tmes the projecton factor K. On the other hand, RP s r tmes snθ, so that r sn θ = K a sn E = e 2 a sne (98) 8

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21 If we square these two expressons and add them, after a lttle manpulaton, we obtan the result r = a ( e cose) (99) So r can be expressed n terms of E just as well as n terms of θ usng eqn(76). If we equate the two expressons for r, we can derve the relaton between the eccentrc and the true anomales: tan ( ) θ 2 = + e ( ) E e tan 2 (00) The eccentrc anomaly s mportant because t s nvolved n the smplest expresson of the poston of the body as a functon of tme. The key s Kepler s 2nd law, whch states that the radus vector sweeps out equal areas of the ellpse n equal tme ntervals. The second fgure ntroduces a unt orbt and a unt crcle, for whch a =. The radus vector FS n the ellpse s just the projecton of the vector FS extendng out to the unt crcle, shown by the dotted lne. The essental pont to realze s ths: as the vector FS sweeps out area, FS sweeps out area on the crcle at a rate proportonal to that on the ellpse, the rato always beng (ellpse)/(crcle)= K, the projecton factor. Now the area of the unt crcle s π, so the area n the sector FS P must amount to π(t t 0 )/P, where t 0 s the tme of perhelon passage (pont P). Now the area of the sector of the crcle FS P s just the area of the sector S OP mnus the area of the trangle OS F. The area of S OP s E/2 (where E s n radans). The area of the trangle s (/2)(base)(heght) = (/2)(e)(snE). Thus the swept area FS P = (E/2) (e sn E/2). Equatng the two expressons for the area FS P then leads to π t t 0 P = E 2 e sn E 2 Let us defne the mean anomaly, M (don t confuse t wth the total mass M!): (0) M = 2π P (t t 0) (02) As the tme goes from 0 to one orbtal perod, M ncreases unformly from 0 to 2π. Thus we fnally arrve at Kepler s equaton: M = E e sn E (03) 2

22 We thus have all we need to fnd where the planet wll be at tme t : From t, t 0, and P, use eqn(02) to fnd M. From M and e, use eqn(03) to fnd E. From E and e, use eqn(00) to fnd θ. From E, e and a, use eqn(99) to fnd r. Or use θ and eqn(76). To carry out ths procedure, we would lke to nvert Kepler s equaton (eqn 03) to gve E as an explct functon of M. Unfortunately, ths s a transcendental equaton whch cannot be nverted to gve a closed expresson for E. Over the years, lterally hundreds of methods have been proposed for the soluton of Kepler s equaton. If e s small, then a seres expansons may be constructed, e.g., E M + e sn M + e2 2 e3 sn 2M + (3 sn 3M sn M) + (04) 8 And such a seres can even be put nto eqn(00) to gve θ drectly as a seres expanson: θ M + 2e sn M e2 sn 2M + 2 e3 (3 sn 3M 3 snm) + (05) Before computers were avalable to solve such problems, volumes of hary math were devoted to ths and related problems. One mportant representaton of the problem s E = M + 2 n= n J n(ne) sn(nm) (06) where J n s the Bessel functon of the frst knd and of order n. Ths s one of the frst places that Bessel functons appeared. If terms of order e 6 or hgher are neglected, the frst few J n n eqn(06) are J (e) = J 2 (2e) = J 3 (3e) = ( 2 e 8 e2 + ) 92 e4 ( 2 e2 3 ) e2 9 6 e3 ( 9 ) 6 e2 The problem wth these expansons s that f e s close to unty (e.g. e = for Halley s comet) the expansons converge very slowly eqn(05) may even dverge. 22

23 A better approach s to fnd the soluton by teraton. We could smply rewrte Kepler s equaton as E = M + e sn E (07) and settng E = M on the rght hand sde, evaluate a new E, and then repeat ths procedure untl the equaton returns the same E. But convergence may be rather slow. The Newton-Raphson method s an teratve procedure whch converges more rapdly (the sgnfcant fgures wll eventually double wth each teraton). The downsde s that sometmes t wll not converge at all. It seems good for ths problem. Suppose we want to fnd the soluton of the equaton f(x) = 0, where f(x) s a functon we can t nvert algebracally. The Newton-Raphson method s to guess a value x 0 for the soluton and then terate accordng to the formula x + = x f(x ) f (x ) where f (x) = df dx. (08) The Newton-Raphson formula appled to Kepler s equaton s just E n+ = E n + M E n + e sn E n e cose n for n = 0,, 2, (09) We pck a frst guess, say E 0 = M, and terate to convergence. 23