The Kepler Problem. 1 Features of the Ellipse: Geometry and Analysis

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1 The Kepler Problem For the Newtonian 1/r force law, a miracle occurs all of the solutions are perioic instea of just quasiperioic. To put it another way, the two-imensional tori are further ecompose into invariant circles. This highly egenerate situation seems unbelievable from the point of view of general theory, yet it is the most interesting feature of the problem. Richar Moeckel, Bull. AMS., 41:1 (003), pp. 11. Review of Classical an Celestial Mechanics, the Recife Lectures, Cabral an Diacu (es.), Princeton University Press, 00. A necessary preliminary to a full unerstaning of the Kepler problem is a full familiarity with the geometric an analytic features of the conics particularly those of the ellipse. 1 Features of the Ellipse: Geometry an Analysis Placing the origin at the center C, with X- an Y -coorinate axes coinciing respectively with the major an minor axes of the ellipse, then in terms of these Cartesian coorinates, the equation of the ellipse reas X a + Y b = 1 (1.1) where a an b measure the semimajor an semiminor axes, respectively. The equation can be characterize parametrically in the form D. Ó Mathúna, Integrable Systems in Celestial Mechanics, oi: / _, Birkhäuser Boston, a part of Springer Science + Business Meia, LLC 008 9

2 30 Ch The Kepler Problem X = a cos E, Y = b sin E. (1.) The line PQ normal to the major axis through an arbitrary point P(X,Y) meets the circumscribe circle at P 0 (X, Y 0 ).WithE enoting the angle subtene at the center C between CP 0 an CQ, the interpretation of (1.) is clear, an furthermore we see that CQ = a cos E, P 0 Q = a sin E, PQ = b sin E. (1.3) For the raius vector CP = R from the center to the arbitrary point P(X,Y) of the ellipse, we have R = X + Y = a (a b ) sin E. (1.4) The eccentricity e of the ellipse may be efine by so that for (1.4), we may write b = a (1 e ) (1.5) R = a [1 e sin E] (1.6a) or R = a[1 e sin E] 1/ (1.6b) as the equation for the ellipse in terms of the eccentric angle E. For the corresponing equation in terms of center-base polar coorinates (R, θ), wenote an equation (1.1) becomes which, on the introuction of (1.5) yiels X = R cos θ, Y = R sin θ (1.7) [ b R ] a cos θ + sin θ = b (1.8) R [1 e cos θ] = a (1 e ) R = (1.9a) a 1 e [1 e cos θ] 1/ (1.9b) as the require equation. The point F(ae,0) is a focus of the ellipse. Moving the origin to the focus through the translation the Cartesian equation (1.1) becomes x = X ae, y = Y (1.10)

3 Sec 1 Features of the Ellipse: Geometry an Analysis 31 (x + ae) Substituting from (1.5) into (1.11) an rearranging yiels a + y b = 1. (1.11) x + y = [ a(1 e ) ex ]. (1.1) We now introuce polar coorinates (r, f ) centere at the focus so that x = r cos f, y = r sin f (1.13) an relation (1.1) may be written [ a(1 e r = a(1 e ] ) ) er cos f = e r cos f. (1.14) e If we consier the line x = a(1 e )/e (parallel to the y-axis), which we call the irectrix, then the factor in square brackets on the right of (1.14) measures the istance from an arbitrary point on the ellipse to the irectrix. Hence equation (1.14) merely states that for an arbitrary point on the curve, the ratio of the istance from the focus to the istance from the irectrix is given by the eccentricity e. This, in fact, can be taken as the general efinition of a conic, which for e<1 is an ellipse, whereas for e>1 it is a hyperbola. Returning to (1.14), we note that it can be put in the neater an possibly more recognizable form r = a(1 e ) 1 + e cos f (1.15) which, with e<1, we take as the stanar equation for the ellipse. For the corresponing relation in terms of the eccentric angle E, r = x + y = (X ae) + Y = X + Y aex + a e (1.16) Introucing R from (1.6) into (1.16) yiels so that = R a e cos E + a e. r = a [1 e sin E e cos E + e ] = a [1 e cos E] (1.17) as the sought-for relation. For the ellipse, therefore, we note the following: r = a[1 e cos E] (1.18) 1. Equation (1.6) relates the center-base raius vector R at the point P to the angle-parameter E, being the angle subtene at the center between the major axis an the raius to the point where the normal to the major axis through P meets the circumscribe circle.

4 3 Ch The Kepler Problem. Equation (1.9) gives the equation of the ellipse in terms of the center-base polar coorinates (R, θ). 3. Equation (1.15) gives the equation of the ellipse in terms of the focus-base polar coorinates (r, f ). 4. Equation (1.18) relates the raial coorinate r of the focus-base system (r, f ) of item 3 above to the angle parameter E referre to in item 1 above. The attractive simplicity of (1.18) must be balance against its mixe nature, involving coorinate systems of ifferent origins. A straightforwar exercise yiels the relation between the angles E an f. Since x = r cos f = r [cos f ] [ 1 = r 1 sin f ] (1.19) we have (1) r cos f = r + x = r + X ae = a(1 e cos E) + a cos E ae = a(1 e)[1 + cos E] = a(1 e) cos E (1.0) an hence r cos f = a(1 e) cos E. (1.1) () r sin f = r x = r X + ae = a(1 e cos E) a cos E + ae = a(1 + e)[1 cos E] = a(1 + e) sin E an hence Diviing (1.) by (1.1) yiels r sin f = a(1 + e) sin E. (1.) tan f = 1 + e 1 e tan E, tan E = 1 e 1 + e tan f. (1.3a,b) This latter relation can now be use to erive the equation for R in terms of f, but its algebraic complexity limits its utility. Returning to the stanar equation (1.15), we see that (with prime enoting ifferentiation with respect to f ) r = r f = ae(1 e ) sin f (1 + e cos f). (1.4) Hence r = 0 for f = 0, ±π,...,±nπ. It can be easily checke that f = 0 is a minimum point for r (as also are f =±nπ) while f = π (as well as

5 Sec The Two-Boy Problem 33 f = ±(n + 1)π) is the maximum point for r. The point f = 0, at which r = a(1 e), we shall call the pericenter; the point f = π, at which r = a(1+e), we shall call the apocenter. At the extremity of the semiminor axis, we have x = ae, y = b = a 1 e (1.5) from which it follows that, at that extremity, r = a, cos f = e (1.6) an hence we have that [a, arccos e] are the focus-base polar coorinates of the extremity of the positive semiminor axis. The Two-Boy Problem We consier the motion of two boies moving uner the influence of their mutual attraction. Denoting the masses of the two boies by m 1 an m, with position vectors r 1 an r, referre to the origin at 0, we write r = r r 1. (.1) In accorance with the inverse square law governing the gravitational attraction of m 1 an m, the equations of motion for m 1 an m are given respectively by m 1 r 1 = Gm 1m r e r = Gm 1m r 3 r, an hence r 1 = Gm r 3 r (.a) m r = Gm 1m r e r = Gm 1m r 3 r, an hence r = Gm 1 r 3 r (.b) where we have use the ot to enote ifferentiation with respect to time t, an where the unit vector e r is efine by r = r e r = r e r. Subtracting (.a) from (.b), we have r r 1 = r = G(m 1 + m ) r 3 r (.3) an as the equation is unaltere by the replacement of r by r, orbythe interchange of m 1 an m, equation (.3) escribes the motion of either boy relative to the other. Moreover, equation (.3) shows that the problem has been

6 34 Ch The Kepler Problem reuce to that of the motion of a particle of unit mass in the gravitational fiel of a boy of mass m, situate at the origin, where anifweset then equation (.3) reas m = m 1 + m (.4) μ = G(m 1 + m ) = Gm (.5) r = μ r 3 r = μ r e r (.6) which is the stanar form. In the case of planetary motion, one may think of m 1 as the Sun an m as the planet. In that case, we may write ( m = m 1 + m = m m ) (.7) an (.6) escribes the motion of the planet in the heliocentric coorinate system. We may also note that the ominance of the mass of the Sun woul permit the approximation m 1 m m 1, μ Gm 1 (.8) when such an approximation is appropriate. At this point, we introuce the gravitational potential. At an arbitrary point P in the gravitational fiels of a mass m at Q, the function U efine by U = Gm PQ = Gm r = μ r (.9) is the potential per unit mass: it has the feature that the force efine by the graient of this function U is in fact the Newtonian gravitational force acting on a particle of unit mass, namely F = U = Gm r e r = Gm r 3 r = μ r 3 r (.10) so that, for the equation of motion of a particle P of unit mass, we have r = Gm r 3 r = μ r 3 r (.11) ientical with (.6). In case of several masses m i, i = 1,...,n, situate respectively at Q i, i = 1,...,n, the potential function per unit mass at P is given by U = n i=1 Gm i PQ i to which we shall have occasion to refer later. (.1)

7 Sec 3 The Kepler Problem: Vectorial Treatment 35 In the next section when we encounter the conservation of energy, we shall see that the potential energy V per unit mass for a particle in the gravitational fiel of a mass m is given by V = Gm r = μ r = U (.13) so that the potential function is the negative of the potential energy. The problem efine by the ifferential equations (.6) with μ given by (.5) is known as the Kepler problem. 3 The Kepler Problem: Vectorial Treatment In the class of problems in Celestial Mechanics, the Kepler problem is istinguishe by several features: it has every possible egeneracy the frequencies associate with all three coorinates coincie so that all boun orbits are perioic (except for collision orbits); but more relevant at this point is the fact that the motion is always planar. This means that it amits a vectorial treatment to which other problems are not amenable. In terms of a (heliocentric) spherical coorinate system (r,θ,ϕ) with unit base vectors e r, e θ,ane ϕ, it follows from that the velocity vector v is given by r = r e r (3.1) v = ṙ = ṙe r + r θe θ + r sin θ ϕe ϕ (3.) where again the ot enotes ifferentiation with respect to time; there follows r ṙ = r ṙ v = v v = ṙ ṙ = ṙ + r θ + r sin θ ϕ. (3.3a) (3.3b) We note that the funamental equation (.6) amits an immeiate first integral which we shall recognize as the energy integral. Taking the scalar prouct of (.6) with the velocity vector ṙ, we fin an so or ṙ r = μ r 3 r ṙ = μ r 3 1 μ (r r) = t r 3 t (ṙ ṙ) = t ( ) μ r t (r ) = μ r ṙ (3.4) (3.5) [ 1 t v μ ] = 0. (3.6) r

8 36 Ch The Kepler Problem Letting E enote the constant of integration, we therefore have the energy integral in the form 1 v μ r =E. (3.7) For a particle of unit mass, the first term is clearly the kinetic energy an the secon term is the potential energy; accoringly, if we use T to enote the kinetic an V the potential energy, then T = 1 v, V = μ, T + V =E (3.8a,b,c) r an the efinition of V is consistent with (.13). Rewriting (3.7) in the form 1 v =E+ μ r (3.9) an noting that the left-han sie is always positive, then if E is negative, relation (3.9) sets the lower limit on μ/r: if we exhibit the case of negative energy by writing an efine a length scale a by setting E= α (3.10) then we have that an hence a = μ α (3.11) μ r α 0 implying μ r α (3.1) r μ = a (3.13) α giving the corresponing upper limit on r : negative energy implies boun orbits, an these shall be the main focus of our attention. Returning to relations (3.1) an (3.) we form the angular momentum vector C by taking the cross prouct of r an v, tofin an we further note that C = r v = r ṙ = r sin θ ϕe θ + r θeϕ (3.14) C t = t (r ṙ) = ṙ ṙ + r r = 0 r μ r 3 r = 0. (3.15) Hence in the central gravitational fiel, the angular momentum vector C is constant. At this point, we observe that

9 t (e r ) = ( ) r = r ṙ ṙ r t r r = r ṙ r ṙ r r 3 (r r)ṙ (r ṙ)r r (ṙ r) = r 3 = r 3 Sec 3 The Kepler Problem: Vectorial Treatment 37 = r C r 3 = C r r 3. (3.16) When C = 0, the above relation implies that, in that case, the unit vector e r is constant hence the motion is rectilinear along the raius vector towar the origin, leaing to collision. When C 0, it follows from (3.14) that r C = r (r ṙ) = 0 (3.17) so that r remains normal to the fixe vector C; hence the motion takes place in the plane efine by the fixe (constant) vector C. It further follows from (3.14) that C = C C = (r ṙ) (r ṙ) = r 4 sin θ ϕ + r 4 θ = r [ r θ + r sin θ ϕ ] = r [v ṙ ] (3.18) an we have a secon integral, this one involving the magnitue of the angular momentum vector C, namely r [v ṙ ] = C. (3.19) Moreover, rewriting the latter as an expression for v, an recalling the energy integral (3.7), we have [ 1 v = 1 C ] r + ṙ =E+ μ (3.0) r giving the relation between the constants C an E. Returning to (3.16) an again applying the gravitational equation (.6) an also noting that Ċ = 0, we fin t (e r ) = C r r 3 = C r = 1 μ μ t (C ṙ) = 1 μ (v C). (3.1) t If we let e enote the arbitrary constant vector introuce by the integration of this latter vector ifferential equation, we have μ(e r + e) = v C = ṙ (r ṙ) = (ṙ ṙ)r (r ṙ)ṙ = v r r ṙ ṙ. (3.) Again, we note in passing that if C = 0, then e = e r,sothate is the unit vector along the raius vector towar the origin. For C 0, we take the scalar prouct with C across (3.), an noting that C is normal to both r an ṙ, we fin e C = 0 (3.3) which implies that the vector e lies in the plane of the motion. Taking the scalar prouct with r across (3.) gives

10 38 Ch The Kepler Problem μ(r + e r) = v r r ṙ = r (v ṙ ) = C (3.4) wherein we have introuce (3.18); we now rewrite (3.4) in the form e e r = C μr 1. (3.5) If, in the plane of the motion, we let the vector e, whose magnitue we enote by e, efine a base axis an if we let f enote the angle in this plane between this base vector an the raius vector r, then(r, f ) constitute a polar coorinate basis in the plane of the motion, an equation (3.5) can be written in the form r[1 + e cos f] = C μ. (3.6) For e = 0, the motion is circular. For e 0, we rewrite (3.6) [in accor with (1.14)] as [ C ] r = e eμ r cos f (3.7) which [referring to equation (1.14) an the subsequent paragraph] efines a conic with a irectrix at a istance C /μe from the origin an with eccentricity e. An for e<1, this conic is an ellipse, an the vector e is the vector base at the focus (origin) irecte at the pericenter an with magnitue e. The vector e is known as the Runge Lenz vector an also the eccentric axis vector. There is one more exercise to be performe on relation (3.). We recall that since C is normal to v, there follows that v C =vc, (v C) = v C. (3.8) Accoringly, if we square both sies of (3.), then on reversing the orer we fin v C = μ (e + e r ) = μ [1 + e + e e r ] ( C = μ [1 + e )] + μr 1 = μ (e 1) + μ C r (3.9) in which we have introuce (3.5) an rearrange. Hence [ μ (1 e ) = C 1 v μ ] = C E (3.30) r from which it is immeiately evient that e 1 correspons to E 0 (3.31) i.e., negative/positive energy correspons to elliptic/hyperbolic orbits as anticipate earlier.

11 Sec 3 The Kepler Problem: Vectorial Treatment 39 Restricting our attention to boun orbits (negative energy), we introuce (3.10) an (3.11) into (3.30), to obtain / 1 e = C μ α = C μ μ α = C μ 1 a (3.3) an hence C μ = a(1 e ) = p (3.33) where we introuce the symbol p to enote the semi latus rectum the value of r at f = π/. In terms of these length parameters, equation (3.6) reas r = p 1 + e cos f = a(1 e ) 1 + e cos f (3.34) as an alternate form for the equation of the orbit, an we write b = a 1 e (3.35) as the length parameter of the semiminor axis. The polar coorinates (r, f ) in the orbit plane together with the axis normal to the plane constitute a cylinrical polar coorinate system. With base unit vectors e r an e f in the orbit plane together with the axial unit vector e k,we may write r = r e r (3.36a) v = ṙ = ṙ e r + r ḟ e f (3.36b) an, for the angular momentum, we have C = r v = r ḟ e k. (3.37) It follows that, for the magnitue of the angular momentum, we have r ḟ = C = μp = μa(1 e ) = μa 1 e (3.38) wherein we have introuce (3.33). If we let τ enote the time for a complete orbit an if we also introuce the mean motion n, measuring the frequency, by the relation n = π τ (3.39) an note that the area trace out in one orbit is πab, we have that the mean areal velocity over an orbit is given by πab τ = πab n π = 1 nab = 1 na 1 e. (3.40)

12 40 Ch The Kepler Problem However, the areal velocity is, in fact, given by one-half the angular momentum of (3.38). Ientifying the quantity in (3.38) with twice the quantity in (3.40) gives, after cancellation of the common 1 e factor, an hence the important relation na = μa (3.41) n a 3 = μ = Gm = G(m 1 + m ) (3.4) whence we have substitute for μ from (.5). We are now in a position to make some observations: 1. The motion takes place in a plane efine by the angular momentum vector, an for negative energy the orbit is the ellipse (3.34); this is Kepler s First Law.. The constancy of the angular momentum (3.38) implies a constant mean areal velocity; this is Kepler s Secon Law. 3. If the approximation (.8) were to be introuce into (3.4), we woul have n a 3 = Gm, a constant for all planets; this is Kepler s Thir Law, more usually state as the square of the orbit perio is proportional to the cube of the semimajor axis. Recalling equation (3.0) for the case of negative energy so that E= α, we rearrange to obtain r ṙ = [ α r μr + C ] [ = α r μr ] α + C α. (3.43) The singularity at r = 0 in this ifferential equation can be regularize by means of a regularizing transformation whereby a new inepenent variable E is introuce through the efining relation E t = α r so that r t = α E (3.44) an, on the introuction of (3.44) an some rearrangement, equation (3.43) becomes ( ) r ] = [r μα C r + E μ μ α = [ r ar + a (1 e ) ] = [ (a r) a e ] (3.45) where we have introuce (3.11) an (3.33). By means of the substitution a r = aez, this immeiately integrates, an we fin r = a[1 e cos E] (3.46)

13 Sec 4 The Kepler Problem: Lagrangian Analysis 41 satisfying the conition that E = 0 when r = a(1 e). Recalling relation (1.18), it is evient that E can be ientifie with the eccentric angle introuce in (1.). It remains to etermine the relation between the angle E an the time t. From the efining relation (3.44), we have α t = r = a[1 e cos E] (3.47) E so that, on integration α (t t 0 ) = a[e e sin E] (3.48) satisfying the requirement that E = 0whent = t 0. From (3.41), we note that an hence n a = μ a = α (3.49) M = n(t t 0 ) = E e sin E, (3.50) known as Kepler s equation. The eccentric angle E efine by (3.44) is, in Celestial Mechanics, calle the eccentric anomaly, an the quantity M = n(t t 0 ) is calle the mean anomaly. The angle f, introuce in equation (3.6), is calle the true anomaly. We postpone to the next section the full treatment of the true anomaly. The vectorial treatment gives a full account of the Kepler orbit in its plane. The fuller picture of the motion in space, incluing the orientation of the orbit plane, is more clearly seen in the Lagrangian analysis, which is the subject of the next section. 4 The Kepler Problem: Lagrangian Analysis In terms of spherical coorinates (r,θ,ϕ) (of the heliocentric system), the three Cartesian coorinates can be expresse as x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ (4.1a) (4.1b) (4.1c) from which it can reaily be euce that the metric coefficients g ij are given by g 11 = 1, g = r, g 33 = r sin θ, g ij = 0, i j. (4.) Then for the kinetic an potential energies per unit mass, we have, respectively,

14 4 Ch The Kepler Problem T = 1 v = 1 [ṙ + r θ + r sin θ ϕ ], V = μ r (4.3) an the Hamiltonian, reflecting the total energy, is H = T + V = 1 [ṙ + r θ + r sin θ ϕ ] μ r (4.4) while, for the Lagrangian, we have L = T V = 1 [ṙ + r θ + r sin θ ϕ ] + μ r. (4.5) From the latter there follows the system of Lagrangian equations, which takes the form t [ṙ] = r θ + r sin θ ϕ μ r t [r θ] = r sin θ cos θ ϕ t [r sin θ ϕ] = 0. (4.6a) (4.6b) (4.6c) As the coorinate ϕ oes not appear explicitly in the Lagrangian (4.5), it is an ignorable coorinate, an the proceure outline in Chapter 1 may be followe; or we may procee irectly. From (4.6c) there follows an immeiate integration yieling r sin θ ϕ = C 3, or ϕ = C 3 r sin θ (4.7a,b) where C 3 is the constant of integration an represents the polar component of angular momentum. The introuction of (4.7) into (4.6a,b) yiels, respectively t [ṙ] = r θ μ r + C3 r 3 sin θ t [r θ] = C cos θ 3 r sin 3 θ. (4.8a) (4.8b) Consiering (4.8b), we multiply across by r θ to obtain r θ t [r θ] = C 3 cos θ θ sin 3 θ which may be rearrange as [ ] t [r θ] = C3 1 t sin θ or alternatively (4.9) (4.10) [ ] (r θ) + C 3 t sin = 0. (4.11) θ

15 Sec 4 The Kepler Problem: Lagrangian Analysis 43 This implies that the expression in square brackets is constant; however, if we substitute for C 3 in terms of ϕ from (4.7a), the expression becomes r [ r θ + r sin θ ϕ ] = r (v ṙ ) (4.1) an if we recall (3.18), we see that this constant is the square of the angular momentum, namely C. Accoringly, the integral of (4.11) may be written r 4 θ + C 3 sin θ = C (4.13) or alternatively r θ = 1 [ ] r 3 C C 3 sin θ (4.14) as the form appropriate for the reuction of (4.8a), which we effect prior to the integration of (4.13). If we substitute for r θ from (4.14) an for ϕ from (4.7b) into equation (4.8a), we see that the terms with C3 cancel an we have C [ṙ] = t r 3 μ r = [ μ r r 1 C ] r. (4.15) If we multiply across by ṙ, we obtain t [ 1 ṙ ] = [ μ t r 1 C ] r or on rearrangement (4.16) [ 1 t ṙ μ r + 1 C ] r = 0 (4.17) an so the expression in square brackets must be constant. Again, recalling (3.19) we see that the expression 1 ṙ μ r + 1 C r = 1 ṙ μ r + 1 (v ṙ ) = 1 v μ r (4.18) is in fact the energy integral, whose constant has alreay been esignate as E (3.7) an for negative energy has been ientifie by α (3.10). Accoringly, the integrate relation reas 1 ṙ μ r + 1 C r = α (4.19) or alternatively [ r ṙ = α r μ ] α r + C α (4.0)

16 44 Ch The Kepler Problem ientical with the previously erive (3.43). The subsequent analysis leaing to the solution (3.46) r = a[1 e cos E] (4.1) follows an ientical pattern. In item 4 below relation (1.18), we have note a shortcoming of this simple form of the equation of the ellipse: for the ynamic problem a secon shortcoming is now coming into view. If one were to apply the transformation (3.44) to equation (4.13) for θ, we woul still have a couple equation, an if one were to substitute for r from (4.1), one has a ifferential equation that is not reaily integrable. In fact, an inspection of equation (4.13) suggests the form of the alternative regularizing transformation that will effect the uncoupling of equations (4.13) an (4.17), whose uncouple form amits a reay integration in the case of each equation. The singularity in the ifferential equation (4.13) can be regularize by means of the [regularizing] transformation f t = C r, C f = r t (4.a,b) an with f as the new inepenent variable, an with prime enoting ifferentiation with respect to f, equation (4.13) becomes C θ + C 3 sin θ = C. (4.3) If we now introuce a new parameter ν, representing the inclination of the orbit plane, an efine by ν = C 3 C (4.4) then equation (4.3) may be written sin θ θ = (1 ν ) cos θ (4.5) which, as we shall see, amits a straightforwar integration. Returning to equation (4.0), we multiply by a further r -factor to obtain r 4 ṙ = r [ α r μr + C ]. (4.6) If we utilize the transformation (4.) to introuce the new inepenent variable f, then after iviing across by C we have [ r = r 1 μ ] C r + α C (4.7a) [ = r 1 p r + 1 ] ap r (4.7b)

17 Sec 4 The Kepler Problem: Lagrangian Analysis 45 where we have introuce the length scales from (3.11) an (3.33) into (4.7a) to obtain (4.7b). The integration of (4.7b) is facilitate by the introuction of an auxiliary epenent variable u, efine by u = 1 r, r = 1 u, r = 1 u u (4.8) an, after a little manipulation, equation (4.7b) becomes [( u = u 1 ) + 1 p ap 1 ] [ e ( p = p u 1 ) ]. (4.9) p By setting u 1 p = e p w (4.30) the ifferential equation for w reas w = 1 w (4.31) with solution w = cos(f + ω 0 ) (4.3) where ω 0 is the constant of integration. It follows from (4.30) that u = 1 p [ 1 + e cos(f + ω0 ) ] (4.33) an hence, noting (4.8), we have r = p 1 + e cos(f + ω 0 ). (4.34a) Except for the factor ω 0, this is ientical with (1.15) for the ellipse, so the variable f has the obvious angular interpretation; moreover, if the angle is measure from the pericenter so that then clearly ω 0 = 0 an we have f = 0 correspons to r = a(1 e) (4.34b) r = p 1 + e cos f (4.35) as the solution for r, ientical with (1.15). Returning to equation (4.5), we note that the integration can be facilitate by setting cos θ = 1 ν S (4.36)

18 46 Ch The Kepler Problem so that equation (4.5) becomes S = 1 S (4.37) with solution S = sin(f + ω) (4.38) where ω is the constant of integration. The point where the orbit crosses the z-plane is calle the noe an the line joining it to the focus is calle the noal line. The crossing of the z-plane correspons to θ = π/, an so noting (4.36) an (4.38), this must correspon to f = ω; henceω measures the angle in the orbit plane subtene at the focus between the major axis an the noal line. An we may write cos θ = 1 ν sin(f + ω) (4.39) as the complete solution for the θ-coorinate. It remains to integrate equation (4.7) for the thir coorinate ϕ. Writing (4.7) in the form r ϕ = C 3 sin θ (4.40) we introuce the regularizing transformation (4.) replacing t as the inepenent variable by f. We then have Cϕ = C 3 sin θ (4.41) an if we ivie across by C an note the efining relation (4.4) for ν, we obtain ϕ = ν sin θ = If we introuce cos θ in terms of f from (4.39), we obtain ν 1 cos θ. (4.4) ϕ ν = 1 (1 ν ) sin (f + ω) ν = cos (f + ω) + ν sin (f + ω) = ν sec (f + ω) 1 + ν tan (f + ω). (4.43) The integration of equation (4.43) is facilitate by the substitution from which we have tan Φ = ν tan(f + ω) (4.44) sec Φ Φ = ν sec (f + ω), sec Φ = 1 + ν tan (f + ω) (4.45a,b) an from (4.43) there follows

19 Sec 4 The Kepler Problem: Lagrangian Analysis 47 ϕ = Φ implying Φ = ϕ + ϕ 0 (4.45c) where ϕ 0 is the constant of integration. Hence (4.44) implies that tan(ϕ + ϕ 0 ) = ν tan(f + ω). (4.46) We have alreay note that at the noal crossing, f = ω; ifwenowletω enote the longitue at this noal line, then from (4.46) there follows an hence, from (4.46), we have tan(ω + ϕ 0 ) = 0, implying ϕ 0 = Ω (4.47) tan(ϕ Ω) = ν tan(f + ω) (4.48) as the solution for the thir coorinate ϕ. The completion of the solution requires the etermination of the time-angle relation connecting the time with the true anomaly f. For this we introuce the expression (4.35) into the inverte form of the efining relation (4.a), an if we substitute for C from (3.33), we fin t f = r C = 1 a (1 e ) μa 1 e (1 + e cos f). (4.49) If we recall from (3.41) that μa = na, it follows that n t f = (1 e ) 3/ (1 + e cos f). (4.50) For the integration of this expression we first note that [ ] e sin f = (1 + e cos f)ecos f + e sin f f 1 + e cos f (1 + e cos f) = e + e cos f (1 + e cos f) = (1 + e cos f) (1 e ) (1 + e cos f) 1 = 1 + e cos f (1 e ) (1 + e cos f) (4.51) an hence, on multiplying by 1 e an rearranging, we have (1 e ) 3/ 1 e (1 + e cos f) = 1 + e cos f [ e 1 e ] sin f. (4.5) f 1 + e cos f For the integration of the first term on the right we note that if we set tan χ = 1 e sin f e + cos f (4.53) there follows

20 48 Ch The Kepler Problem sec χ = Taking the erivative of (4.53), we fin (1 + e cos f) e + cos f, cos χ = (e + cos f) 1 + e cos f, (4.54a,b) 1 e sin χ = sin f 1 + e cos f. (4.54c) sec χ χ (1 + e cos f) = (e + cos f) χ = which, with (4.54a), yiels 1 e (e + cos f)cos f + 1 e sin f (e + cos f) (1 + e cos f) = 1 e (e + cos f) (4.55) 1 e χ = 1 + e cos f (4.56) an hence, noting (4.53), we have 1 e [ 1 e sin f f = χ = arctan 1 + e cos f e + cos f ]. (4.57) Accoringly, the integration of (4.50) is accomplishe by combining (4.5) an (4.57) to yiel [ 1 e M = n(t t 0 ) = arctan ] sin f e 1 e sin f (4.58) e + cos f 1 + e cos f where t 0, reflecting the constant introuce by the integration, is the time of the pericenter passage, i.e., t = t 0 correspons to f = 0.

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