The two-body Kepler problem

Transcription

1 The two-body Kepler problem set center of mass at the origin (X = 0) ignore all multipole moments (spherical bodies or point masses) define r := r 1 r 2,r:= r,m:= m 1 + m 2,µ:= m 1 m 2 /m reduces to effective one-body problem a = Gm r 2 n Energy and angular momentum conserved: E = 1 2 m 1v m 2v 2 2 G m 1m 2 r 1 r 2 = 1 2 µv2 G µm r L = m 1 r 1 v 1 + m 2 r 2 v 2 = µr v orbital plane is fixed

2 Effective one-body problem Make orbital plane the x-y plane r v = r 2 d dt := he z v = dr dt =ṙn + r From energy conservation: ε = E/µ Reduce to quadratures (integrals) t t i = ± ṙ 2 =2[" V e (r) = h2 r 2 i = h Z r dr 0 p r i 2[" Ve (r 0 )] Z t t i dt 0 r(t 0 ) 2 V e (r)] Gm r y r φ λ=dn/dφ n x

3 Radial acceleration, or d/dt of energy equation: r h 2 r 3 = Gm r 2 Find the orbit in space: convert from t to φ: d/dt = d/d d 2 d 2 Keplerian orbit solutions 1 r + 1 r = Gm h 2 =(h/r 2 )d/d 1 r = 1 p (1 + e cos f) f :=! true anomaly p := h 2 /Gm semilatus rectum Elliptical orbits (e < 1, a > 0) r peri = p 1+e, =! r apo = p 1 e, =! + a := 1 2 (r peri + r apo )= p 1 e 2 Hyperbolic orbits (e > 1, a < 0) in out = 2 arcsin(1/e)

4 Keplerian orbit solutions Useful relationships ṙ = he p sin f v 2 = Gm p (1 + 2e cos f + e2 )=Gm E = Gµm 2a e 2 =1+ 2h2 E µ(gm) 2 a 3 P =2 Gm 1/2 2 r for closed orbits 1 a Alternative solution r = a(1 e cos u) n(t T )=u e sin u tan f 2 = r 1+e 1 e tan u 2 n =2 /P u = eccentric anomaly f = true anomaly n = mean motion

5 Dynamical symmetry in the Kepler problem a and e are constant (related to E and h) orbital plane is constant (related to direction of h) ω is constant -- a hidden, dynamical symmetry Runge-Lenz vector A := v h n Gm = e cos! e x +sin! e y ) = constant Comments: responsible for the degeneracy of hydrogen energy levels added symmetry occurs only for 1/r and r 2 potentials deviation from 1/r potential generically causes dω/dt

6 Six orbit elements: Keplerian orbit in space i = inclination relative to reference plane: Ω = angle of ascending node ω = angle of pericenter e = A cos = ĥ e Z cos = ĥ e Y sin sin! = A e z e sin a = h^2/gm(1-e 2 ) T = time of pericenter passage z n h f λ ω x Ω i y T = t Z f 0 r 2 h df Comment: equivalent to the initial conditions x 0 and v 0

7 Osculating orbit elements and the perturbed Kepler problem a = Gmr r 3 + f(r, v,t) Define: r := rn, r := p/(1 + e cos f), p = a(1 e 2 ) osculating orbit he sin f v := n + h, h := p Gmp p r n := cos cos(! + f) cos sin sin(! + f) e X + sin cos(! + f) + cos cos sin(! + f) e Y +sin sin(! + f) e Z Same x & v actual orbit := cos sin(! + f) cos sin cos(! + f) e X + sin sin(! + f) + cos cos cos(! + f) e Y new osculating orbit +sin cos(! + f) e Z ĥ := n =sin sin e X sin cos e Y + cos e Z e, a, ω, Ω, i, T may be functions of time

8 A = v h Gm Decompose: dh dt Gm da dt Example: ḣ cos Perturbed Kepler problem a = Gmr r 3 h = r v =) dh dt = r f n =) Gm da dt f = Rn + S = rw + rs ĥ + f(r, v,t) = f h + v (r f) + Wĥ =2hS n hr + rṙs rṙw ĥ. ḣ = rs d dt (h cos ) =ḣ e Z h d dt sin = rw cos(! + f)sin + rs cos

9 Lagrange planetary equations r dp p dt =2 3 1 Gm 1+ecos f S, r apple de p dt = sin f R + 2 cos f + e(1 + cos2 f) Gm 1+ecos f d dt = sin d dt = d! dt = 1 e Perturbed Kepler problem r p cos(! + f) Gm 1+ecos f W, r p sin(! + f) Gm 1+ecos f W, r apple p cos f R + Gm An alternative pericenter angle: 2+e cos f 1+e cos f S, sin f S e cot sin(! + f) 1+e cos f W $ :=! + cos d$ dt = 1 r apple p e Gm cos f R + 2+e cos f 1+e cos f sin f S

10 Perturbed Kepler problem Comments: these six 1 st -order ODEs are exactly equivalent to the original three 2 nd -order ODEs if f = 0, the orbit elements are constants if f << Gm/r 2, use perturbation theory yields both periodic and secular changes in orbit elements can convert from d/dt to d/df using df dt = h r 2 d! dt + cos d dt Drop if working to 1st order

11 Secular variations of orbit elements dx dt = Q (X,t) df dt = h r 2 = m1/2 (1 + e cos f)2 p3/2 =) dx df = Q (X,f) First order perturbation theory: set the X β = const in the RHS and integrate Z f X (f) = 0 Q (X,f 0 )df 0 There could be a net change in X α over one orbit - called a secular effect X = Z 2 0 Q (X,f 0 )df 0

12 Worked example: perturbations by a third body a = Gmr r 3 Perturbed Kepler problem r 12 a 1 = Gm 2 r12 3 r 12 a 2 =+Gm 1 Gm 3 r R 3 r 3 12 h n r 13 Gm 3 r13 3, r 23 Gm 3 r23 3 i 3(n N)N + O(Gm 3 r 2 /R 4 ) r 23 3 r 13 R := r 23,N:= r 23 / r 23,m:= m 1 + m 2 R := f n = S := f Gm 3r R 3 1 3(n N) 2, =3 Gm 3r R 3 (n N)( N), W := f ĥ =3 Gm 3r R 3 (n N)(ĥ N) Put third body on a circular orbit r df G(m + N = e X cos F + e Y sin F, dt = m3 ) R 3 2 df dt r=r 12 1

13 Integrate over f from 0 to 2π holding F fixed, then average over F from 0 to 2π: h ai =0 h ei = 15 2 h!i = a e(1 e 2 ) 1/2 sin 2 sin! cos! m R 3 a (1 e 2 ) 1/2h 5 cos 2 sin 2! +(1 e 2 )(5 cos 2! 3)i m 3 m 3 m h i = 15 2 h i = 3 2 Perturbed Kepler problem Worked example: perturbations by a third body R 3 a e 2 (1 e 2 ) 1/2 sin cos sin! cos! m R 3 a (1 e 2 ) 1/2 (1 5e 2 cos 2! +4e 2 ) cos R m 3 m 3 m Also: h $i = 3 2 m 3 m 3 a (1 e 2 ) 1/2h 1+sin 2 (1 5sin!)i 2 R

14 Perturbed Kepler problem Worked example: perturbations by a third body Case 1: coplanar 3 rd body and Mercury s perihelion (i = 0) h $i = 3 2 m 3 m a R 3 (1 e 2 ) 1/2 For Jupiter: d$/dt = 154 as per century (153.6) For Earth d$/dt = 62 as per century (90)

15 Mercury s Perihelion: Trouble to Triumph 1687 Newtonian triumph 1859 Leverrier s conundrum 1900 A turn-of-the century crisis 575 per century Planet Advance Venus Earth 90.0 Mars 2.5 Jupiter Saturn 7.3 Total Discrepancy 42.9 Modern measured value ± General relativity prediction 42.98

16 Case 2: the Kozai-Lidov mechanism h ai =0 h ei = 15 2 h!i = a e(1 e 2 ) 1/2 sin 2 sin! cos! m R 3 a (1 e 2 ) 1/2h 5 cos 2 sin 2! +(1 e 2 )(5 cos 2! 3)i m 3 m 3 m h i = 15 2 Perturbed Kepler problem Worked example: perturbations by a third body m 3 m R a R 3 e 2 (1 e 2 ) 1/2 sin cos sin! cos! Stationary point:! c = 2 or3 2 A conserved quantity: e cos h ei +sin h i =0 1 e2 =) p 1 e 2 cos =constant L Z! 1 e 2 c = 3 5 cos2 c

17 Perturbed Kepler problem Worked example: perturbations by a third body Case 2: the Kozai-Lidov mechanism Eccentricity Inclination Pericenter

18 Kozai oscillations and resonant relaxation Capture by BH Merritt, Alexander, Mikkola & Will, PRD 84, (2011) Animations courtesy David Merritt

19 Worked example: body with a quadrupole moment a = Perturbed Kepler problem Gmr 3 r 3 2 J GmR 2 n 5(e 2 n) 2 r 4 1 o n 2(e n)e, a =0, e =0, =0 2 R 5! =6 J 2 1 p 4 sin2 = 3 J 2 R p 2 cos For Mercury (J 2 = 2.2 X 10-7 ) d$ dt =0.03 as/century For Earth satellites (J 2 = 1.08 X 10-3 ) 7/2 d R dt = 3639 cos deg/yr a LAGEOS (a=1.93 R, i = 109 o.8): 120 deg/yr! Sun synchronous: a= 1.5 R, i = 65.9

20 Curved spacetime tells matter how to move Continuous matter, stress energy tensor Perfect fluid: T 1st law of Thermodynamics Relativistic Euler equation Compare with Newton =( c p)u u /c 2 + pg j = u r T =0, r j =0 u r T =0= d" d (µ + p) Du d dv dt ru = ρ = rest mass density ε = energy density p = pressure u α = four velocity +(" + p)r ~u d("v)+pdv =0 = c 2 g + u u /c 2 r p rp

21 Matter tells spacetime how to curve Riemann tensor + µ µ µ Ricci tensor R = R µ µ Ricci scalar R = g R Einstein tensor G = R 1 2 g R Bianchi identities Action S = Einstein s equations: r G =0 c3 16 G G Z p = 8 G c 4 grd 4 x + S matter T