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Osculating Orbit Elements and the Perturbed Kepler Problem

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Osculating Orbit Elements and the Perturbed Kepler Problem Osculating orbit elements and the perturbed Kepler problem Gmr a = 3 + f(r, v,t) Same − r x & v Define: r := rn ,r:= p/(1 + e cos f) ,p= a(1 e2) − osculating orbit he sin f h v := n + λ ,h:= Gmp p r n := cos ⌦cos(! + f) cos ◆ sinp ⌦sin(! + f) e − X +⇥ sin⌦cos(! + f) + cos ◆ cos ⌦sin(! + f⇤) eY actual orbit +sin⇥ ◆ sin(! + f) eZ ⇤ λ := cos ⌦sin(! + f) cos ◆ sin⌦cos(! + f) e − − X new osculating + sin ⌦sin(! + f) + cos ◆ cos ⌦cos(! + f) e orbit ⇥ − ⇤ Y +sin⇥ ◆ cos(! + f) eZ ⇤ hˆ := n λ =sin◆ sin ⌦ e sin ◆ cos ⌦ e + cos ◆ e ⇥ X − Y Z e, a, ω, Ω, i, T may be functions of time Perturbed Kepler problem Gmr a = + f(r, v,t) − r3 dh h = r v = = r f ⇥ ) dt ⇥ v h dA A = ⇥ n = Gm = f h + v (r f) Gm − ) dt ⇥ ⇥ ⇥ Decompose: f = n + λ + hˆ R S W dh = r λ + r hˆ dt − W S dA Gm =2h n h + rr˙ λ rr˙ hˆ. dt S − R S − W Example: h˙ = r S d (h cos ◆)=h˙ e dt · Z d◆ h˙ cos ◆ h sin ◆ = r cos(! + f)sin◆ + r cos ◆ − dt − W S Perturbed Kepler problem “Lagrange planetary equations” dp p3 1 =2 , dt rGm 1+e cos f S de p 2 cos f + e(1 + cos2 f) = sin f + , dt Gm R 1+e cos f S r d◆ p cos(! + f) = , dt Gm 1+e cos f W r d⌦ p sin(! + f) sin ◆ = , dt Gm 1+e cos f W r d! 1 p 2+e cos f sin(! + f) = cos f + sin f e cot ◆ dt e Gm − R 1+e cos f S− 1+e cos f W r An alternative pericenter angle: $ := ! +⌦cos ◆ d$ 1 p 2+e cos f = cos f + sin f dt e Gm − R 1+e cos f S r Perturbed Kepler problem Comments: § these six 1st-order ODEs are exactly equivalent to the original three 2nd-order ODEs § if f = 0, the orbit elements are constants § if f << Gm/r2, use perturbation theory § yields both periodic and secular changes in orbit elements § can convert from d/dt to d/df using df df d! d⌦ = + cos ◆ dt dt − dt dt ✓ ◆Kepler ✓ ◆ Drop if working to 1st order Perturbed Kepler problem Worked example: perturbations by a third body 3 r12 r13 a1 = Gm2 3 Gm3 3 , − r12 − r13 r12 r23 a2 =+Gm1 3 Gm3 3 r12 − r23 r r13 Gmr Gm r 23 a = 3 n 3(n N)N + O(Gm r2/R4) r3 − R3 − · 3 h i R := r ,N:= r / r ,m:= m + m | 23| 23 | 23| 1 2 Gm3r 2 := f n = 1 3(n N) , r=r12 3 2 R · − R − · 1 Gm3r ⇥ ⇤ := f λ =3 (n N)(λ N), S · R3 · · Gm r := f hˆ =3 3 (n N)(hˆ N) W · R3 · · Put third body on a circular orbit dF G(m + m3) df N = eX cos F + eY sin F, = 3 dt r R ⌧ dt Perturbed Kepler problem Worked example: perturbations by a third body Integrate over f from 0 to 2π holding F fixed, then average over F from 0 to 2π: ∆a =0 h i 15⇡ m a 3 ∆e = 3 e(1 e2)1/2 sin2 ◆ sin ! cos ! h i 2 m R − ✓ ◆ 3 3⇡ m3 a 2 1/2 2 2 2 2 ∆! = (1 e )− 5 cos ◆ sin ! +(1 e )(5 cos ! 3) h i 2 m R − − − ✓ ◆ 3 h i 15⇡ m3 a 2 2 1/2 ∆◆ = e (1 e )− sin ◆ cos ◆ sin ! cos ! h i − 2 m R − ✓ ◆ 3 3⇡ m3 a 2 1/2 2 2 2 ⌦ = (1 e )− (1 5e cos ! +4e ) cos ◆ h i − 2 m R − − ✓ ◆ Also: 3⇡ m a 3 ∆$ = 3 (1 e2)1/2 1+sin2 ◆(1 5sin2 !) h i 2 m R − − ✓ ◆ h i Perturbed Kepler problem Worked example: perturbations by a third body Case 1: coplanar 3rd body and Mercury’s perihelion (i = 0) 3⇡ m a 3 ∆$ = 3 (1 e2)1/2 h i 2 m R − ✓ ◆ For Jupiter: d$/dt = 154 as per century (153.6) For Earth d$/dt = 62 as per century (90) Mercury’s Perihelion: Trouble to Triumph • 1687 Newtonian triumph 575 “ • 1859 Leverrier’s conundrum per century • 1900 A turn-of-the century crisis Planet Advance Venus 277.8 Earth 90.0 Mars 2.5 Jupiter 153.6 Saturn 7.3 Total 531.2 Discrepancy 42.9 Modern measured value 42.98 0.001 General relativity prediction± 42.98 Perturbed Kepler problem Worked example: perturbations by a third body Case 2: the Kozai-Lidov mechanism ∆a =0 h i 15⇡ m a 3 ∆e = 3 e(1 e2)1/2 sin2 ◆ sin ! cos ! h i 2 m R − ✓ ◆ 3 3⇡ m3 a 2 1/2 2 2 2 2 ∆! = (1 e )− 5 cos ◆ sin ! +(1 e )(5 cos ! 3) h i 2 m R − − − ✓ ◆ 3 h i 15⇡ m3 a 2 2 1/2 ∆◆ = e (1 e )− sin ◆ cos ◆ sin ! cos ! Stationary point: h i − 2 m R − ✓ ◆ ⇡ 3⇡ ! = or c 2 2 A conserved quantity: 3 e 1 e2 = cos2 ◆ cos ◆ ∆e +sin◆ ∆◆ =0 − c 5 c 1 e2 h i h i − = 1 e2 cos ◆ =constant LZ ! ) − p Perturbed Kepler problem Worked example: perturbations by a third body Case 2: the Kozai-Lidov mechanism Eccentricity Inclination Pericenter Incorporating post-Newtonian effects in N-body dynamics Capture by BH Merritt, Alexander, Mikkola & Will, PRD 84, 044024 (2011) Animations courtesy David Merritt Perturbed Kepler problem Worked example: body with a quadrupole moment Gmr 3 GmR2 a = J 5(e n)2 1 n 2(e n)e , r3 − 2 2 r4 · − − · n⇥ ⇤ o ∆a =0, ∆e =0, ∆◆ =0 2 R 5 2 -7 ∆! =6⇡J 1 sin ◆ For Mercury (J2 = 2.2 X 10 ) 2 p − 4 ✓ ◆ ✓ ◆ d$ R 2 =0.03 as/century ⌦ = 3⇡J cos ◆ dt − 2 p ✓ ◆ -3 For Earth satellites (J2 = 1.08 X 10 ) d⌦ R 7/2 = 3639 cos ◆ deg/yr dt − a ✓ ◆ § LAGEOS (a=1.93 R, i = 109o.8): 120 deg/yr ! § Sun synchronous: a= 1.5 R, i = 65.9 .
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