Osculating orbit elements and the perturbed
Kepler problem
a=
Gmr
+ f (r, v, t)
3
r
Same
x&v
Define:
r := rn , r := p/(1 + e cos f ) , p = a(1 e2 )
osculating orbit
p
he sin f
h
v :=
n+
, h := Gmp
p
r
⇥
⇤
n := cos ⌦ cos(! + f ) cos ◆ sin ⌦ sin(! + f ) eX
⇥
⇤
+ sin ⌦ cos(! + f ) + cos ◆ cos ⌦ sin(! + f ) eY
+ sin ◆ sin(! + f ) eZ
⇥
⇤
:=
cos ⌦ sin(! + f ) cos ◆ sin ⌦ cos(! + f ) eX
⇥
⇤
+
sin ⌦ sin(! + f ) + cos ◆ cos ⌦ cos(! + f ) eY
+ sin ◆ cos(! + f ) eZ
ĥ := n ⇥
= sin ◆ sin ⌦ eX
sin ◆ cos ⌦ eY + cos ◆ eZ
e, a, ω, Ω, i, T may be functions of time
actual orbit
new osculating
orbit
Perturbed Kepler problem
a=
Gmr
+ f (r, v, t)
3
r
dh
=r⇥f
dt
dA
n =) Gm
= f ⇥ h + v ⇥ (r ⇥ f )
dt
h = r ⇥ v =)
A=
v⇥h
Gm
Decompose:
f = Rn + S + W ĥ
dh
= rW + rS ĥ
dt
dA
Gm
= 2hS n
hR + rṙS
dt
Example:
rṙW ĥ.
ḣ = rS
d
(h cos ◆) = ḣ · eZ
dt
d◆
ḣ cos ◆ h sin ◆ = rW cos(! + f ) sin ◆ + rS cos ◆
dt
Perturbed Kepler problem
“Lagrange planetary equations”
r
dp
p3
1
=2
S,
dt
Gm 1 + e cos f
r

de
p
2 cos f + e(1 + cos2 f )
=
sin f R +
S ,
dt
Gm
1 + e cos f
r
d◆
p cos(! + f )
=
W,
dt
Gm 1 + e cos f
r
d⌦
p sin(! + f )
sin ◆
=
W,
dt
Gm 1 + e cos f
r

d!
1
p
2 + e cos f
sin(! + f )
=
cos f R +
sin f S e cot ◆
W
dt
e Gm
1 + e cos f
1 + e cos f
An alternative pericenter angle:
$ := ! + ⌦ cos ◆
r

d$
1
p
2 + e cos f
=
cos f R +
sin f S
dt
e Gm
1 + e cos f
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
=
dt
✓
df
dt
◆
Kepler
✓
d!
d⌦
+ cos ◆
dt
dt
◆
Drop if working to
1st order
Perturbed Kepler problem
Worked example: perturbations by a third body
r12
a1 = Gm2 3
r12
r12
a2 = +Gm1 3
r12
Gmr
a=
r3
Gm3 r h
n
R3
r13
Gm3 3 ,
r13
r23
Gm3 3
r23
i
3(n · N )N + O(Gm3 r2 /R4 )
3
r13
r23
R := |r23 | , N := r23 /|r23 | , m := m1 + m2
⇤
Gm3 r ⇥
2
R := f · n =
1 3(n · N ) ,
R3
Gm3 r
(n · N )( · N ),
S := f · = 3
3
R
Gm3 r
W := f · ĥ = 3
(n · N )(ĥ · N )
R3
Put third body on a circular orbit
N = eX cos F + eY sin F ,
dF
=
dt
r
2
r=r12
G(m + m3 )
df
⌧
R3
dt
1
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:
h ai = 0
h ei =
h !i =
h ◆i =
h ⌦i =
✓ ◆3
15⇡ m3 a
e(1 e2 )1/2 sin2 ◆ sin ! cos !
2 m R
✓ ◆3
h
3⇡ m3 a
(1 e2 ) 1/2 5 cos2 ◆ sin2 ! + (1 e2 )(5 cos2 !
2 m R
✓ ◆3
15⇡ m3 a
e2 (1 e2 ) 1/2 sin ◆ cos ◆ sin ! cos !
2 m R
✓ ◆3
3⇡ m3 a
(1 e2 ) 1/2 (1 5e2 cos2 ! + 4e2 ) cos ◆
2 m R
Also:
3⇡ m3
h $i =
2 m
✓
a
R
◆3
(1
h
e2 )1/2 1 + sin2 ◆(1
5 sin2 !)
i
3)
i
Perturbed Kepler problem
Worked example: perturbations by a third body
Case 1: coplanar 3rd body and Mercury’s perihelion (i = 0)
3⇡ m3
h $i =
2 m
✓
a
R
◆3
(1
e2 )1/2
For Jupiter:
d$/dt = 154 as per century (153.6)
For Earth
d$/dt = 62 as per century (90)
Mercury’s Perihelion: Trouble to Triumph
•  1687 Newtonian triumph
575 “
per
century
•  1859 Leverrier’s conundrum
•  1900 A turn-of-the century crisis
Planet
Venus
Earth
Mars
Jupiter
Saturn
Total
Discrepancy
Modern measured value
General relativity prediction
Advance
277.8
90.0
2.5
153.6
7.3
531.2
42.9
42.98 ± 0.001
42.98
Perturbed Kepler problem
Worked example: perturbations by a third body
Case 2: the Kozai-Lidov mechanism
h ai = 0
✓ ◆3
15⇡ m3 a
h ei =
e(1 e2 )1/2 sin2 ◆ sin ! cos !
2 m R
✓ ◆3
h
i
3⇡ m3 a
2
2
1/2
2
2
2
h !i =
(1 e )
5 cos ◆ sin ! + (1 e )(5 cos ! 3)
2 m R
✓ ◆3
15⇡ m3 a
h ◆i =
e2 (1 e2 ) 1/2 sin ◆ cos ◆ sin ! cos !
Stationary point:
2 m R
A conserved quantity:
e
cos ◆h ei + sin ◆h ◆i = 0
1
p
=) 1 e2 cos ◆ = constant
e2
1
LZ !
⇡ 3⇡
!c = or
2
2
3
2
ec = cos2 ◆c
5
Perturbed Kepler problem
Worked example: perturbations by a third body
Case 2: the Kozai-Lidov mechanism
Eccentricity
Inclination
Pericenter
Incorporating post-Newtonian effects
in N-body dynamics
Capture by BH
Merritt, Alexander, Mikkola &
Will, PRD 84, 044024 (2011)
Animations courtesy David Merritt
Perturbed Kepler problem
Worked example: body with a quadrupole moment
Gmr
a=
r3
a = 0,
3 GmR2 n⇥
2
J2
5(e
·
n)
2
r4
e = 0, ◆ = 0
✓ ◆2 ✓
◆
R
5
! = 6⇡J2
1
sin2 ◆
p
4
✓ ◆2
R
⌦ = 3⇡J2
cos ◆
p
⇤
1 n
2(e · n)e ,
For Mercury (J2 = 2.2 X 10-7)
d$
= 0.03 as/century
dt
For Earth satellites (J2 = 1.08 X 10-3)
✓ ◆7/2
d⌦
R
= 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
o