Astro 6570

Physics of the Planets

Planetary Perturbation Theory

Keplerian Orbital Elements

i = inclinationΩ = longitude of the (ascending) nodeω = argument of pericenter(ν = true anomaly)a = semi-major axise = eccentricityT = time of pericenter passage

Also, !ω ≡ Ω +ω , the longitude of pericenter L= !ω + ν , the true longitude

For the 2-body problem, all of these elements (except v and L) are constants of the motion.

Perturbation TheoryPurpose: calculate deviations from Keplerian 2-body

motion due to external influences.Examples:• Gravitational effects of a 3rd body (e.g., another planet or

satellite)• Non-spherical primary body (J2)• Atmospheric drag or thruster firings• Radiation forces (e.g., light pressure on small particles)• EM forces (e.g., charged particles in a planet�s

magnetosphere)• Relativistic corrections (e.g., Mercury�s orbit)

From a numerical integration of the Solar System.

Note: In this plot and the next, the reference plane is the Invariable Plane of the solar system, i.e, the plane perpendicular to its total orbital angular momentum, H.

This differs by ~1.5o from the Ecliptic plane, which is not fixed over long time intervals.

The invariable plane is largely determined by the orbits & masses of Jupiter & Saturn.

Note: The variations for Jupiter and Saturn are almost perfectly anti-correlated, as they exchange angular momentum periodically. The dominant period for this exchange is ~50,000 yrs.

Perturbation Theory (cont’d.):Approaches: at least 3 are commonly employed:1. Calculate perturbing forces directly, resolve in radial,

azimuthal & normal directions (R, B, N) and derive the effect of each on the orbital elements (a, e, etc.)• Simple, but inelegant & generally messy…

2. Write perturbation in terms of a potential (the “Disturbing Function”, R) and derive the effect on the orbital elements.

• Conservative forces only, but generally simpler and more direct.

3. Rewrite Kepler problem in terms of Hamiltonian (canonical coordinates & momenta) and apply canonical perturbation theory.

• Elegant & simple, but non-intuitive.

Perturbation Theory: Lagrange’s Planetary Equations

(Danby pp 238-252)

u

Method 1: the perturbing force approach:

p = a(1-e2) = h2/GM

u = w + v

Notes on perturbation equations:

• The equations for di/dt and dW/dt involve only the normal force, N and the argument of latitude, u = w + v.

• The equation for da/dt involves only the radial and transverse forces, R and B, but mostly depends on B for small e, as this directly affects the orbital energy.

• The equations for de/dt and dw/dt involve both the radial and transverse forces, R and B, as well as (indirectly) the normal force, N via dW/dt .

• R strongly affects de/dt at v = 90 and 270o, while B strongly affects de/dt at v = 0 and 180o.

• The effects of R and B on dw/dt are opposite those on de/dt.• The dependence on N is stronger for dw/dt, as it is measured from

the node.

Note on longitude perturbationsThe most troublesome orbital element is the 6th, the time of pericenter

passage (T) or equivalently the longitude at t = 0. This is most often written as the �mean longitude at epoch�, ε:

λ t( ) = Ω +ω + M

≡ !ω + n t − T( ) = ε + ntFrom Kepler's equation, E − esin E = M , we get ε = !ω + E − esin E − nt and when the orbit is perturbed

"ε = !"ω + 1− ecos E( ) "E − "esin E − n − "nt

Even if !"ω , "E, "e, and "n remain small, the last term gets large as t →∞.This is circumvented by introducing a different quantity, which depends on the history of the perturbation:

λ t( ) ≡ ε1 + n t( )0

t

∫ dt

so that "ε1 satisfies the same equation, minus the "nt term. See Danby's text on Celestial Mechanics for more details.

Note that, since n2a3 = GM exactly for the unperturbed Kepler problem, we have

"nn ≡ −32"aa( ).

Method 2: the Disturbing Function approach:

Danby, p. 251

Note: See J. Burns, Amer. J. Phys. 44, 944 for a simplified derivation of the perturbation equations

Perturbations for small e or i

Note: In many papers h & k are reversed, and (h2, k2) à (p, q)

Note that the 2nd and3rd terms can often be neglected if h1 and k1 are small.

Ditto for 2nd term here.

Satellite Orbits Around an Oblate PlanetAs an example of perturbation techniques, we calculate the effects of a planet�s oblateness on a close

satellite (natural or artificial). The planet�s gravitational potential can be expanded in a series of multipole terms; we keep only the first 2 terms in this expansion:

Where R is the planet�s equatorial radius, J2 is a dimensionless constant of order 10-3 to 10-2, and P2is the 2nd Legendre polynomial:

( is the usual spherical polar co-ordinate.)

The 2nd term in V is our perturbing potential, , which we must express in terms of a, e, i, … !

V r,θ( ) = − µ

r 1− J2Rr( )2

P2 cosθ( ){ } (17)

P2 cosθ( ) = 1

23cos2θ −1( ).

θ

i.e.,! r,θ( ) = − µJ2 R2

2r33cos2θ −1( )

From geometry, we have cosθ = sin isin ω +υ( ), so

!= −µJ2 R2

4r33sin2 i 1− cos2 ω +υ( )⎡⎣ ⎤⎦ − 2{ } and

r −3 = a 1− e2( )⎡⎣

⎤⎦−3

1+ ecosυ( )3

Since the perturbation equations involve ε (i.e., mean longitude) rather than the true anomaly, υ, it is necessary to expand the cosine

terms in terms of the mean anomaly, M = n t − T( ). We can further simplify the problem, however, by considering only the cumulative

perturbations over many orbits and neglecting the various short-period perturbations.

Accordingly, we average ! over one orbital period as follows:

! = −µJ2 R2

4i2π

3sin2 i 1− cos2 ω +υ( )⎡⎣ ⎤⎦ − 2

r30

2π

∫ dM

the integrals involved are

cos2υr3

dM0

2π

∫ ,sin2υ

r3dM

0

2π

∫ ,1r3

dM0

2π

∫ ,

of which the 1st and 2nd are equal to zero (verification left as an exercise to the student). To evaluate the 3rd integral, we employ Kepler's 2nd law again:

1r3

dM0

2π

∫ =1r3

dυdM

⎛

⎝⎜⎞

⎠⎟

−1

dυ = n 1r3"υ( )−1

0

2π

∫0

2π

∫ dυ = nh

dυr0

2π

∫ , since r 2 "υ = h

(Note that R is conventionally defined as thenegative of the perturbing potential.)

∴1r3

dM0

2π

∫ =n

ha 1− e2( )1+ ecosυ( )dυ = 2π

a3 1− e2( )32

0

2π

∫

∴! = −µJ2 R2

2a3 1− e2( )32

32sin2 i −1⎡⎣ ⎤⎦ (18)

Note the ! is a function of a,e, and i only, so Lagrange's Equations ⇒

"a = "e = didt= 0

i.e., there are no long-term perturbations in a, e or i. (There are short - term perturbations due to the parts of ! that averaged to zero over one orbit, however.) Applying Lagrange's Equation, we obtain

dΩdt

= − 32

n J2Ra( )2 cos i

1− e2( )⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪ (19)

d #ωdt

= 32

n J2Ra( )2

1− 32 sin2 i − 1

2sin2i tan i

2( )1− e2( )2

⎧

⎨⎪⎪

⎩⎪⎪

⎫

⎬⎪⎪

⎭⎪⎪

(20)

and instead of dε1

dt, we calculate

dMdt

=ddt

ε1 + n dt − #ω∫( ) =

dε1

dt−

d #ωdt

+ n

= n 1+ 32

J2Ra( )2 1− 3

2 sin2 i

1− e2( )32

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

(21)

Note that all factors enclosed in { }

are ! 1 for small e and small i.

These results have considerable practical significance. Since J2 > 0 for any planet

flattened by rotation, we see that

(i) the line of nodes (i.e., Ω) regresses for prograde orbits, except for i = 90o , so the plane of the obit rotates backwards in inertial space, relative to a fixed direction.

(ii) for all i less than some critical value (46.o5), the line of apsides (i.e., !ω ) advances, so the pericenter rotates forwards around the orbit, relative to a fixed direction.

(iii) for all i < sin−1 23= 54.o7, the mean anomaly increases at a rate greater than n,

the Keplerian mean motion; i.e., the particle's orbital period (pericenter → pericenter) is less than that of a Keplerian orbit about a spherical planet of the same mass, M .

(i) and (ii) provide the means to determine J2 for planets with satellites in eccentric and/or

inclined orbits (E, Mars, J, S, U, N), as well as make possible the orbits of sun-synchronous Earth satellites such as LANDSAT. (iii) means that one must be careful in calculating planetary masses from observed values of a and "M for satellites.

In physical terms, (i) arises because an oblate planet exerts a torque on an inclined orbit, which causesits angular momentum vector to precess (see later lectures on planetary shapes). Items (ii) and (iii) are less obvious, but involve the breakdown of spherical symmetry in the Kepler problem.

{Nodal and apsidal precession

For i << π2

and e << 1, we have

!"ω # 32

n J2Ra( )2

= − "Ω

and

"M # n 1+ 32

J2Ra( )2{ }

∴ the sidereal average mean motion is "λ = n + "ε1

= "M + !"ω

# n 1+ 3J2Ra( )2{ }

≡ nG , the observed mean motion,

where

n ≡ GMa3 is the osculating, Keplerian

mean motion.

The osculating semi-major axis, a, is determined by the actual orbital energy:

EK =12

nG2 aG

2 − GMaG

where aG is the observed mean radius.

≡ − GM2a

∴ GM2a3 1+ 6 J2

Ra( )2{ }aG

2 − GMaG! − GM

2a

12

aG

a( )3 1+ 6 J2Ra( )2{ } + 1

2

aG

a = 1

writing aG

a = 1+ δ , we have 1+ 3δ + 6 J2Ra( )2

+1+ δ − 2 ! 0

∴δ ! − 32

J2Ra( )2

i.e., aG ! a 1− 32

J2Ra( )2{ }

ASIDE: osculating vs. �geometric� elements

or

The osculating eccentricity is determined by noting that, for eG = 0, the particle must be at pericenter

(since nG > n) :

∴aG = a 1− e( ) ∴e ! 3

2J2

Ra( )2

for eG = 0

Note that nG and aG do NOT satisfy Kepler's 3rd law:

nG2 aG

3 ! 1+ 6 J2Ra( )2

− 92

J2Ra( )2{ }

= GM 1+ 32

J2Ra( )2{ } ≠ constant

This is due, of course, to the additional radial gravity provided by J2 :

V r( ) = − GMr 1+ 1

2J2

Ra( )2{ } @ θ = π

2 so

−∂V∂r

= − GMr2 −

3GM J2 R2

2 R4 = −nG2 r

∴nG2 r3 = GM 1+ 3

2J2

Ra( )2( )… as above

If J2 were to be abruptly "turned off", the particle would be travelling faster than the circular

velocity at r = aG , and thus would be at the pericenter of an elliptical orbit with a > aG . This

is the true meaning of the osculating orbit.