alberto abad, antonio elipe grupo de mec´anica espacial...

Analytical Model for Lunar Orbiter

Alberto Abad, Antonio Elipe

Grupo de Mecanica Espacial, University of Zaragoza, Spain

Juan Felix San-Juan, Montserrat San-Martın

Dpt. Maths & Computation, University of La Rioja, Spain

A. Abad, A. Elipe, J.F. San-Juan & M. San-Martın CADE 2007. Turku. 2

Moon orbiter

• Theoretical interest:

– Normal forms in perturbed Hamiltonians

– Periodic orbits

– Bifurcations, . . .

• Practical interest:

– Science missions to the Moon, Jovian satellites, . . .


Frozen orbits

Definition: Those whose average e and I are constant.

They are the equilibria of the zonal problem, once short period terms

are removed. That is, mean orbits with fixed e, I and ω.

Example: Critical inclination in the main problem

Interest: avoiding maneuvers

repeated elevation at each pass

maintaining relative configuration

. . .


How are they determined?

• Find equilibria in the Gauss (Lagrange) equations

• Normal forms

• Continuation of families of periodic orbits

• . . .


Normal forms Coffey, Deprit & Deprit (1994)

Zonal problem in the Artificial Satellite

H = H(ℓ, g, h, L, G, H)

Lie transformations

H′ = H′(−, g′,−, L′, G′, H ′)

One degree of freedom.

Find equilibria (families when parameters vary)

(Mercator map or more convenient representation)


Families of periodic orbits

Broucke; Deprit, Elipe & Lara, . . .

Consider the system

x = 2Ay + Wx,

y = −2Ay + Wy,

with the integral

C = 2W − (x2 + y2)

and

A = A(x, y; σ), W = W (x, y; σ). σ a parameter


Let (x0, y0, x0, y0) be the initial conditions for t = 0

Let us assume the solution

x(t; x0, y0, x0, y0), y(t; x0, y0, x0, y0), C0,

be periodic

x(t; ξ0) = x(t + T0; ξ0), y(t; ξ0) = y(t + T0; ξ0).


Problem: Find ∆x0, ∆y0, ∆x0, ∆y0 for the new orbit be periodic

(T = T0 + ∆T0) and (C = C0 + ∆C0)

Periodicity condition: ξ0 + ∆ξ0 = ξ(T0 + ∆T0; ξ0 + ∆ξ0).

By Poincare method of continuity, ∃J (C0 ∈ J ) and initial conditions

X0 = x0 +∑

k≥1

x0,k(C − C0)k, Y0 = y0 +

∑

k≥1

y0,k(C − C0)k,

X0 = x0 +∑

k≥1

x0,k(C − C0)k, Y0 = y0 +

∑

k≥1

y0,k(C − C0)k,

such that the solution of the ODE system is periodic

T = T0 +∑

k≥1

T0,k(C − C0)k,


Natural Family

X(t) = x(t) +∑

k≤1

xk(t)(C − C0)k

Y (t) = y(t) +∑

k≤1

yk(t)(C − C0)k

Actually, it should be numerically computed !!!


Forces involved?

• Moon gravitational potential

• Earth gravitational potential (3th body attraction)

• Other

Spin-orbit resonance 1:1


Moon gravitational potential

Series expansion; e.g. in spherical coordinates

U = −µ

r

1 +

∑

n≥2

(α

r

)nJnPn(cosβ)

+∑

1≤m≤n

(Cmn cos mλ + Sm

n sin mλ)Pmn (cosβ)

where

Pmn Legendre Polynomials.

Jn zonal harmonics; Cmn , Sm

n tesseral harmonics.


λ = λ(t), hence, t appears explicitly !

Alternative: Formulate the problem in a synodic frame.

Consequences:

1) t no longer appears.

2) There is a new term in the kinetic energy −w Ω

which may cause difficulties in the tesseral case.


For axial symmetry, we have the Zonal Problem

U = −µ

r

1 +

∑

n≥2

(α

r

)n

JnPn(cosβ)

.

We will restrict to the Zonal problem !

J2 and J7 are the dominant terms (Elipe & Lara, 2003).


Third body attraction (Earth)

Ve = −µe

1

‖xe − x‖−

xe · x

‖xe‖3

But1

‖xe − x‖=

1

ae

1 +

(r

ae

)2

− 2r

aecos α

−1/2

=1

ae

∑

n≥0

(r

ae

)n

Pn

(x

r

),

with

P0 = 1, P1 = cos α, P2 =1

2(−1 + 3 cos2 α), . . .

Ve =ω2

2(r2 − 3x2)


In a rotating frame (ω), the Hamiltonian is

H = HK + HC + HZ + H3b

HK =1

2X · X −

µ

r=

1

2

R2 +

Θ2

r2

−

µ

r

HC = −ω · (x × X) = −ω N,

HZ =∑

n≥2

µ

r

(α

r

)n

JnPn(sin i sin(f + g)),

H3b =ω2r2

2

(1 − 3 [ cos h cos(f − g) − sinh sin(f − g) cos i]

2)

.


Scaling Hamiltonian

Low altitude (< 100 km)

Hl = H0 + ǫH1

with H0 = HK and ǫH1 = HZ, ǫ = J2.

High altitude

Hh = H0 + ǫH1 +ǫ2

2H2

with H0 = HK, ǫ = ω/n, ǫH1 = HC, and ǫ2 H2/2 = H3b + HZ.


Normal forms

Transformation (ℓ, g, h, L, G, H) −→ (ℓ′, g′, h′, L′, G′, H ′), then

H −→ K

such that K safisfies some properties


Lie Transform

Is a canonical transformation χ : (q, Q, ǫ) −→ (p, P ), solution of

dp

dǫ= (p ; W) =

∂W

∂P,

dP

dǫ= (P ; W) = −

∂W

∂p,

with the initial conditions

p(q, Q; 0) = q, P (q, Q; 0) = Q,

the generator W being

W ≡ W(p, P , ǫ) =∑

n≥0

ǫn

n!Wn+1(p, P ).


Given a function

F (p, P , ǫ) =∑

n≥0

ǫn

n!Fn,0(p, P ),

What is the action of the Lie-transformation onto it?

χ∗F (q, Q, ǫ) is a Taylor series:

χ∗F =∑

n≥0

ǫn

n!

dnF

dǫn|ǫ=0

∑

n≥0

ǫn

n!F0,n(q, Q)


Problem:

The functions Fn,0 and Wn given, determine F0,n.

Homological equation (Deprit, 1969)

Fi,j = Fi+1,j−1 +∑

0≤n≤i

i

n

(Fi−n,j−1; Wn+1)


The Lie–Deprit Triangle:

F0,0

ւ

F1,0 −→ F0,1

ւ ւ

F2,0 −→ F1,1 −→ F0,2

ւ ւ ւ

F3,0 −→ F2,1 −→ F1,2 −→ F0,3

ւ ւ ւ ւ

For the computing of each term, we need the previous ones in the row, and

the elements above the diagonal of this one. For example,

F2,1 = F3,0 + (F2,0 ; W1) + 2(F1,0 ; W2) + (F0,0 ; W3).


Elimination of the Parallax (Deprit, 1981)

The main objective of this simplification is to reduce the factors

µ

r

(p

r

)2n

toµ

p

(p

r

)2

,

while eliminating the explicit appearance of θ


Elimination of the Parallax

• Basics.

H0,0 =1

2

R2 +

Θ2

r

−

µ

r

State functions:

p =Θ2

µ, C =

(Θ

r−

Θ

p

)cos θ + R sin θ,

N = Θ cos I, S =

(Θ

r−

Θ

p

)sin θ − R cos θ,

∗ L0(p) = L0(N) = L0(C) = L0(S) = 0.

∗∗ F (p, N, C, S) ∈ ker(L0)


Elimination of the Parallax

Given a function

F (r, θ, R, Θ) =∑

j≥0

(Cj cos jθ + Sj sin jθ), Cj, Sj ∈ ker(L0),

the PDE in W , F ′:

L0(W ) + F ′ =Θ

r2F

is satisfied by choosing

F ′ =Θ

r2C0

W =∑

j≥1

1

j(Cj sin jθ − Sj cos jθ)


Delaunay Normalization

L0 =µ2

L3

∂

∂ℓ

The new Hamiltonian is chosen to be the averaged over ℓ

Kn =1

2π

∫ 2π

0Hn dℓ, Wn =

1

n

∫ (Hn −Kn

)dℓ


Simplified Hamiltonian

K = KZ + δ K3b,

where

K3b =ω2a2

16[(3c2 − 1)(3η2 − 5) − 15e2s2 cos 2g]),

KZ =1

2π

∫ 2π

0HZ dℓ,

and

δ =

0 for low altitude

1 for high altitude


J2 and J7 are the dominant terms.

KZ =µJ2α

2

4

3s2 − 2

a3η3

+3µJ7α

7

16384

es

a8η13[231e4s4(13s2 − 12) sin 5g

−105e2s2(3e2 + 8)(143s4 − 220s2 + 80) sin 3g

+70(5e4 + 20e2 + 8)(429s6 − 792s4 + 432s2 − 64) sin g].


Equations of motion

dg

dt= m00 + m01 sin g + m02 sin 2g + m03 sin 3g + m04 sin 5g,

dG

dt= m10 cos g + m11 cos 2g + m12 cos 3g + m13 cos 5g,

where mi,j depends on a, e, i, µ, α, ω, J2 and J7.

For δ = 0, m02 = m11 = 0


Low altitude obiter (δ = 0)

dg

dt= m00 + m01 sin g + m03 sin 3g + m04 sin 5g,

dG

dt= m10 cos g + m12 cos 3g + m13 cos 5g.

Frozen orbits ⇐⇒ Equilibria

m00 + m01 sin g + m03 sin 3g + m04 sin 5g = 0

m10 cos g + m12 cos 3g + m13 cos 5g = 0


Low altitude frozen orbits

m00 + m01 sin g + m03 sin 3g + m04 sin 5g = 0,

m10 cos g + m12 cos 3g + m13 cos 5g = 0,

g = π/2, 3π/2

0

0.5

1

1.5

00.050.10.1511.051.11.151.2

11.051.1

a

e

i


a = α + 100 km Impact orbits

0 30 ° 60 ° 90 °i

0.2

0.4

0.6

0.8

1

e


a = α + 100 km (small eccentricity)

0 30 ° 60 ° 90 °i

0.02

0.04

0.06

0.08

0.1

0.12

0.14

e


Previous results by numerical continuation (Elipe & Lara, 2003)

0 20 40 60 80

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

Log

. ecc

entr

icity

Inclination

o o o o o

S

S

S

0

1

2


How frozen are the orbits?

⋆ Initial conditions from the averaged equations

⋆ Propagate the orbit for 20 days (e cos g, e sin g)

0.00530.00540.00550.00560.00570.0058

-0.0001

0.0001

0.0002


⋆ Propagate the orbit for 5 years (e cos g, e sin g)

0.0046 0.0048 0.0052 0.0054 0.0056 0.0058

-0.0004

-0.0002

0.0002

0.0004


How frozen are the orbits?

⋆ Initial conditions from the averaged equations

⋆ Inverse Lie transformations =⇒ i.c. original problem

⋆ Propagate the orbit for 20 days (e cos g, e sin g)

0.00480.0049 0.00510.00520.00530.00540.0055

-0.00015

-0.0001

-0.00005

0.00005

0.0001

0.00015


⋆ Propagate the orbit for 5 years (e cos g, e sin g)

0.0048 0.0049 0.0051 0.0052 0.0053 0.0054 0.0055

-0.0002

-0.0001

0.0001

0.0002


-10000

1000

-1000

0

1000

-1000

0

1000

-10000

1000

-1000

0

1000


Low altitude frozen orbits (cont.)

m00 + m01 sin g + m03 sin 3g + m04 sin 5g = 0,

m10 cos g + m12 cos 3g + m13 cos 5g = 0,

g 6= π/2, 3π/2

Put x = cos g, y = sin g. Use Tchebyshev polynomials properties

dg

dt= m00 + k10y + k12yx2,

dG

dt= x(k00 + k02x

2 + k04x4).

bi-quadratic equation


0 30 ° 60 ° 90 °

0

0.2

0.4

0.6

0.8

1

There are no more frozen orbits !


High altitude frozen orbits Third body

m00 + m01 sin g + m02 sin 2g + m03 sin 3g + m04 sin 5g = 0,

m10 cos g + m11 cos 2g + m12 cos 3g + m13 cos 5g = 0,


0 20 ° 40 ° 60 ° 80 °

0.2

0.4

0.6

0.8

1Α+100

0 20 ° 40 ° 60 ° 80 °

0.2

0.4

0.6

0.8

1Α+1000

0 20 ° 40 ° 60 ° 80 °

0.2

0.4

0.6

0.8

12Α

0 20 ° 40 ° 60 ° 80 °

0.2

0.4

0.6

0.8

13Α

alberto abad, antonio elipe grupo de mec´anica espacial...

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