basics of orbital mechanics i - ocw...

Basics of Orbital Mechanics I

Modeling the Space Environment

Manuel Ruiz Delgado

European Masters in Aeronautics and SpaceE.T.S.I. Aeronauticos

Universidad Politecnica de Madrid

April 2008

Basics of Orbital Mechanics I – p. 1/20

Basics of Orbital Mechanics I

Two-Body ProblemInertial FormulationCenter of Mass FormulationRelative FormulationSimplifications→ Kepler Problem

Kepler ProblemIntegralsTrajectoryEnergy and PeriodVelocityTime-Law: Kepler EquationClassical Orbital Elements


Two-Body Problem: Inertial Formulation

r1

r2

S1

m1, m2 point masses;S1 inertial reference



r1

r2

S1

F12

F21

m1, m2 point masses;S1 inertial reference

F12, F21 gravitational attraction



r1

r2

S1

F12

F21

P1

P2m1, m2 point masses;S1 inertial reference


P1, P2 other forces (“Perturbations”)



r1

r2

S1

F12

F21

P1




m1 r1 =G m1 m2

|r2 − r1|3(r2 − r1) + P1

m2 r2 = − G m1 m2

|r2 − r1|3(r2 − r1) + P2



r1

r2

S1

F12

F21

P1




m1 r1 =G m1 m2

|r2 − r1|3(r2 − r1) + P1

m2 r2 = − G m1 m2

|r2 − r1|3(r2 − r1) + P2

Numerical integration:r1(t), r2(t)


Two-Body Problem:COM Formulation

G

r1

r2

S1

G Center of mass ofm1, m2



G

r1

r2

S1


(m1 + m2) rG = m1 r1 + m2 r2



G

r1

r2

S1

S0


(m1 + m2) rG = m1 r1 + m2 r2

S0 Non-rotating reference, origin inG



G

r1

r2

S1

r′1

r′2 S0


(m1 + m2) rG = m1 r1 + m2 r2


r′1 = r1 − rG = m2

m1+m2(r1 − r2)

r′2 = r2 − rG = m1

m1+m2

(r2 − r1) = −m1

m2

r′1



G

r1

r2

S1

r′1

r′2 S0


(m1 + m2) rG = m1 r1 + m2 r2


r′1 = r1 − rG = m2

m1+m2(r1 − r2)

r′2 = r2 − rG = m1

m1+m2

(r2 − r1) = −m1

m2

r′1

(m1 + m2) rG = m1 r1 + m2 r2 = P1 + P2



G

r1

r2

S1

r′1

r′2 S0


(m1 + m2) rG = m1 r1 + m2 r2


r′1 = r1 − rG = m2

m1+m2(r1 − r2)

r′2 = r2 − rG = m1

m1+m2

(r2 − r1) = −m1

m2

r′1

(m1 + m2) rG = m1 r1 + m2 r2 = P1 + P2

m1r′1 = m1r1 − m1rG =



G

r1

r2

S1

r′1

r′2 S0


(m1 + m2) rG = m1 r1 + m2 r2


r′1 = r1 − rG = m2

m1+m2(r1 − r2)

r′2 = r2 − rG = m1

m1+m2

(r2 − r1) = −m1

m2

r′1

(m1 + m2) rG = m1 r1 + m2 r2 = P1 + P2

m1r′1 = m1r1 − m1rG =

= G m1 m2

|r2−r1|3(r2 − r1) + P1 − m1

m1+m2

(P1 + P2) ⇒



G

r1

r2

S1

r′1

r′2 S0


(m1 + m2) rG = m1 r1 + m2 r2


r′1 = r1 − rG = m2

m1+m2(r1 − r2)

r′2 = r2 − rG = m1

m1+m2

(r2 − r1) = −m1

m2

r′1

(m1 + m2) rG = m1 r1 + m2 r2 = P1 + P2

m1r′1 = m1r1 − m1rG =

= G m1 m2

|r2−r1|3(r2 − r1) + P1 − m1

m1+m2

(P1 + P2) ⇒

m1r′1 = −Gm1 m2

(1 + m1

m2

)−2r′

1

|r′1|3 + m1m2

m1+m2

(P1

m1

− P2

m2

)


Two-Body Problem:Primary Formulation

S2

m1

m2

S2 Non-rotatingNon-inertial, origin in m2



r

S2

m1

m2


r = r1 − r2 Relativeposition vector



r

S2

m1

m2



COM motion: (m1 + m2) rG = P1 + P2



r

S2

m1

m2




Relativemotion: r = r1 − r2



r

S2

m1

m2





r =[

G m2

|r2−r1|3(r2 − r1) + P1

m1

]−

[− G m1

|r2−r1|3(r2 − r1) + P2

m2

]⇒



r

S2

m1

m2





r =[

G m2

|r2−r1|3(r2 − r1) + P1

m1

]−

[− G m1

|r2−r1|3(r2 − r1) + P2

m2

]⇒

r = −G (m2 + m1)r

|r|3+ P1

m1

− P2

m2



r

S2

F12

m1

m2

P1





r =[

G m2

|r2−r1|3(r2 − r1) + P1

m1

]−

[− G m1

|r2−r1|3(r2 − r1) + P2

m2

]⇒

r = −G (m2 + m1)r

|r|3+ P1

m1

− P2

m2

Direct terms



r

S2

F12

F21

m1

m2

P1

P2





r =[

G m2

|r2−r1|3(r2 − r1) + P1

m1

]−

[− G m1

|r2−r1|3(r2 − r1) + P2

m2

]⇒

r = −G (m2 + m1)r

|r|3+ P1

m1

− P2

m2

Indirect terms



r

S2

F12

−F21

m1

m2

P1

−P2

m1





r =[

G m2

|r2−r1|3(r2 − r1) + P1

m1

]−

[− G m1

|r2−r1|3(r2 − r1) + P2

m2

]⇒

r = −G (m2 + m1)r

|r|3+ P1

m1

− P2

m2

Indirect terms asinertia forces → m1 + m2


Two-Body Problem: Formulations

2

1

G

Inertial motion

2

1

r′2

r′1

Relative toCOM

2

1

G

r

Relative toPrimary


Two-Body Problem: Simplifications

r = −G (m2 + m1)r

|r|3︸︷︷︸Kepler Problem

+P1

m1− P2

m2︸︷︷︸Perturbation



r = −G (m2 + m1)r


+P1

m1− P2


Isolated System: P1 = P2 = 0 → Kepler problem



r = −G (m2 + m1)r


+P1

m1− P2



Small mass:m1 ≪ m2 → G(m2 + m1) ≃ G m2 = µ Gravitational constant



r = −G (m2 + m1)r


+P1

m1− P2



Small mass:m1 ≪ m2 → G(m2 + m1) ≃ G m2 = µ Gravitational constant

Close pair: Moon/Earth, satellite/Earth Third-Body perturbations:P1

m1

− P2

m2

≃ 0 → Kepler problem


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M

Autonomous:C6 tied to initial time:t − t0


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M


Point masses or spherical symmetry


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M



Potential force: F = −∇V V = −µr


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M




Central force:F = −f(r) r ⇒ r ∧ r = 0 ⇒ r ∧ v = Const


Kepler Problem

F = m r = −G M mr

r3= −µm

r

r3

d

dt

r

v

=

v

−µ r

r3

r = r (t, C1, C2, C3, C4, C5, C6)

rF

m

M




Central force:F = −f(r) r ⇒ r ∧ r = 0 ⇒ r ∧ v = Const

Solution: Search forintegrals(conserved magnitudes)


Integrals of Motion

Potential: Specific Energy conservationv2

2− µ

r=

E

m


Integrals of Motion


2− µ

r=

E

m

Central: Specific Angular Momentum r ∧ v = h


Integrals of Motion


2− µ

r=

E

m


h

rv

⇒ Plane motion

Color code:h Constant magnituder Fast variable


Integrals of Motion


2− µ

r=

E

m


h

rv

⇒ Plane motion

F ∝ − 1r2 → Laplace/Runge-Lenz vector −r

r− h ∧ v

µ= e


Integrals of Motion


2− µ

r=

E

m


h

rv

⇒ Plane motion

F ∝ − 1r2 → Laplace/Runge-Lenz vector −r

r− h ∧ v

µ= e

d

dt(h ∧ r) = h ∧ r + h ∧ r = (r ∧ r) ∧ −µr

r3=

µr

r3∧ (r ∧ r) =

=µ

r3

(r r r − r2 r

)= µ

(r r

r2− r

r

)= −µ

d

dt

(r

r

)


Integrals of Motion: Dependencies

r = r(t, C1, C2, C3, C4, C5 , C6

), C6 → t − t0

7 Constants:E (1), h (3), e (3) → Only 5 can be independent



r = r(t, C1, C2, C3, C4, C5 , C6

), C6 → t − t0


h = r ∧ v ⊥ e = −r

r− h ∧ v

µh

e

rv



r = r(t, C1, C2, C3, C4, C5 , C6

), C6 → t − t0


h = r ∧ v ⊥ e = −r

r− h ∧ v

µh

e

rv

E =µ2m

2h2

(e2 − 1

)



r = r(t, C1, C2, C3, C4, C5 , C6

), C6 → t − t0


h = r ∧ v ⊥ e = −r

r− h ∧ v

µh

e

rv

E =µ2m

2h2

(e2 − 1

)

e2 = e·e =(r

r

)2+

(⊥︷︸︸︷

h ∧ v

µ

)2+2

r · (h ∧ v)

rµ= 1+

h2v2

µ2+2

h ·−h︷︸︸︷

v ∧ r

rµ=

= 1 +h2v2

µ2− 2h2

rµ= 1 +

2h2

µ2

( v2

2− µ

r︸︷︷︸E/m

)= 1 +

2h2E

µ2m


Trajectory

r

m

θ

eM

Polar coordinates in the plane of motion (⊥ h)

Main axis: direction ofe


Trajectory

r

m

θ

eM



r·e = r e cos θ = −r− r · h ∧ v

µ= −r+

h · r ∧ v

µ=

= −r +h2

µ⇒ r =

h2/µ

1 + e cos θ


Trajectory

r

m

θ

eM

θ : True Anomaly



r·e = r e cos θ = −r− r · h ∧ v

µ= −r+

h · r ∧ v

µ=

= −r +h2

µ⇒ r =

h2/µ

1 + e cos θ

Polar equation of aConic Section


Trajectory

r

m

θ

eM

θ : True Anomaly



r·e = r e cos θ = −r− r · h ∧ v

µ= −r+

h · r ∧ v

µ=

= −r +h2

µ⇒ r =

h2/µ

1 + e cos θ


e Eccentricity→ e Eccentricity vector

e = 0 Circlee < 1 Ellipsee = 1 Parabolae > 1 Hyperbola


Trajectory

r

m

θ

eM

θ : True Anomaly



r·e = r e cos θ = −r− r · h ∧ v

µ= −r+

h · r ∧ v

µ=

= −r +h2

µ⇒ r =

h2/µ

1 + e cos θ


e Eccentricity→ e Eccentricity vector

e = 0 Circlee < 1 Ellipsee = 1 Parabolae > 1 Hyperbola

p = h2/µ Parameteror semilatus rectum: radius at 90o


Trajectory

p

FF ′

θ

r

c

b

a e

rpra

e < 1p

FF ′

θ

e

rra∞

e = 1

eF F ′

θ

pr

e > 1

Ellipse Parabola Hyperbola

Parameter p h2/µ h2/µ h2/µ

Eccentricity e < 1 1 > 1

Pericenter rpp

1+e

p

2p

1+e

Apocenter rap

1−e∞ ∄

Major semiaxis a p

(1−e2) ∞ −p

(e2−1)

Minor semiaxis b p√1−e2

∞ p√e2−1

Focal distance c ae ∞ ae


Energy

EnergyE is related toh ande as:E

m=

µ2

2h2

(e2 − 1

)


Energy


m=

µ2

2h2

(e2 − 1

)

From the trajectory we can identify:µ(e2 − 1)

h2= −1

a


Energy


m=

µ2

2h2

(e2 − 1

)


h2= −1

a

Therefore:

E

m= − µ

2aOrbit size dependsonly onE


Energy


m=

µ2

2h2

(e2 − 1

)


h2= −1

a

Therefore:

E

m= − µ

2aOrbit size dependsonly onE

E also shows the type of Conic:

E < 0 EllipseE = 0 ParabolaE > 0 Hyperbola


Period of the Elliptic Orbit

Kepler’s 2nd Law: Areas are swept at aconstantrate

dA = 12r2dθ → dA

dt = 12r2 dθ

dt = 12h

dθ

dA

r




dA = 12r2dθ → dA

dt = 12r2 dθ

dt = 12h

dθ

dA

r

dAdt = Area

Period = πabT = 1

2h

T · 1

2h = π a b = π a

√a p = π a

√

ah

2

µ⇒ T = 2π

√a3

µ




dA = 12r2dθ → dA

dt = 12r2 dθ

dt = 12h

dθ

dA

r

dAdt = Area

Period = πabT = 1

2h

T · 1

2h = π a b = π a

√a p = π a

√

ah

2

µ⇒ T = 2π

√a3

µ

Kepler’s 3rd Law: Period squared is proportional to semiaxis cubed.




dA = 12r2dθ → dA

dt = 12r2 dθ

dt = 12h

dθ

dA

r

dAdt = Area

Period = πabT = 1

2h

T · 1

2h = π a b = π a

√a p = π a

√

ah

2

µ⇒ T = 2π

√a3

µ

Kepler’s 3rd Law: Period squared is proportional to semiaxis cubed.

Mean angular rate: n =2π

T=

√µ

a3


Energy and Eccentricity

V

D

B

C

A

r =, v ↑E ↑, h ↑a ↑, e l

E

m= − µ

2a=

v2

2− µ

rBasics of Orbital Mechanics I – p. 15/20

Energy and Eccentricity

V

D

B

C

A

r =, v ↑E ↑, h ↑a ↑, e l

E

m= − µ

2a=

v2

2− µ

r

–1

–0.5

0

0.5

1

–1 –0.5 0.5 1 1.5 2

r =, v =, r, v ↑E =, h ↑a =, e ↓

e2 = 1 +2h2E

µ2mBasics of Orbital Mechanics I – p. 15/20

Velocity

Computev(θ) through

Trajectory: r = h2/µ

1+e cos θ

Area Law: h = r2θ uz

Polar coordinates: r = r ur, v = r ur + rθ uθ


Velocity



1+e cos θ



rθ =h

r=

µ

h(1 + e cos θ)

(µ

h=

√µp

)


Velocity



1+e cos θ



rθ =h

r=

µ

h(1 + e cos θ)

(µ

h=

√µp

)

r = drdθ · θ = h2/µ

(1+e cos θ)2

e sin θ · µ2

h3(1 + e cos θ)2 =

µ

he sin θ


Velocity



1+e cos θ



rθ =h

r=

µ

h(1 + e cos θ)

(µ

h=

√µp

)

r = drdθ · θ = h2/µ

(1+e cos θ)2

e sin θ · µ2

h3(1 + e cos θ)2 =

µ

he sin θ

v =µ

h[e sin θ ur + (1 + e cos θ) uθ]

v =µ

h[uθ + e j] =

µ

h[− sin θi + (e + cos θ) j]

rrθv

θ

j

uθ

ur


Time Law (Elliptic): Kepler Equation

O Fae

Æ θ

r

b

a

Q

Q′

P

Constant area rate: Area=12 h · (t − τ)

Circle/Ellipse affinity:

Area FPQ = baArea FPQ′



O Fae

Æ θ

r

b

a

Q

Q′

P




Æ : Eccentric Anomaly

M = n (t − τ) : Mean Anomaly

τ : Time at pericenterP



O Fae

Æ θ

r

b

a

Q

Q′

P







h

2(t − τ) = Area FPQ =

b

a

(AreaOPQ′ − AreaOFQ′

)=

=b

a

(1

2a2Æ− 1

2ae a sin Æ

)=

h

ab(t − τ) = Æ− e sin Æ



O Fae

Æ θ

r

b

a

Q

Q′

P







h

2(t − τ) = Area FPQ =

b

a

(AreaOPQ′ − AreaOFQ′

)=

=b

a

(1

2a2Æ− 1

2ae a sin Æ

)=

h

ab(t − τ) = Æ− e sin Æ

hab = h

a√

ah2/µ= n ⇒ n (t − τ) = Æ− e sin Æ = M


Kepler Equation

n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)

True Anomalyθ ↔ ÆEccentric Anomaly


Kepler Equation

n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)


cos θ = cos Æ−e1−e cos Æ cos Æ = e+cos θ

1+e cos θ

sin θ =√

1−e2 sin Æ1−e cos Æ sin Æ =

√1−e2 sin θ1+e cos θ

tanθ

2=

√(1 + e)

(1 − e)tan

Æ2


Kepler Equation

n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)


cos θ = cos Æ−e1−e cos Æ cos Æ = e+cos θ

1+e cos θ

sin θ =√

1−e2 sin Æ1−e cos Æ sin Æ =

√1−e2 sin θ1+e cos θ

tanθ

2=

√(1 + e)

(1 − e)tan

Æ2

Implicit Equation. Simplest method:Iteration

t → M → Æ→ θ ( · · · + k 2π)

Æ1 = M

Æ2 = M + e sin Æ1

Æ3 = M + e sin Æ2

. . . Fast except fore → 1


Classical Orbital Elements

Line of nodes Peric.

Sat.

h

eω

Ω

Ω

θ

i

i

x1

y1

z1

i InclinationΩ Longitud of ascending nodeω Argument of pericentera Semimajor axise Eccentricityτ Time of pericenter passage

= Ω + ω Longitude of pericenter Ascending nodeL = + M Mean longitude Descending node↔ Line of Nodes Aries Point

Ω from to

i ∈ [0, 180o] Ω ∈ [0, 360o] ω ∈ [0, 360o]


Reference Systems and Time

Line of nodes Peric.

Sat.

h

eω

Ω

Ω

θ

uN

i

i

x1

y1

z1

Equatorial i1() j1 k1

Nodal uN () h ∧ uN h

Perifocal e (Per) h ∧ e h

Orbital ur uθ h

Julian Date (JD): Days from Jan 01, 4713BC, 12:00 noon

Modified Julian Date (MJD): JD-2,400,000.5

J2000=JD 2,451,545.0 Epoch 1 Jan 2000 12:00 TT

J2000=MJD 51,544.5 Epoch 1 Jan 2000 12:00 TT


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