basics of orbital mechanics i - ocw...
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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
Basics of Orbital Mechanics I – p. 2/20
Two-Body Problem: Inertial Formulation
r1
r2
S1
m1, m2 point masses;S1 inertial reference
Basics of Orbital Mechanics I – p. 3/20
Two-Body Problem: Inertial Formulation
r1
r2
S1
F12
F21
m1, m2 point masses;S1 inertial reference
F12, F21 gravitational attraction
Basics of Orbital Mechanics I – p. 3/20
Two-Body Problem: Inertial Formulation
r1
r2
S1
F12
F21
P1
P2m1, m2 point masses;S1 inertial reference
F12, F21 gravitational attraction
P1, P2 other forces (“Perturbations”)
Basics of Orbital Mechanics I – p. 3/20
Two-Body Problem: Inertial Formulation
r1
r2
S1
F12
F21
P1
P2m1, m2 point masses;S1 inertial reference
F12, F21 gravitational attraction
P1, P2 other forces (“Perturbations”)
m1 r1 =G m1 m2
|r2 − r1|3(r2 − r1) + P1
m2 r2 = − G m1 m2
|r2 − r1|3(r2 − r1) + P2
Basics of Orbital Mechanics I – p. 3/20
Two-Body Problem: Inertial Formulation
r1
r2
S1
F12
F21
P1
P2m1, m2 point masses;S1 inertial reference
F12, F21 gravitational attraction
P1, P2 other forces (“Perturbations”)
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)
Basics of Orbital Mechanics I – p. 3/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
G Center of mass ofm1, m2
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
r′1 = r1 − rG = m2
m1+m2(r1 − r2)
r′2 = r2 − rG = m1
m1+m2
(r2 − r1) = −m1
m2
r′1
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
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
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
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 =
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
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) ⇒
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
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) ⇒
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:COM Formulation
G
r1
r2
S1
r′1
r′2 S0
G Center of mass ofm1, m2
(m1 + m2) rG = m1 r1 + m2 r2
S0 Non-rotating reference, origin inG
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
)
Basics of Orbital Mechanics I – p. 4/20
Two-Body Problem:Primary Formulation
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
r =[
G m2
|r2−r1|3(r2 − r1) + P1
m1
]−
[− G m1
|r2−r1|3(r2 − r1) + P2
m2
]⇒
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
m1
m2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
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
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
F12
m1
m2
P1
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
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
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
F12
F21
m1
m2
P1
P2
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
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
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem:Primary Formulation
r
S2
F12
−F21
m1
m2
P1
−P2
m1
S2 Non-rotatingNon-inertial, origin in m2
r = r1 − r2 Relativeposition vector
COM motion: (m1 + m2) rG = P1 + P2
Relativemotion: r = r1 − r2
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
Basics of Orbital Mechanics I – p. 5/20
Two-Body Problem: Formulations
2
1
G
Inertial motion
2
1
r′2
r′1
Relative toCOM
2
1
G
r
Relative toPrimary
Basics of Orbital Mechanics I – p. 6/20
Two-Body Problem: Simplifications
r = −G (m2 + m1)r
|r|3︸ ︷︷ ︸Kepler Problem
+P1
m1− P2
m2︸ ︷︷ ︸Perturbation
Basics of Orbital Mechanics I – p. 7/20
Two-Body Problem: Simplifications
r = −G (m2 + m1)r
|r|3︸ ︷︷ ︸Kepler Problem
+P1
m1− P2
m2︸ ︷︷ ︸Perturbation
Isolated System: P1 = P2 = 0 → Kepler problem
Basics of Orbital Mechanics I – p. 7/20
Two-Body Problem: Simplifications
r = −G (m2 + m1)r
|r|3︸ ︷︷ ︸Kepler Problem
+P1
m1− P2
m2︸ ︷︷ ︸Perturbation
Isolated System: P1 = P2 = 0 → Kepler problem
Small mass:m1 ≪ m2 → G(m2 + m1) ≃ G m2 = µ Gravitational constant
Basics of Orbital Mechanics I – p. 7/20
Two-Body Problem: Simplifications
r = −G (m2 + m1)r
|r|3︸ ︷︷ ︸Kepler Problem
+P1
m1− P2
m2︸ ︷︷ ︸Perturbation
Isolated System: P1 = P2 = 0 → Kepler problem
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
Basics of Orbital Mechanics I – p. 7/20
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
Basics of Orbital Mechanics I – p. 8/20
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
Basics of Orbital Mechanics I – p. 8/20
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
Point masses or spherical symmetry
Basics of Orbital Mechanics I – p. 8/20
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
Point masses or spherical symmetry
Potential force: F = −∇V V = −µr
Basics of Orbital Mechanics I – p. 8/20
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
Point masses or spherical symmetry
Potential force: F = −∇V V = −µr
Central force:F = −f(r) r ⇒ r ∧ r = 0 ⇒ r ∧ v = Const
Basics of Orbital Mechanics I – p. 8/20
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
Point masses or spherical symmetry
Potential force: F = −∇V V = −µr
Central force:F = −f(r) r ⇒ r ∧ r = 0 ⇒ r ∧ v = Const
Solution: Search forintegrals(conserved magnitudes)
Basics of Orbital Mechanics I – p. 8/20
Integrals of Motion
Potential: Specific Energy conservationv2
2− µ
r=
E
m
Basics of Orbital Mechanics I – p. 9/20
Integrals of Motion
Potential: Specific Energy conservationv2
2− µ
r=
E
m
Central: Specific Angular Momentum r ∧ v = h
Basics of Orbital Mechanics I – p. 9/20
Integrals of Motion
Potential: Specific Energy conservationv2
2− µ
r=
E
m
Central: Specific Angular Momentum r ∧ v = h
h
rv
⇒ Plane motion
Color code:h Constant magnituder Fast variable
Basics of Orbital Mechanics I – p. 9/20
Integrals of Motion
Potential: Specific Energy conservationv2
2− µ
r=
E
m
Central: Specific Angular Momentum r ∧ v = h
h
rv
⇒ Plane motion
F ∝ − 1r2 → Laplace/Runge-Lenz vector −r
r− h ∧ v
µ= e
Basics of Orbital Mechanics I – p. 9/20
Integrals of Motion
Potential: Specific Energy conservationv2
2− µ
r=
E
m
Central: Specific Angular Momentum r ∧ v = h
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
)
Basics of Orbital Mechanics I – p. 9/20
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
Basics of Orbital Mechanics I – p. 10/20
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
h = r ∧ v ⊥ e = −r
r− h ∧ v
µh
e
rv
Basics of Orbital Mechanics I – p. 10/20
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
h = r ∧ v ⊥ e = −r
r− h ∧ v
µh
e
rv
E =µ2m
2h2
(e2 − 1
)
Basics of Orbital Mechanics I – p. 10/20
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
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
Basics of Orbital Mechanics I – p. 10/20
Trajectory
r
m
θ
eM
Polar coordinates in the plane of motion (⊥ h)
Main axis: direction ofe
Basics of Orbital Mechanics I – p. 11/20
Trajectory
r
m
θ
eM
Polar coordinates in the plane of motion (⊥ h)
Main axis: direction ofe
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
Basics of Orbital Mechanics I – p. 11/20
Trajectory
r
m
θ
eM
θ : True Anomaly
Polar coordinates in the plane of motion (⊥ h)
Main axis: direction ofe
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
Basics of Orbital Mechanics I – p. 11/20
Trajectory
r
m
θ
eM
θ : True Anomaly
Polar coordinates in the plane of motion (⊥ h)
Main axis: direction ofe
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
e Eccentricity→ e Eccentricity vector
e = 0 Circlee < 1 Ellipsee = 1 Parabolae > 1 Hyperbola
Basics of Orbital Mechanics I – p. 11/20
Trajectory
r
m
θ
eM
θ : True Anomaly
Polar coordinates in the plane of motion (⊥ h)
Main axis: direction ofe
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
e Eccentricity→ e Eccentricity vector
e = 0 Circlee < 1 Ellipsee = 1 Parabolae > 1 Hyperbola
p = h2/µ Parameteror semilatus rectum: radius at 90o
Basics of Orbital Mechanics I – p. 11/20
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
Basics of Orbital Mechanics I – p. 12/20
Energy
EnergyE is related toh ande as:E
m=
µ2
2h2
(e2 − 1
)
Basics of Orbital Mechanics I – p. 13/20
Energy
EnergyE is related toh ande as:E
m=
µ2
2h2
(e2 − 1
)
From the trajectory we can identify:µ(e2 − 1)
h2= −1
a
Basics of Orbital Mechanics I – p. 13/20
Energy
EnergyE is related toh ande as:E
m=
µ2
2h2
(e2 − 1
)
From the trajectory we can identify:µ(e2 − 1)
h2= −1
a
Therefore:
E
m= − µ
2aOrbit size dependsonly onE
Basics of Orbital Mechanics I – p. 13/20
Energy
EnergyE is related toh ande as:E
m=
µ2
2h2
(e2 − 1
)
From the trajectory we can identify:µ(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
Basics of Orbital Mechanics I – p. 13/20
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
Basics of Orbital Mechanics I – p. 14/20
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
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
Basics of Orbital Mechanics I – p. 14/20
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
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.
Basics of Orbital Mechanics I – p. 14/20
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
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=
õ
a3
Basics of Orbital Mechanics I – p. 14/20
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
Basics of Orbital Mechanics I – p. 16/20
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
rθ =h
r=
µ
h(1 + e cos θ)
(µ
h=
õp
)
Basics of Orbital Mechanics I – p. 16/20
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
rθ =h
r=
µ
h(1 + e cos θ)
(µ
h=
õp
)
r = drdθ · θ = h2/µ
(1+e cos θ)2
e sin θ · µ2
h3(1 + e cos θ)2 =
µ
he sin θ
Basics of Orbital Mechanics I – p. 16/20
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
rθ =h
r=
µ
h(1 + e cos θ)
(µ
h=
õp
)
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
Basics of Orbital Mechanics I – p. 16/20
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′
Basics of Orbital Mechanics I – p. 17/20
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′
Æ : Eccentric Anomaly
M = n (t − τ) : Mean Anomaly
τ : Time at pericenterP
Basics of Orbital Mechanics I – p. 17/20
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′
Æ : Eccentric Anomaly
M = n (t − τ) : Mean Anomaly
τ : Time at pericenterP
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
Basics of Orbital Mechanics I – p. 17/20
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′
Æ : Eccentric Anomaly
M = n (t − τ) : Mean Anomaly
τ : Time at pericenterP
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
Basics of Orbital Mechanics I – p. 17/20
Kepler Equation
n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)
True Anomalyθ ↔ ÆEccentric Anomaly
Basics of Orbital Mechanics I – p. 18/20
Kepler Equation
n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)
True Anomalyθ ↔ ÆEccentric Anomaly
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
Basics of Orbital Mechanics I – p. 18/20
Kepler Equation
n (t − τ) = Æ− e sin Æ = M ( · · · + k 2π)
True Anomalyθ ↔ ÆEccentric Anomaly
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
Basics of Orbital Mechanics I – p. 18/20
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]
Basics of Orbital Mechanics I – p. 19/20
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
Basics of Orbital Mechanics I – p. 20/20