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16.346 Astrodynamics
Fall 2008
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Lecture 9
Lambert's Theorem and the Lagrange Time Equation
The Theorem of Johann Heinrich Lambert
√
#6.6
µ (t2 − t1 ) = F (a, r1 + r2 , c)
Developing Lagrange’s Equations
3
√
• Kepler’s Equation
µ (t − τ ) = a 2 (E − e sin E)
3
√
√
µ (t2 − τ ) − µ (t1 − τ ) = a 2 [(E2 − e sin E2 ) − (E1 − e sin E1 )
3
√
µ (t2 − t1 ) = 2a
2 [ 12 (E2 − E1 ) − e sin 12 (E2 − E1 ) cos 12 (E1 + E2 )]
• Equation of Orbit
r = a(1 − e cos E)
r1 + r2 = a(1 − e cos E1 ) + a(1 − e cos E2 )
= 2a[1 − e cos 12 (E2 − E1 ) cos 12 (E1 + E2 )]
• Chord
c=
�
r12 + r22 − 2r1 r2 cos θ
c2 = r12 + r22 + 2r1 r2 (1 − 2 cos2 12 θ)
= (r1 + r2 )2 − 4r1 r2 cos2 12 θ
Recall the relations between eccentric and true anomalies.
�
�
√
√
Using
r sin 12 f = a(1 + e) sin 12 E
r cos 12 f = a(1 − e) cos
√
E
√
r1 r2 cos 12 (f2 − f1 ) = r1 r2 (cos 12 f2 cos 12 f1 + sin 12 f2 sin 12 f1 )
√
√
√
√
= r2 cos 12 f2 r1 cos 12 f1 + r2 sin 12 f2 r1 sin 12 f1
r1 r2 cos 12 θ =
√
1
2
= a(1 − e) cos 12 E2 cos 12 E1 + a(1 + e) sin 12 E2 sin 12 E1
= a cos 12 (E2 − E1 ) − ae cos 12 (E2 + E1 )
• Lagrange Parameters
ψ = 12 (E2 − E1 )
cos φ = e cos 12 (E1 + E2 )
• Lagrange Equations
√
Note: We also have
16.346 Astrodynamics
3
µ(t2 − t1 ) = 2a 2 (ψ − sin ψ cos φ)
r1 + r2 = 2a(1 − cos ψ cos φ)
c = 2a sin ψ sin φ
√
r1 r2 cos
1
2
θ = a(cos ψ − cos φ)
Lecture 9
Lagrange’s Time Equation
Define
Then
α=φ+ψ
β =φ−ψ
ψ = 12 (α − β)
=⇒
φ = 12 (α + β)
3
√
µ(t2 − t1 ) = 2a 2 (ψ − sin ψ cos φ)
3
= a
2 [α − β − 2 sin 12 (α − β) cos 12 (α + β)]
3
= a
2 [(α − sin α) − (β − sin β)]
Also:
r1 + r2 + c = 2a[1 − cos(φ + ψ)] = 2a(1 − cos α) = 4a sin2 12 α
r1 + r2 − c = 2a[1 − cos(φ − ψ)] = 2a(1 − cos β) = 4a sin2 12 β
Hence, Lagrange’s analytic form of Lambert’s theorem is
�
µ
(t − t1 ) = (α − sin α) − (β − sin β)
a3 2
sin2
where
1
2
α=
s
2a
sin2
1
2
β=
s−c
2a
in terms of the semiperimeter of the triangle:
s=
1
2
(r1 + r2 + c)
Euler’s Equation for Parabolic Orbits
Since
� α
3
�
3
3
α
5
2
2
a (α − sin α) = a
+
+ ···
3!
5!
and
�
� s
� 12
1 � s �
32
s
α = 2 arcsin
=2
+
+
· · · 2a
2a
3 2a
√
� 1 �
3
2
3
a
2 (α − sin α) = s 2 + O
2
Then
3
a
√
� 1 �
3
3
2
2
2
a
(β − sin β) = (s − c)
+ O
2
Similarly,
3
a
√
3
2 3
√
[s 2 ∓ (s − c)
2 ]
Therefore:
µ (t2 − t1 ) =
3
◦
The choice of sign is minus for θ < 180 and plus for θ > 180 ◦ .
Alternately,
3
3
√
6 µ (t2 − t1 ) = (r1 + r2 + c) 2 ∓ (r1 + r2 − c) 2
16.346 Astrodynamics
Lecture 9
The Orbital Parameter
From Page 1 of Lecture 8
� p �2
p
− 2D
+1=0
pm
pm
where
D=
r r
r1 + r2
− 1 2 cos2
c
ac
1
2
θ
Use the Lagrange equations
sin2 ψ + sin2 φ
2a(1 − cos ψ cos φ) a2 (cos ψ − cos φ)2
=
−
2a sin ψ sin φ
2a2 sin ψ sin φ
2 sin ψ sin φ
�
sin2 ψ − sin2 φ
D2 − 1 =
2 sin ψ sin φ
D=
so that
⎧
sin 12 (α + β)
sin φ
⎪
⎪
⎪
⎨ sin ψ = sin 1 (α − β)
p
2
=
⎪
pm
sin 12 (α − β)
sin ψ
⎪
⎪
=
⎩
sin φ
sin 12 (α + β)
Note: The orbital parameter equations were not developed by Lagrange.
Skewed-Velocity Components in Terms of the Semimajor Axis
From
p=
#6.8
sin φ
r r
sin φ
pm =
× 1 2 (1 − cos θ)
sin ψ
sin ψ
c
and
α=φ+ψ
c = 2a sin ψ sin φ
√
we obtain
r1 r2 cos 12 θ = a(cos ψ − cos φ)
β =φ−ψ
�
�
√
c µp
µ
µ
sin φ
vc =
=
=
(cot 12 β + cot 12 α)
r1 r2 sin θ
a cos ψ − cos φ
4a
�
�
�
µ 1 − cos θ
µ
µ
sin ψ
vρ =
=
=
(cot 12 β − cot 12 α)
p sin θ
a cos ψ − cos φ
4a
Hence:
��
v1 =
µ
µ
+
−
2(s − c) 4a
16.346 Astrodynamics
�
µ
µ
−
2s 4a
�
��
ic +
µ
µ
−
−
2(s − c) 4a
Lecture 9
�
µ
µ
−
2s 4a
�
ir1