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16.346 Astrodynamics
Fall 2008
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Lecture 27
Variation of Parameters
Chapter 10
Lagrange’s Variational Methods for Linear Equations
Consider the equation
d2 y
+ y = sec t
dt2
dy1
= y2
dt
=⇒
dy2
+ y1 = sec t
dt
� � �
�� � �
�
d y1
0 1
y1
0
=
+
−1 0
sec t
y2
dt y2
=⇒
which is equivalent to
dy
= Fy + g
dt
where
�
y
y= 1
y2
�
�
0 1
and F =
−1 0
Now the Wronskian matrix W
�
sin t
W =
cos t
�
− cos t
sin t
�
�
0
and g =
sec t
satisfies
�
dW
= FW
dt
and the solution of the homogeneous equation is
� �
c
yh = Wc where c = 1
c2
We now seek a solution of the general equation
dy
= Fy + g of the form y = Wc(t)
dt
Substitute and obtain
dc
dW
c+W
= FWc + g which reduces to
dt
dt
W
dc
=g
dt
Hence
dc
= W−1 g
dt
which is solved by quadratures to obtain
�
��
�
dc1
dc
sin t
cos t
0
=
or
= 1 and
− cos t sin t
sec t
dt
dt
dc2
= tan t
dt
Hence
c1 (t) = t + c1
and c2 (t) = log(sec t + tan t) + c2
so that the general solution is simply
y = Wc(t)
16.346 Astrodynamics
or y(t) = c1 (t) sin t − c2 (t) cos t
Lecture 27
Derivation of the Variational Equations
⎡ ⎤ ⎡
r⎥ ⎢
O
d ⎢
⎢ ⎥=⎢
µ
dt ⎣ ⎦ ⎣
v
− 3I
r
⎤⎡ ⎤ ⎡ ⎤
I ⎥⎢ r ⎥
0⎥
⎥⎢ ⎥ + ⎢
⎦⎣ ⎦ ⎣ ⎦
ad
O
v
Two-Body Motion:
ds
= F s + η
dt
⇐⇒
where
�
r(t, α)
s = s(t, α) =
v(t, α)
∂s
= Fs
∂t
Disturbed Motion:
ds
= Fs + η
dt
Seek solutions of the form s = s(t, α1 , α2 , α3 , α4 , α5 , α6 )
where, for example,
α
T
= [Ω i
ω
a
e
λ = −nτ ]
�
and
�
0
η =
ad
�
Differentiate
ds
∂s
∂s d α
=
+
= Fs + η
dt
∂t
∂ α dt
Since
then
=⇒
∂s d α
=η
∂ α dt
∂s ∂ α
∂ α ∂s
=
=I
∂ α ∂s
∂s ∂ α
�∂α ∂α � � � ∂α
∂α
∂ α ∂s d α
0
=
η =
=
a
ad
∂s ∂ α dt
∂s
∂r
∂v
∂v d
so that
∂α
dα
=
a
dt
∂v d
� � �
�� �
x = f (ξ, η)
dx
fξ fη
dξ
=
y = g(ξ, η)
dy
gξ gη
dη
��
�� �
� � �
F x Fy
dx
dx
fξ fη
=
dy
gξ gη
Gx Gy
dy
� � �
�� �
ξ = F (x, y)
dξ
Fx Fy
dx
=
η = G(x, y)
dη
G x Gy
dy
�
��
�
fξ fη
F x Fy
=⇒
=I
gξ gη
Gx Gy
Variation of the Classical Elements
�2 1�
µ
−
= v2 = v · v = v T v
r
a
h = r × v = Sr v
16.346 Astrodynamics
=⇒
µ ∂a
= 2v T
a2 ∂v
=⇒
2a2
da
=
v · ad
dt
µ
∂h
∂v
= Sr
= Sr I = Sr
∂v
∂v
=⇒
dh
= r × ad
dt
=⇒
Lecture 27
h2 = h T h
=⇒
2h
∂h
= 2h T Sr
∂v
dh
= ih · r × ad = ih × r · ad = r i θ · ad
dt
=⇒
or, alternately,
h2 = (r × v) · (r × v) = r T rv T v − r T vr T v
p=
h2
= a(1 − e2 )
µ
µ
=⇒
=⇒
2µae
1
dh
= r T (rv T − vr T )ad
h
dt
de
da
dh
− 2h
= µ(1 − e2 )
dt
dt
dt
de
= ad × (r × v) + (ad × r) × v
dt
Variation of i and Ω
From Page 84 in the textbook
h = h i h = h(sin Ω sin i i x − cos Ω sin i i y + cos i i z )
(2.6)
dh
dΩ
di
dh
= h sin i
in − h im +
i
dt
dt
dt
dt h
Then
where
in = cos Ω ix + sin Ω iy
im = ih × in = − sin Ω cos i ix + cos Ω cos i iy + sin i iz
(2.5)
(2.8)
Hence
1
r sin θ
dΩ
=
in × r · ad =
i ·a
dt
h sin i
h sin i h d
1
r cos θ
di
= − im × r · ad =
ih · ad
dt
h
h
where θ = ω + f is the argument of latitude.
Variation of the true anomaly f
r(1 + e cos f ) =
h2
µ
=⇒
re sin f
∂f
∂e
2h ∂h
= r cos f
−
∂v
∂v
µ ∂v
µ
∂f
∂e
r · v ∂h h T
re sin f = r · v =⇒ re cos f
= −r sin f
+
+ r
h
∂v
∂v
µ ∂v µ
Multiply the first by cos f , the second by sin f and add to obtain
From Eq. (3.29)
reh
∂f
∂h
= (p cos f ) r T − (p + r) sin f
∂v
∂v
16.346 Astrodynamics
to be used in
Lecture 27
h
∂f
df
= 2+
a
dt
r
∂v d
Variation of ω
in = cos Ω i x + sin Ω i y
=⇒
cos θ = in · ir = cos Ω (ix · ir ) + sin Ω (iy · ir )
Then
− sin θ
∂θ
∂Ω
= [− sin Ω (ix · ir ) + cos Ω (iy · ir )]
∂v
∂v
since
=⇒
∂θ
∂Ω
= − cos i
∂v
∂v
ix · ir = cos Ω cos θ − sin Ω sin θ cos i
iy · ir = sin Ω cos θ + cos Ω sin θ cos i
This gives the perturbative derivative of θ , i.e., the change in θ due to the change in in
from which the angle θ is measured. The total time rate of change of θ is the sum
∂θ
∂θ
h
dΩ
dθ
=
+
ad = 2 − cos i
dt
∂t
∂v
r
dt
Since θ = ω + f , then
∂f
dΩ
dω
=−
ad − cos i
dt
∂v
dt
Gauss’ form of Lagrange’s variational equations in polar coordinates
dΩ
=
dt
di
=
dt
dω
=
dt
da
=
dt
de
=
dt
df
=
dt
r sin θ
a
h sin i dh
r cos θ
adh
h
1
r sin θ cos i
[−p cos f adr + (p + r) sin f adθ ] −
adh
he
h sin i
�
p
2a2 �
e sin f adr + adθ
h
r
1
{p sin f adr + [(p + r) cos f + re]adθ }
h
h
1
+
[p cos f adr − (p + r) sin f adθ ]
2
eh
r
Gauss’ form of the variational equations in tangential-normal coordinates
dω
dt
da
dt
de
dt
df
dt
�
�
� r sin θ cos i
1 �
r
2 sin f adt + 2e + cos f adn −
adh
ev
a
h sin i
2a2 v
=
a
µ
dt
�
1�
r
=
2(e + cos f )adt − sin f adn
v
a
�
�
�
h
1 �
r
= 2−
2 sin f adt + 2e + cos f adn
ev
a
r
=
16.346 Astrodynamics
Lecture 27