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16.346 Astrodynamics
Fall 2008
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Lecture 13 Gauss' Method for the Time-Constrained BVP
Lagrange’s equations for the boundary-value problem
3
√
µ (t2 − t1 ) = 2a 2 (ψ − sin ψ cos φ)
r1 + r2 = 2a(1 − cos ψ cos φ)
√
r1 r2 cos 12 θ = a(cos ψ − cos φ)
# 7.3
(1)
(2)
(3)
Gauss’s Equation for the Semimajor Axis
Eliminate cos φ between (2) and (3):
2 sin2 ψ
1
=
√
a
r1 + r2 − 2 r1 r2 cos
1
2
θ cos ψ
When ψ and θ are very small (of the order of 2 or 3 degrees), then r1 and r2 will have
almost the same value. The denominator will be determined as the difference between two
almost equal terms resulting in a severe loss of accuracy. To prevent this, Gauss wrote
sin2 ψ
1
= √
a
2 r1 r2 cos 12 θ (� + sin2
1
2
ψ)
(4)
where � is defined as
�
�=
�
r2
r1
+
r1
r2
1
−
1
2
4 cos 2 θ
The problem of subtracting two almost equal quantities still exists but Gauss had a different
method for calculating � which avoided any subtraction:
� � 14
sin2 14 θ + tan2 2ω
r2
1
where tan( 4 π + ω) =
�=
1
r1
cos 2 θ
An alternate method, which does not require any inverse trigonometric function is to
express
r2 = r1 (1 + �)
The quantity � is simply the fractional part resulting when r2 is divided by r1 (assuming,
of course, that r2 exceeds r1 ). The result is
tan 2ω = �
2
16.346 Astrodynamics
r
r2
+ 2
r1
r1
1 2
4��
�
2+
Lecture 13
r2
r1
�
Gauss’s Time Equation
Eliminate cos φ between (1) and (3):
�
√
2 r1 r2 cos
µ
(t − t1 ) = 2ψ − sin 2ψ +
a3 2
a
1
2
θ
sin ψ
(5)
Next using Eq. (4) for 1/a he obtained
√
µ(t2 − t1 ) sin3 ψ
m sin3 ψ
sin3 ψ
≡
=
2ψ
−
sin
2ψ
+
3
3
3
√
� + sin2 12 ψ
(2 r1 r2 cos 12 θ)
2 (� + sin2 12 ψ)
2
(� + sin2 12 ψ)
2
√
m=
where he defined
which requires that
√
µ(t2 − t1 )
(2 r1 r2 cos
1
2
3
θ) 2
0 < θ < 180 ◦ .
Finally, Gauss defined
m2
y =
� + sin2
2
1
2
ψ
so that the time equation (5) can be written as
y2
2ψ − sin 2ψ
y3
+ 2
m× 3 =
m
m
sin3 ψ
or
y 3 − y 2 = m2
2ψ − sin 2ψ
sin3 ψ
which are Gauss’ equations, to be solved for y and ψ .
The Orbital Parameter and the Significance of y
From Lecture 9 on Page 3
2r1 r2 sin2
sin φ
sin φ
p=
p =
×
sin ψ m
sin ψ
c
Then from
a=
we have
√
2 r1 r2 cos
1
2
1
2
θ (� + sin2
θ
1
2
2r1 r2 sin2 12 θ
r1 r2 sin2 12 θ
sin φ
=
×
=
sin ψ
2a sin ψ sin φ
a sin2 ψ
ψ)
and
sin2 ψ
√
2m2 r1 r2 cos
a sin ψ =
y2
2
1
2
θ
m2
y =
� + sin2
2
where
m2 =
from which
h2
r12 r22 y 2 sin2 θ
=
µ
µ(t2 − t1 )2
1
2 h(t2 − t1 )
1
2 r1 r2 sin θ
16.346 Astrodynamics
= y=
Area of sector
Area of triangle
Lecture 13
ψ
µ(t2 − t1 )2
√
(2 r1 r2 cos 12 θ)3
so that
p=
1
2
Changing the independent variable from ψ to x = sin2
Define
Q=
3Q sin2 ψ cos ψ + sin3 ψ
dx
= sin
dψ
so that
1
2
ψ
2ψ − sin 2ψ
sin3 ψ
then
Now
1
2
dQ
= 2 − 2 cos 2ψ = 4 sin2 ψ
dψ
ψ cos
3Q cos ψ +
1
2
ψ=
1
sin ψ
2
1
dQ
sin2 ψ
=4
2
dx
Since
cos ψ = 1 − 2 sin2 12 ψ = 1 − 2x
sin2 ψ = 4 sin2 12 ψ(1 − sin2 12 ψ) = 4x(1 − x)
2x(1 − x)
then
dQ
= 4 − (3 − 6x)Q
dx
4
(1 + q1 x + q2 x2 + q3 x3 + q4 x4 + · · ·)
3
Substitute and equate like powers of x to obtain:
Write
Q=
q1 =
6
5
4
Q(x) = F (3, 1;
3
5
2
q2 =
8
q
7 1
q3 =
10
q
9 2
q4 =
12
q
11 3
etc.
resulting in
F (3, 1;
5
2
; x) =
�
4
�
6
6 ·
8 2 6 ·
8 ·
10 3 6 ·
8 ·
10 ·
12 4
; x) = 1 + x +
x
+
x
+
x
+ · · · 3
5
5 ·
7
5 ·
7 ·
9
5 ·
7 ·
9 ·
11
1
γ1 x
1 −
γ2 x
1−
γ x
1 −
3
1 − ..
⎧
(n + 2)(n + 5) ⎪
⎪
⎪
⎨ (2n + 1)(2n + 3) γn =
⎪
n(n − 3)
⎪
⎪
⎩
(2n + 1)(2n + 3) n odd
n even
.
a. The series converges for −1 < x < 1.
b. The continued fraction converges for −∞ < x < 1.
Note: The function F (α, β; γ; x) is Gauss’ Hypergeometric Function which we will
exam in some detail in the next Lecture.
16.346 Astrodynamics
Lecture 13
The Universal Form of Gauss’ Method
We can extend the definition of x so that
⎧
2 1
⎪
⎨ sin 4 (E2 − E1 )
x=
0
⎪
⎩
− sinh2 14 (H2 − H1 )
ellipse
parabola
hyperbola
The range of x is −∞ < x < 1 . The series representation of Q(x) will not converge when
x < −1. However the continued fraction does converge over the full range.
Possible Algorithm
Gauss’ equations are:
y2 =
m2
�+x
and y 3 − y 2 = m2 Q(x)
in terms of x . The following is a recursive algorithm for the solution:
1. Set x = 0
2. Solve of cubic y 3 − y 2 = m2 Q(x)
Note: Solution of the cubic:
3. Obtain new x from
y =1+
4
3
sinh2
x=
1
3
z
where
3
2
sinh z =
�
3m2 Q
m2
−�
y2
and repeat until the process converges.
Gauss’ Successive Substitution Algorithm
√
µ (t2 − t1 )
1
r
+ r
2
−
� =
√ 1
1
4 r
1 r2 cos 2 θ 2
r 1 , r2 , θ ,
1. Given
2. Compute
3. Initialize
0<θ<π
µ(t2 − t1 )2
m
=
√
(2 r
1 r2 cos 12 θ)3
2
and
x=0
4. Calculate
2
35
ξ(x) =
1+
2
35
h=
5
6
m
+ � + ξ(x)
5. Solve the cubic
y 3 − y 2 − hy −
h
=0
9
x=
16.346 Astrodynamics
x
4
99
1−
1−
x
70
143 x
18
195 x
108
255
x
1 − ...
and repeat until x no longer changes.
m2
−�
y2
7. Calculate the orbital elements:
8r r y 2 x(1 − x)(1 + cos θ)
1
= 1 2
a
µ(t2 − t1 )2
6. Determine new
40
63
x−
1−
2
and
x2
Lecture 13
1−
r12 r22 y 2 sin2 θ
p=
µ(t2 − t1 )2
Avoiding the Continued Fraction When ψ is Not Small
Instead of the continued fraction (which we shall learn more about in the next lecture),
we can use the closed form expression
sin3 ψ −
ξ(ψ) = (2ψ − sin 2ψ)(1 − 65 sin2 12 ψ)
9
10 (2ψ − sin 2ψ)
√
Since x = sin2 12 ψ , then ψ = 2 arcsin( x ).
“The numerator of this expression is a quantity of the seventh order, the denominator of
the third order, and ξ , therefore, of the fourth order, if ψ is regarded as a quantity of
the first order. Hence it is inferred that this formula is not suited to the exact numerical
computation of ξ when ψ does not denote a very considerable angle.”
3
4
Karl Friedrich Gauss
Solving the Cubic Equation
Pages 321 & 54
The solution of the cubic equation
y
3 − y
2 − hy −
19 h = 0 using the method developed on Page 321 of your textbook, is
√
y = 13 (1 + w 1 + 3h )
where w is the solution of
w
3 − 3w = 2
Note:
1 + 6h
3
(1 + 3h) 2
= 2b
Barker’s Equation is w3 + 3w = 2b
We must address the cases b < 1 and b ≥ 1 separately:
b < 1 Write w = 2 cos
23 x = 2(1 − 2 sin2 13 x) and b = cos 2x = 1 − 2 sin2 x
Then the cubic equation becomes
4 cos2 23 x = 3 cos
23 x = cos 2x which is an iden­
tity for cosine functions. Hence:
w = 2 cos(
13 arccos b)
b ≥ 1 Define w = 2 cosh
23 x = 2(1 + 2 sinh2 x)
and
b = cosh 2x = 1 + 2 sinh2 x
Then the cubic equation becomes 4 cosh2 23 x − 3 cosh
23 x = cosh 2x
identity for hyperbolic cosines. Hence:
w = 2 cosh(
13 arccosh b)
16.346 Astrodynamics
Lecture 13
which is an