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16.346 Astrodynamics
Fall 2008
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Lecture 11
Hyperbolic Orbits
Hyperbolic Orbits
x = a sec ζ
y = b tan ζ
⇐⇒
y2
x2
−
=1
b2
a2
r = a − ex
=⇒
r = a(1 − e sec ζ)
To understand the geometrical significance of ζ , write the equation of orbit as
r + re cos f = a(1 − e2 )
a(1 − e sec ζ) + re cos f = a(1 − e2 )
−a sec ζ + r cos f = −ae = F C
positive negative
and relate the terms of the last equation to the diagram
Fig. 4.12 from An Introduction to
the Mathematics and Methods of
Astrodynamics. Courtesy of AIAA.
Used with permission.
Also
x = a cosh H
⇐⇒
y = b sinh H
Then
tan
1
2
f=
y2
x2
−
=1
a2
b2
e+1
tan
e−1
r = a − ex
=⇒
1
2
ζ
or
tan
1
4
or
1
2
f=
r = a(1 − e cosh H)
e+1
tanh
e−1
1
2
H
and the analogs of Kepler’s equation are
N = e tan ζ − log tan( 12 ζ +
where
16.346 Astrodynamics
N=
π)
µ
(t − τ )
(−a)3
Lecture 11
N = e sinh H − H
Lagrange’s Equations for Hyperbolic Orbits
For hyperbolic orbits, ψ and φ are defined as
1
2
ψ=
(H2 − H1 )
1
2
cosh φ = e cosh
(H1 + H2 )
and the basic equations are
3
√
µ(t2 − t1 ) = 2(−a) 2 (sinh ψ cosh φ − ψ)
r1 + r2 = 2a(1 − cosh ψ cosh φ)
√
c = −2a sinh ψ sinh φ
r1 r2 cos
= a(cosh ψ − cosh φ)
1
2θ
The Lagrange parameters are defined as for the ellipse. Then
3
√
µ(t2 − t1 ) = (−a) 2 [(sinh α − α) − (sinh β − β)]
where
sinh2
1
2
α=−
s
2a
sinh2
1
2
β=−
s−c
2a
Hyperbolic Injection Velocity
Page 294
Recall to the velocity vector in terms of the semimajor axis :
µ
µ
µ
µ
µ
µ
µ
µ
+
−
ic +
−
−
ir1
−
−
v1 =
2(s − c) 4a
2s 4a
2(s − c) 4a
2s 4a
Then
lim
θ= const.
r2 →∞
Now
1 + cos θ
= cos2
2
1
2
θ=
i c = i∞
s(s − c)
r 1 r2
so that
lim
θ= const.
r2 →∞
and
lim
θ= const.
r2 →∞
1
=0
s
lim
θ= const.
r2 →∞
(s − c) =
2 1
−
v2 = µ
r
a
Therefore, in the limit:
v1 = (D +
where
D=
v2
v◦2
+ ∞
1 + cos θ
4
θ= const.
r2 →∞
s
=1
r2
1
s(s − c)
=
×1×
r1 r2
r1
1
2
=⇒
2
v∞
=
v∞ ) i ∞ + (D −
and
1
2
v∞ ) i r1
Note: v◦ is the circular speed at the pericenter radius r1 , i.e., v◦2 =
16.346 Astrodynamics
Lecture 11
µ
−a
cos θ = ir1 · i∞
µ
.
r1
lim
θ= const.
r2 →∞
r1
(1 + cos θ)
2
Also, from the vis-viva integral:
lim
(s − c)
Injection from Pericenter of a Hyperbolic Orbit
For injection from pericenter of the hyperbola
0 = ir1 · v1 = (D +
1
2
v
∞ ) cos θ + (D −
21 v
∞ )
so that
(1 + cos θ)D =
1
2
v∞ (1 − cos θ)
Hence,
v1 =
v∞
(i − cos θ ir1 )
1 + cos θ ∞
To determine cos θ , first square both sides of the equation for D
2
(1 − cos θ)2
(1 + cos θ)2 D2 =
14 v∞
Then, from the previous equation for D ,
D=
v
2
v
◦2
+ ∞
1 + cos θ
4
we also have
2
(1 + cos θ)2 D2 = v
◦2 (1 + cos θ) + 14 v∞
(1 + cos θ)2
Therefore:
2
2
v
◦2 (1 + cos θ) + 14 v
∞
(1 + cos θ)2 =
41 v
∞
(1 − cos θ)2
Then
cos θ = cos( 12 π + ν) = − sin ν
=⇒
sin ν =
1
v2
1+ ∞
v◦2
where ν is the angle between the hyperbolic asymptote and the minor axis.
Out-of-Plane Injection from Pericenter of a Hyperbolic Orbit
The vector v∞ = v∞ i∞ is in the orbital transfer plane in which both P1 and P2 lie. The
orientation of this plane can be specified by the vector iN defined as iN = Unit(r1 × v∞ ).
where r1 is the vector position of point P1
Having determined cos θ from
2
cos θ = 0
v
◦2 (1 + cos θ) + v
∞
or
v
2
cos θ = −
2 o 2
v
o + v
∞
and knowing i∞ , we can obtain the pericenter direction irm using the vector rotation
calculation developed in Lecture 12 on Page 2.
Rotate i∞ , clockwise, through the angle θ to obtain irm . Specifically:
irm = i∞ − sin θ(iN × i∞ ) + (1 − cos θ)iN × (iN × i∞ )
16.346 Astrodynamics
Lecture 11