MIT OpenCourseWare
http://ocw.mit.edu
16.346 Astrodynamics
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 4
The Initial-Value Problem
Polar Coordinates and Orbital Plane Coordinates
From Lecture 2
⎤ ⎡
⎡
1
r
⎢ r ⎥ ⎣ 1
⎣h ⎦=
e sin f
v
µ
⎤⎡
0
⎦⎣
1 + e cos f
ir
⎤
⎡
⎦
⎣
iθ
ir
iθ
⎤
⎡
⎦=⎣
cos f
sin f
− sin f
cos f
⎤⎡
⎦⎣
ie
⎤
⎦
ip
Representation in the Hodograph Plane
Fig. 3.5 (a) and (b) from An Introduction to the Mathematics and
Methods of Astrodynamics. Courtesy of AIAA. Used with permission.
Physical Plane
Hodograph Plane
Flight-Direction Angle
From the position and velocity equations in polar coordinates at the top of this page:
µre sin f
h
In terms of the flight-direction angle γ (shown in the figures above),
r·v=
r = r ir
v = v cos γ ir + v sin γ iθ
=⇒
r · v = rv cos γ
Also
h = |r × v| = rv sin γ
Therefore:
def r · v
rv cos γ
h cot γ
σ = √
= √
= √
µ
µ
µ
σ=
16.346 Astrodynamics
√
or, since p =
p cot γ
Lecture 4
h2
µ
we have
The Initial-Value Problem using Lagrange Coefficients F , G, Ft , Gt
⎡ ⎤ ⎡
r
r cos f
⎣ ⎦=⎣ µ
v
− sin f
h
⎤⎡
r sin f
ie
⎤
⎦⎣ ⎦
µ
(e + cos f )
ip
h
r = F r0 + G v0
v = Ft r0 + Gt v0
or
⎡
⎡ µ
(e + cos f0 )
⎣ ⎦ = ⎣ h2 µ
ip
sin f0
h2
ie
⎤
r
r
=Φ 0
v
v0
F
with Φ =
Ft
G
Gt
#3.6
⎤⎡ ⎤
r
r0
− 0 sin f0
h
⎦
⎣
⎦
r0
cos f0
v0
h
and |Φ| = 1
The value of the determinant |Φ| = F Gt − GFt = 1 follows from the conservation of
angular momentum
r × v = (F Gt − GFt ) r0 × v0 = r0 × v0
Lagrange Coefficients in Terms of the True Anomaly Difference
rr
G = √ 0 sin θ
µp
r
Gt = 1 − 0 (1 − cos θ)
p
r
F = 1 − (1 − cos θ)
p
√
µ
√
σ0 (1 − cos θ) − p sin θ
Ft =
r0 p
where
p
r
=
√
r0
r0 + (p − r0 ) cos θ − p σ0 sin θ
with
θ = f − f0
and σ0 =
Lagrange Coefficients in Terms of the Eccentric Anomaly Difference
Define
r0 · v0
√
µ
Page 162
ϕ = E − E0 . Then:
a
F = 1 − (1 − cos ϕ)
r
√ 0
µa
sin ϕ
Ft = −
rr0
where
(3.42)
√
√
µ G = aσ0 (1 − cos ϕ) + r0 a sin ϕ
√
r = a + (r0 − a) cos ϕ + σ0 a sin ϕ
Kepler’s Equation is then
M − M0 =
(4.41)
a
Gt = 1 − (1 − cos ϕ)
r
and
σ0 =
r0 · v0
√
≡ p cot γ0
√
µ
µ
(t − t0 ) = (E − e sin E) − (E0 − e sin E0 )
a3
or, in terms of ϕ = E − E0
σ0
r0 µ
√
sin ϕ
(t
−
t
)
=
ϕ
+
(1
−
cos
ϕ)
−
1
−
0
a3
a
a
16.346 Astrodynamics
Lecture 4
(4.43)
Lagrangian Coefficients for Parabolic Orbits
Pages 155–156
Since tan 12 f is obtained directly as the solution of Barker’s equation, it is more convenient
to express all trigonometric functions in terms of this function of the true anomaly.
Thus, the position and velocity vectors for the parabola in orbital plane coordinates
are
p
(1 − tan2 12 f ) ie + p tan 12 f ip
2√
√
µp
µp
1
i
v=−
tan 2 f ie +
r
r p
r=
r·v √
σ = √ = p tan
µ
Define
Therefore, with
χ = σ − σ0 ,
1
2
f=
√
p cot γ
we have
χ2
F =1−
2r
√ 0
µχ
Ft = −
rr0
χ
G = √ (2r0 + σ0 χ)
2 µ
Gt = 1 −
χ2
2r
together with Barker’s equation and the equation of orbit:
√
6 µ(t − t0 ) = 6r0 χ + 3σ0 χ2 + χ3
r = r0 + σ0 χ + 21 χ2
σ = σ0 + χ
Solving the Generalized Form of Barker’s Equation
The solution is
χ=
√
p z − σ0
where z is obtained by solving the cubic equation
z 3 + 3z = 2B
with
B=
1
p
3
2
√
[σ0 (r0 + p) + 3 µ(t − t0 )]
Therefore, all the solution methods developed for Barker’s equation are applicable without
modification provided that B > 0 .
16.346 Astrodynamics
Lecture 4
Construction of a Parabola and its Tangents
Pages 156–157
Fig. 4.7 from An Introduction to the Mathematics and Methods of
Astrodynamics. Courtesy of AIAA. Used with permission.
16.346 Astrodynamics
Lecture 4