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 15
The Fundamental Ellipse of the BVP
#6.3
Fig. 6.13 from An Introduction to the Mathematics and Methods of
Astrodynamics. Courtesy of AIAA. Used with permission.
Recall:
Equation of orbit
e · r1 = p − r1
e · r2 = p − r2
e·r=p−r
=⇒
at P1 and P2 so that
e · (r2 − r1 ) = r1 − r2
or
−e · ic =
r 2 − r1
c
Orbital Elements of the Fundamental (Minimum Eccentricity) Ellipse
eF =
Since
|r2 − r1 |
c
pF = aF (1 − e2F )
aF =
1
2
(r1 + r2 )
pF
r + r2
= 1
pm
c
Page 258
and
c2 − (r2 − r1 )2
1
= 2 (c + r2 − r1 )(c + r1 − r2 )
2
c
c
4
4r r
2p
= 2 (s − r1 )(s − r2 ) = 12 2 sin2 12 θ = m
c
c
c
From the figure at the top of this page, we see that sin φF = eF . Also the angle ωp
is the complement of φF so that cos ωp = eF . Therefore:
1 − e2F =
The axes of the conjugate parabolic orbits coincide with the lines through the
focus F and the extremities of the minor axis of the fundamental ellipse.
16.346 Astrodynamics
Lecture 15
Locus of Mean Points
#6.4
Definition of mean point:
At the point r0 , the velocity v0 is parallel to the chord.
Using the eccentricity vector at this point, we have v0 × h · (r2 − r1 ) = 0 Therefore:
µ µ
µe · (r2 − r1 ) = v0 × h − r0 · (r2 − r1 ) = 0 − r0 · (r2 − r1 )
r0
r0
Hence:
1
e · (r2 − r1 ) = − r0 · (r2 − r1 )
r0
= r1 − r2
=⇒
r0 · (r2 − r1 ) = r0 (r2 − r1 )
The loci of all mean points are the lines through the focus F and the
extremities of the minor axis of the fundamental ellipse.
The Line Segment FS:
Page 266
The line F S is the distance along mean point locus from the focus F to intersection
with chord. The flight direction angle at P0 is γ0 and δ is the angle opposite the line
segment SP1 . Use the law of sines for the triangles:
F P1 S :
FS
r1
=
sin(γ0 + δ)
sin γ0
F P1 P2 :
c
r2
=
sin θ
sin(γ0 + δ)
and use the calculation on the previous page for 1 − e2F :
b
2√
F P0 C : sin γ0 = F = 1 − e2F =
r r sin
aF
c 1 2
Therefore:
FS =
Recall the proposition:
√
r1 r2 cos
1
2
1
2
θ
θ
Lecture 8, Page 2
The line connecting the focus and the point of intersection of the orbital tangents at the
terminals bisects the transfer angle.
⎧
⎨ F N cos 12 (E2 − E1 )
√
r1 r2 = F N
⎩
F N cosh 12 (H2 − H1 )
Fig. 6.7 from An Introduction to the
Mathematics and Methods of
Astrodynamics. Courtesy of AIAA.
Used with permission.
16.346 Astrodynamics
Lecture 15
ellipse
parabola
hyperbola
Eccentric Anomaly of the Mean-Point
For the ellipse
(E 0 − E 1 ) =
F N1 p = r1 r0 p
F N1 cos
and for the parabola
1
2
Pages 267–270
√
r 1 r0
F N2 p
F N2 cos 12 (E2 − E0 ) =
= r 0 p r2
√
r 0 r2
Triangle ∆F N1 p N2 p is similar to triangle ∆F N1 N2 . Therefore
F N1 p
F N2 p
=
F N1
F N2
=⇒
cos
1
2
(E0 − E1 ) = cos
1
2
(E2 − E0 )
Fig. 6.16 from An Introduction to the
Mathematics and Methods of
Astrodynamics. Courtesy of AIAA.
Used with permission.
Hence
E0 =
1
2
(E1 + E2 )
6.73
The eccentric anomaly of the mean point of an orbit connecting two termini is
the arithmetic mean between the eccentric anomalies of those termini.
16.346 Astrodynamics
Lecture 15
Mean-Point Radius of the Parabolic Orbit
Lecture 8, Page 1
The parameter of the parabola is obtained from
p 2
pp
r + r2
s(s − c)
r + r2
p
− 2D
+ 1 = 0 where D = 1
−
= 1
pm
pm
c
ac
c
so that
pp
1
√
= (r1 + r2 + 2 r1 r2 cos
pm
c
1
2
θ) =
2
√
1 √
s+ s−c
c
The mean point radius of the parabola is
pp
pp c
pp
pp
r0p =
=
=
=
2
2
2 cos φF
2(1 − eF )
4pm
1 + cos 2φF
Hence
r0p =
1
4
√
(r1 + r2 + 2 r1 r2 cos
1
2
θ) =
1
2
(aF + F S)
The mean point radius of the parabola extends to the midpoint between
the chord and the extremity of the minor axis of the fundamental ellipse.
Mean-Point Radius of Ellipic and Hyperbolic Orbits
Page 270
From the derivation of the eccentric anomaly of the mean point, we have
√
F N2 cos 12 (E2 − E0 ) = F N2 cos 12 [E2 − 12 (E1 + E2 )] = F N2 cos 14 (E2 − E1 ) = r0 r2
F N 2 p = r0 p r2
so that
F N2
cos
F N2 p
1
4
(E2 − E1 ) =
r0
r0 p
But, from similar triangles,
r
F N2
= 0
F N2 p
r0 p
Therefore, we have the truly elegant expression
r0 = r0 p sec2
1
4
(E2 − E1 ) = r0 p sec2
1
2
ψ = r0 p (1 + tan2
1
2
ψ)
and, as we might expect,
r0 = r0 p sech2
1
4
(H2 − H1 )
obtains also for hyperbolic orbits.
r0p (1 + tan2 12 ψ)
a[1 − e cos 12 (E1 + E2 )] = a(1 − cos φ)
=
r0 =
a[1 − e cosh 12 (H1 + H2 )]
r0p (1 − tanh2 14 (H2 − H1 )
16.346 Astrodynamics
Lecture 15