Final Exam (SOLUTION)
ASE 366K Fall 2004
Name:_______________________________________________________
There are four problems and one extra credit problem on this examination. Please answer problem 1
on this page and the remaining problems on separate sheets of paper. It should go without saying, but the
work you show must be clear and legible for it to be credited. For problems 2,3, and 4, …rst outline
a procedure on how you are going to solve the problem; then implement the procedure and
highlight key steps and results.
1. (10 points) True or False
(a) F_____The zero sphere of in‡uence patched conic model approximates well the motion of a
very small particle ‡ying between two very large equal masses that are in circular orbits about
each other.
(b) T_____Lambert’s Theorem states that the time of ‡ight on an elliptical orbit between two
points on the orbit is only a function of the semi-major axis, the sum of the magnitudes of the
position vectors of the two points, and the distance between the two points.
(c) F_____The components of the angular momentum vector and the eccentricity vector uniquely
represent a complete set of independent integrals or constants of motion for the two body problem
with three degrees of freedom.
(d) T_____The lower and upper limiting values for the turning angle of a hyperbola are 0 and
180 respectively.
(e) F_____The eccentricity vector is not de…ned for a circular orbit.
(f) F_____A velocity impulse along the angular momentum vector may change the magnitude
of the angular momentum vector but never its direction.
(g) F_____If the earth’s rotation rate about its polar axes increases, then the radius of geostationary orbits would also have to increase.
(h) T_____Elliptical, inclined orbits about the Earth can be designed to be geosynchronous.
(i) F_____Changing the ‡yby periapsis radius of a ‡yby orbit about Jupiter will always only
change the turning angle
(j) F_____The CW equations model exactly the relative motion between a chase vehicle and a
target vehicle that is in a circular orbit about a point mass central body.
1
2. (30 points) Consider the circular and elliptical orbits shown in the sketch. Spacecraft B is at the point
shown at t0 , and all motion is counter-clockwise. At what true anomaly should spacecraft A be at
t0 = 0 so that it intercepts spacecraft B at the second intersection point as shown? The central body
has = 1 du3 =tu2 , spacecraft B is in a circular orbit of radius 1 du, and spacecraft A is in an elliptical
orbit with e = 72 and a = 1 du. Note that the sketch is generic and it will not necessarily correspond
to the actual values for all quantities.
(t0 ) = __________ deg
Solution:
Procedure:
1. …nd the true anomalies at which the two orbits intersect; observe that we are interested in the one
that has a value greater than 180 degrees; call this f
2. compute the eccentric anomaly at this point for the elliptical orbit, Ef
3. compute the corresponding mean anomaly, Mf
4. determine the time it takes for spacecraft B to get to this point; note that spacecraft B travels an
angular distance f 90 , since it is already on the y-axis at t0
5. compute the mean anomaly for spacecraft A at t0
6. solve Kepler’s equation to get E0
7. compute 0 knowing E0
The semi-latus parameter is
p = a(1 e2 ) = :918367 du
To get the intersection true anomaly use
r=
p
1 + e cos
The value we need is
f
= 253:398
Use
tan
to get Ef
2
=
r
1+e
E
tan
1 e
2
Ef = 270
2
The time it takes S/C B to move from where it is at t0 the intersection point is
r
n=
= 1 rad/tu
a3
90
=2
f
=
n
n
Using Kepler’s equation we …nd the time from periapsis to this intersection point
t=
f
ntf = Ef
e sin Ef
tf = 4:99810 tu
The initial time is
t0 = t f
t = 2:14626 tu
We solve Kepler’s equation iteratively to get E0 ;
M0
= E0
nt0
= E0
e sin E0
2
sin E0
7
E0 = 2:349617
We again use the true anomaly to eccentric anomaly equation to get
0
0
= 2:537441 rad
= 145:385
3.
3
t; its mean motion is n
3. (30 points) Consider a parabolic orbit of a comet relative to the Sun. The comet encounters Jupiter
for a ‡yby. The perihelion of this orbit before the ‡yby coincides with Jupiter’s mean semi-major axes about
the Sun (ajupiter = 5:2 AU): The ‡yby will change the orbit of the comet as shown in the sketch. The
Sun’s gravitational parameter is sun = 1 AU3 =tu2 and Jupiter’s gravitational parameter is approximated
1
AU3 =tu2 . The perijove ‡yby radius at Jupiter is rp_f lyby = 0:0052 du. Assume a patched
as jup = 1000
conic model applies for your analysis. Note that the sketch is just symbolic and the comet’s new orbit may
not be the nice simple ellipse shown. What is the new ‡ight path angle, perihelion, and aphelion of the
comet’s new orbit with respect to the Sun? Based on your results attempt to sketch the real post Jupiter
‡yby orbit of the comet about the Sun in the sketch below.
new
rp
ra
= ____ deg
= _____du
= _____du
Solution:
Procedure:
1. …nd the perihelion velocity of the comet when it arrives at Jupiter’s orbit
2. …nd the velocity of Jupiter at this point
3. compute v1 , note that at t (time just before the ‡yby) the velocities are parallel
4. we know the energy of the hyperbola w/r to Jupiter; we know the required radius of the ‡yby at
Jupiter...with this compute vp the perijove velocity
5. compute the magnitude of the angular momentum vector of the ‡yby hyperbola
6. compute the eccentricity of the hyperbola and then the turning angle
7. with this information we compute the post ‡yby velocity of the comet w/r to the Sun and the ‡ight
path angle (who’s sign depends on which side of Jupiter the comet ‡ies by; this is not important since the
problem does not ask this, so either +; value is OK; call this f
8. we can now compute the post ‡yby energy of the comet w/r to the Sun, and get the needed quantities,
namely ra ; rp
Implementation
The perihelion velocity is found from
vp2
sun
E=
2
r
4
where E = 0, and r = ajup
vp = :620173 au/tu
the velocity of Jupiter is
vc = :438520 au/tu
so
v1 = :181644
The energy of the hyperbola is
2
v1
= :016497 au2 =tu2
2
The velocity at perijove is found from the energy equation; we know rp_f lyby and
Ehyperbola =
jup
vp_f lyby = :646228 au/tu
Angular momentum at perijove is
h
= rp vp
= :003360 au2 =tu
The eccentricity of the hyperbola is
e =
=
s
1+
2Eh2
2
1:171573
The turning angle is
1
=
2
e
= 117:201
sin
From the velocity ‡yby diagram we can get v + ; which is the post ‡yby velocity w/r to the Sun
v+
2
2
= vc2 + v1
2vc v1 cos(
)
note the angle......verify via the velocity ‡yby diagram.
v + = :390486 au/tu
The ‡ight path angle is the angle b/n the vectors v+ and vc (which is normal to the radius vector from the
Sun) (we use the law of cosines)
=
=
2
cos 1 ((v1
24:440
v+
2
vc2 )=( 2=v + =vc ))
The post ‡yby energy and semi-major axes is
E+ =
v +2
2
E =
a =
SU N
ajup
=
sun
2a
:116068 au2 =tu2
4:307821 au
The angular momentum after the ‡yby is
h
= rv + cos
= 1:84859 au2 /tu
5
The eccentricity is
e =
s
1+
2Eh2
2
sun
= :45467
and …nally
rp
ra
= a(1 e) = 2:249180 au
= a(1 + e) = 6:266462 au
We observe that had the perijove radius been smaller, then it would have been possible to have rp
making a near Earth crossing comet.
6
1 au,
4. 4. (30 points) Consider an astronaut orbiting the international space station (ISS) as shown in the
sketch. The space station is in a circular 407 km altitude orbit about the Earth (rearth = 6378 km,
105 km3 =s2 ). The initial conditions for the astronaut in the CW frame at t0 are
earth = 3:986
x0 = 0; y0 = 1 km; y_ 0 = 0, and x_ 0 is the required value to produce a periodic orbit (one without
secular drift). What is x_ 0 ? The period of the astronaut’s orbit about the ISS is Tp ; the same as the
period of the ISS’s orbit about the Earth. At t0 the astronaut determines she needs to return and
rendezvous with the station. She needs to do this transfer in a time period Tp =2 as follows. She
…rst coasts along the orbit to ti (intersection with the x-axis), and then does a maneuver to put her
on an intercept path with the ISS. Then a …nal rendezvous maneuver is made at tf . We know that
t0 = 0; ti = Tp =4; tf = Tp =2. What is the sum of the two maneuvers, vtotal ?
Solution:
Procedure:
1. compute the mean motion of the ISS and compute the initial velocity in the x direction of the
astronaut to generate a zero drift orbit
2. propagate the orbit to the x-axis so we know the position and velocity just before the maneuver at ti
3. compute the 2 impulse transfer from this point to the origin; the transfer time is tf ti = Tp =2 Tp =4
4. we know all the velocities; and hence compute the total delta-v
Implementation
The semi-major axis of the ISS is
a = 6788 km
and
n = 0:0011296 rad/sec.
yielding a time period of the ISS orbit (and also for the astronaut around the ISS) of
Tp = 5562:064 sec
We use the 2-D form of the CW phi matrix to compute all quantities:
0
1 0
2(1 cos nt)
sin nt
x
4 3 cos nt
0
n
n
2(1 cos nt)
(4 sin nt 3nt)
B y C B 6(sin nt nt)
1
B
C=B
n
n
@ x_ A @
3n sin nt
0
cos nt
2 sin nt
y_
6n(1 cos nt) 0
2 sin nt
4 cos nt 3
7
10
1
x0
C B y0 C
CB
C
A @ x_ 0 A
y_ 0
But …rst we note that we must null out the secular terms by using the second row of this matrix; we have
the expression that contains secular terms
6(sin nt
nt)x0 + 1y0
2(1
cos nt)
(4 sin nt
x_ 0 +
n
n
3nt)
y_ 0
we remove the sinusoidal terms and isolate the constant and secular terms
2
x_ 0
n
6ntx0 + y0
3ty_ 0
the initial conditions for our problem are of the form
x0
y0
x_ 0
y_ 0
= 0
= given
= unknown
= 0
with this we have a relationship b/n y0 and x_ 0 which we set to zero so cancel out the drift
2
x_ 0 = 0
n
y0
so that
x_ 0 =
n
y0
2
we know y0 (=1km) so we get x_ 0
x_ 0 = 0:0005648 km/s
We evaluate
at t = ti = Tp =4 and multiply is by the initial condition vector to get
1
1
0
0
0:5
x
C
B
B y C
0
C km/s
C
B
B
A
@ x_ A (ti ) = @
0
0:0011296
y_
We evaluate from ti to tf so that we can work the rendezvous part. If we partition
matrices we compute
rf = Mri + Nv+
i
into the M; N; S; T
so
vi+ = N
1
(rf
Mri )
where
rf = 0
we have
vi+ =
0:0006872
0:0007860
km/s
We use this know velocity to propagate the new trajectory to the …nal state to get the …nal velocity just
before the …nal impulse
0:00012239
vf =
km/s
0:00034361
we know that
vi =
0
0:0011296
and
vf+ =
0
0
8
km/s
km/s
We have now all the information needed
vi = vi+
vi
vf+
vf
vf =
with this we have
vi
vf
=
=
0:000768 km/s
0:000365 km/s
and
vtotal = 0:001133 km/s
5.
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Extra Credit Problem (20 points)
Consider the following …gure:
Two position vectors are given with respect to a central body with gravitational parameter . It is required
to determine the transfer orbit that connects these two position vectors in a given transfer time t. Discuss
and describe as many aspects of the problem as you can, highlighting the key issues, and using sketches if
you think they will clarify your discussion.
Solution:
this problem is open ended and its solution is based on the notes handed out in class; it is key to
identify the four possible transfer paths and discuss the nature of these; as such the solution is based on the
information provided in the summary notes and are not repeated here. We agree this was an easy problem
since it was just conceptual; however, in your solution do not say anything that is incorrect....this will show
me whether you understand the problem or not.
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