Transcript ASEN 5050 SPACEFLIGHT DYNAMICS

ASEN 5050
SPACEFLIGHT DYNAMICS
Mid-Term Review
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 19: Mid-Term Review
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Announcements
• Alan’s office hours are on FRIDAY this week! 1:00 pm.
• No Concept Quiz active after this lecture
• STK LAB 2, due 10/17
• Homework #6 will be due Friday 10/24 (2014 not 2013)
– CAETE by Friday 10/31
• Mid-term Exam will be handed out Friday, 10/17 and will be due
Wed 10/22. (CAETE 10/29)
– Take-home. Open book, open notes.
– Once you start the exam you have to be finished within 24 hours.
– It should take 2-3 hours.
• Today: review. Friday: GRAIL and more perturbation fun.
Lecture 19: Mid-Term Review
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Final Project
• Get started on it!
• Worth 20% of your grade, equivalent to 6-7 homework assignments.
• Find an interesting problem and investigate it – anything related to
spaceflight mechanics (maybe even loosely, but check with me).
• Requirements: Introduction, Background, Description of
investigation, Methods, Results and Conclusions, References.
• You will be graded on quality of work, scope of the investigation,
and quality of the presentation. The project will be built as a
webpage, so take advantage of web design as much as you can
and/or are interested and/or will help the presentation.
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Final Project
•
Instructions for delivery of the final project:
•
Build your webpage with every required file inside of a directory.
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•
Name your main web page “index.html”
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Name the directory “<LastName_FirstName>”
there are a lot of duplicate last names in this class!
You can link to external sites as needed.
i.e., the one that you want everyone to look at first
Make every link in the website a relative link, relative to the directory structure
within your named directory.
–
We will move this directory around, and the links have to work!
•
Test your webpage! Change the location of the page on your computer and make
sure it still works!
•
Zip everything up into a single file and upload that to the D2L dropbox.
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Space News
• MAVEN’s first scientific announcement!
• UV views of Mars’ escaping atmosphere
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Concept Quiz 12
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Concept Quiz 12
Honestly, I didn’t make this clear enough.
There IS indeed correlation that the geoid follows mass
(strictly speaking it is defined by the gravity of matter!)
But it IS NOT perfectly correlated and sometimes it
appears quite the opposite.
Lecture 19: Mid-Term Review
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ASEN 5050
SPACEFLIGHT DYNAMICS
Mid-Term Review
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 19: Mid-Term Review
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Mid-Term Logistics
• Please write your name on the exam
• Carry “infinite” precision in your math at all steps.
At the very end you are welcome to round your
answers, but don’t round too much!
• Document which equations / process you use, and
then verify that your math is correct using a
computer.
• Try to learn something  I’m not JUST doing this
exam to torture you.
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Course Topics
• Two-Body Problem
–
–
–
–
Newton’s Law of Gravitation
Orbital elements
Converting Cartesian to/from Keplerian Orbital Elements
Using Eccentric Anomaly to determine when an orbit is at some
radius from the central body and, given that, estimate how much
time has elapsed since periapse, etc.
– Vis-Viva Equation
• Coordinate Systems
– IJK, XYZ, Perifocal, RSW, SEZ, VNC, etc.
• Time Systems
– UT1, UTC, TAI, GPS, ET
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Course Topics
• Orbital Maneuvering
– Changing each orbital element
– Plane Changes are expensive
• Orbital Transfers
– Hohmann Transfer (2 tangential burns)
– Phase Changing
• Intercepting an object and rendezvousing with it
– Combining maneuvers, such as an optimal LEO – GEO transfer
• Where do you perform the inclination change?
• Do you perform all of the inclination change there?
– One-tangent burn
• To speed up the transfer
– Lambert’s Problem
• Minimum-energy orbital transfer between two arbitrary position vectors.
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Course Topics
• Proximity Operations
– Clohessy-Wiltshire / Hill’s Equations
– We’ll talk about this more in a few minutes
• Groundtracks
– Plotting the sub-point of a satellite over time
– Repeat groundtracks and other practical groundtracks
• Build me an orbit that exactly repeats its groundtrack every 12
days, after 121 revolutions.
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Course Topics
• Additional Topics
– The shape of the Earth
• Geocentric latitude vs. geodetic latitude
– Solar day vs. Sidereal day
• 86400 seconds in a solar day, 86164.1 ish in a sidereal day. Be
careful!
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Review Requests (first 43)
•
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•
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•
•
•
•
•
CW/H x14
Groundtracks x2
Coordinate systems x5
Transformations x7
f and g series x4
Canonical units x2
Odd orbital elements
Geocentric/geodetic
Kepler’s Equation derivation
Lambert’s Problem x3
Synodic period
Plane changes / maneuvers not covered in HW x5
EVERYTHING x4
NOTHING x5
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Coordinate Systems
Geocentric Coordinate System (IJK)
- aka: Earth Centered Inertial (ECI), or the Conventional
Inertial System (CIS)
- J2000 – Vernal equinox on Jan 1, 2000 at noon
- non-rotating
Intersection of ecliptic and celestial eq
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Coordinate Systems
Earth-Centered Earth-Fixed Coordinates (ECEF)
Topocentric Horizon Coordinate System (SEZ)
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Coordinate Systems
Perifocal Coordinate System (PQW)
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Coordinate Systems
Satellite Coordinate Systems:
RSW – Radial-Transverse-Normal
NTW – Normal-Tangent-Normal; VNC is a rotated version
V
R
C
S
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Coordinate Transformations
Coordinate rotations can be accomplished through rotations about
the principal axes.
0 ù
sin a ú
ú
cos a úû
- sin a ù
0 ú
ú
cos a ûú
0
é1
cos a
ROT 1( a ) = ê0
ê
- sin a
êë0
0
écos a
1
ROT 2( a ) = ê 0
ê
0
ëê sin a
é cos a
ROT 3( a ) = ê- sin a
ê
êë 0
Lecture 19: Mid-Term Review
sin a
cos a
0
0ù
0ú
ú
1úû
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Coordinate Transformations
To convert from the ECI (IJK) system to ECEF, we simply rotate
around Z by the GHA:
rECEF = ROT 3( q GST )rIJK
GST = Greenwich Sidereal Time
or rIJK = ROT 3( -q GST )rECEF
ignoring precession, nutation, polar motion, motion of equinoxes.
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Coordinate Transformations
To convert from ECEF to SEZ:
rSEZ = ROT 2( 90° - f )ROT 3( l )rECEF
= ROT 2( 90° - f )ROT 3( q LST = l + q GST )rIJK
To set up a SYSTEM you also
need to specify a center,
which can be anything but is
usually the reference site.
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Coordinate Transformations
• One of the coolest shortcuts for building
transformations from one system to any other,
without building tons of rotation matrices:
The unit vector in the S-direction,
expressed in I,J,K coordinates
Lecture 19: Mid-Term Review
(sometimes this is
easier, sometimes not)
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Latitude/Longitude
Geocentric latitude
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(Vallado, 1997)
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Latitude/Longitude
For geodetic latitude use:
tan f gc
tan f gd =
1 - eÅ2
where e=0.081819221456
(Vallado, 1997)
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Latitude/Longitude
Rotate into ECEF
rECEF = ROT 3( q GST )rIJK
æ ry ö
l = tan çç ÷÷
è rx ø
-1 æ rz ö
f gc = sin ç ÷
èrø
-1
rECEF
ér cos f cos lù érx ù
ê
ú ê ú
= ê r cos f sin l ú = êry ú
êë r sin f úû êërz úû
r = rx2 + ry2 + rz2
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Coordinate Transformations
To convert between IJK and PQW:
rIJK = ROT 3( -W )ROT 1( -i )ROT 3( -w )rPQW
rPQW = ROT 3( w )ROT 1( i )ROT 3( W )rIJK
To convert between PQW and RSW:
R
S
rRSW = ROT 3(n )rPQW
rPQW = ROT 3( -n )rRSW
Thus, RSW  IJK is:
rIJK = ROT 3( -W )ROT 1( -i )ROT 3( -u )rRSW
where
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P
u =n + w
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CW / H
• Useful to answer questions like:
– If I deploy a satellite from my current position in orbit, and
the deployment imparts some small Delta-V, where does
the satellite go, relative to me?
– If I’m approaching a space station, what Delta-V should I
execute to rendezvous with the station after 10 minutes?
– Also helps evaluate trajectories rapidly, since you don’t
have to numerically integrate them.
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Coordinate Systems
Satellite Coordinate System (RSW) -- (Radial-Transverse-Normal)
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Coordinate Transformations
To convert between IJK and PQW:
rIJK = ROT 3( -W )ROT 1( -i )ROT 3( -w )rPQW
rPQW = ROT 3( w )ROT 1( i )ROT 3( W )rIJK
To convert between PQW and RSW:
rRSW = ROT 3(n )rPQW
rPQW = ROT 3( -n )rRSW
Thus, RSW  IJK is:
rIJK = ROT 3( -W )ROT 1( -i )ROT 3( -u )rRSW
where
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u =n + w
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
Use RSW coordinate system (may be different from NASA)
Target satellite has two-body motion:
m rtgt
rtgt = - 3
rtgt
The interceptor is allowed to have thrusting
mr
rint = - 3int + F
rint
Then rrel = rint - rtgt Þ rrel = rint - rtgt
So, rrel = -
m rint
Lecture 19: Mid-Term Review
3
int
r
+F+
m rtgt
rtgt3
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
Need more information to solve this:
rrel = -
m rint
3
int
r
+F+
m rtgt
rtgt3
Lecture 14 (Slide 27+) takes you through
all of the steps needed to convert
this acceleration into one that is
only dependent on the relative
vector and omega:
{
}
rrel R = -w 2 xRˆ + ySˆ + zWˆ - 3xRˆ + F + 2w yRˆ - 2w xSˆ + w 2 xRˆ + w 2 ySˆ
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
m
We assume circular motion:
{
3
tgt
r
= w2 ,
w = 0,
}
rrel R = -w 2 xRˆ + ySˆ + zWˆ - 3xRˆ + F + 2w yRˆ - 2w xSˆ + w 2 xRˆ + w 2 ySˆ
Thus,
x - 2wy - 3w 2 x = f x
y + 2w x = f y
CW or Hill’s Equations
z + w2 z = fz
Assume F = 0 (good for impulsive DV maneuvers, not for
continuous thrust targeting).
Equations
also assume circular orbits and close proximity!
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
These equations can be solved (see book: Algorithm 48) leaving:
æ
æ
2y ö
2y ö
sin w t - ç 3x0 + 0 ÷ cosw t + ç 4x0 + 0 ÷
è
è
w
w ø
w ø
æ
æ
4y ö
2x
2x ö
y ( t ) = ç 6x0 + 0 ÷ sin w t + 0 cosw t - ( 6w x0 + 3y0 ) t + ç y0 - 0 ÷
è
è
w ø
w
w ø
z
z ( t ) = z0 cosw t + 0 sin w t
x (t ) =
x0
w
x ( t ) = x0 cosw t + ( 3w x0 + 2y0 ) sin w t
y ( t ) = ( 6w x0 + 4y0 ) cosw t - 2x0 sin w t - ( 6w x0 + 3y0 )
z ( t ) = -z0w sin w t + z0 cosw t
So, given x0 , y0 , z0 , x0 , y0 , z0 of interceptor, can determine x , y , z ,
x , y , z of interceptor at future time.
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Hubble’s Drift from Shuttle
• RSW Coordinate Frame
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
æ
æ
2y ö
2y ö
sin w t - ç 3x0 + 0 ÷ cosw t + ç 4x0 + 0 ÷
è
è
w
w ø
w ø
æ
æ
4y ö
2x
2x ö
y ( t ) = ç 6x0 + 0 ÷ sin w t + 0 cosw t - ( 6w x0 + 3y0 ) t + ç y0 - 0 ÷
è
è
w ø
w
w ø
z
z ( t ) = z0 cosw t + 0 sin w t
x (t ) =
x0
w
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
We can also determine DV needed for rendezvous. Given x0, y0, z0, we
want to determine x0 , y0 , z0 necessary to make x = y = z = 0.
Set first 3 equations to zero, and solve for x0 , y0 , z0 .
Assumptions:
1. Satellites only a few km apart
2. Target in circular orbit
3. No external forces (drag, etc.)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
NOTE: This is not the Delta-V,
this is the new required relative
velocity!
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CW/H Examples
• Scenario 1:
Use eqs to compute x(t), y(t), and z(t)
RSW coordinates, relative to deployer.
Note: in the CW/H equations,
the reference state doesn’t
move – it is the origin!
Deployment: x0, y0, z0 = 0,
• Scenario 1b:
Lecture 19: Mid-Term Review
Delta-V = non-zero
Use eqs to compute x(t), y(t), and z(t)
RSW coordinates, relative to reference.
Initial state: non-zero
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CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
Initial state in the
RSW frame
Lecture 19: Mid-Term Review
Target: some constant
values in the RSW frame
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CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
This is easy if the targets are zero (Eq. 6-66)
This is harder if they’re not!
Initial state in the
RSW frame
Lecture 19: Mid-Term Review
Velocity needed to
get onto transfer
Target: some constant
values in the RSW frame
46
CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
This is easy if the targets are zero (Eq. 6-66)
This is harder if they’re not!
Initial state in the
RSW frame
Velocity needed to
get onto transfer
The Delta-V is the
difference of these
Lecture 19: Mid-Term Review
velocities
Target: some constant
values in the RSW frame
47
CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
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Shuttle: reference frame
48
CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Result of deployment
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Shuttle: reference frame
49
CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Rendezvous trajectory (Eq 6-66)
Result of deployment
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Shuttle: reference frame
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CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Delta-V is the difference
of these velocities.
Rendezvous trajectory (Eq 6-66)
Result of deployment
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Shuttle: reference frame
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Example from last year’s mid-term!
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Mid-Term Review
• Problem 4
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Mid-Term Review
• Problem 4
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Mid-Term Review
• Problem 4
• First, we need the velocity of the shuttle relative to the
experiment before the maneuver:
– Experiment’s velocity relative to shuttle, from Algorithm 48 of
part (a)
– Shuttle’s velocity relative to experiment is just the opposite of
that:
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Mid-Term Review
• Problem 4
• Second, we need the velocity that the shuttle must obtain
to perform the rendezvous.
– Equation 6-66 in Vallado’s 4th edition
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Mid-Term Review
• Problem 4
• The initial conditions are the state of the shuttle relative to
the experiment (opposite signs of the experiment’s
position relative to the shuttle from part (a))
• t = 10*60 sec = 600 sec (you can keep omega the same
as before, or update it; doesn’t make much difference)
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Mid-Term Review
• Problem 4, part (b)
• Velocity required to achieve the transfer:
• Delta-V for the rendezvous:
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Mid-Term Review
• Problem 4, part (c)
• Use Algorithm 48 once again and now plug in the initial
position and velocity of the shuttle relative to the
experiment.
• The position of the shuttle after 10 minutes (15 minutes
after the deployment) should be zero (GOOD CHECK!)
• The velocity of the shuttle after 10 minutes (15 minutes
after the deployment) will not be zero.
• The Delta-V is that which will remove the relative
velocity of the shuttle relative to the experiment.
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Mid-Term Review
• Problem 4, part (c)
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Orbit Maneuvering
•
•
•
•
•
Hohmann Transfers
Bi-elliptic Transfers
Circular Rendezvous
Coplanar Rendezvous
Changing orbital elements
– a, e, rp, ra, P, M, AOP are all coplanar
– i, RAAN are plane changes
• Lambert’s Problem
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Changing Orbital Elements
•
•
•
•
•
•
Δa  Hohmann Transfer
Δe  Hohmann Transfer
Δi  Plane Change
ΔΩ  Plane Change
Δω  Coplanar Transfer
Δν  Phasing/Rendezvous
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Changing Inclination
• Δi  Plane Change
• Inclination-Only Change vs. Free Inclination Change
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Changing Inclination
• Let’s start with circular orbits
Vf
V0
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Changing Inclination
• Let’s start with circular orbits
Vf
V0
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Changing Inclination
• Let’s start with circular orbits
Are these vectors the
same length?
Vf
Δi
V0
What’s the ΔV?
Is this more expensive
in a low orbit or a high
orbit?
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Changing Inclination
• More general inclination-only maneuvers
Where do you perform
the maneuver?
How do V0 and Vf
compare?
What about the FPA?
Line of Nodes
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Changing Inclination
• More general inclination-only maneuvers
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Changing The Node
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Changing The Node
Where is the
maneuver located?
Neither the max latitude
nor at any normal
feature of the orbit!
There are somewhat
long expressions for
how to find uinitial and
ufinal in the book for
circular orbits.
Lambert’s Problem
gives easier solutions.
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Changing Argument of Perigee
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Changing Argument of Perigee
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Changing Argument of Perigee
Which ΔV is
cheaper?
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Maneuver Combinations
• General rules to consider:
– Energy changes are more efficient when traveling FAST
• Periapse
– Plane changes are more efficient when traveling SLOW
• Apoapse
– Combinations take advantage of vector addition
•
•
•
•
3+4 = 5 not 7 
Some inclination change at periapse is optimal
Some energy change at apoapse is optimal (or necessary)
Delta-V vector = V final vector – V initial vector
– One’s initial and final orbit do not always intersect; if they
don’t you have to build a transfer orbit.
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Questions
• These are the sorts of questions I hope you can
answer:
–
–
–
–
–
Given orbit X, when does the satellite reach radius R?
Where in orbit Y is the satellite at its maximum latitude?
Compute element XYZ given element ABC
And all of those concept quiz type questions.
There will be math and concepts.
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ASEN 5050
SPACEFLIGHT DYNAMICS
GRAIL
Prof. Jeffrey S. Parker
University of Colorado – Boulder
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ASEN 5050
SPACEFLIGHT DYNAMICS
Perturbations
Prof. Jeffrey S. Parker
University of Colorado – Boulder
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Perturbation Discussion Strategy
✔• Introduce the 3-body and n-body problems
– We’ll cover halo orbits and low-energy transfers later
✔• Numerical Integration
✔• Introduce aspherical gravity fields
– J2 effect, sun-synchronous orbits
• Solar Radiation Pressure
• Introduce atmospheric drag
– Atmospheric entries
• Other perturbations
• General perturbation techniques
• Further discussions on mean motion vs. osculating
motion.
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