Transcript ASEN 5050 SPACEFLIGHT DYNAMICS - CCAR | CU

ASEN 5050
SPACEFLIGHT DYNAMICS
Intro to STK, More 2-Body
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 6: The Two Body Problem
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Announcements
• Friday: no class at 9:00. STK Lab 1
–
–
–
–
–
ITLL 1B10, 2B10, or Visions Lab – or on your own computer.
Groups of 1-3.
Intro to STK today
Alan will be in ITLL 2B10 to help answer questions from 1-3.
Due Friday 9/26 at 9:00 am
• Homework #2 is due Friday 9/12 at 9:00 am
– D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or
9:10 so don’t be late!
• Homework #3 is due Friday 9/19 at 9:00 am
• Concept Quiz #5 will be available at 10:00 am, due Monday morning at
8:00 am.
• Reading: Chapters 1 and 2
Lecture 6: The Two Body Problem
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Quiz 4
Lecture 6: The Two Body Problem
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Quiz 4
Lecture 6: The Two Body Problem
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Quiz 4
Lecture 6: The Two Body Problem
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Quiz 4
Lecture 6: The Two Body Problem
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Challenge #3
• We examined Pluto’s and Neptune’s orbits last time.
• Question: since Pluto sometimes travels interior to
Neptune’s orbit, could they ever collide?
– If so, what sort of order of duration do we need to wait
until it may statistically happen? Years? Millennia? Eons?
Lecture 6: The Two Body Problem
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Derivation of Kepler’s Equation
If n is given:
If t is given:
M = n(t - tp)
Solve E-e sin E = M
for E
r = a( 1-e cos E)
and
cosn =
sinn =
=
a cos E - ae
r
r
p
1 + e cos n
cos E =
sinE =
b
sin E
r
a 1-e
r=
r cos n + ae
a
r sinn
b
Solve E-e sin E = M
2
for t (or M)
sin E
Another Useful Relation:
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
• Given M, solve for E
• What would a plot
of this look like?
• Circular orbit?
• Elliptical?
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
• Given M, solve for E
• To see what we have to do, here are a few plots of
this relationship:
Near-circular
Lecture 6: The Two Body Problem
~ parabolic
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
What is E that produces
M=100?
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Subtract M from all values
Find zero-crossing
What is E that produces
Value=0?
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Step 1: Initial guess (E0 = M)
Step 1: Value = -5.64 deg
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Step 2: Modify guess and find the zero
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Newton-Raphson
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Newton-Raphson
Guess #2
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Solving Kepler’s Equation
• Newton Raphson Method applied to Kepler’s
Equation
Xn+1 = Xn + dn = Xn -
E - esin E = M
Þ
f (Xn )
f ¢(Xn )
f (E) = M - E + esin E = 0
f '(E) = -1+ ecos E
En+1
Lecture 6: The Two Body Problem
M - En + e sin En
= En +
1 - e cos En
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Solving Kepler’s Equation
•Let’s say e = 0.1 and M = 100 deg (1.745 rad)
•Guess 1: E0 = 100 deg
– Error 1: -5.64 deg
En+1
M - En + e sin En
= En +
1 - e cos En
•Guess 2: E1 = 105.546 deg
– Error 2: 0.0263 deg
•Guess 3: E2 = 105.521 deg
– Error 3: 0.0000 deg
•Guess 4: E3 = 105.521 deg
– Error 4: 0.0000 deg
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
• Let’s say e = 0.1
• Any tricks or challenges? No.
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
• Let’s say e = 0.95
• Any tricks or challenges? Not really.
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Solving Kepler’s Equation
• Let’s say e = 0.95, M = 180 deg
Iteration History:
E0 = 180.0000 deg
Error = 0.0000 deg
Lecture 6: The Two Body Problem
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Solving Kepler’s Equation
• Let’s say e = 0.95, M = 300 deg
Iteration History:
E0 = 300.0000 deg
Value = 47.1386 deg
E1 = 210.2122 deg
Error = -62.3980 deg
E2 = 244.4787 deg
Error = -6.4014 deg
E3 = 249.0209 deg
Error = -0.1562 deg
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Lecture 6: The Two Body Problem
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3
0
E4 = 249.1375 deg
Error = -0.0001 deg
E5 = 249.1376 deg
Error = -0.0000 deg
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Solving Kepler’s Equation
• Let’s say e = 0.95, M = 350 deg
Iteration History:
E0 = 350.0000 deg
Value = 9.4518 deg
E1 = 203.3066 deg
Error = -125.1577 deg
E2 = 270.1472 deg
Error = -25.4220 deg
E3 = 295.6314 deg
Error = -5.2939 deg
E4 = 304.6185 deg
Error = -0.5873 deg
E5 = 305.8945 deg
Error = -0.0111 deg
Lecture 6: The Two Body Problem
E6 = 305.9195 deg
Error = -0.0000 deg 23
Solving Kepler’s Equation
• How many iterations does it take?
• e=0
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Solving Kepler’s Equation
• How many iterations does it take?
• e = 0.5
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Solving Kepler’s Equation
• How many iterations does it take?
• e = 0.9
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Solving Kepler’s Equation
• How many iterations does it take?
• e = 0.95
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Solving Kepler’s Equation
• How many iterations does it take?
• e = 0.99
Not good!
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Kepler’s Equation
• Algorithm 2 in Vallado:
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Orbital Elements
(Vallado, 1997)
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Orbital Elements
Now, let’s define our other orbital elements.
The inclination, i, refers to the tilt of the orbit plane. It is the
angle between Kˆ and h , and varies from 0-180°.
0 < i < 90
90 < i < 180
i = 90
Lecture 6: The Two Body Problem
Prograde orbit (w/Earth' s rotation)
Retrograde orbit (against Earth rotation)
Polar Orbit
Kˆ × h
cos i =
Kˆ h
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Orbital Elements
The right ascension of the ascending node, W, is the
angle in the equatorial plane from Iˆ to the ascending
node. The ascending node is the point on the equator
where the satellite passes from South to North
(opposite for the descending node).
The line of nodes connects the ascending and
descending nodes. The node vector, n , points
towards the ascending node and is denoted:
n = Kˆ ´ h
The node lies between 0° and 360°.
Iˆ × n nx
cos W =
=
If ny < 0, then W = 360 - W
Iˆ n n
Lecture 6: The Two Body Problem
hx
sin W =
hsini
cos W =
-hy
hsini
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Orbital Elements
The argument of periapse, w, measured from the ascending
node, locates the closest point of the orbit (periapse) and is the
angle between n and e .
n ×e
cos w =
ne
Lecture 6: The Two Body Problem
If (ez < 0 ), w = 360 - w
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Orbital Elements
The true anomaly, n, is the angle between periapse and the
satellite position; thus:
e ×r
cosn =
er
If (r × v) < 0, n = 360 -n
( r × v is positive going away from periapse, negative coming
towards periapse.)
Lecture 6: The Two Body Problem
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Special Cases
Elliptical Equatorial Orbits – W is undefined, so we use true
~ ,
longitude of periapse, w
true
cos w~true
Iˆ × e
=
Iˆ e
If (e y < 0) , then w~true = 360 - w~true
~,
This is equivalent to astronomers’ longitude of periapse, w
~ = W +w
where w
Lecture 6: The Two Body Problem
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Special Cases
Circular Orbits – w is undefined, use argument of latitude, u,
where:
n×r
cos u =
nr
Lecture 6: The Two Body Problem
If (rz < 0) , then u = 360 - u
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Special Cases
Circular Equatorial – w and W undefined
Iˆ × r
cos( ltrue ) =
Iˆ r
Lecture 6: The Two Body Problem
If ( rj < 0 ), ltrue = 360 - ltrue
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Two Line Element Sets
(Vallado, 1997)
Available on class web page. (Can be read by many programs
including STK.)
n= m
n
2
n
6
a
3
e
i
W
1 CD A
B* =
r0
2 m
w
M
UTC
a , n are “Kozai” means. B* is a drag parameter.
Lecture 6: The Two Body Problem
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Two Line Element Sets
Example
1 16609U 86017A 93352.53502934 .00007889 00000 0 10529-3 34
2 16609 51.6190 13.3340 0005770 102.5680 257.5950 15.59114070 44786
Epoch: Dec 18, 1993 12h 50min 26.5350 sec UTC
n = 15.59114070 rev / day Þ a = 1.06118087 ER = 6768.357 km
n
= 7.889 ´ 10 -5 rev / day 2
2
B* = 1.0529 ´ 10 - 4
W = 13.3340
n
= 0.0 rev / day 3
6
e = 0 .0005770
w = 102.5680
i = 51.6190
M = 257.5950
Errors can be as large as a km or more.
Lecture 6: The Two Body Problem
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Orbital Elements from r and v (and t)
p. 112 - 116
Algorithm 9 in the book
First compute the following vectors
h = r ´v
n = Kˆ ´ h
h= h
n= n
v2 m
Compute the energy: - = e
2 r
2
ö
m æ
p
h
a=çor p = m Þ a =
2 ÷
(1e
)ø
è
2e
e=
v´h
m
-
r
r
e= e
2
é
ù
e= e
or e = ê1+ 2eh 2 ú
m û
ë
h
i = cos-1 ( z h )
12
Lecture 6: The Two Body Problem
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Orbital Elements from r and v (and t)
Test using Example 2-5 in book
Also,
n
1+ e
E
tan( ) =
tan( )
2
1- e
2
M = E - esin E
(E - esin E)
tp = t n
n=
Lecture 6: The Two Body Problem
m
a3
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STK
• STK! Systems Tool Kit version 10.0
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Welcome Splash
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STK Boot-up
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Save
• Save frequently
• Save each scenario in its own directory
• Please note: each scenario includes many files, for
each object. If two scenarios are in the same
directory and share objects, they may not behave
properly.
Lecture 6: The Two Body Problem
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Lecture 6: The Two Body Problem
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A satellite
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View
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ASEN 5050
SPACEFLIGHT DYNAMICS
Coordinate and Time Systems
Prof. Jeffrey S. Parker
University of Colorado - Boulder
Lecture 6: The Two Body Problem
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Coordinate Systems
• Given a full state, with position and velocity known.
• Or, given the full set of coordinate elements.
• What coordinate system
is this state represented
in?
• Could be any nonrotating coordinate
system!
• Earth J2000 or ecliptic
J2000 or Mars, etc.
Lecture 6: The Two Body Problem
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Coordinate Systems
Celestial Sphere
– Celestial poles intersect
Earth’s rotation axis.
– Celestial equator extends
Earth equator.
– Direction of objects
measured with right
ascension (a) and
declination (d).
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Coordinate Systems
The Vernal Equinox defines the reference
direction. A.k.a. The Line of Aries
The ecliptic is defined as the mean plane of
the Earth’s orbit about the Sun.
The angle between the Earth’s mean
equator and the ecliptic is called the
obliquity of the ecliptic, e~23.5.
Lecture 6: The Two Body Problem
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Coordinate Frames
• Inertial: fixed orientation in space
– Inertial coordinate frames are typically tied to hundreds of
observations of quasars and other very distant near-fixed
objects in the sky.
• Rotating
– Constant angular velocity: mean spin motion of a planet
– Osculating angular velocity: accurate spin motion of a
planet
Lecture 6: The Two Body Problem
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Coordinate Systems
• Coordinate Systems = Frame + Origin
– Inertial coordinate systems require that the system be nonaccelerating.
• Inertial frame + non-accelerating origin
– “Inertial” coordinate systems are usually just non-rotating
coordinate systems.
• Is the Earth-centered J2000 coordinate system
inertial?
Lecture 6: The Two Body Problem
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Useful Coordinate Systems
• ICRF
• International Celestial Reference Frame, a realization of the ICR System.
• Defined by IAU (International Astronomical Union)
• Tied to the observations of a selection of 212 well-known quasars and other
distant bright radio objects.
– Each is known to within 0.5 milliarcsec
• Fixed as well as possible to the observable universe.
• Motion of quasars is averaged out.
– Coordinate axes known to within 0.02 milliarcsec
• Quasi-inertial reference frame (rotates a little)
• Center: Barycenter of the Solar System
Lecture 6: The Two Body Problem
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Useful Coordinate Systems
• ICRF2
• Second International Celestial Reference Frame, consistent with the first
but with better observational data.
• Defined by IAU in 2009.
• Tied to the observations of a selection of 295 well-known quasars and other
distant bright radio objects (97 of which are in ICRF1).
– Each is known to within 0.1 milliarcsec
• Fixed as well as possible to the observable universe.
• Motion of quasars is averaged out.
– Coordinate axes known to within 0.01 milliarcsec
• Quasi-inertial reference frame (rotates a little)
• Center: Barycenter of the Solar System
Lecture 6: The Two Body Problem
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Useful Coordinate Systems
• EME2000 / J2000 / ECI
• Earth-centered Mean Equator and Equinox of J2000
– Center = Earth
– Frame = Inertial (very similar to ICRF)
• X = Vernal Equinox at 1/1/2000 12:00:00 TT (Terrestrial Time)
• Z = Spin axis of Earth at same time
• Y = Completes right-handed coordinate frame
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Useful Coordinate Systems
• EMO2000
• Earth-centered Mean Orbit and Equinox of J2000
– Center = Earth
– Frame = Inertial
• X = Vernal Equinox at 1/1/2000 12:00:00 TT (Terrestrial Time)
• Z = Orbit normal vector at same time
• Y = Completes right-handed coordinate frame
– This differs from EME2000 by ~23.4393 degrees.
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Useful Coordinate Systems
• Note that J2000 is very similar to ICRF and ICRF2
– The pole of the J2000 frame differs from the ICRF pole by ~18 milliarcsec
– The right ascension of the J2000 x-axis differs from the ICRF by 78 milliarcsec
• JPL’s DE405 / DE421 ephemerides are defined to be consistent with the
ICRF, but are usually referred to as “EME2000.” They are very similar,
but not actually the same.
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Useful Coordinate Systems
• ECF / ECEF / Earth Fixed / International Terrestrial
Reference Frame (ITRF)
• Earth-centered Earth Fixed
– Center = Earth
– Frame = Rotating and osculating (including precession,
nutation, etc)
• X = Osculating vector from center of Earth toward the equator
along the Prime Meridian
• Z = Osculating spin-axis vector
• Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem
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Useful Coordinate Systems
• Earth Rotation

The angular velocity vector ωE is
not constant in direction or
magnitude
◦ Direction: polar motion
 Chandler period: 430 days
 Solar period: 365 days
◦ Magnitude: related to length of day
(LOD)

Lecture 6: The Two Body Problem
Components of ωE depend on
observations; difficult to predict
over long periods
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Useful Coordinate Systems
• Principal Axis Frames
• Planet-centered Rotating System
– Center = Planet
– Frame:
• X = Points in the direction of the minimum moment of inertia, i.e.,
the prime meridian principal axis.
• Z = Points in the direction of maximum moment of inertia (for
Earth and Moon, this is the North Pole principal axis).
• Y = Completes right-handed coordinate frame
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Useful Coordinate Systems
• IAU Systems
• Center: Planet
• Frame: Either inertial or fixed
• Z = Points in the direction of the spin axis of the body.
– Note: by convention, all z-axes point in the solar system North
direction (same hemisphere as Earth’s North).
– Low-degree polynomial approximations are used to compute the
pole vector for most planets wrt ICRF.
• Longitude defined relative to a fixed surface feature for rigid
bodies.
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Useful Coordinate Systems
• Example:
– Lat and Lon of Greenwich, England, shown in EME2000.
– Greenwich defined in IAU Earth frame to be at a constant
lat and lon at the J2000 epoch.
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Useful Coordinate Systems
• Synodic Coordinate Systems
• Earth-Moon, Sun-Earth/Moon, Jupiter-Europa, etc
– Center = Barycenter of two masses
– Frame:
• X = Points from larger mass to the smaller mass.
• Z = Points in the direction of angular momentum.
• Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem
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Announcements
• Friday: no class at 9:00. STK Lab 1
–
–
–
–
–
ITLL 1B10, 2B10, or Visions Lab – or on your own computer.
Groups of 1-3.
Intro to STK today
Alan will be in ITLL 2B10 to help answer questions from 1-3.
Due Friday 9/26 at 9:00 am
• Homework #2 is due Friday 9/12 at 9:00 am
– D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or
9:10 so don’t be late!
• Homework #3 is due Friday 9/19 at 9:00 am
• Concept Quiz #5 will be available at 10:00 am, due Monday morning at
8:00 am.
• Reading: Chapters 1 and 2
Lecture 6: The Two Body Problem
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