Transcript Slide 1

with your host….
Dr. Hy, the rocket scientist guy
AERO 426, Lecture #5
Spacecraft Dynamics- Questions Addressed
How can we tell where our spacecraft is ?
What are some simple ways to estimate the motion of spacecraft in the
vicinity of a NEA?
How can we plan space trajectories and estimate propulsion system
requirements?
Regarding available and future launch systems, what are the
implications for cost versus payload size, weight, etc.?
Suggested reading:
L&W, Chap.5 intro or P&M, Sect. 3.3 (coordinate systems),
L&W, Sect. 6.1.1 - 6.1.3 or P&M, Sect. 3.6 (Keplerian orbits),
L&W, Sect. 6.3 (orbit maneuvering),
L&W, Sect. 17.2 or P&M, Sect. 4.2.1 and 4.3 (rocket propulsion
and motion),
L&W, Sect. 17.3 (types of rockets),
L&W, Sect. 18.2 (launch system data)
What’s our coordinates? Use Nature’s Gyros!
Jorbit ~ RESX MEVE ~ Const
 Orbit plane is fixed
Jspin~
Constant
RES
VE
So, we have two axes that are fixed: The perpendicular to the orbit
plane and the axis of rotation of the Earth (which actually nutates once
every 26,000 years)
Actually, in the Ecliptic coordinate system;
We use the normal to the orbit plane (called the
Ecliptic Pole) as the Z-axis
Y-axis
In the position
of the vernal
equinox, the
rotation axis
vector is
perpendicular
to the SunEarth vector
and Northern
Hemisphere
spring
commences
Ecliptic Pole,
Z-axis
X-axis
Orbit
Plane
Coordinate systems used in space applications
Coordinate
Name
Fixed with
respect to
Center
Z-axis or
Pole
X-axis or
Ref. Point
Applications
Celestial
(Inertial)
Inertial
space
Earth or
spacecraft
Celestial
Pole
Vernal
equinox
Orbit analysis,
astronomy,
inertial motion
Earth-fixed
Earth
Earth
Earth
pole=celestial
pole
Greenwich
meridian
Geolocation,
apparent
satellite motion
Spacecraftfixed
Spacecraft
Defined by
engineering
drawings
Spacecraft
axis toward
nadir
Spacecraft
axis in
direction of
velocity vector
Position and
orientation of
spacecraft
instruments
Ecliptic
Inertial
space
Sun
Ecliptic pole
Vernal
equinox
Solar system
orbits,
lunar/solar
ephemerides
Lunar
The Moon
Moon
Lunar North
pole
Average
center of
Lunar Disk
Locating lunar
features
Locating Events in Time
The Julian day or Julian day number (JDN) is the integer number of days that have elapsed
since the initial epoch defined as noon Universal Time (UT) Monday, January 1, 4713 BC in
the Julian calendar.
The Julian date (JD) is a continuous count of days and fractions elapsed since the same
initial epoch. The integral part gives the Julian day number. The fractional part gives the
time of day since noon UT as a decimal fraction of one day with 0.5 representing midnight
UT.
Example: A Julian date of 2454115.05486 means that the date and Universal Time is Sunday
14 January 2007 at 13:18:59.9.
The decimal parts of a Julian date:
0.1 = 2.4 hours or 144 minutes or 8640 seconds
0.01 = 0.24 hours or 14.4 minutes or 864 seconds
0.001 = 0.024 hours or 1.44 minutes or 86.4 seconds
0.0001 = 0.0024 hours or 0.144 minutes or 8.64 seconds
0.00001 = 0.00024 hours or 0.0144 minutes or 0.864 seconds.
The Julian day system was introduced by astronomers to provide a single system of dates
that could be used when working with different calendars. Also, the time separation
between two events can be determined with simple subtraction.
To make conversions, several handy web-sites are available; e.g.,
http://aa.usno.navy.mil/cgi-bin/aa_jdconv.pl
Orbital Dynamics - Made Simple
Most of the time (with many important exceptions) spacecraft orbital dynamics
involves bodies that are either (1) very, very small relative to inter-body distances, or (2)
are nearly spherically symmetric -- then:
Bodies behave (attract and are attracted) as if they are point masses.
Motion can be described by keeping track of the centers of mass.
Also, most of the time (with many important exceptions) spacecraft orbital dynamics is
a two-body problem (the s/c and the Earth, or the s/c and the sun, or, etc.) - so we have
two gravitationally attracting point masses, and:
Both bodies move in a plane (the same plane)
Both trace out conic sections with one focus at the total center of mass.
Each body moves periodically on its conic section, tracing and retracing the same curve
forever.
Finally,
most of the time (with many important exceptions), one of the bodies is
much more massive than the other ( the Earth versus a s/c, or the sun versus the Earth,
etc.). Then in addition to the above:
 The smaller body moves on a conic section with a focus on the larger body's center
of mass, which is also approximately the total center of mass.
 The motion of the smaller body does not depend on its mass.
 The smaller body's motion depends on the gravitational constant, G, and the larger
body's mass only through the combination:
 = "The Gravitational Parameter"
= GM
G = 6.673 x 10-11 m3/ kg-s2
M = Mass of the larger body
Euler Angle Description of the Orbit Plane

ˆi
z
i
Orbit Plane
i
h
Longit ude of t he ascending node
Inclinat ion angle

Argument of t he perihelion
f
T rue anomaly
r
Equitorial
Ecliptic plane
ˆi
p
f
ˆi
h
ˆi
e


i
Periapsis
ˆi
x
ˆi
y
h
angular moment um vect or
ˆi
h
Unit vect or along h
ˆi
e
Unit vect or point ing t o perihelion
ˆi
p
ˆi  ˆi
h
e
Orbital Dynamics - Briefly Summarized

d
Parabola:
parabola
vmax = (2e)1/2 =vescape
E=0
v0
hyperbola
v0 = vf
d  d(v0)2/
Hyperbola:
vmax = v0 [ 1 + (1 + d2)1/2] / d,
rmin = dv0 / vmax
sin() = d /(1 + d2 )1/2
E0
rmax
vf
ellipse
hyperbola
Ellipse:
rmax = rmin (vmax)2 / (2 e - (vmax)2)
0  E  E0
circle
Circle:
v = vmax = e1/2
E = E0 = - e /2
For all orbits:
e =  / rmin
E = v2/2 -  / r = - / 2a
a = (rmax + rmin)/ 2
rmin
For bound orbits:
P  2
vmax
a3

Location of a Body in its Orbit as a Function of Time
K epler 's Equat ion :
M  E - e sin E  E  K e, M 
M
Mean anomaly  M 0  n t - t 0 
M   0, 2 
b
n
mean angular mot ion 
P
Orbit period  2
r
F2
E
r 
a
f
h2 
1  e cos f
F1
ae
a3

 1e
E
f  2 t an -1 
t an 
2 
 1 - e
p
r 
1  e cos f
p  a 1 - e2


2
P
Getting from Earth to a NEA - Patched Conics Method
When S/C crosses
asteroid’s activity sphere
boundary, subtract the
asteroid’s velocity relative to
the sun.This gives initial
conditions for the asteroiddominated portion of the
rendezvous
Asteroid
Sphere of Influence of the
asteroid:
S/C acceleration due to asteroid >
Perturbing acceleration due to the
Earth. SI radius given by:
RSI  RA-E (Masteroid /MEarth)2/5
(Masteroid = 4.6X1010 kg
MEarth=5.9737X1024 kg )
(Masteroid /MEarth)2/5= 2.2626X10-6
Within SI and ref. frame moving
with the asteroid, S/C approx.
interacts only with the asteroid.
Sun
Earth
When outside the
Earth’s activity
sphere, calculate
only the S/C orbit
around the Sun.
(which follows a
conic section).
V (km/s) Topography
Mars
Sun
Low Mars orbit
Phobos
0.5
4.1
0.9
Phobos transfer
0.3
Deimos transfer
0.7
0.2
30
Mars C3
Deimos
0.9
Mars transfer
0.6
Optional
Aerobrake
Earth C3
0.7
Orbital
location
GTO
2.5
1.6
1.6
0.7
GEO
LEO
3.8
4.1
1.7
L4/5
9.3 - 10
0.7
Earth
Lunar orbit
1.6
Moon
Low Thrust Transfer Maneuvers
Suppose we have a very low thrust
engine that provides constant acceleration , A. It's most efficient to
direct the thrust along the velocity
vector of an initially circular orbit.
In this case, the orbit semimajor axis
is slowly varying and approximately
satisfies:
da 2 A 3 2

a
dt

Then :
1
1
A
 t - t 0  , a  a0
a0
a 
1- 4
a 02


So the time required to go from the initial semimajor axis to the final one is:
t - t0 

A
a
-1 2
0
- a -1 2 
Low Thrust Transfer Maneuvers - Continued
In the previous chart we considered the case of constant acceleration. Now considering constant
thrust, we use the relation between mass flow rate and thrust (see next two charts):
dm
F
, m  t 0   m0
dt
gI sp
where m is the vehicle mass, and F is the constant thrust. Also, substituting A  F m into the
previous equation for the semimajor axis, we get:
da
2F 3 2

a
dt m 
These two equations can be integrated to obtain:
t - t0 
  1
gI sp m 0 
1  
1 - exp 

  
F 
gI
a 0   
 sp  a

F
m0 - m 
t - t 0 
gI sp
Planar Circular Restricted 3-Body Problem
(PCR3BP)
• “Restricted” = Gravitational field is determined by two
massive bodies (The “primaries”). The third body is too
small to affect the primaries.
• “Circular” = The primaries are in circular orbits about the
total center of mass
• “Planar” = All three bodies move in the same plane.
• Normalized Units:
–
–
–
–
Unit of mass = m1+m2
Unit of length = constant separation between m1and m2
Unit of time: Orbital period of m1and m2 is 2 (G = 1)
The only parameter in the system is  = m1/(m1+m2)
Unit of distance: L = distance between m1 and m2 (km)
Unit of Velocity: V = orbital velocity of m2 (km/s)
Unit of time = orbital period of the primaries (s)
Equations of Motion (In the rotating frame)
U
px  p y - x x
U
p y  - px - y y
x  px  y ,
y  p y - x,
where:
1-  
U  - x  y r1
r2
1
2
2
2
r   x     y , r   x -1    y2
2
1
2
2
2
2
2
  m2  m1  m2 
Energy Integral:
E= 12  x 2  y 2   U ,
 C  -2 E  Jacobi Integral 
Planar Circular Restricted Three Body Problem
(PCR3BP)
Effective Potential:
The Open Realms and the Forbidden Zone
Five Cases of Possible Motions
Types of Orbits in the “Neck” Region
Tangled Trajectories in the Neck Region
Structure of the Neck Region
Global Orbit Structure: Homoclinic/Heteroclinic Chains
Patched 3-Body Method:
The Interplanetary Super Highway
Patched 3-Body Method:
LL1 to EL2 in 40 days with a single 14m/s V
Patched 3-Body Method:
Space Mission Application
Rotational Dynamics of Axisymmetric Rigid Bodies
H  angular momentum
  spin rate
Z axis inertia = C
z

  precession
angle
T  kinetic energy
C  C  2


1
cos



A  A





C-A
  Precession rate 

A
H2

2C
For C  A we have a problem. When there
are any moveable bodies within the interior,
x
y
x and y axes moment of
inertia = A
H  C  cos    constant
the precession will excite their motion, and
the rigid body kinetic energy will be drained.
continually. A state of steady axial spin,   0
is actually unstable.
Rotational Dynamics of Axisymmetric Rigid Bodies
If we start with a pure z-axis spin (  0). then T  t  0   H 2 2C . Since H remains constant:
sin 2 
2C
1
T  0  - T  t  
2
H 1- C A 
Thus as T  t  decreases,  increases. Ultimately,    2 and the system moves in a flat tumble.
z


x
y
This does not happen when C >A
Propulsion Function
Comments/ Typical Requirements
Launch and injection into LEO
Really in the domain of "Launch Systems" - which we discuss
separately below
DV for raising the orbit from LEO to a higher orbit
60 to 1500 m/s, Use kick motor
Acceleration to escape velocity from LEO parking orbit
3600 to 4000 m/s for injection into an interplanetary trajectory
Interplanetary trajectory - From Earth escape to in-mission parking
orbit.
Depends heavily on the trajectory design - Have a wide choice
among min energy maneuvers, swing-by maneuvers, etc.
In-Mission Operations
Orbit correction V
15 to 75 m/s per year, for Earth orbits
Stationkeeping V
Up to 45 to 55 m/s per year, Earth orbits
"Formation Flying" V's
Could be relevant to stand-off mode of operation.
Attitude control
Acquisition of Sun, Earth, Star - for navigational and target
acquisition purposes
< 5000 N-s total impulse, 1K to 10K pulses, 0.01 to 5.0 s pulse width
In-mission pointing control, 3-axis stabilization
100K to 200K pulses, min impulse bit of 0.01N-s, 0.01 to 0.25s pulse
width.
Propulsion Systems - Key Parameters
Oxidizer
Fuel
dm/dt
F  Thrust
 Ve (dm/dt)
Ve =exhaust
velocity
Combustion
Chamber
F
Nozzle
dm/dt = propellant
and oxidizer
mass flow
rate
Isp  Specific Impulse
= F / (g dm/dt)
-- depends on propulsion type
(liquid, solid, chemical, electric,
Ve
etc.) , energetics of chemical
reactions, etc.
Key Propulsion Parameters Related to
Important Trajectory Parameters
Suppose we have a thruster burn event with constant thrust (maybe
to inject the spacecraft into a higher orbit, etc.).
Define: m0  Total mass of vehicle before burn event
mp Mass of propellant (& oxidizer) used in burn event
Trajectory Parameters/ Propulsion System Relations:
ΔV = Total change in vehicle speed
= g Isp ln (m0/( m0 - mp))
Δt = Time elapsed during burn event
= g Isp mp/ F
•Trajectory Requirements
Needed V and t
•Use above relations to estimate total mass of propellant
•Select propulsion system (F & Isp) and design trajectory to minimize total propulsion system
mass
Determining Propulsion System Requirements
- For Transport of S/C to its Mission Station
Lay out the entire trajectory and itemize the V maneuvers.
Start from the last V maneuver and use the V/ mp equation to determine mp
(where here, m0 - mp = the final S/C mass), for several values of Isp
From considerations of the t desired, or other practical constraints,
determine any thrust level requirements. Now narrow the selection of propulsion
systems to those consistent with required thrust levels.
Now, carry out the above process for all the V
maneuvers, working back along the trajectory. Get a range of
values for mp and F.
Finally, obtain the total
propulsion system masses
corresponding to different
propulsion system options.
Select option with smallest
cost and/or launch weight.
Launch Systems
Key Parameters are:
Mass of payload that can be injected into LEO
or GTO or GEO
Fairing diameter and length
Data for Systems with Fairing Diameters >3.0 m
Fairing Envelopes
Launch System
Upper Stage (if
any)
LEO
(kg)
GTO
(kg)
GEO
(kg)
Diam
(m)
Length (m)
ATLAS II
Cent-2
6395
2680
570
4.2
SHUTTLE
IUS
TOS
PAM-D
24,400
----
-5900
5900
1300
-2360
---
4.6
18.3
TITAN III
NUS
PAM-D2
TRAN
TOS
14,400
----
-1850
4310
5000
-1360
1360
--
3.6
12.4
15.5
16.0
TITAN IV
NUS
Cent
IUS
17,700
---
-5760
6350
-4540
2380
4.5
17.0
20.0
23.0,26.0
ARIANE 40
(France)
42P
42L
44P
44LP
44L
H-10
H-10
H-10
H-10
H-10
H-10
EPS
4900
6100
7400
6900
8300
9600
18,000 at 550
1900
2600
3200
3000
3700
4200
6800
--------
3.6
8.6 to 12.4
km
H-2 (Japan)
--
10,500
4000
2200
3.7
10.0
LONG MARCH (China)
Star 63F
9265
3370
1500
3.8
7.5
D1
D1e
EUS, RCS
Block D
20,000
90,000
13,740
-5500
-4300
-2200
18,000
4100
3.3
4.1
5.5
3.3
4.2-7.5
19-37
5.8-9
CZ2E
PROTON
(Russia)
ENERGIA
ZENIT 2
and watch out for those irate Romulans!