with your host….

Dr. Hy, the rocket scientist guy

AERO 426, Lecture #5 Spacecraft Dynamics- Questions Addressed

Regarding available and future launch systems, what are the implications for cost versus payload size, weight, etc.?

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?

Suggested reading:Suggested reading:L&W, Chap.5 intro or P&M, Sect. 3.3 (coordinate systems),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. 6.1.1 - 6.1.3 or P&M, Sect. 3.6 (Keplerian orbits),L&W, Sect. 6.3 (orbit maneuvering),L&W, Sect. 6.3 (orbit maneuvering),L&W, Sect. 17.2 or P&M, Sect. 4.2.1 and 4.3 (rocket L&W, Sect. 17.2 or P&M, Sect. 4.2.1 and 4.3 (rocket

propulsion propulsion and motion),and motion),

L&W, Sect. 17.3 (types of rockets),L&W, Sect. 17.3 (types of rockets),L&W, Sect. 18.2 (launch system data)L&W, Sect. 18.2 (launch system data)

What’s our coordinates? Use Nature’s Gyros!

RES

Jorbit ~ RESX MEVE ~ Const

Orbit plane is fixed

VE

Jspin~ Constant

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)

Ecliptic Pole, Z-axis

Orbit Plane

Actually, in the Ecliptic coordinate system;We use the normal to the orbit plane (called the

Ecliptic Pole) as the Z-axis

In the position of the vernal equinox, the rotation axis

vector is perpendicular

to the Sun-Earth vector and Northern Hemisphere

spring commences

X-axis

Y-axis

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

Spacecraft-fixed

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 seconds0.01 = 0.24 hours or 14.4 minutes or 864 seconds0.001 = 0.024 hours or 1.44 minutes or 86.4 seconds0.0001 = 0.0024 hours or 0.144 minutes or 8.64 seconds0.00001 = 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 SimpleMost 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

xi

zi

yi

i

h

ei

hi

pi

r

i

f

Equitorial Ecliptic plane

Orbit Plane

Periapsis

Longitude of the ascending node

Inclination angle

Argument of the perihelion

True anomaly

i

f

angular momentum vector

ˆ Unit vector along

ˆ Unit vector pointing to perihelion

ˆ ˆ ˆ

e

p e

h

h

h

i h

i

i i i

Orbital Dynamics - Briefly Summarized

vf

d

v0

rmax

parabola

ellipsehyperbola

v0 = vf d(v0)2/

circle

rmin

vmax

For all orbits:

= / rmin

E = v2/2 / r = / 2a a = (rmax + rmin)/ 2Circle:

v = vmax = 1/2 E = E0 = - /2

Ellipse:

rmax = rmin (vmax)2 / (2 - (vmax)2)0 E E0

Parabola: vmax = (2)1/2 =vescape

E = 0

Hyperbola: vmax = v0 [ 1 + (1 + 2)1/2] / , rmin = dv0 / vmax

sin() = /(1 + 2 )1/2

E 0

hyperbola

For bound orbits:3

2a

P

Location of a Body in its Orbit as a Function of Time

2

1 cos

hr

e f

b

f

a F1

E

ae

F2

r

0 0

3

sin ,

Mean anomaly

0,2

2mean angular motion

Orbit period 2

M E e E E K eM

M M n t t

M

nP

aP

Kepler's Equation :

1

2

12tan tan

1 2

1 cos1

e Ef

e

pr

e fp a e

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 asteroid-dominated portion of the

rendezvous

When outside the Earth’s activity

sphere, calculate only the S/C orbit around the Sun. (which follows a conic section).

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.

Asteroid

Earth

Sun

V (km/s) Topography

Mars

Phobos

Earth

Low Mars orbit

Phobos transfer

Deimos transfer

Deimos

Mars transfer

Mars C3

L4/5

GEO

GTO

Earth C3

Sun

LEO

Moon

Lunar orbit9.3 - 10

4.1

4.1

2.5

0.7

1.73.8

1.6

1.61.6

0.2

0.9

0.6

30

0.7

0.7

0.7

0.90.5

0.3

Optional Aerobrake

Orbital location

Low Thrust Transfer ManeuversSuppose we have a very low thrust

engine that provides constant accel-

eration , . It's most efficient to

direct the thrust along the velocity

vector of an initially circular orbit.

In this

A

32

case, the orbit semimajor axis

is slowly varying and approximately

satisfies:

2da A

adt

12 120 0

20

0 00

:

So the time required to go from the initial semimajor axis to the final one is:

1 1, 1 4

Then

t t a aA

aAt t a a

a a

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):

0 0 ,

where is the vehicle mass, and is the constant thrust. Also, substituting into the

previous equation for the semimajor axis, we g

sp

dm Fm t m

dt gI

m F A F m

3 2

00

et:

2

These two equations can be integrated to obtain:

1 1 1 expsp

sp

da Fa

dt m

gI mt t

F gI a a

0

0 0 sp

Fm m t t

gI

Low Thrust Transfer Maneuvers - Continued

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)

2 212

1 2

2 22 2 2 21 2

2 1 2

2 212

,

,

where:

1

, 1

Energy Integral:

E= , 2 Jacobi Integral

x x y

y y x

Ux p y p p x

xU

y p x p p yy

U x yr r

r x y r x y

m m m

x y U C E

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

x

y

z

angular momentumH

precession

angle

x and y axes moment of inertia = A

Z axis inertia = C

2

2

cos onstant

kinetic energy

1 cos2

Precession rate

H C c

T

H C C

C A A

C A

A

spin rate

For we have a problem. When there

are any moveable bodies within the interior,

the precession will excite their motion, and

the rigid body kinetic energy will be drained.

continually. A sta

C A

te of steady axial spin, 0

is actually unstable.

Rotational Dynamics of Axisymmetric Rigid Bodies

2

22

If we start with a pure z-axis spin ( 0). then 0 2 . Since remains constant:

2 1 sin 0

1

Thus as decreases, increases. Ultimately, 2 an

T t H C H

CT T t

C AH

T t

d the system moves in a flat tumble.

x

y

z

This does not happen when C >A

Propulsion Function Comments/ Typical RequirementsLaunch 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 OperationsOrbit correction V

Stationkeeping V

"Formation Flying" V's

15 to 75 m/s per year, for Earth orbits

Up to 45 to 55 m/s per year, Earth orbits

Could be relevant to stand-off mode of operation.

Attitude control

Acquisition of Sun, Earth, Star - for navigational and target acquisition purposes

In-mission pointing control, 3-axis stabilization

< 5000 N-s total impulse, 1K to 10K pulses, 0.01 to 5.0 s pulse width

100K to 200K pulses, min impulse bit of 0.01N-s, 0.01 to 0.25s pulse width.

Propulsion Systems - Key Parameters

F

CombustionChamber

Ve

Oxidizer Fuel

dm/dt

Nozzle

F Thrust Ve (dm/dt)

Ve =exhaust velocity

dm/dt = propellant and oxidizer mass

flow rate

Isp Specific Impulse

= F / (g dm/dt)

-- depends on propulsion type (liquid, solid, chemical, electric, 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 eventmp Mass of propellant (& oxidizer) used in burn event

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

Trajectory Parameters/ Propulsion System Relations:

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 SystemsKey Parameters are:

Mass of payload that can be injected into LEO or GTO or GEO

Fairing diameter and length

Launch SystemUpper Stage (if

any)LEO(kg)

GTO(kg)

GEO(kg)

Fairing Envelopes

Diam(m) Length (m)

ATLAS II Cent-2 6395 2680 570 4.2

SHUTTLE IUSTOSPAM-D

24,400------

--590059001300

--2360----

4.6 18.3

TITAN III NUSPAM-D2TRANTOS

14,400------

--185043105000

--13601360--

3.6 12.415.516.0

TITAN IV NUSCentIUS

17,700----

--57606350

--45402380

4.5 17.020.023.0,26.0

ARIANE 40(France)42P42L44P44LP44L

H-10H-10H-10H-10H-10H-10EPS

49006100740069008300960018,000 at 550

km

1900260032003000370042006800

--------------

3.6 8.6 to 12.4

H-2 (Japan) -- 10,500 4000 2200 3.7 10.0

LONG MARCH (China)CZ2E

Star 63F 9265 3370 1500 3.8 7.5

PROTON(Russia)ENERGIAZENIT 2

D1D1eEUS, RCSBlock D

20,00090,00013,740

--5500--4300

--220018,0004100

3.34.15.53.3

4.2-7.519-375.8-9

Data for Systems with Fairing Diameters >3.0 m

and watch out for those irate Romulans!