5b. orbital maneuvers - uliege motivation coplanar transfers have wide applicability (e.g., phasing...

Gaëtan Kerschen

Space Structures &

Systems Lab (S3L)

5B. Orbital Maneuvers

Astrodynamics (AERO0024)

2

Previous Lecture: Coplanar Maneuvers

5.1 INTRODUCTION

5.1.1 Why ?

5.1.2 How ?

5.1.3 How much ?

5.1.4 When ?

5.2 COPLANAR MANEUVERS

5.2.1 One-impulse transfer

5.2.2 Two-impulse transfer

5.2.3 Three-impulse transfer

5.2.4 Nontangential burns

5.2.5 Phasing maneuvers

3

Motivation

Coplanar transfers have wide applicability (e.g., phasing

maneuvers and orbit raising). They alter 4 orbital elements:

1. Semi-major axis.

2. Eccentricity.

3. True anaomaly.

4. Argument of perigee.

Noncoplanar transfers requires applying v out of the orbit

plane and can therefore alter the last two elements:

1. Inclination.

2. RAAN.

4

5. Orbital Maneuvers

5.3 Noncoplanar transfers

5.4 Rendez-vous

http://upload.wikimedia.org/wikipedia/commons/d/d2/Gemini_6_Views_Gemini_7.jpg

5


5.3 Noncoplanar transfers

5.3.1 Inclination change

5.3.2 RAAN change


6

Gaussian VOP (Lecture 4)

2

2cos(1 )

1 cos

Na ei

e

2

2sin(1 )

sin 1 cos

Na e

i e

ˆ ˆ ˆR R T NT N F e e e

2


7

Where ?

Cranking maneuver: with a single v maneuver, one

wants to change only the inclination of the orbit plane.

Where should we apply this maneuver ?


8

Where ? At an Equator Crossing

The two orbit planes intersect in the equatorial plane. We

can therefore change the inclination at an equator

crossing, by a simple rotation of the velocity vector.

vrv

v

View down the line of intersection

of the two orbital planes


9

ΔV Computation for a Circular Orbit

2 1 2 1 2 2 1 1ˆ ˆ ˆ

r r rv v v v v v v u u u

22 2 2

2 1 1 2 1 12 cosr rv v v v v v v

1 2 1 2,r rv v v v

2 sin2

v v


10

Inclination Changes Are Very Expensive

0 10 20 30 40 50 60 70 80 900

0.5

1

1.5

, degrees

v

(%

of

v)

(24º , 41.6%)

(60º , 100%)


How to minimize ΔV ?

Is this cranking maneuver

the minimum-fuel transfer ?

11

Real-Life Examples: Space Shuttle and Hubble

The Space Shuttle is capable of a plane change in orbit of

only about 3º, a maneuver which would exhaust its entire

fuel capacity.

What is Hubble space telescope’s inclination ?

28.46 (HST) vs. 28.35 (KSC)…


12

What’s Unique with this Picture ?


Too costly ! Instead, NASA has

chosen to have another shuttle

ready to lift off to retrieve the

astronauts if needed.

13

Minimizing ΔV

1. Do not modify the radial velocity.

2. The maneuver should occur where the azimuth

component is smallest, which is at apogee.


14

Minimum-Fuel Transfer: Bi-Elliptic Transfer


One can always save fuel over a one-impulse maneuver by

using three-impulses: noncoplanar bi-elliptic transfer.

J.E. Prussing, B.A. Conway, Orbital Mechanics, Oxford University Press

15

Motivation for RAAN Change

5.3.2 RAAN change

Constraints on the line of nodes may be imposed to a

spacecraft (e.g., for rendez-vous):

Launch window to match up the line of nodes.

If we wait too long for such a window, the satellite can be launched

sooner and thrust can be applied to change the line of nodes.

16

Why is RAAN Change Complicated ?

A change in the node affects the argument of perigee

(except for circular and polar orbits) !

5.3.2 RAAN change

3

22 Resin 1 cos

1

aa T e

e

2(1 )

sin cos cosa e

e R T E

2

2cos(1 )

1 cos

Na ei

e

2 sin 2 cos1 (1 )cos cos

1 cos

T ea ei R

e e

2 2(1 ) 2 cos cos sin 2 cos, with

1 cos

e R e e T eaM nt

e e

2

2sin(1 )

sin 1 cos

Na e

i e

17

RAAN Change: Circular Orbits

One-burn is sufficient. But where ?

Applying a change in velocity at any other points than the

common points of intersection of the orbital planes will

change both inclination and RAAN.

5.3.2 RAAN change

18


5.4 Rendez-vous

5.4.1 Hill – Clohessy – Wiltshire equations

5.4.2 Gemini and Space Shuttle


19

Rendez-vous: Two Successive Phases

1. Long-distance navigation: basic maneuvers bring the

spacecraft in a vicinity (<100 km) of the target in the

same orbit plane.

2. Terminal rendez-vous: the interest is in the relative

motion, i.e. the motion of the maneuvering spacecraft in

relation to the target (spacecraft or neighboring orbit).

Focus of the present discussion.

5.4.1 H-C-W equations

Rescue of Intelsat-6: new

perigee kick motor (STS-49)

ISS & ATV

STS & HST

http://upload.wikimedia.org/wikipedia/commons/5/53/Three_Crew_Members_Capture_Intelsat_VI_-_GPN-2000-001035.jpg

21

Basic Assumption

The rendez-vous can be solved using the “classical” 2-

body problem. But nonlinear equations of motion have to

be solved and one should resort to numerical methods

(e.g., Lambert’s problem).

Can we use approximate equations considering that

distances are small relative to the dimensions of the orbit ?

Yes ! Use linearization procedures.


22

Problem Statement

0r

r

0

0

, 1r

r

r r r

target

(known

orbit)

What is the equation governing the relative motion r ?

A rendez-vous occurs when r =0.

r

chase


23

Mathematical Developments

3r

rr

0 r r r

00 3r

r rr r

?

2 2 2

0 0 0 0( ).( ) 2 .r r r r r r r r r

3

23 3 0

0 2

0

2 .1r r

r

r r

Binomial

theorem

3 3 00 2

0

3 .1r r

r

r r

Relative

motion as

well !


24

Mathematical Developments

00 03 5

0 0

3 .1

r r

r rr r r r

0 00 03 3 2

0 0 0

3 .

r r r

r r rr r r r

Higher-order terms

neglected

00 3r

r

r

003 2

0 0

3 .

r r

r rr r r

Linear ODE.

Observer in an

inertial frame


25

Co-Moving Frame & Circular Orbit

0r

r

r

i

jk

0 02 rel rel

d

dt

Ωr r r Ω Ω r Ω v a r r

2. 2 rel rel r Ω Ω r r Ω v a

Local vertical

coordinate

frame

Our focus is on a

rotating observer in

the local frame (e.g.,

an ISS astronaut

watching the ATV)


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2. 2 rel rel r Ω Ω r r Ω v a

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ

rel

rel

x y z

x y z

x y z

n

r i j k

v i j k

a i j k

Ω k

Centrifugal Coriolis

2 2ˆ ˆ ˆ( 2 ) ( 2 )n x n y x n y n x y z r i j k

Interpretation

of the terms ?


27


2 2ˆ ˆ ˆ( 2 ) ( 2 )n x n y x n y n x y z r i j k

20 00 03 2 2

0 0 0

3 . 3 .ˆ ˆ ˆˆ r xn x y z r

r r r

r rr r r i j k i

2

2

3 2 0

2 0

0

x n x n y

y n x

z n z

Hill – Clohessy – Wiltshire

equations: acceleration relative

to a rotating observer fixed in

the moving frame.


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An Obvious Solution

2

2

3 2 0

2 0

0

x n x n y

y n x

z n z

?

?

?

x

y

z

x=0, y=constant.

z=harmonic motion (the out-of plane motion is

decoupled from the in-plane motion).

Physical interpretation ?

Two spacecraft on the same orbit.


29

In Matrix Form

2

2

0 2 0 3 0 0 0

2 0 0 0 0 0 0

0 0 0 0 0 0

x n x n x

y n y y

z z n z

Linear ODEs with constant coefficient: we can solve them !

Gyroscopic damping:

imposes a rotation if

there is a relative velocity

Centrifugal +

gravitational terms: move

away from the target if

not on the same orbit


30

Interpretation

Impulse to the left

rotation due to

Coriolis

Forward impulse

backward

motion !

Backward

impulse

forward motion !


31

Solution

0 02 0 2 2y n x y n x Cte y n x

2 2 2

0 03 2 0 2 4x n x n y x n x n y n x

0 0

2sin cos 4x A nt B nt y x

n

…


32

Solution: Overall

H. Curtis, Orbital Mechanics for Engineering Students, Elsevier.

0 0

0 0

0 0

2 1 cossin0

4 3cos 0 02 cos 1 4sin 3

6 sin 1 0 0

0 0 cossin

0 0

ntnt

n nx nt x x

nt nt nty nt nt y y

n nz nt z z

nt

n

0 0

0 0

0 0

3 sin 0 0 cos 2sin 0

6 cos 1 0 0 2sin 4cos 3 0

0 0 sin 0 0 cos

x n nt x nt nt x

y n nt y nt nt y

z n nt z nt z

( )rr tΦ ( )rv tΦ

( )vr tΦ ( )vv tΦ

( )tr 0 ( )tr 0 ( )t v

( )t v 0 ( )tr 0 ( )t v


33

Key Feature ?

0 0

0 0

0 0

2 1 cossin0

4 3cos 0 02 cos 1 4sin 3

6 sin 1 0 0

0 0 cossin

0 0

ntnt

n nx nt x x

nt nt nty nt nt y y

n nz nt z z

nt

n

Secular terms ! What do they mean ?


34

Example: HST and STS

The HST is released from the Space Shuttle, which is in a

circular orbit of 590 km altitude. The relative velocity of

ejection is 0.1 m/s down, 0.04 m/s backwards and 0.02 m/s

to the right. Find the position of HST after 5,10,20 minutes.

0 0

0 0

0 0

2 1 cossin0

4 3cos 0 02 cos 1 4sin 3

6 sin 1 0 0

0 0 cossin

0 0

ntnt

n nx nt x x

nt nt nty nt nt y y

n nz nt z z

nt

n

(-0.1;-0.04;-0.02)T

3

3986000.0011 rad/s

6968n


35


5 minutes: (-33.345 ; -1.473 ; -5.894)T

10 minutes: (-70.933 ; 20.357 ; -11.170)T

20 minutes: (-143.000 ; 137.279 ; -17.766)T


36


-200 -150 -100 -50 0 50-500

0

500

1000

1500

2000

2500

3000

x (m)

y

(m

)


37

Further Reading on the Web Site

They use Hill – Clohessy –

Wiltshire equations !


38

Solution: Overall

0 0

0 0

( ) ( ) ( )

( ) ( ) ( )

rr rv

vr vv

t t t

t t t

r Φ r Φ v

v Φ r Φ v

Assumptions:

1. The two satellites are within a few kms of each other.

2. The target is in a circular orbit.

3. There are no external forces on the interceptor.

Starting from t=0, how to exploit these equations for

rendez-vous after a time tf ?

Linear analog of

Kepler’s equation


39

Two-Impulse Rendez-vous Maneuvers

0 00 , and knownt r v

0 00 ( ) ( )rr f rv ft t Φ r Φ v

0Compute for rendez-vous in a time f

t v

1

0 0( ) ( )rv f rr ft t v Φ Φ r

0 0

1

0 0

Compute ( ) ( )

( ) ( ) ( ) ( )

f vr f vv f

vr f vv f rv f rr f

t t

t t t t

v Φ r Φ v

Φ r Φ Φ Φ r

Impose 0f v f f v v

0 0 0 v v v


40

Physical Interpretation

The initial velocity change places the vehicle on a

trajectory which will intercept the origin at time tf.

The final velocity change cancels the arrival velocity of

the vehicle at the origin and completes the rendez-vous.


41


After 10 minutes, the STS astronauts want to retrieve HST.

Compute the v0+ to rendez-vous in 5 and 15 minutes.

5 minutes: (0.2742 ; 0.0135 ; 0.0359)T

15 minutes: (0.1356 ; 0.0753 ; 0.0082)T


42

Real Data: Gemini 10 and Agena 8

A 4-orbit rendez-vous plan similar to that of Gemini 6-7:

O1: no maneuvers to allow spacecraft checkout.

O2: height adjustment at perigee (0.2 m/s for orbital decay)

phase adjustment maneuver at apogee (17 m/s).

O3: switch to a circular orbit in the same plane as Agena

(16 m/s).

O4: terminal phase initiation (10 m/s)

velocity matching (13 m/s)

Total: 56 m/s

NH

NCI

NSR

TPI

TPF


Nc: nominal corrective burn

Nh: nominal height adjust burn


44



45



46


47

Date Time UTC Name Impulsion Orbite

(m/s) (km x km)

23 15:46:44 MECO ---- 229 x 56

23 16:16:38 OMS-2 70,74 296 x 229

23 18:32:45 NC1 26,06 319 x 296

24 08:50:56 NC2 6,13 319 x 316

24 19:27:09 NC3 0,91 319 x 317

25 07:30:35 NH 7,01 344 x 319

25 08:24:45 NC4 2,74 345 x 327

25 09:55:43 Ti 2,71 347 x 332

25 11:12:45 MC1 0,64 347 x 334

25 11:26:20 MC2 0,64 348 x 334

STS-120 (Rendez-vous with ISS, Oct. 07)

Pour STS123: http://www.obsat.com/sts123tle.htm 5.4.2 Gemini and Space Shuttle

48

STS-120 (Rendez-vous with ISS, Oct. 07)


49

Maneuvers can find many practical applications and may

drive spacecraft design (fuel consumption).

Focus on impulsive maneuvers (chemical propulsion).

Most important maneuvers in Earth’s orbit have been

described (plane change, circularization, phasing, rendez-

vous).

In Summary

50

Combined maneuvers (e.g., circularization and inclination

change for a GEO satellite).

Fixed V maneuvers.

Other maneuvers

Gaëtan Kerschen

Space Structures &

Systems Lab (S3L)


Astrodynamics (AERO0024)

5b. orbital maneuvers - uliege motivation coplanar transfers have wide applicability (e.g., phasing...

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