Aerospace Astronautics

ARO309 -‐ Astronau0cs and Spacecra6 Design

Winter 2013

Try Lam CalPoly Pomona Aerospace Engineering

Earth Orbiters

Pork Chop Plot

High Thrust Interplanetary Transfer

Low-‐Thrust Interplanetary Transfer

Low-‐Thrust Europa End Game

Stable for > 100 days

Orbit Stability

Enceladus Orbit

Other Missions

Lecture 01 and 02: Two-‐Body Dynamics: Conics

Chapter 2

Equa0ons of Mo0on

!

where r = R 2 " R 1

Equa0ons of Mo0on

•  Fundamental Equa0ons of Mo0on for 2-‐Body Mo0on

!

˙ r = " µr3 r

!

˙ r = "µr3 r =

˙ x ˙ y ˙ z

#

$

% % %

&

'

( ( (

= "µr3

xyz

#

$

% % %

&

'

( ( (

!

where µ = G M + m( ) and ˙ r = ˙ R 2 " ˙ R 1

Conic Equa0on

!

˙ r = " µr3 r

!

r =h2 /µ

1+ ecos"=

p1+ ecos"

!

h = r " v

!

˙ r " h = # µr3 r " h

From 2-body equation to conic equation

Angular Momentum

!

v" =hr

=µh1+ ecos#( )

Other Useful Equations

!

h = r " v and h = rv cos#

!

vr =µhesin"

!

rp =h2 /µ1+ e

!

tan" =vrv#

=esin$

1+ ecos$

Energy

!

" = K + P =v 2

2#

µr

= constant

" = " p =vp2

2#

µrp

= #µ2 /h2

21# e2( )

NOTE: ε = 0 (parabolic), ε > 0 (escape), ε < 0 (capture: elliptical and circular)

Conics

!

r =p

1+ ecos"=a 1# e2( )1+ ecos"

Circular Orbits

!

e = 0 and " < 0

!

vcircular =µr

!

r =p

1+ 0" cos#= p =

h2

µ

!

h = rv

!

Tcircular =2"r

µr

= 2" r3

µ

!

" = #µ2 /h2

2= #

µ2r

Ellip0cal Orbits

!

0 < e <1 and " < 0

Ellip0cal Orbits

!

r =p

1+ ecos"# ra =

p1$ e

and rp =p

1+ e

!

since 2a = rp + ra " a =p

1# e2

!

" ra = a 1+ e( ) and rp = a 1# e( )

!

NOTE : b = a 1" e2

!

" = #µ2 /h2

21# e2( ) = #

µ2a

!

Telliptical =2"abh

=2"ab

µa 1# e2( )= 2" a3

µ

Ellip0cal Orbits

!

e =ra " rpra + rp

!

h = µa 1" e2( )va = h /ra and vp = h /rp

!

tan" =vrv#

=esin$

1+ ecos$

Parabolic Orbits

•  Parabolic orbits are borderline case between an open hyperbolic and a closed ellip0cal orbit

!

e =1 and " = 0

!

" = 0 =v 2

2#

µr$ vparabolic =

2µr

= vescape

NOTE: as v à 180°, then r à ∞

!

" parabolic = # /2

!

tan" =vrv#

=esin$

1+ ecos$=1 % sin$1+1 % cos$

Hyperbolic Orbits

!

e >1 and " > 0

!

sin"# =e2 $1e

!

" = cos#11e$

% & '

( )

!

r = "a 1" e2( )1+ ecos#

!

" = a e2 #1

Hyperbolic Orbits

!

" =µ2a

!

" =v 2

2#

µr

=µ2a

!

v" =µa

=µhesin#" =

µh

e2 $1Hyperbolic excess speed

!

" =v 2

2#

µr

=v$2

2#

µr$

=v$2

2

!

v 2 = v"2 +2µr

= v"2 + vescape

2

!

C3 = v"2

Proper0es of Conics

0 < e < 1

Conic Proper0es

Vis-‐Viva Equa0on

!

" =v 2

2#

µr

= #µ2a

!

v 2 = µ2r"1a

#

$ %

&

' ( Vis-viva equation

Mean Mo0on

!

n =2"T

=µa3

Perifocal Frame “natural frame” for an orbit centered at the focus with x-‐axis to periapsis and z-‐axis toward the angular momentum vector

!

r = x ˆ p + y ˆ q and ˆ w = h /h

r = rcos" ˆ p + rsin" ˆ q

v = ˙ r = ˙ x ˆ p + ˙ y ˆ q

v = ˙ r cos" # r ˙ " sin"( ) ˆ p

+ ˙ r sin" + r ˙ " cos"( ) ˆ q

Perifocal Frame

FROM

!

˙ r = vr =µh

esin" and r ˙ " = v# =µh

1+ ecos"( )

!

˙ x = "µh

sin# and ˙ y =µh

e + cos#( )THEN

!

v = ˙ r = ˙ x ˆ p + ˙ y ˆ q

v =µh

"sin#( ) ˆ p + e + cos#( ) ˆ q [ ]

!

h = r " v =ˆ p ˆ q ˆ w x y 0˙ x ˙ y 0

= x ˙ y # y ˙ x ( ) ˆ w and h = x ˙ y # y ˙ x ( )

Lagrange Coefficients •  Future es0mated state as a func0on of current state

!

r = x ˆ p + y ˆ q

v = ˙ x ˆ p + ˙ y ˆ q

!

r = f r 0 + g v 0v = ˙ f r 0 + ˙ g v 0

Solving unit vector based on initial conditions

!

f =1"µ rh2 1" cos#$( ); ˙ f =

µh

1" cos#$sin#$

%

& '

(

) *

µh2 1" cos#$( ) " 1

r0

"1r

+

, -

.

/ 0

g =r r0

hsin#$; ˙ g =1"

µ r0

h2 1" cos#$( )

!

r =h2 /µ

1+h2

µ r0"1

#

$ % %

&

' ( ( cos)* "

h vr0µ

#

$ %

&

' ( sin)*

!

vr0 =r 0 " v 0

r0and Where

Lagrange Coefficients •  Steps finding state at a future Δθ using Lagrange Coefficients

1.  Find r0 and v0 from the given posi0on and velocity vector 2.  Find vr0 (last slide) 3.  Find the constant angular momentum, h

4.  Find r (last slide) 5.  Find f, g, fdot, gdot 6.  Find r and v !

h = r0v" 0 = r0 v02 # vr0

2

Lagrange Coefficients •  Example (from book)

•  Example (from book)

Lagrange Coefficients

ALSO

Since Vr0 is < 0 we know that S/C is approaching periapsis (so 180°<θ<360°)

CR3BP •  Circular Restricted Three Body Problem (CR3BP)

!

" = µ /r123 = GM /r12

3

M = m1 +m2

#1 =m1

m1 +m2

# 2 =m2

m1 +m2

!

m1x1 +m2x2 = 0x1 + r12 = x2

!

x1 = "# 2r12x2 = #1r12

!

"1 =1#"2

CR3BP

!

r1 = x +"2r12( )ˆ i + yˆ j + z ˆ k

r2 = x #"1r12( )ˆ i + yˆ j + z ˆ k

r = xˆ i + yˆ j + z ˆ k

!

˙ r = v inertialCM +"# r + v rel

˙ r = ˙ a inertialCM + ˙ " # r +"# " # r ( ) + 2"# v rel + a rel

where v rel = ˙ x i + ˙ y j + ˙ z k

where a rel = ˙ x i + ˙ y j + ˙ z k

Kinematics (LHS):

!

m˙ r = F1 +F2

!

˙ r = ˙ x " 2#˙ y "#2x( )ˆ i + ˙ y + 2#˙ x "#2y( ) ˆ j + ˙ z k

CR3BP

!

F1 = "µ1mr13 r 1 and F2 = "

µ2mr23 r 2

Kinematics (RHS):

!

m˙ r = F1 +F2

!

˙ x " 2#˙ y "#2x = "µ1

r13 x +$ 2r12( ) " µ2

r23 x "$1r12( )

˙ y + 2#˙ x "#2y = "µ1

r13 y "

µ2

r23 y

˙ z = "µ1

r13 z "

µ2

r23 z

CR3BP

CR3BP Plots are in the rotating frame

Tadpole Orbit Horseshoe Orbit

Lyapunov Orbit

DRO

CR3BP: Equilibrium Points

!

"#2x = "µ1r13 x +$ 2r12( ) " µ2

r23 x "$1r12( )

"#2y = "µ1r13 y "

µ2r23 y

0 = "µ1r13 z "

µ2r23 z

Equilibrium points or Libration points or Lagrange points

!

˙ x = ˙ y = ˙ z = ˙ x = ˙ y = ˙ z = 0

!

z = 0L1 L2 L3

L4

L5

!

JC =12v 2 "

12#2 x 2 + y 2( ) " µ1

r1"

µ2r2

Jacobi Constant

Lecture 03 and 04: Two-‐Body Dynamics:

Orbit Posi0on as a Func0on of Time

Chapter 3

Introduc0ons

•  Chapter 2 (Lec0on 1 and 2) relates posi0on as a func0on of θ (true anomaly) but not 0me

•  Time was only introduced when referring to orbit period

•  Here we acempt to find the rela0ons between posi0on of the S/C and 0me à Kepler’s Equa4on

Time versus True Anomaly

•  Recall from Chapter 2

!

v" =hr

= r ˙ # therefore h = r2 ˙ #

!

d"dt

=hr2

!

r =h2 /µ

1+ ecos"Since Then

!

µ2

h3dt =

d"1+ ecos"( )2

Integrating from 0 (assuming tp = 0) to t and from 0 to θ

!

µ2

h3t =

d"1+ ecos"( )20

#

$


!

µ2

h3t =

d"1+ ecos"( )20

#

$


!

µ2

h3t =

d"1+ ecos"( )20

#

$

Simple Case: Circular Orbits (e=0)

If e = 0, then

!

µ2

h3t = d" = #

0

#

$ therefore

!

t =h3

µ2 "

Since for a circular orbit we have

!

r = h2 /µ then

!

t =r3

µ"

FOR CIRCULAR ORBIT OR

!

t ="2#

Tcir


!

µ2

h3t =

d"1+ ecos"( )20

#

$

Elliptical Orbits (0<e<1)

Then a = 1 and b = e, therefore we have b < a

!

µ2

h3t =

d"1+ ecos"( )20

#

$ =1

1% e2( )3 / 22tan%1

1% e1+ e

tan#2

&

' (

)

* + %

e 1% e2 sin#1+ ecos#

,

- . .

/

0 1 1

Me = Mean anomaly for the ellipse


!

µ2

h31" e2( )3 / 2 t = Me = 2tan"1

1" e1+ e

tan#2

$

% &

'

( ) "

e 1" e2 sin#1+ ecos#

*

+ , ,

-

. / /

Therefore we have

From the orbit period of an ellipse we know (or can derive) that

!

Tellipse =2"abh

=2"µ2

h1# e2

$

% &

'

( )

3

= 2" a3

µ

Therefore we can solve for me as function orbit period as

!

Me =2"Tellipse

t OR

!

Me = n t where n = mean motion = 2π/Te



!

Me = n t = 2tan"11" e1+ e

tan#2

$

% &

'

( ) "


*

+ , ,

-

. / /

We need to fine out Me still ?

Let’s introduce another variable E = eccentric anomaly

!

acosE = ae + rcos"

= ae +a 1# e2( )1+ ecos"

cos"

!

cosE =e + cos"1+ ecos"



!

cosE =e + cos"1+ ecos"

This relates E and θ, but it leaves the quadrant of the solution unknown and you get two values of E for the equation. To eliminate this ambiguity we use the following identity

!

sinE =1" e2 sin#1+ ecos#

OR

!

tan2E2

=sin2 E /2( )cos2 E /2( )

=1" cosE( ) /21+ cosE( ) /2

=1" cosE1+ cosE

tan2 E2

=1" e1+ e

1" cos#1+ cos#$

% &

'

( ) =1" e1+ e

tan2 #2

!

tan E2

=1" e1+ e

tan#2Therefore

!

E = 2tan"1 1" e1+ e

tan#2

$

% &

'

( ) or



!

Me = n t = 2tan"11" e1+ e

tan#2

$

% &

'

( ) "


*

+ , ,

-

. / /

We need to fine out Me still

E

!

esinE

!

Me = E " esinE

This is Kepler’s Equation


To find t given Δθ

•  Given orbital parameters, find e and h (assume θ = 0 deg)

•  Find E:

•  Find T (orbit period):


!

h = µ rp 1+ e( )

!

tan E2

=1" e1+ e

tan#2

!

Telliptical = 2" a3

µ=2"µ2

h1# e2

$

% &

'

( )

3


To find t given Δθ

•  Fine Me:

•  Find t:


!

t =Me

2"Telliptical

!

Me = E " esinE

Question: What if you are going from a θ = θa to θ = θb? Answer: Find the time from θ = 0 to θ = θa and the time from θ = 0 to θ = θb. Then subtract the differences.


To find θ given Δt

•  Given orbital parameters, find e and h (assume θ = 0 deg

•  Find T (orbit period):

•  Find Me:

•  Find E using Newton’s method (or a transcendental solver)


!

Telliptical = 2" a3

µ=2"µ2

h1# e2

$

% &

'

( )

3

!

Me = 2" t /Telliptical

!

Me = E " esinE



•  Using Newton’s Method:

–  Ini0alize E = Eo:

–  Find f(E):

–  Find f’(E):

–  If abs( f(E) / f’(E) ) > TOL, then repeat with

–  Else Econverged = En


!

Ei+1 = Ei " f Ei( ) / f / Ei( )!

f E( ) = E " esinE "Me

!

f / E( ) =1" ecosE!

E0 = Me + e /2 or E0 = Me " e /2


For Me < 180 deg For Me > 180 deg


•  A6er finding the converged E, then find θ


!

tan E2

=1" e1+ e

tan#2



!

µ2

h3t =

d"1+ ecos"( )20

#

$

Parabolic Orbits (e = 1)

Then a = 1 and b = e, therefore we have b = a

!

µ2

h3t =

d"1+ ecos"( )20

#

$ =12tan

#2

+16tan3

#2

= MP

MP = Parabolic Mean Anomaly

Time versus True Anomaly Parabolic Orbits (e = 1)

!

16tan"2

#

$ %

&

' ( 3

+12tan"2)MP = 0

Thus given t or Δt we can find MP

!

MP =µ2

h3t

To fine θ we can find the root of the below equation

Which has one real root

!

tan"2

= 3MP + 9MP2 +1( )

1/ 3# 3MP + 9MP

2 +1( )#1/ 3

STEPS: Find h Find MP Find θ


!

µ2

h3t =

d"1+ ecos"( )20

#

$

Hyperbolic Orbits (e > 1)

Then a = 1 and b = e, therefore we have b > a

!

µ2

h3t =

1e2 "1

esin#1+ ecos#

"1e2 "1

lne +1 + e "1tan # /2( )e +1 " e "1tan # /2( )

$

% &

'

( )

*

+ , ,

-

. / /

Time versus True Anomaly Hyperbolic Orbits (e > 1)

!

Mh =e e2 "1sin#1+ ecos#

" lne +1 + e "1tan # /2( )e +1 " e "1tan # /2( )

$

% &

'

( )

Where the Hyperbolic mean anomaly is

Thus we have

!

Mh =µ2

h3e2 "1( )3 / 2 t

Similar with Ellipse we will intro a new variable, F, the hyperbolic eccentric anomaly to help solve for the Mean Hyperbolic anomaly, Mh.


!

sinhF = y /b and coshF = x /a

Hyperbolic eccentric anomaly for the Hyperbola

Since:

!

sinh x = ex " e"x( ) /2cosh x = ex + e"x( ) /2

y = rsin#

r =a e2 "1( )1+ ecos#

b = a e2 "1


!

sinhF =e2 "1sin#1+ ecos#We now have

Solving for F and since

!

sin" =2tan " /2( )1+ tan2 " /2( )

and cos" =1# tan2 " /2( )1+ tan2 " /2( )

!

sinh"1 x = ln x + x 2 +1( ) we now have

Using the following trig identities for sine and cosine

!

F = lne2 "1sin# + cos# + e

1+ ecos#

$

% & &

'

( ) )


We now have

!

F = ln e +1 + e "1tan # /2( )e +1 " e "1tan # /2( )

$

% &

'

( )

Therefore we now have:

!

Mh = esinhF " F

This is Kepler’s Equation for Hyperbola

Similar to Elliptical orbits we can solve for F as a function of θ, which is found to be. Thus given θ we can find F, and Mh, and finally t.

!

tanh F2

=e "1e +1

tanh#2

STEPS TO FIND θ (given t) •  Set ini0al F0 = Mh where

•  Find f and f’

•  If abs( f / f’ ) > TOL, repeat steps with updated F

•  Else, Fconverged = Fi. Now find θ


If time, t, was given and θ is to be found then we have to solve for Kepler’s equation for hyperbola iteratively using Newton’s method

!

f F( ) = esinhF " F "Mh

!

f / F( ) = ecoshF "1

!

Fi+1 = Fi " f Fi( ) / f / Fi( )

!

Mh =µ2

h3e2 "1( )3 / 2 t

!

tanh"2

=e +1e #1

tanh F2

Universal Variables •  What happens if you don’t know what type of orbit you are

in? Why use 3 set of equa0ons?

•  Kepler’s equa0on can be wricen in terms of a universal variable or universal anomaly, Χ, and Kepler’s equa0on becomes the universal Kepler’s equa4on.

!

µ "t =r0vr0

µ#2 C $#2( ) + 1%$ r0( )#3 S $#2( ) + r0#

Where

!

" =1/aIf α < 0, then orbit is hyperbolic If α = 0, then orbit is parabolic If α > 0, then orbit is elliptical

Universal Variables

•  Stumpff func0ons

or for z = αΧ2

!

S z( ) =

z " sin z( ) z( )3

(z > 0)

sinh "z " "z( ) "z( )3(z < 0)

1/6 (z = 0)

#

$

% %

&

% %

!

C z( ) =

1" cos z( ) z (z > 0)

cosh "z "1( ) "z (z < 0)1/2 (z = 0)

#

$ % %

& % %

,

Universal Variables

•  To use Newton’s method we need to define the following func0on and it’s deriva0ve

•  Iterate with the following algorithm

!

f / "( ) =r0vr0

µ" 1#$"2 S $"2( )[ ] + 1#$ r0( )"2 C $"2( ) + r0

!

f "( ) =r0vr0

µ"2 C #"2( ) + 1$# r0( )"3 S #"2( ) + r0"$ µ %t

!

"0 = µ # $twith

!

"i+1 = "i #f "( )f / "( )

Universal Variables

•  Rela0on ship between X and the orbits

!

" =

h tan # /2( ) $ tan #0 /2( )( ) µ parabolaa E $ E0( ) ellipse$a F $ F0( ) hyperbola

%

& '

( '

!

" =

h tan # /2( ) µ parabolaaE ellipse$aF hyperbola

%

& '

( '

For t0 = 0 at periapsis

Universal Variables

•  Example 3.6 (Textbook: Cur0s’s)

Find h and e

!

h = µ r0 1+ ecos"0( ) = 63135 1+ 0.866e

v# 0 = h /r0 = 6.314 1+ 0.866e

vr0 =µhesin"0 = 3.16 e

1+ 0.866e

Since , then

!

v02 = v" 0

2 + vr02

!

3.16 e1+ 0.866e

"

# $

%

& ' 2

+ 6.314 1+ 0.866e( )2

=102

!

e =1.47

Universal Variables


Therefore

!

vr0 = 3.075 km2 /s

!

" =1a

=1

h2 /µ1# e2

=1

#19,655 km= #5.09E # 05 km#1

!

"0 = µ # $t =115.6

So X0 is the initial X to use for the Newton’s method to find the converged X

Universal Variables


Universal Variables


Thus we accept the X value of X = 128.5

!

" = #a F # F0( )

!

tanh F02

=e "1e +1

tan#02

= 0.1667

!

F0 = 0.23448 rad

where

!

"F =1.15

!

tanh"2

=e +1e #1

tan F2

=1.193

!

" =100°

Lagrange Coefficients II •  Recall Lagrange Coefficients in terms of f and g coefficients

•  From the universal anomaly X we can find the f and g coefficients

!

r = f r 0 + g v 0v = ˙ f r 0 + ˙ g v 0

!

f =1" #2

r0C $#2( )

g = %t " 1µ#3S $#2( )

!

˙ f =µ

r r0

"#3S "#2( ) $ #[ ]

˙ g =1$ #2

rC "#2( )

Lagrange Coefficients II

!

vr0 =r 0 " v 0

r0and Where

!

" =1a

=2r0#v02

µ

•  Steps finding state at a future Δθ using Lagrange Coefficients

1.  Find r0 and v0 from the given posi0on and velocity vector 2.  Find vr0 and α 3.  Find X 4.  Find f and g 5.  Find r, where r = f r0 + g v0 6.  Find fdot and gdot 7.  Find v

Lecture 05 and 06: Two-‐Body Dynamics:

Orbits in 3D

Chapter 4

Introduc0ons

•  So far we have focus on the orbital mechanics of a spacecra6 in 2D.

•  In this Chapter we will now move to 3D and express orbits using all 6 orbital elements

Geocentric Equatorial Frame

!

r = X 2 +Y 2 + Z 2

" = sin#1 Z /r( )

$ =cos#1 X /r

cos"%

& '

(

) * (Y /r > 0)

2+ # cos#1 X /rcos"%

& '

(

) * (Y /r , 0)

-

. / /

0 / /

Orbital Elements

•  Classical Orbital Elements are: a = semi-‐major axis (or h or ε) e = eccentricity i = inclina0on Ω = longitude of ascending node ω = argument of periapsis θ = true anomaly

Orbital Elements

!

vr = r" v /r

h = r # v =

ˆ i ˆ j ˆ k X Y ZVx Vy Vz

n = ˆ k # h =

ˆ i ˆ j ˆ k 0 0 1hx hy hz

Orbital Elements

!

i = cos"1hzh

#

$ %

&

' (

!

" =cos#1

nxn

$

% &

'

( ) (ny * 0)

2+ # cos#1nxn

$

% &

'

( ) (ny < 0)

,

- . .

/ . .

!

e =1µ

v 2 "µr

#

$ %

&

' ( r " rvrv

)

* +

,

- . and e = e/ e or e = 1+

h2

µ2 v 2 "2µr

#

$ %

&

' ( !

" =cos#1

n$ ene

%

& '

(

) * (ez + 0)

2, # cos#1n$ ene

%

& '

(

) * (ez < 0)

-

. / /

0 / /

Orbital Elements

!

" =cos#1

e$ rer

%

& '

(

) * (vr + 0)

2, # cos#1e$ rer

%

& '

(

) * (vr < 0)

-

. / /

0 / /

=

cos#1 1eh2

µr#1

%

& '

(

) *

%

& '

(

) * (vr + 0)

2, # cos#1 1eh2

µr#1

%

& '

(

) *

%

& '

(

) * (vr < 0)

-

.

/ /

0

/ /

Coordinate Transforma0on •  Answers the ques0on of “what are the parameters in another coordinate frame”

Q

Transformation (or direction cosine)

matrix

x

y

z

x’

y’

z’

!

Q[ ]T Q[ ] = 1[ ] =

1 0 00 1 00 0 1

"

#

$ $ $

%

&

' ' '

Q is a orthogonal transformation matrix

!

Q[ ] Q[ ]T = 1[ ]

Coordinate Transforma0on

!

x '[ ] = Q[ ] x[ ]

x[ ] = Q[ ]T x'[ ]Where

!

Q[ ] =

Q11 Q12 Q13

Q21 Q22 Q23

Q31 Q32 Q33

"

#

$ $ $

%

&

' ' '

=

ˆ i / ( ˆ i ˆ i / ( ˆ j ˆ i / ( ˆ k ˆ j / ( ˆ i ˆ j / ( ˆ j ˆ j / ( ˆ k ˆ k / ( ˆ i ˆ k / ( ˆ j ˆ k / ( ˆ k

"

#

$ $ $

%

&

' ' '

And

Where is made up of rotations about the axis {a, b, or c} by the angle {θd, θe, and θf}

!

Q[ ] = Ra "d( )[ ] Rb " e( )[ ] Rc " f( )[ ]

1st rotation 2nd rotation 3rd rotation


!

Q[ ] = R3 "( )[ ] R1 #( )[ ] R3 $( )[ ]

For example the Euler angle sequence for rotation is the 3-1-3 rotation

!

0 " # < 360°( ) 0 " $ "180°( ) 0 " % < 360°( )

where you rotate by the angle α along the 3rd axis (usually z-axis), then by β along the 1st axis, and then by γ along the 3rd axis.

!

Q[ ]313 =

"sin# cos$sin% + cos# cos% cos# cos$sin% + sin# cos% sin$sin%"sin# cos$cos% " cos# sin% cos# cos$cos% " sin# sin% sin$cos%

sin# sin$ "cos# sin$ cos$

&

'

( ( (

)

*

+ + +

!

tan" =Q31

#Q32

cos$ =Q33 tan% =Q13Q23

Thus, the angles can be found from elements of Q

Coordinate Transforma0on Classic Euler Sequence from xyz to x’y’z’


!

Q[ ] = R1 "( )[ ] R2 #( )[ ] R3 $( )[ ]

For example the Yaw-Pitch-Roll sequence for rotation is the 1-2-3 rotation

!

0 " # < 360°( ) $90 < % < 90°( ) 0 " & < 360°( )

where you rotate by the angle α along the 3rd axis (usually z-axis), then by β along the 2nd axis, and then by γ along the 1st axis.

!

Q[ ]123 =

cos" cos# sin" cos# $sin#cos" sin#sin% $ sin" cos% sin" sin#sin% + cos" cos% cos#sin%cos" sin#cos% + sin" sin% sin" sin#cos% $ cos" sin% cos#cos%

&

'

( ( (

)

*

+ + +

!

tan" =Q12Q11

sin# = $Q13 tan% =Q23

Q33

Thus, the angles can be found from elements of Q

Coordinate Transforma0on Yaw, Pitch, and Roll Sequence from xyz to x’y’z’

Transforma0on between Geocentric Equatorial and Perifocal Frame

Transferring between pqw frame and xyz

!

r{ }pqw =h2 /µ

1+ ecos"

cos"sin"0

#

$ %

& %

'

( %

) %

v{ }pqw =µh

*sin"e + cos"0

#

$ %

& %

'

( %

) %

!

Q[ ]xyz" pqw = R3 #( )[ ] R1 $( )[ ] R3 %( )[ ] = R3 &( )[ ] R1 i( )[ ] R3 '( )[ ]

Transformation from geocentric equatorial to perifocal frame

Transforma0on between Geocentric Equatorial and Perifocal Frame

Transformation from perifocal to geocentric equatorial frame is then

Therefore

!

Q[ ]pqw" xyz = Q[ ]xyz" pqwT

!

r{ }xyz = Q[ ]xyz" pqwT r{ }pqw

v{ }xyz = Q[ ]xyz" pqwT v{ }pqw

!

r{ }pqw= Q[ ]xyz" pqw r{ }xyz

v{ }pqw= Q[ ]xyz" pqw v{ }xyz

!

Q[ ]xyz" pqw =

cos# sin# 0$sin# cos# 00 0 1

%

&

' ' '

(

)

* * *

1 0 00 cosi sini0 $sini cosi

%

&

' ' '

(

)

* * *

cos+ sin+ 0$sin+ cos+ 00 0 1

%

&

' ' '

(

)

* * *

Perturba0on to Orbits

•  Planets are not perfect spheres Oblateness

!

oblateness =Req " Rpole

Req

Perturba0on to Orbits Oblateness

!

˙ r = " µr3 r + p

!

p = pr ˆ u r + pt ˆ u t + phˆ h

!

pr = "1.5 µr2J2

Rr

#

$ %

&

' ( 2

1" 3sin2 i( ) sin2 ) +*( )[ ]

pt = "1.5 µr2J2

Rr

#

$ %

&

' ( 2

sin2 i( ) sin2 2 ) +*( )( )

ph = "1.5 µr2J2

Rr

#

$ %

&

' ( 2

sin 2i( ) sin2 ) +*( )


!

˙ " = #1.5µa7

J2R2

1# e2( )2

$

%

& &

'

(

) ) cosi

!

if 0 " i < 90°( ) # ˙ $ < 0

if 90° < i "180°( ) # ˙ $ > 0


!

˙ " = #1.5µa7

J2R2

1# e2( )2

$

%

& &

'

(

) )

52

sin2 i # 2$

% &

'

( )

= ˙ * 5 /2( )sin2 i # 2

cosi$

% &

'

( )

!

if 0 " i < 63.4°( ) or 116.6 < i "180°( ) # ˙ $ > 0

if 63.4° < i "116.6°( ) # ˙ $ < 0

Sun-‐Synchronous Orbits Orbits where the orbit plane is at a fix angle α from the Sun-planet line

Thus the orbit plane must rotate 360° per year (365.25 days) or 0.9856°/day

!

˙ " SunSync =2#$ rad

365.25$ 86400$ sec˙ " SunSync =1.991E % 07 rad /s

Finding State of S/C w/Oblateness •  Given: Ini0al State Vector •  Find: State a6er Δt assuming oblateness (J2)

•  Steps finding updated state at a future Δt assuming perturba0on

1.  Compute the orbital elements of the state 2.  Find the orbit period, T, and mean mo0on, n 3.  Find the eccentric anomaly 4.  Calculate 0me since periapsis passage, t, using Kepler’s equa0on

!

Me = nt = E " esinE

Finding State of S/C w/Oblateness 5.  Calculate new 0me as tf = t + Δt 6.  Find the number of orbit periods elapsed since original periapsis passage

7.  Find the 0me since periapsis passage for the final orbit

8.  Find the new mean anomaly for orbit n

9.  Use Newton’s method and Kepler’s equa0on to find the Eccentric anomaly (See slide 57)

!

np = t f /T

!

torbit _ n = np " floor np( )[ ] T

!

Me )orbit _ n = n torbit _ n

Finding State of S/C w/Oblateness 10.  Find the new true anomaly

11.  Find posi0on and velocity in the perifocal frame

!

r{ }pqw =h2 /µ

1+ ecos"

cos"sin"0

#

$ %

& %

'

( %

) %

v{ }pqw =µh

*sin"e + cos"0

#

$ %

& %

'

( %

) %

!

tan"orbit _ n2

=1+ e1# e

tanEorbit _ n

2

Finding State of S/C w/Oblateness 12.  Compute the rate of the ascending node

13.  Compute the new ascending node for orbit n

14.  Find the argument of periapsis rate

15.  Find the new argument of periapsis

!

"orbit _ n ="0 + ˙ " #t

!

" orbit _ n =" 0 + ˙ " #t

!

˙ " = #1.5µa7

J2R2

1# e2( )2

$

%

& &

'

(

) ) cosi

!

˙ " = ˙ # 5 /2( )sin2 i $ 2

cosi%

& '

(

) *

Finding State of S/C w/Oblateness 16.  Compute the transformation matrix [Q] using the inclination, the

UPDATED argument of periapsis, and the UPDATED longitude of ascending node

17.  Find the r and v in the geocentric frame

!

r{ }xyz = Q[ ]pqw" xyz r{ }pqw

v{ }xyz = Q[ ]pqe" xyz v{ }pqw

!

Q[ ]pqw" xyz =

#sin $( )cos i( )sin %( ) + cos $( )cos %( ) #sin $( )cos i( )cos %( ) # cos $( )sin %( ) sin $( )sin i( )cos $( )cos i( )sin %( ) + sin $( )cos %( ) cos $( )cos i( )cos %( ) # sin $( )sin %( ) #cos $( )sin i( )

sin i( )sin %( ) sin i( )cos %( ) sin i( )

&

'

( ( (

)

*

+ + +

Ground Tracks Projection of a satellite’s orbit on the planet’s surface


!

"E #15.04 deg/hr


Ground Tracks reveal the orbit period

!

T ="LONequator deg15.04 deg/hr

Ground Tracks reveal the orbit inclination

!

i = LATmax ormin

If the argument of perispais, ω, is zero, then the shape below and above the equator are the same.

Lecture 07: Preliminary Orbit Determina0on

Chapter 5

Introduc0ons

•  This chapter only covers the basic concept of determining an orbit from some observa0on

•  In prac0ce, this is not referred to as orbit determina0on

•  Space OD is actually a sta0s0cal es0ma0on or filtering method (example: Kalman Filter)

•  We will only cover Lambert’s problem (Sec0on 5.3) from this Chapter

Lambert’s Problem

•  Given 2 posi0ons on an orbit r1 and r2 and Δt, what are the veloci0es at those two points, v1 and v2.

Lambert Fit •  Steps to find v1 and v2:

1.  Find the magnitude of r1 and r2 2.  Decide if the orbit is prograde or retrograde

3.  Compute the following

4.  Compute Δθ

!

r1 " r2( )z= r1 " r2( )# ˆ k

!

"# =

cos$1 r1 % r2r1 r2

&

' ( (

)

* + + if r1 , r2( )Z - 0

2. $ cos$1r1 % r2r1 r2

&

' ( (

)

* + + if r1 , r2( )Z < 0

/

0

1 1

2

1 1

!

"# =

cos$1 r1 % r2r1 r2

&

' ( (

)

* + + if r1 , r2( )Z < 0

2- $ cos$1r1 % r2r1 r2

&

' ( (

)

* + + if r1 , r2( )Z . 0

/

0

1 1

2

1 1

for prograde for retrograde

Lambert Fit 5.  Compute the func0on

6.  Find z by itera0ng using Newton’s method un0l convergence

you can start with z0 = 0 (or posi0ve z0 if an ellip0cal orbit), where

!

A = sin"#r1 r2

1$ cos"#

!

zi+1 = zi "F zi( )F ' zi( )

!

F z( ) =y z( )C z( )"

# $

%

& '

3 / 2

S z( ) + A y z( ) ( µ)t

!

F / z( ) =

y z( )C z( )"

# $

%

& '

3 / 2 C z( ) ( 3S z( )2C z( )2z

+3S z( )2

4C z( )

"

#

$ $ $ $

%

&

' ' ' '

+A83 S z( )C z( )

y z( ) + A C z( )y z( )

"

# $ $

%

& ' '

if z ) 0

240

y 0( )3 / 2 +A8

y 0( ) + A 12y 0( )

"

# $

%

& ' if z = 0

*

+

, , ,

-

, , ,

Lambert Fit where

7.  Note: the sign of the converged z tells you the orbit type:

z < 0 à Hyperbolic Orbit z = 0 à Parabolic Orbit z > 0 à Elliptical Orbit

!

y z( ) = r1 + r2 + Az S z( ) "1C z( )

#

$ % %

&

' ( (

!

S z( ) =

z " sin zz3 if z > 0

sinh "z " "z"z

3 if z < 0

16

if z = 0

#

$

% % % %

&

% % % %

!

C z( ) =

1" cos zz

if z > 0

cosh "z "1"z

if z < 0

12

if z = 0

#

$

% % % %

&

% % % %

Lambert Fit 8.  Compute the func0on y(z) using the converged z

9.  Compute f, g, fdot, gdot

10.  Compute

!

y z( ) = r1 + r2 + Az S z( ) "1C z( )

#

$ % %

&

' ( (

!

f =1" y z( )r1

; ˙ f =µ

r1 r2

y z( )C z( )

z S z( ) "1[ ]

g = A y z( )µ

; ˙ g =1" y z( )r2

!

v1 =1gr2 " f r1( ); v2 =

1g

˙ g r2 " r1( )

Lecture 08 and 10: Orbital Maneuvers

Chapter 6

Maneuvers

•  We assume our maneuvers are “instantaneous” •  ΔV changes can be applied by changing the magnitude “pump” or by changing the direc0on “crank”

•  Rocket equa0on: rela0ng ΔV and change in mass

where g0 = 9.806 m/s2

⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ

final

initialSP m

mIgV ln0

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ=

SP

initialfinal

IgV

mm

0

expor

Isp •  Isp is the specific impulse (units of seconds) and measures the performance of the rocket

ΔV •  ΔV (“delta”-‐V) represents the instantaneous change in

velocity (from current velocity to the desired velocity).

12 VV −=ΔVCircular Orbit Velocity

Elliptical Orbit Velocity

RVC

µ=

⎟⎠

⎞⎜⎝

⎛ −=aR

VE12

µ

Tangent Burns

Your Spacecraft

Your Initial Orbit

R1

•  Ini0ally you are in a circular orbit with radius R1 around Earth

11 R

VCµ

=

Tangent Burns

ΔV Your new orbit after the ΔV burn

R1 R2


•  You perform a burn now which

puts in into the red-‐doced orbit. So you perform a ΔV.

11 R

VCµ

=

11 CE VVV −=Δ

Tangent Burns

R1 R2


•  You perform a burn now which

puts in into the red-‐doced orbit. So you perform a ΔV.

!V = VE1"VC1

where

VE1= µ

2R1

"1a

#

$%

&

'( and a = 1

2R1 + R2( )

11 R

VCµ

=

Hohmann Transfer

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=Δ

Δ+Δ=Δ

21

1

22

21

2

11

21

21

12

RRR

RV

RRR

RV

VVV

µ

µ

ΔV1

R1 R2

ΔV2

The most efficient 2-burnmaneuver to transfer between 2 co-planar circular orbit

µπ

3

21 aPTOF ==

Hohmann Transfer (non-‐circular)

•  Another way to view it is using angular momentum

h = 2µrarpra + rp

!

"##

$

%&&

!Vtotal _3 = !VA +!VB!Vtotal _3' = !VA ' +!VB '

Hohmann Transfer (non-‐circular)

h1 = 2µ rArA 'rA + rA '

!

"#

$

%&

h2 = 2µ rBrB 'rB + rB '

!

"#

$

%&

h3 = 2µ rArBrA + rB

!

"#

$

%&

h3' = 2µ rA 'rB 'rA ' + rB '

!

"#

$

%&

VA 1 = h1 / rAVB 1 = h2 / rBVA ' 1 = h1 / rA 'VB ' 1 = h2 / rB '

VA 3= h3 / rA

VB 3= h3 / rB

VA ' 3' = h3' / rA 'VB ' 3' = h3' / rB '

!VA = abs VA 3"VA 1( )

!VA ' = abs VA ' 3 "VA ' 1( )!VB = abs VB 2

"VB 3( )!VB ' = abs VB ' 2 "VB ' 3'( )

Bi-‐Ellip0c Transfer

Bi-‐Ellip0c Transfer If rc/ra < 11.94 then Hohmann is more efficient If rc/ra > 15.58 then Bi-Elliptic is more efficient

Non-‐Hohmann Transfers w/Common Line of Aspides


etransfer = !rA ! rB

rA cos!A ! rB cos!B

htransfer = µrArBcos!A ! cos!B

rA cos!A ! rB cos!B

"

#$

%

&'

!VA = V12 +Vtrans

2 " 2V1Vtrans cos!!A

A

tan! = !vr@A

!v"@A

=vr@A orbit2

# vr@A orbit1

v"@A orbit2# v"@A orbit1


!1 = tan!1 µ / h1( )e1 sin!1

h1 / rA

"

#$

%

&'

" trans = tan!1 µ / htrans( )etrans sin!1

htrans / rA

"

#$

%

&'

!!A = ! trans "!1( )@A

! i = tan"1 VrV#

$

%&

'

()i @A

!VA = V12 +Vtrans

2 " 2V1Vtrans cos!!A

Apse Line Rota0on Opportunity to transfer from one orbit to another using a single maneuver

occurs at intersection points of the orbits

! ="1 !"2

Apse Line Rotation Angle

rorbit1@I = rorbit2@I

h12 /µ

1+ e1 cos!1=

h22 /µ

1+ e2 cos!2

NOTE:



!2 =!1 !"Setting and using the following identity

cos !1 !"( ) = cos!1 cos" + sin!1 sin"

Then has two solution corresponding to point I and J !1

!1 =" ± cos!1 h1

2 ! h22

e1h22 ! e2h1

2 cos#cos"

"

#$

%

&'

" = tan!1 !e2h12 sin#

e1h22 ! e2h1

2 cos#"

#$

%

&'



Given the orbit information of the two orbits Next, compute r, v!i, vri, ! i using "1 and "2 ="1 "#

r = h12 /µ

1+ e1 cos!1v!1 = h1 / rvr1 = µ / h1( )e1 sin!1"1 = tan

"1 vr1 / v!1( )

v1 = vr12 + v!1

2

v!2 = h2 / rvr2 = µ / h2( )e2 sin!2"2 = tan

"1 vr2 / v!2( )

v2 = vr22 + v!2

2

#v = v12 + v2

2 " 2v1v2 cos !2 "!1( )



tan! = !vr@I

!v"@I

=vr@I orbit2

# vr@I orbit1

v"@I orbit2# v"@I orbit1

If you want to find the effect of a !v on orbit 1 at !1

tan!2 =v!1 +"v!( ) vr1 +"vr( )

v!1 +"v!( )2 e1 cos!1 + 2v!1 +"v!( )"v!v!12

µ / r( )

If !1 = vr = !v" = 0 then tan! = ! rv"1µe1

#vr (at periapsis)



Another useful equation:

e2 =h1 + r1!v"( )2 e1 cos!1 + 2h1 + r1!v"( )r1!v"

h12 cos!2

Plane Change Maneuvers In general a plane change maneuver changes the orbit plane

!

v1

v2

!v = v2 " v1 = vr2 " vr1( ) ur + v#2u#2 " v#1u#1!v = !v " !v

!v = vr2 # vr1( )2 + v$12 + v$12 # 2v$2v$1 cos!

or

!v = v122 + v2

2 " 2v1v2 cos!! " cos!1 cos!2 1" cos"( )#$ %&

where !! = !2 "!1

Plane Change Maneuvers In general a plane change maneuver changes the orbit plane

!v = v122 + v2

2 " 2v1v2 cos!!

If there is no plane change, then ! = 0

and we get the same equation as from slide 130.

If then vr1 = vr2 = 0 and v!1 = v1 and v!2 = v2

!v = v122 + v2

2 " 2v1v2 cos!

No plane change

Pure plane change (common line of apisdes)

Plane Change !vPC = 2vC sin

!2"

#$%

&'= 2vC sin

!i2

"

#$

%

&'

Pure plane change (circular to circular)

!v(a) = v2 " v1( )2 + 4v1v2 sin2 ! / 2( )

!v(b) = 2v1 sin ! / 2( )+ v2 " v1 !v(c) = v2 " v1 + 2v2 sin ! / 2( )

Rendezvous Catching a moving target: Hohmann transfer assume the target will be there

independent of time, but for rendezvousing the target body is always moving.

target

spacecraft

TIME = 0

φ0

motionmean

where

32

2

20

≡=

⋅−=

an

TOFn

µ

πφΔV

Rendezvous

target

spacecraft

TIME = TOF

φ0

motionmean

where

3target

target

target0

≡=

⋅−=

an

TOFn

µ

πφΔV

Catching a moving target: Hohmann transfer assume the target will be there independent of time, but for rendezvousing the target body is always moving.

Wait Time (WT) How long do you wait before you can perform (or initiate) the transfer?

target

spacecraft

TIME = -WT

φ0

scnnW

−=

target

0-T φφ

φ

ΔV

Co-‐orbital Rendezvous

target

spacecraft

Chasing your tail

φ0

ΔV

Reverse thrust to speed up

Forward thrust to

slow down

Phasing orbit size

3/12

target

initialphasing 2

2⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅

−=

na

πφπ

µ

Patched-‐Conic An approximation breaks the interplanetary trajectory into

regions where conic approximation is applicable

Sphere of Influence 5/2

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Sun

planetplanetSOI m

maR



•  Computational Steps –  PATCH 1: Compute the Hohmann transfer ΔVs –  PATCH 2: Compute the Launch portion –  PATCH 3: Compute the Arrival portion

•  PATCH 1 –  The ΔV1 from the interplanetary Hohmann is V∞ at Earth –  The ΔV2 from the interplanetary Hohmann is V∞ at target

body arrival



ΔVDEP

VEarth V∞ @Earth

RC@DEP DEP

DEP

DEP

DEP

C

DEPC

C

departureCDEP

RV

VV

VV

µ=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟

⎟⎠

⎞⎜⎜⎝

⎛+=Δ ∞

where

12 2

@

Earth Departure (assume circular orbit)

RSOI



ΔVCAP

VMars V∞ @Mars

RC@CAP

CAP

CAP

CAP

CAP

C

CAPC

C

arrivalCCAP

RV

VV

VV

µ=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟

⎟⎠

⎞⎜⎜⎝

⎛+=Δ ∞

where

12 2

@

Target Body Capture (assume circular orbit)

RSOI



CAPDEPMISSION VVV Δ+Δ=Δ

Total Mission ΔV for a simple 2 burn transfer

ΔVCAP

ΔVDEP

Planetary Flyby

•  Gravity-‐assists or flybys are used to –  Reduce or increase the heliocentric (wrt. Sun) energy of the spacecra6

(orbit pumping), or –  Change the heliocentric orbit plane of the spacecra6 (orbit cranking),

or both

Getting a boost

Planetary Flyby Getting a boost


Stationary Jupiter Moving Jupiter


Vplanet

Fg Fg

Decrease energy wrt. Sun Increases energy wrt. Sun

Planetary Flyby

•  V∞ leveraging is the use of deep space maneuvers to modify the V∞ at a body. A typical example is an Earth launch, ΔV maneuver (usually near apoapsis), followed by an Earth gravity assist (ΔV-‐EGA).

Getting a boost

ΔV

If no maneuver

post maneuver

Lecture 11and 12: Rela0ve Mo0on

Chapter 7

Rela0ve Mo0on and Rendezvous

•  In this chapter we will look at the rela0ve dynamics between 2 objects or 2 moving coordinate frames, especially in close proximity

•  We will also look at the linearized mo0on, which leads to the Clohessy-‐Wiltshire (CW) equa4ons

Co-‐Moving LVLH Frame (7.2)

Local Vertical Local Horizontal (LVLH) Frame i = rA

rA, j= k! i, k = hA

hA

TARGET

CHASER (or observer)

•  The target frame is moving at an angular rate of Ω

where and

•  Chapter 1: Rela0ve mo0on in the INERTIAL (XYZ) frame

Co-‐Moving LVLH Frame

hA = rA ! vA = rAvA"k = rA2# k = rA

2!

! =rA ! vArA2

!! = hAddt

1rA2

!

"#

$

%&= '2

vA ( rArA2 !

rrel = rB ! rAvrel = vB ! vA !! " rrelarel = aB ! aA ! !! " rrel !! " ! " rrel( )! 2! " vrel

•  We need to find the mo0on in the non-‐iner4al rota0ng frame

where Q is the rota0ng matrix from


rrel _ rot =Qiner _ to_ rot rrel _ inervrel _ rot =Qiner _ to_ rot vrel _ inerarel _ rot =Qiner _ to_ rot arel _ iner

Qiner _ to_ rot =

ij

k

!

"

####

$

%

&&&&

=

rA / rAk' ihA / hA

!

"

####

$

%

&&&&

•  Steps to find the rela0ve state given the iner0al state of A and B.


1.  Compute the angular momentum of A, hA

2.  Compute the unit vectors

3.  Compute the rotating matrix Q

4.  Compute

5.  Compute the inertial acceleration of A and B

i, j, and k

! and !!

aA = !µ rA / rA3, and aB = !µ rB / rB

3

•  Steps to find the rela0ve state given the iner0al state of A and B.


6.  Compute the relative state in inertial space

7.  Compute the relative state in the rotating coordinate system

rrel _ rot =Qiner _ to_ rot rrel _ inervrel _ rot =Qiner _ to_ rot vrel _ inerarel _ rot =Qiner _ to_ rot arel _ iner


Rotating Frame

Lineariza0on of the EOM (7.3)

r =R+!r

!r / R <<1

!!!r = ! !!R!µ R+!rR+!r 3

!!!r = ! µR3

!r! 3R2

R "!r( ) R#

$%&

'(

neglecting higher order terms

Lineariza0on of the EOM

!!!r = ! µR3

!2 !x!y!z

"

#

$$$

%

&

'''= !!rB ! !!rA = aB ! aA

Assuming R = R i

Acceleration of B relative to A in the inertial frame

!arel = !!!r! !! "!r!! " ! "!r( )! 2! "!vrel

! =hR2k !! = !

2 V "R( )hR4

k


After further simplification we get the following EOM

!!!x ! 2µR3

+h2

R4"

#$

%

&'!x +

2 V (R( )hR4

!y! 2hR2! !y = 0

!!!y+ µR3

!h2

R4"

#$

%

&'!y!

2 V (R( )hR4

!x + 2hR2! !x = 0

!!!z + µR3!z = 0

Thus, given some initial state R0 and V0 we can integrate the above EOM (makes no assumption on the orbit type)


e = 0.1

e = 0

Clohessy-‐Whiltshire (CW) Equa0ons (7.4)

Assuming circular orbits: Then EOM becomes

!!!x !3n2!x ! 2n ! !y = 0!!!y+ 2n ! !x = 0!!!z + n2!z = 0

where

V !R( ) = 0 and h = µR

n = µR3

=V / R

Clohessy-‐Whiltshire (CW) Equa0ons

Where the solution to the CW Equations are:

!x = 4!x0 +2n! !y0 +

1n! !x0 sin nt( )! 3!x0 +

2n! !y0

"

#$

%

&'cos nt( )

!y = !y0 !2n! !x0 !3 2n !x0 +! !y0( ) t + 2 3!x0 + 2n! !y0

"

#$

%

&'sin nt( )+ 2

n! !x0 cos nt( )

!z = 1n! !z0 sin nt( )+!z0 cos nt( )

Maneuvers in the CW Frame (7.5) The position and velocity can be written as where

!r t( ) =!rr!r0 +!rv!v0!v t( ) =!vr!r0 +!vv!v0

!rr =

4"3cos nt( ) 0 0

6 sin nt( )" nt( ) 1 0

0 0 cos nt( )

#

$

%%%%

&

'

((((

!rv =

1/ n( )sin nt( ) 2 / n( ) 1" cos nt( )( ) 0

2 / n( ) cos nt( )"1( ) 1/ n( ) 4sin nt( )"3nt( ) 0

0 0 1/ n( )sin nt( )

#

$

%%%%%

&

'

(((((

Maneuvers in the CW Frame

!vv =

cos nt( ) 2sin nt( ) 0

"2sin nt( ) 4cos nt( )"3 0

0 0 cos nt( )

#

$

%%%%

&

'

((((

!vr =

3nsin nt( ) 0 0

6n cos nt( )"1( ) 0 0

0 0 "nsin nt( )

#

$

%%%%

&

'

((((

and

Maneuvers in the CW Frame

Two-Impulse Rendezvous: from Point B to Point A

Maneuvers in the CW Frame Two-Impulse Rendezvous: from Point B to Point A where

where is the relative velocity in the Rotating frame, i.e., If the target and s/c are in the same circular orbits then

!v = !v@B t = 0( ) + !v@A t = t f( )

!v@B = " #rv t f( )$% &'"1#rr t f( )$% &'( )!r0 "!v0"

!v@A = "vr t f( )#$ %&' "vv t f( )#$ %& "rv t f( )#$ %&'1"rr t f( )#$ %&( )!r0

!v0!

!v0! =Qinertial _ to_ rotating!v =QXx vs/c ! vtar !! tar!r( )

!v0! = 0

Maneuvers in the CW Frame Two-Impulse Rendezvous example:

Lecture 13: Rigid Body Dynamics Aytude Dynamics

Chapter 9-‐10

Rigid Body Mo0on RB =RA +RB/A

RB/A = constant

vB = vA +! !RB/A

Note:

dRB/A / dt =! !RB/A

aB = aA +! !RB/A +! ! ! !RB/A( )

Position, Velocity, and Acceleration of points on a rigid body, measure in the same inertial frame of reference.

Angular Velocity/Accelera0on

•  When the rigid body is connected to and moving rela0ve to another rigid body, (example: solar panels on a rota0ng s/c) computa0on of its iner0al angular velocity (ω) and the angular accelera0on (α) must be done with care.

•  Let Ω be the iner0al angular velocity of the rigid body

! =d"dt

+! !"

Note: if ! =d"dt

! =!

Example 9.2

Angular Velocity of Body

! = N k

Angular Velocity of Panel

! = ! !! j + N k

rA/O = !w2sin! i + d j + w

2cos! k

vA/O =! ! rA/O = "w2!! cos! + Nd

#

$%

&

'( i "

w2N sin! j " w

2!! sin! k

Example 9.2 (con0nues)

aA/O =! ! rA/O +! ! ! ! rA/O( )

aA/O =w2N 2 + !! 2( )sin! i ! N Nd +w !! cos!( ) j ! w

2!! 2 cos! k

! =d"dt

+! !! =ddt

" !! j + N k( )+ N k( )! " !! j + N k( ) = !!N i

0

Example: Gimbal

! = !! k + N sin! j + N cos! k

!rotor = !! i + N sin! j + N cos! +"spin( ) k

Equa0ons of Mo0on

•  Dynamics are divided to transla0onal and rota0onal dynamics

Translational:

Ftrans =m !!RG

Equa0ons of Mo0on

•  Dynamics are divided to transla0onal and rota0onal dynamics

Rotational: MPnet

= r! !!R dmm"

MPnet= r! dFnet"

MPnet= !HP + v p !mvG

If then where v p = vG MGnet= !HG HG = ! ! " !!( ) dm"

Angular Momentum

HG = ! ! " !!( ) dm"

?

! ! " !!( ) =" ! "!( )# ! " "!( )

! ! " !!( ) =

y2 + z2( )"x # xy"y # xz"z

#yx"x + x2 + z2( )"y # yz"z

#zx"x # zy"y + x2 + y2( )"z

$

%

&&&&&

'

(

)))))

Angular Momentum

Since: HG =

Hx

Hy

Hz

!

"

####

$

%

&&&&

= ?[ ]!x

!y

!z

!

"

####

$

%

&&&&

HG = I!

I =

Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz

!

"

####

$

%

&&&&

=

y2 + z2( )' dm ( xy' dm ( xz' dm

( yx' dm x2 + z2( )' dm ( yz' dm

( zx' dm ( zy' dm x2 + y2( )' dm

!

"

#####

$

%

&&&&&

Note: Iyx = Ixy, Izx = Ixz, and Iyz = Izy

Angular Momentum

If has 2 planes of symmetry then I

I =Ixx 0 00 Iyy 0

0 0 Izz

!

"

####

$

%

&&&&

=A 0 00 B 00 0 C

!

"

###

$

%

&&&

therefore Hx = A !x, Hy = A !y, Hz = A !z

Moments of Iner0a

Euler’s Equa0ons

•  Rela0ng M and for pure rota0on. Assuming body fixed coordinate is along principal axis of iner0a

•  Therefore

!

Mnet = !H = !Hrelative +!"H

H = Hx Hy Hz( )T

= A!x B!y C!z( )T

Mnet = A !!x B !!y C !!z( )T

+

i j k!x !y !z

A!x B!y C!z

Euler’s Equa0ons

•  Assuming that moving frame is the body frame, then this leads to Euler’s Equa4ons:

Mxnet= A !!x + C !B( )!y!z

Mynet= B !!y + A!C( )!z!x

Mznet=C !!z + B! A( )!x!y

! =!

Kine0c Energy

T = 12

v2m! dm =

12v "v

m! dm =

12mvG

2 +12! "HG = Ttrans +Trot

Trot =12!xHx +!yHy +!zHz( ) = 12!

TI!

Spinning Top •  Simple axisymmetric top spinning at point 0

Introduces the topic of 1.  Precession 2.  Nutation 3.  Spin

Assumes:

Ixx = Iyy = A and Izz =C

! p = !" !n = !#

Notes: If A < C (oblate) If C < A (prolate)

Spinning Top

! p = !" !n = !#

From the diagram we note 3 rotations: where therefore:

! =!ni +! pK +!sk

K = sin! j + cos! k

! =

!x

!y

!z

!

"

####

$

%

&&&&

=

!n

! p sin"

!s +! p cos"

!

"

####

$

%

&&&&

Spinning Top

! p = !" !n = !#

From the diagram we note the coordinate frame rotation therefore:

! =!ni +! pK

! =

!x

!y

!z

"

#

$$$$

%

&

''''

=

!n

! p sin"

! p cos"

"

#

$$$$

%

&

''''

! M0net =

A !!x

A !!y

C !!z

"

#

$$$$

%

&

''''

+

i j k(x (y (z

A!x A!y C!z

= d k ) *mg( ) K =mgd sin! i

Spinning Top •  Some results for a spinning top

–  Precession and spin rate are constant –  For precession two values exist (in general) for

–  If spin rate is zero then

•  If A > C, then top’s axis sweeps a cone below the horizontal plane •  If A < C, then top’s axis sweeps a cone above the horizontal plane

!! p = !!s = 0

! p !s=0= ±

mgdC ! A( )cos!

if C ! A( )cos! > 0

! p =C

2 A!C( )cos"!s ± !s

2 !4mgd A!C( )cos"

C2

"

#$$

%

&''

! ! 90°

Spinning Top •  Some results for a spinning top

–  If then

•  If , then precession occurs regardless of 0tle angle •  If , then precession occurs 0tle angle 90 deg

–  If then a minimum spin rate is required for steady precession at a constant 0lt

–  If then

A!C( )cos! = 0

! p =mgdC!s

if A!C( )cos" = 0

A =CA !C

A!C( )cos! > 0

!s_MIN =2C

mgd A!C( )cos" if A!C( )cos" > 0

A!C( )cos! < 0 !s ! 0, and ! p !! p !s=0

Axisymmetric Rotor on Rota0ng Pla|orm

! = 90° ! =! p j +!sk

! =! p j

MGnet =! !H =

i j k0 ! p 0

0 A! p C!s

=! p j !C!sk = ! p"# $%!Hs

Thus, if one applies a torque or moment (x-axis) it will precess, rotating spin axis toward moment axis

Euler’s Angles (revisited) •  Rota0on between body fixed x,y,z to rota0on angles using Euler’s angles (313 rota0on)

Qiner _ to_body =R3 !( )R1 !( )R3 !( )

!body =

!x

!y

!z

!

"

####

$

%

&&&&

=

! p sin" sin# +!n cos#

! p sin" cos# '!n sin#

!s +! p cos"

!

"

####

$

%

&&&&

!body =Qiner _ to_body!inertal

! p

!n

!s

!

"

####

$

%

&&&&

=

!"!#!$

!

"

####

$

%

&&&&

=

1sin#

!x sin$ +!y cos$( )!x cos$ '!y sin$

'1tan#

!x sin$ +!y cos$( )+!z

!

"

######

$

%

&&&&&&

Euler’s Angles (revisited)

!body =

!x

!y

!z

!

"

####

$

%

&&&&

=

! p sin" sin# +!n cos#

! p sin" cos# '!n sin#

!s +! p cos"

!

"

####

$

%

&&&&

!inertal =QTiner _ to_body !!body

Mxnet= A !!x + C !B( )!y!z

Mynet= B !!y + A!C( )!z!x

Mznet=C !!z + B! A( )!x!y

Satellite Aytude Dynamics

•  Torque Free Mo0on MG _net = !HG = 0 = !HG _ rel +! !HG

cos! = HG

HG

! k

!! ="n = !A!B( )"x"y

HG sin!

Euler’s Equa0on for Torque Free Mo0on

0 = A !!x + C ! A( )!y!z

0 = B !!y + A!C( )!z!x

0 =C !!z + B! A( )!x!y

A = B A !!x + C ! A( )!y!z = 0

A !!y + A!C( )!z!x = 0C !!z = 0

!z =!0 = constant!n = 0

!!x !"!y = 0!!y +"!x = 0

! =A!CA

"0

!!!x +"!x = 0


! =!! +!0 =!! +!0kFor Then:

!s = !" =A!CA

!0

!xy =CA!0 tan"

! p = !" =C

A!C!s

cos#

If A > C (prolate), ωp > 0 If A < C (oblate), ωp < 0


HG =

Hx

Hy

Hz

!

"

####

$

%

&&&&

=

A!x

A!y

C!z

!

"

####

$

%

&&&&

=

A!x

A!y

C!0

!

"

####

$

%

&&&&

tan! = ACtan"

If A > C (prolate), γ < θ If A < C (oblate), γ > θ

HG = A! p

Stability of Torque-‐Free S/C Assumes: ! =!0k

! !!"x,y + k!"x,y = 0

0 = A !!x + C ! A( )!y!z

0 = B !!y + A!C( )!z!x

0 =C !!z + B! A( )!x!y

k =A!C( ) B!C( )

AB!02

Stability of Torque-‐Free S/C ! !!"x,y + k!"x,y = 0

•  If k > 0, then solution is bounded •  A > C and B > C or A < C and B < C •  Therefore, spin is the major axis (oblate) or minor

axis (prolate)

•  If k < 0, then solution is unstable •  A > C > B or A < C < B •  Therefore, spin is the intermediate axis

!"x,y = c1ei kt + c2e

!i kt

!"x,y = c1ekt + c2e

! kt

Stability of Torque-‐Free S/C •  With energy dissipa0on ( )

Trot =12!TI! =

12A!!

2 +12C!z

2

!!z =1

C!z

!Trot !12A d!"

2

dt#

$%

&

'(

d!!2

dt= 2C

A

!TrotC " A#

$%

&

'(

!Trot < 0

d!!2

dt< 0 if C > A (oblate)" asymtotically stable

d!!2

dt> 0 if C < A (prolate)" unstable

Stability of Torque-‐Free S/C •  Kine0c Energy rela0ons

Trot =12A!!

2 +12C!z

2 =12HG

2

A+12A"CA

#

$%

&

'(C!z

2

Trot =12HG

2

A1+ A!C

Acos2!

"

#$

%

&'

Trot =

12HG

2

C for major axis spinner

12HG

2

A for minor axis spinner

!

"

##

$

##

Conning Maneuvers •  Maneuver of a purely spinning S/C with fixed angular momentum magnitude

! =!0k

HG,0 =C!0k

!HG = !HG1 +!HG2

!HG = MG dt0

t f

"

Conning Maneuvers Before the Maneuver

During the Maneuver

!s =!0 HG =C!0

!s =A!CA

"

#$

%

&'!0

!P =CA

!0

cos " / 2( )

!

"##

$

%&&

HG = A! p =C!0

cos " / 2( )

Another maneuver is required ΔHG2 after precession 180 deg

Conning Maneuvers Another maneuver is required ΔHG2 after precession 180 deg.

At the 2nd maneuver we want to stop the precession (normal to the spin axis):

!s =! p

!HG1 = !HG2

! = 2cos!1 CA!C"

#$

%

&'

Required deflection angle to precess 180 deg for a single coning mnvr

t = !" p

=!AC!0

cos " / 2( )

!HG = !HG1 + !HG2

= 2 HG0 tan ! / 2( )( )" HG0 !

Gyroscopic Aytude Control

•  Momentum exchange gyros or reac0on wheels can be used to control S/C aytude without thrusters.

•  The wheels can be fixed axis (reac0on wheels) or gimbal 2-‐axis (cmg)


HG = IGbus + IG

i!( )! + IGi!rel

i!= IG

s/c! + IGi!rel

i!MG,net,ext =

dHG

dt+! !HG

Example: HG = I p + Iw( )! + Iw!rel

If external torque free then therfore

HG t = 0( ) = HG t = !t( )

!!rel = " 1+ I p / Iw( )!!


HG = IB + I1 + I2 + I3( )! + I1!1 + I2!2 + I3!3

Example II: S/C with three identical wheels with their axis along the principal axis of the S/C bus, where the wheels spin axis moment of inertial is I and other axis are J. Also, the bus moment of inertia are diagonal elements (A, B, C).

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