Selected Problems in Dynamics:

Stability of a Spinning Body-and-

Derivation of the Lagrange PointsJohn F. Fay

June 24, 2014

Outline

Stability of a Spinning Body

Derivation of the Lagrange Points

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Stability of a Spinning Body

Euler’s Second Law:

(good for an inertial coordinate system)Problem: Inertia matrix “I” of a rotating body changes in an inertial coordinate system

Idt

dM

Moments of Torque

Inertia MatrixAngular Velocity

Euler’s Second Law (2)

In a body-fixed coordinate system:

Inertia Matrix: Describes distribution

of mass on the body

IIdt

dM body

Momentsof Torque

Inertia MatrixAngular Velocity

zzyzxz

yzyyxy

xzxyxx

III

III

III

I

Euler’s Second Law (3)

Planes of Symmetry: x-y plane: Ixz = Iyz = 0 x-z plane: Ixy = Iyz = 0 y-z plane: Ixz = Ixz = 0

For a freely-spinning body:

0 IIdt

dbody

zz

yy

xx

I

I

I

I

00

00

00

0M

Euler’s Second Law (4)

Rigid body: “I” matrix is constant in body coordinates

Do the matrix multiplication:

0 IIdt

dbody

zzz

yyy

xxx

z

y

x

zz

yy

xx

I

I

I

I

I

I

I

00

00

00

0 I

dt

dI

Euler’s Second Law (5)

Do the cross product:

Result:

xxxyyyyx

zzzxxxxz

yyyzzzzy

zzz

yyy

xxx

z

y

x

II

II

II

I

I

I

I

0

00

00

00

xxxyyyyx

zzzxxxxz

yyyzzzzy

z

y

x

zz

yy

xx

II

II

II

dt

d

I

I

I

Euler’s Second Law (6)

As scalar equations:

0 zyyyzzx

xx IIdt

dI

0 xzzzxxy

yy IIdt

dI

0 yxxxyyz

zz IIdt

dI

0

00

00

00

xxxyyyyx

zzzxxxxz

yyyzzzzy

z

y

x

zz

yy

xx

II

II

II

dt

d

I

I

I

Small Perturbations:

Assume a primary rotation about one axis:

Remaining angular velocities are small

xx '

yy '

zz ' zyx ',','

Small Perturbations (2)

Equations:

0'''

zyyyzzx

xx IIdt

dI

0'''

xzzzxxy

yy IIdt

dI

0'''

yxxxyyz

zz IIdt

dI

Small Perturbations (3)

Linearize: Neglect products of small quantities

0'

dt

dI xxx

0''

zzzxxy

yy IIdt

dI

0''

yxxyyz

zz IIdt

dI

Small Perturbations (4)

Combine the last two equations:

z

yy

zzxxy

I

II

dt

d'

'

0''

yxxyyz

zz IIdt

dI

0''

zzzxxy

yy IIdt

dI

y

zz

yyxxz

I

II

dt

d'

'

switched

Small Perturbations (4)

Combine the last two equations:

dt

d

I

II

dt

dz

yy

zzxxy ''2

2

0''

yxxyyz

zz IIdt

dI

0''

zzzxxy

yy IIdt

dI

dt

d

I

II

dt

d y

zz

yyxxz''

2

2

Small Perturbations (4)

Combine the last two equations:

0'

' 22

2

yzzyy

yyxxzzxxy

II

IIII

dt

d

0''

yxxyyz

zz IIdt

dI

0''

zzzxxy

yy IIdt

dI

0'

' 22

2

zzzyy

yyxxzzxxz

II

IIII

dt

d

Small Perturbations (5)

Case 1: Ixx > Iyy, Izz or Ixx < Iyy, Izz:

Sinusoidal solutions for , Rotation is stable

02

zzyy

yyxxzzxx

II

IIII

y' z'

0'

' 22

2

yzzyy

yyxxzzxxy

II

IIII

dt

d

0'

' 22

2

zzzyy

yyxxzzxxz

II

IIII

dt

d

Small Perturbations (6)

Case 2: Izz > Ixx > Iyy or Izz < Ixx < Iyy:

Hyperbolic solutions for , Rotation is unstable

02

zzyy

yyxxzzxx

II

IIII

y' z'

0'

' 22

2

zzzyy

yyxxzzxxz

II

IIII

dt

d

0'

' 22

2

yzzyy

yyxxzzxxy

II

IIII

dt

d

Numerical Solution

Ixx = 1.0Iyy = 0.5Izz = 0.7

Body Coordinates

Numerical Solution (2)

Ixx = 1.0Iyy = 0.5Izz = 0.7

Global Coordinates

Numerical Solution (3)

Ixx = 1.0Iyy = 1.2Izz = 2.0

Body Coordinates

Numerical Solution (4)

Ixx = 1.0Iyy = 1.2Izz = 2.0

Global Coordinates

Numerical Solution (5)

Ixx = 1.0Iyy = 0.8Izz = 1.2

Body Coordinates

Numerical Solution (6)

Ixx = 1.0Iyy = 0.8Izz = 1.2

Global Coordinates

Spinning Body: Summary

Rotation about axis with largest inertia:Stable

Rotation about axis with smallest inertia:Stable

Rotation about axis with in-between inertia:Unstable

Derivation of the Lagrange Points

Two-Body Problem: Lighter body orbiting around a heavier

body(Let’s stick with a circular orbit)

Actually both bodies orbit around the barycenter

x

y W

mM

D

Two-Body Problem (2)

Rotating reference frame: Origin at barycenter of two-body system Angular velocity matches orbital velocity

of bodies Bodies are stationary in the rotating

reference frame

W

Mx

y

m

D

Two-Body Problem (3)

Force Balance:Body “M” Body “m”

02

2 D

GMmMxM 0

22

D

GMmmxm

W

x

y

m

D

Mm xxD

xmxM(<0)

M

CentrifugalForceGravitational Force

Two-Body Problem (4)

Parameters: M – mass of large body m – mass of small body D – distance between bodies

Unknowns: xM – x-coordinate of large body xm – x-coordinate of small body W – angular velocity of coordinate system

Three equations, three unknowns

Mm xxD

02

2 D

GMmMxM

02

2 D

GMmmxm

In case anybody wants to see the math …

mmm xM

mMx

M

mxD

02

2 D

GMmMxM

02

2 D

GMmmxm

Mm xxD 0

22 D

GmxM 0

22

D

GMxm

222

Dx

GM

Dx

Gm

mM

mM x

M

x

m Mm Mxmx

DmM

Mxm

DmM

mxM

32

2

D

mMG

Dx

GM

m

Two-Body Problem (5)

Solutions:

DmM

Mxm

DmM

mxM

3

2

D

mMG

W

x

y

m

DxmxM(<0)

M

Lagrange Points

Points at which net gravitational and centrifugal forces are zero

Gravitational attraction to body “M” Gravitational attraction to body “m” Centrifugal force caused by rotating

coordinate system

W

x

y

mM

(x,y)

Lagrange Points (2)

Sum of Forces: particle at (x,y) of mass “ m ”

x-direction:

y-direction:

02

23

2223

22

x

yxx

xxGm

yxx

xxGM

m

m

M

M

02

23

2223

22

y

yxx

yGm

yxx

yGM

mM

Lagrange Points (3)

Simplifications: Divide by mass “m” of particle Substitute for “W2” and divide by “G”

0

32

3222

322

D

mMx

yxx

xxm

yxx

xxM

m

m

M

M

03

23

2223

22

D

mMy

yxx

my

yxx

My

mM

Lagrange Points (4)

Qualitative considerations: Far enough from the system, centrifugal force will

overwhelm everything else Net force will be away from the origin

Close enough to “M”, gravitational attraction to that body will overwhelm everything else

Close enough to “m”, gravitational attraction to that body will overwhelm everything else

mM

Lagrange Points (5)

Qualitative considerations (2) Close to “m” but beyond it, the attraction to “m”

will balance the centrifugal force search for a Lagrange point there

On the side of “M” opposite “m”, the attraction to “M” will balance the centrifugal force search for a Lagrange point there

Between “M” and “m” their gravitational attractions will balance search for a Lagrange point there

mM

Some Solutions: Note that “y” equation is uniquely

satisfied ify = 0 (but there may be nonzero solutions also)

03

23

2223

22

D

mMy

yxx

my

yxx

My

mM

Lagrange Points (6)

Lagrange Points (7)

Some Solutions: “x” equation when “y” is zero:

Note:

0

32

322

32

D

mMx

xx

xxm

xx

xxM

m

m

M

M

3323

2MMM xxxxxx ?

0

33

MMmm

MMmm

xxxxxxxxmMx

xxxxmDxxxxMD

Lagrange Points (8)

Oversimplification: Let “m” = 0 xM = 0

Solutions: Anywhere on the orbit of “m”

0

32

322

D

Mx

yx

Mx

0

32

322

D

My

yx

My

323

22 Dyx

0

32

3222

322

D

mMx

yxx

xxm

yxx

xxM

m

m

M

M

03

23

2223

22

D

mMy

yxx

my

yxx

My

mM

Lagrange Points (9)

Perturbation:

For the y = 0 case:

Mm 10 DxM Dxm 1

0111

113

3

DxDxDxDxx

DxDxD

DxDxD

0

33

MMmm

MMmm

xxxxxxxxmMx

xxxxmDxxxxMD

Lagrange Points (10)

For the y = 0 case (cont’d):– Lagrange Points L1,L2: near x = D:

0111

11 33

DxDxDxDxx

DxDxDDxDxD

1Dx

0111

125

255

D

DD

0

33

MMmm

MMmm

xxxxxxxxmMx

xxxxmDxxxxMD

L1, L2: y = 0, x = D(1 + )d

Collecting terms:

Multiplying things out:

01

11112

2

22

332222

3232

2221

253

57333

0111

125

255

D

DD

L1, L2 (2): y = 0, x = D(1 + )d

Balancing terms:

Linearizing:

22

332222

3232

2221

253

57333

Order of magnitude d3

Order of magnitude e

33 3

3

1

Lagrange Points (11)

Lagrange Point L3: y = 0, near x = –D

Since | e |, | d | << 1 < 2:

0112211

1122

1Dx

01211

1222

22

0111

11 33

DxDxDxDxx

DxDxDDxDxD

L3: y = 0, x = –D(1 + )d

Multiply out and linearize:

12

5

02214441

444

01211

1222

22

08844444

444

0125

Lagrange Points (12)

Summary for a Small Perturbation and y = 0:

Lagrange Points:

L1:

L2:

L3:

Mm 10 DxM Dxm 1

3

3

11 Dx

12

51Dx

3

3

11 Dx

Lagrange Points (13)

What if y is not zero? Mm 10 DxM Dxm 1

x

y

2q

1DR

Lagrange Points (14)

Polar Coordinates:

Mm 10 DxM

Dxm 1

x

y

2q

2sin,2cos DDR

0,RF

1DR

2sin1

,2cos1

D

DR

ereq

Lagrange Points (15)

Consider the forces in the R-direction:

From M:

From m:

Centrifugal force:

2

322 2sin2cos1

2cos1

DDD

DDGMFM

2

322 2sin12cos11

2cos11

DDD

DDGmFm

12DFC

R-direction Forces

Simplify: From M:

From m:

Centrifugal force:

2

3222 2sin2cos1

2cos1

D

GMFM

2

3222 2sin12cos11

2cos11

D

MGFm

1

12D

GMFC

R-direction Forces (2)

Linearize in d and e: From M:

From m:

Centrifugal force:

2cos2212cos331

2cos122

D

GM

D

GMFM

2

232 22212cos1221

2cos11

D

GM

D

MGFm

1

2D

GMFC

R-direction Forces (3)

Balance the Forces:

0122

2cos221222

D

GM

D

GM

D

GM

0122

2cos221

22

2cos2211

22

2cos23

3

2cos2122

1

Lagrange Points (16)

Consider the forces in the q-direction:

From M:

From m:

Centrifugal force is completely radial

2

322 2sin2cos1

2sin

DDD

DGMFM

2

322 2sin12cos11

2sin1

DDD

DGmFm

0CF

q-Direction Forces

Simplify:

2

322 2sin2cos1

2sin

DDD

DGMFM

2

322 2sin12cos11

2sin1

DDD

DGmFm

2

3222 2sin2cos1

2sin

D

GMFM

2

3222 2sin12cos11

2sin1

D

GMFm

q-Direction Forces (2)

Linearize in d and e:

22

2sin

2cos331

2sin

D

GM

D

GMFM

2

3222 2sin2cos1

2sin

D

GMFM

2

3222 2sin12cos11

2sin1

D

GMFm

232

232

2cos22

2sin

212cos1221

2sin1

D

GM

D

GMFm

q-Direction Forces (3)

Simplifying further:

2

2sin

D

GMFM

322

322

23

222

sin8

2sin

sin222

2sin

sincos122

2sin

D

GM

D

GM

D

GMFm

2

2sin

D

GMFM

232 2cos22

2sin

D

GMFm

q-Direction Forces (4)

Balance the two forces:

0

sin8

2sin2sin322

D

GM

D

GM

2

2sin

D

GMFM

32 sin8

2sin

D

GMFm

322 sin8

2sin2sin

D

GM

D

GM

1sin8 3

2

1sin

306

60

32

Lagrange Points (17)

Solving for our radius:

3

2cos2122

1

1DR

603

2

12

281

32

121

22

11

32cos21

22

11

DD

DR

Lagrange Points (13 redux)

What if y is not zero? Mm 10 DxM Dxm 1

x

y

602

12

281DR

Lagrange Points (18):

Summary for zero y: Between M and m: L1:

On the opposite side of m from M: L2:

On the opposite side of M from m: L3:

Mm 10 DxM Dxm 1

3

3

11 Dx

3

3

11 Dx

12

51Dx

Lagrange Points (19):

Summary for nonzero y: Ahead of m in orbit: L4:

Behind m in orbit: L5:

602

602

Mm 10 DxM Dxm 1

12

281DR

12

281DR

Summary of Lagrange Points

(courtesy of Wikipedia, “Lagrange Points”)

(What is wrongwith this picture?)

Potential Energy Contours

Stability of Lagrange Points

Requires saving higher-order terms in linearized equations

Results:– L1, L2, L3 unstable

Body perturbed from the point will move away from the point

– L4, L5 neutrally stable Body perturbed from the point will drift

around the point

Conclusions

Stability of Spinning Bodies

Derivation of Lagrange Points