john f. fay june 24, 2014
DESCRIPTION
Selected Problems in Dynamics: Stability of a Spinning Body -and- Derivation of the Lagrange Points. John F. Fay June 24, 2014. Outline. Stability of a Spinning Body Derivation of the Lagrange Points. TEAS Employees: Please log this time on the Training Tracker. - PowerPoint PPT PresentationTRANSCRIPT
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
TEAS Employees: Please log this time on the Training Tracker
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
IdtdM
Moments of TorqueInertia Matrix
Angular Velocity
Euler’s Second Law (2)
In a body-fixed coordinate system:
Inertia Matrix: Describes distribution
of mass on the body
IIdtdM body
Momentsof Torque
Inertia MatrixAngular Velocity
zzyzxz
yzyyxy
xzxyxx
IIIIIIIII
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 IIdtd
body
zz
yy
xx
II
II
000000
0M
Euler’s Second Law (4)
Rigid body: “I” matrix is constant in body coordinates
Do the matrix multiplication:
0 IIdtd
body
zzz
yyy
xxx
z
y
x
zz
yy
xx
III
II
II
00
0000
0 IdtdI
Euler’s Second Law (5)
Do the cross product:
Result:
xxxyyyyx
zzzxxxxz
yyyzzzzy
zzz
yyy
xxx
z
y
x
IIIIII
III
I
000
0000
xxxyyyyx
zzzxxxxz
yyyzzzzy
z
y
x
zz
yy
xx
IIIIII
dtd
II
I
Euler’s Second Law (6)
As scalar equations: 0 zyyyzz
xxx IIdtdI
0 xzzzxxy
yy IIdtd
I
0 yxxxyyz
zz IIdtdI
000
0000
xxxyyyyx
zzzxxxxz
yyyzzzzy
z
y
x
zz
yy
xx
IIIIII
dtd
II
I
Small Perturbations:
Assume a primary rotation about one axis:
Remaining angular velocities are smallxx '
yy '
zz ' zyx ',','
Small Perturbations (2)
Equations: 0'''
zyyyzzx
xx IIdtdI
0'''
xzzzxxy
yy IIdtd
I
0''' yxxxyy
zzz IIdtdI
Small Perturbations (3)
Linearize: Neglect products of small quantities
0'
dtdI x
xx
0''
zzzxxy
yy IIdtd
I
0'' yxxyy
zzz IIdtdI
Small Perturbations (4)
Combine the last two equations:
z
yy
zzxxy
III
dtd
''
0'' yxxyy
zzz IIdtdI
0''
zzzxxy
yy IIdtd
I
y
zz
yyxxz
III
dtd ''
switched
Small Perturbations (4)
Combine the last two equations:
dtd
III
dtd z
yy
zzxxy ''2
2
0'' yxxyy
zzz IIdtdI
0''
zzzxxy
yy IIdtd
I
dtd
III
dtd y
zz
yyxxz''
2
2
Small Perturbations (4)
Combine the last two equations:
0'
' 22
2
yzzyy
yyxxzzxxy
IIIIII
dtd
0'' yxxyy
zzz IIdtdI
0''
zzzxxy
yy IIdtd
I
0'' 2
2
2
zzzyy
yyxxzzxxz
IIIIII
dtd
Small Perturbations (5)
Case 1: Ixx > Iyy, Izz or Ixx < Iyy, Izz:
Sinusoidal solutions for , Rotation is stable
02
zzyy
yyxxzzxx
IIIIII
y' z'
0'
' 22
2
yzzyy
yyxxzzxxy
IIIIII
dtd
0'' 2
2
2
zzzyy
yyxxzzxxz
IIIIII
dtd
Small Perturbations (6)
Case 2: Izz > Ixx > Iyy or Izz < Ixx < Iyy:
Hyperbolic solutions for , Rotation is unstable
02
zzyy
yyxxzzxx
IIIIII
y' z'
0'' 2
2
2
zzzyy
yyxxzzxxz
IIIIII
dtd
0'
' 22
2
yzzyy
yyxxzzxxy
IIIIII
dtd
Numerical SolutionIxx = 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
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
Mx
y
m
D
Two-Body Problem (3)
Force Balance:Body “M” Body “m”
022
DGMmMxM 02
2 DGMmmxm
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 – angular velocity of coordinate system
Three equations, three unknowns
Mm xxD
022
DGMmMxM
022
DGMmmxm
In case anybody wants to see the math …
mmm xMmMx
MmxD
022
DGMmMxM
022
DGMmmxm
Mm xxD 02
2 DGmxM 02
2 DGMxm
222
DxGM
DxGm
mM
mM xM
xm
Mm Mxmx
DmM
Mxm
DmMmxM
32
2
DmMG
DxGM
m
Two-Body Problem (5)
Solutions:D
mMMxm
DmM
mxM
3
2
DmMG
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
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
xyxx
xxGm
yxx
xxGM
m
m
M
M mmm
02
23
2223
22
yyxx
yGm
yxx
yGM
mM
mmm
Lagrange Points (3)
Simplifications: Divide by mass “m” of particle Substitute for “2” and divide by “G”
03
23
2223
22
DmMx
yxx
xxm
yxx
xxM
m
m
M
M
03
23
2223
22
DmMy
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
DmMy
yxx
my
yxx
My
mM
Lagrange Points (6)
Lagrange Points (7) Some Solutions:
“x” equation when “y” is zero:
Note:
032
32232
DmMx
xx
xxm
xx
xxM
m
m
M
M
33232
MMM xxxxxx ?
0
33
MMmm
MMmm
xxxxxxxxmMx
xxxxmDxxxxMD
Lagrange Points (8)
Oversimplification: Let “m” = 0 xM = 0
Solutions: Anywhere on the orbit of “m”
03
2322
DMx
yx
Mx
03
2322
DMy
yx
My
32322 Dyx
032
3222
322
DmMx
yxx
xxm
yxx
xxM
m
m
M
M
03
23
2223
22
DmMy
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 +
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 + Balancing terms:
Linearizing:
22
332222
3232
2221
253
57333
Order of magnitude 3
Order of magnitude
33 3
31
Lagrange Points (11)
Lagrange Point L3: y = 0, near x = –D
Since | |, | | << 1 < 2:
0112211
1122
1Dx
01211
1222
22
0111
11 33
DxDxDxDxx
DxDxDDxDxD
L3: y = 0, x = –D(1 +
Multiply out and linearize:
125
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
311 Dx
1251Dx
3
311 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
qq 2sin,2cos DDR
0,RF
1DR
q
q2sin1
,2cos1D
DR
ereq
Lagrange Points (15) Consider the forces in the R-
direction: From M:
From m:
Centrifugal force:
2
322 2sin2cos1
2cos1
qm
DDD
DDGMFM
2
322 2sin12cos11
2cos11
qm
DDD
DDGmFm
m 12DFC
R-direction Forces Simplify:
From M:
From m:
Centrifugal force:
2
3222 2sin2cos1
2cos1
qm
D
GMFM
2
3222 2sin12cos11
2cos11
qm
D
MGFm
m
11
2DGMFC
R-direction Forces (2) Linearize in and :
From M:
From m:
Centrifugal force:
qmqqm 2cos221
2cos3312cos1
22
DGM
DGMFM
m
q
qm22
32 22212cos1221
2cos11D
GM
D
MGFm
m 12D
GMFC
R-direction Forces (3) Balance the Forces:
0122
2cos221 222 mmqmDGM
DGM
DGM
0122
2cos221 q
22
2cos2211 q
22
2cos23 q
32cos21
221 q
Lagrange Points (16)
Consider the forces in the q-direction:
From M:
From m:
Centrifugal force is completely radial
2
322 2sin2cos1
2sin
qm
DDD
DGMFM
2
322 2sin12cos11
2sin1
qm
DDD
DGmFm
0CF
q-Direction Forces
Simplify:
2
322 2sin2cos1
2sin
qm
DDD
DGMFM
2
322 2sin12cos11
2sin1
qm
DDD
DGmFm
2
3222 2sin2cos1
2sin
qm
D
GMFM
2
3222 2sin12cos11
2sin1
qm
D
GMFm
q-Direction Forces (2)
Linearize in and :
qm
qqm
22
2sin2cos331
2sinD
GMD
GMFM
2
3222 2sin2cos1
2sin
qm
D
GMFM
2
3222 2sin12cos11
2sin1
qm
D
GMFm
q
qmq
qm
232
232
2cos22
2sin212cos1221
2sin1
D
GMD
GMFm
q-Direction Forces (3)
Simplifying further: qm
2
2sinD
GMFM
qqm
q
qm
qm
322322
23222
sin82sin
sin222
2sinsincos122
2sin
DGM
D
GMD
GMFm
qm2
2sinD
GMFM
q
qm2
32 2cos22
2sin
D
GMFm
q-Direction Forces (4)
Balance the two forces:
0
sin82sin2sin
322
qqmqm
DGM
DGM
qm2
2sinD
GMFM
qqm
32 sin82sin
DGMFm
qqmqm
322 sin82sin2sin
DGM
DGM
1sin8 3 q
21sin q
306
q 60
32
q
Lagrange Points (17)
Solving for our radius:
32cos21
221 q
1DR
603
2 q
q
12281
32121
2211
32cos21
2211
DD
DR
Lagrange Points (13 redux) What if y is not zero? Mm 10
DxM Dxm 1
x
y
602 q
12281DR
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
311 Dx
3
311 Dx
1251Dx
Lagrange Points (19):
Summary for nonzero y: Ahead of m in orbit: L4:
Behind m in orbit: L5:
602 q
602 q
Mm 10 DxM Dxm 1
12281DR
12281DR
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