Selected Problems in Dynamics: Stability of a Spinning Body -and- Derivation of the Lagrange Points John F. Fay June 24, 2014

Outline v Stability of a Spinning Body v Derivation of the Lagrange Points TEAS Employees: Please log this time on the Training Tracker

Stability of a Spinning Body v Euler’s Second Law: (good for an inertial coordinate system) Problem: Inertia matrix “ I ” of a rotating body changes in an inertial coordinate system Moments of Torque Inertia Matrix Angular Velocity

Euler’s Second Law (2) v In a body-fixed coordinate system: v Inertia Matrix: u Describes distribution of mass on the body Moments of Torque Inertia Matrix Angular Velocity

Euler’s Second Law (3) v Planes of Symmetry: u x-y plane: I xz = I yz = 0 u x-z plane: I xy = I yz = 0 u y-z plane: I xz = I xz = 0 v For a freely-spinning body:

Euler’s Second Law (4) v Rigid body: u “ I ” matrix is constant in body coordinates v Do the matrix multiplication:

Euler’s Second Law (5) v Do the cross product: v Result:

Euler’s Second Law (6) v As scalar equations:

Small Perturbations: v Assume a primary rotation about one axis: u Remaining angular velocities are small

Small Perturbations (2) v Equations:

Small Perturbations (3) v Linearize: u Neglect products of small quantities

Small Perturbations (4) v Combine the last two equations: switched

Small Perturbations (4) v Combine the last two equations:

Small Perturbations (4) v Combine the last two equations:

Small Perturbations (5) v Case 1: I xx > I yy, I zz or I xx < I yy, I zz : u Sinusoidal solutions for, u Rotation is stable

Small Perturbations (6) v Case 2: I zz > I xx > I yy or I zz < I xx < I yy : u Hyperbolic solutions for, u Rotation is unstable

Numerical Solution I xx = 1.0 I yy = 0.5 I zz = 0.7 Body Coordinates

Numerical Solution (2) I xx = 1.0 I yy = 0.5 I zz = 0.7 Global Coordinates

Numerical Solution (3) I xx = 1.0 I yy = 1.2 I zz = 2.0 Body Coordinates

Numerical Solution (4) I xx = 1.0 I yy = 1.2 I zz = 2.0 Global Coordinates

Numerical Solution (5) I xx = 1.0 I yy = 0.8 I zz = 1.2 Body Coordinates

Numerical Solution (6) I xx = 1.0 I yy = 0.8 I zz = 1.2 Global Coordinates

Spinning Body: Summary v Rotation about axis with largest inertia: Stable v Rotation about axis with smallest inertia: Stable v Rotation about axis with in-between inertia: Unstable

Derivation of the Lagrange Points v Two-Body Problem: u Lighter body orbiting around a heavier body (Let’s stick with a circular orbit) u Actually both bodies orbit around the barycenter x y  m M D

Two-Body Problem (2) v Rotating reference frame: u Origin at barycenter of two-body system u Angular velocity matches orbital velocity of bodies u Bodies are stationary in the rotating reference frame  M x y m D

Two-Body Problem (3) v Force Balance: Body “M”Body “m”  x y m D xmxm x M (<0) M Centrifugal Force Gravitational Force

Two-Body Problem (4) v Parameters: u M – mass of large body u m – mass of small body u D – distance between bodies v Unknowns: u x M – x-coordinate of large body u x m – x-coordinate of small body   – angular velocity of coordinate system v Three equations, three unknowns

In case anybody wants to see the math …

Two-Body Problem (5) v Solutions:  x y m D xmxm x M (<0) M

Lagrange Points v Points at which net gravitational and centrifugal forces are zero u Gravitational attraction to body “ M ” u Gravitational attraction to body “ m ” u Centrifugal force caused by rotating coordinate system  x y m M (x,y)

Lagrange Points (2)  Sum of Forces: particle at ( x, y ) of mass “  ” u x-direction: u y-direction:

Lagrange Points (3) v Simplifications:  Divide by mass “  ” of particle  Substitute for “   ” and divide by “ G ”

Lagrange Points (4) v Qualitative considerations: u Far enough from the system, centrifugal force will overwhelm everything else  Net force will be away from the origin u Close enough to “ M ”, gravitational attraction to that body will overwhelm everything else u Close enough to “ m ”, gravitational attraction to that body will overwhelm everything else m M

Lagrange Points (5) v Qualitative considerations (2) u Close to “ m ” but beyond it, the attraction to “ m ” will balance the centrifugal force  search for a Lagrange point there u On the side of “ M ” opposite “ m ”, the attraction to “ M ” will balance the centrifugal force  search for a Lagrange point there u Between “ M ” and “ m ” their gravitational attractions will balance  search for a Lagrange point there m M

v Some Solutions: u Note that “ y ” equation is uniquely satisfied if y = 0 (but there may be nonzero solutions also) Lagrange Points (6)

Lagrange Points (7) v Some Solutions: u “ x ” equation when “ y ” is zero: u Note: ?

Lagrange Points (8) v Oversimplification: Let “ m ” = 0 u x M = 0 v Solutions: u Anywhere on the orbit of “ m ”

Lagrange Points (9) v Perturbation: v For the y = 0 case:

Lagrange Points (10) v For the y = 0 case (cont’d): –Lagrange Points L1,L2: near x = D :

L1, L2: y = 0, x = D(1 +  v Collecting terms: v Multiplying things out:

L1, L2 (2): y = 0, x = D(1 +  v Balancing terms: v Linearizing: Order of magnitude  3 Order of magnitude 

Lagrange Points (11) v Lagrange Point L3: y = 0, near x = – D  Since | , |  << 1 < 2:

L3: y = 0, x = –D(1 +  v Multiply out and linearize:

Lagrange Points (12) v Summary for a Small Perturbation and y = 0: v Lagrange Points: u L1: u L2: u L3:

Lagrange Points (13) v What if y is not zero? x y 22

Lagrange Points (14) v Polar Coordinates: x y 22 erer ee

Lagrange Points (15) v Consider the forces in the R -direction: u From M : u From m : u Centrifugal force:

R-direction Forces v Simplify: u From M : u From m : u Centrifugal force:

R-direction Forces (2)  Linearize in  and  : u From M : u From m : u Centrifugal force:

R-direction Forces (3) v Balance the Forces:

Lagrange Points (16)  Consider the forces in the  -direction: u From M : u From m : u Centrifugal force is completely radial

 -Direction Forces v Simplify:

 -Direction Forces (2)  Linearize in  and  :

 -Direction Forces (3) v Simplifying further:

 -Direction Forces (4) v Balance the two forces:

Lagrange Points (17) v Solving for our radius:

Lagrange Points (13 redux) v What if y is not zero? x y

Lagrange Points (18): v Summary for zero y : u Between M and m : L1: u On the opposite side of m from M : L2: u On the opposite side of M from m : L3:

Lagrange Points (19): v Summary for nonzero y : u Ahead of m in orbit: L4: u Behind m in orbit: L5:

Summary of Lagrange Points (courtesy of Wikipedia, “Lagrange Points”) (What is wrong with this picture?)

Potential Energy Contours

Stability of Lagrange Points v Requires saving higher-order terms in linearized equations v Results: –L1, L2, L3 unstable u Body perturbed from the point will move away from the point –L4, L5 neutrally stable u Body perturbed from the point will drift around the point

Conclusions v Stability of Spinning Bodies v Derivation of Lagrange Points