John F. Fay June 24, 2014

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

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

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

4. Euler’s Second Law (2) Momentsof Torque Angular Velocity Inertia Matrix • In a body-fixed coordinate system: • Inertia Matrix: • Describes distributionof mass on the body

5. 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:

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

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

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

9. Small Perturbations: • Assume a primary rotation about one axis: • Remaining angular velocities are small

10. Small Perturbations (2) Equations:

11. Small Perturbations (3) • Linearize: • Neglect products of small quantities

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

13. Small Perturbations (4) Combine the last two equations:

14. Small Perturbations (4) Combine the last two equations:

15. Small Perturbations (5) • Case 1: Ixx > Iyy, Izz or Ixx < Iyy, Izz: • Sinusoidal solutions for , • Rotation is stable

16. Small Perturbations (6) • Case 2: Izz > Ixx > Iyy or Izz < Ixx < Iyy: • Hyperbolic solutions for , • Rotation is unstable

17. Numerical Solution Body Coordinates Ixx = 1.0Iyy = 0.5Izz = 0.7

18. Numerical Solution (2) Global Coordinates Ixx = 1.0Iyy = 0.5Izz = 0.7

19. Numerical Solution (3) Body Coordinates Ixx = 1.0Iyy = 1.2Izz = 2.0

20. Numerical Solution (4) Global Coordinates Ixx = 1.0Iyy = 1.2Izz = 2.0

21. Numerical Solution (5) Body Coordinates Ixx = 1.0Iyy = 0.8Izz = 1.2

22. Numerical Solution (6) Global Coordinates Ixx = 1.0Iyy = 0.8Izz = 1.2

23. 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

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

25. Two-Body Problem (2) y W D x m M • 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

26. Two-Body Problem (3) CentrifugalForce Gravitational Force y W xM(<0) xm D x m M Force Balance:Body “M” Body “m”

27. 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

28. In case anybody wants to see the math …

29. Two-Body Problem (5) y W xM(<0) xm D x m M Solutions:

30. Lagrange Points y (x,y) W x m M • 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

31. Lagrange Points (2) • Sum of Forces: particle at (x,y) of mass “m ” • x-direction: • y-direction:

32. Lagrange Points (3) • Simplifications: • Divide by mass “m” of particle • Substitute for “W2” and divide by “G”

33. Lagrange Points (4) m M • 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

34. Lagrange Points (5) m M • 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

35. Lagrange Points (6) • Some Solutions: • Note that “y” equation is uniquely satisfied ify = 0 (but there may be nonzero solutions also)

36. Lagrange Points (7) ? • Some Solutions: • “x” equation when “y” is zero: • Note:

37. Lagrange Points (8) • Oversimplification: Let “m” = 0 • xM = 0 • Solutions: • Anywhere on the orbit of “m”

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

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

40. L1, L2: y = 0, x = D(1 + d) Collecting terms: Multiplying things out:

41. L1, L2 (2): y = 0, x = D(1 + d) Order of magnitude d3 Order of magnitude e Balancing terms: Linearizing:

42. Lagrange Points (11) Lagrange Point L3: y = 0, near x = –D Since |e |, |d | << 1 < 2:

43. L3: y = 0, x = –D(1 + d) Multiply out and linearize:

44. Lagrange Points (12) • Summary for a Small Perturbation and y = 0: • Lagrange Points: • L1: • L2: • L3:

45. Lagrange Points (13) y 2q x What if y is not zero?

46. Lagrange Points (14) er eq y 2q x Polar Coordinates:

47. Lagrange Points (15) • Consider the forces in the R-direction: • From M: • From m: • Centrifugal force:

48. R-direction Forces • Simplify: • From M: • From m: • Centrifugal force:

49. R-direction Forces (2) • Linearize in d and e: • From M: • From m: • Centrifugal force:

50. R-direction Forces (3) Balance the Forces: