2. T. K. Ng, HKUST

3. Lecture III: (1)Reference frame problem: Coriolis force (2)Circular motion and angular momentum (3)Planetary motion (Kepler ’ s Laws)

4. Recall Newton ’ s First Law: An object remains in its state of rest or uniform motion (moving with uniform velocity) in a straight line if there is no net force acting on it. The first part of the first law is probably easy to accept. However, the second part is not so trivial. F = 0 (1)Reference frame problem, coriolis force.

5. An astronaut at an orbit around the earth (although the astronaut is accelerating, he/she does not feel force!). It is trickier. It is related to the problem of so- called inertial reference frame “Non-trivial” example that seems to violate Newton’s first law.

6. Reference frame problem: Newton’s first law cannot be simultaneously correct for all observers (with different motions).

7. Question: Suppose you are sitting in a rotating “Merry-go-around”. Do you think that Newton’s law will be correct? For example, is the net force acting on a body zero when it is not moving? I have a question.

8. Problem: Motion is defined only relative to a coordinate system

9. You are on the platform. You can see the passengers like this…

10. Now you are on the train. You will see the passengers…

11. Like this.

12. Two observers moving with acceleration with respect to each other and looking at an object at rest with the first observers.

14. a To the first observer, the force acting on the object is zero according to Newton’s Law since it is at rest.

15. a To the second observer, the object is accelerating, so there must be force acting on the object according to Newton’s Law!

16. a

17. Who is correct!?

18. Inertial frame? The reference frames where Newton’s first law holds are called inertial frames. Roughly speaking, inertial frames are those which are not accelerating (or decelerating) themselves.

19. Transformation between 2 reference frames Let Coriolis force

20. In the example of astronaut going around the earth, people on the earth see that the astronaut is subjected to gravitational force and is accelerating towards the center of earth.

23. However, the astronaut himself/herself is accelerating and is not in an inertial frame. Therefore, he/she does not find Newton’s first law to be correct! ? ? ?

24. In solving problems: (i) we can freely choose any inertial frame (Q.5 and Q.7 in assignment 1) (ii)we can transform to an non-inertial frame if we introduce Coriolis force correctly (Q. 4 in assignment 1).

25. II. Circular motion & Angular Momentum: circular motion (car turning in circle, planets around sun with constant speed)

26. Newton’s Law  a radial attractive force between two body is needed to sustain circular motion of one body around another.

27. Why?

28. now later Direction of acceleration in circular motion – pointing towards center of circle  radial attractive force needed from F = ma !

29. Mathematics of Circular motion (constant speed): x(t)=rcos(  t), y(t) = rsin(  t); show that v x (t) = -v o sin(  t), v y (t) = v o cos(  t) a x (t) = -a o cos(  t), a y (t) = -a o sin(  t). What are v o, a o ?

30. Mathematics of Circular motion (constant speed): Show that to sustain circular motion, there must exist a force F=mr  2 pointing towards center of circle according to Newton ’ s Law.

31. An example of circular motion Earth moving around sun under gravitational force Exercise: Derive the relation between velocity and r in this case.

32. Some other examples:

33. (1) Objects on an Merry-go-around tend to “fly outside” unless bound by a radial force (or motion stopped by friction).

34. (2) Bicycles in circular motion don’t fall even if it is not “standing straight”

35. There are two forces acting on the person: gravitational force (acting down) reaction force ( along the bicycle ) = radial force that supports circular motion of bicycle. +

36. Angular Momentum:

37. Cross-Product: Or + Right Hand Rule

38. Cross-Product:

39. For Circular motion: y x

40. y x Notice, angular momentum is a constant vector for circular motion

41. Angular Momentum: This is an example of conservation of angular momentum.

42. Angular Momentum: In general, angular momentum is nonzero when an object is not moving along a straight line, i.e. it is a measure of rotation!

43. Angular Momentum: How about the following trajectory?

44. Energy in circular motion In general: Total Energy=P.E.+ K.E. For gravitational force,

45. Energy in circular motion Notice that both P.E. and K.E. are constants of motion for circular motion (r remains unchanged)

46. Actually Circular motion is just one possible realization of motion of objects under central force

47. Actually Circular motion is just one possible realization of motion of objects under central force In general, objects can move in many different ways under central force,

48. We shall consider one particular example: planetary motion in the following. In general, objects can move in many different ways under central force,

49. II. Planetary Motion. (Motion of (point) object under gravitational force)

50. a. Conservation of angular momentum for objects moving under central force. (polar coordinate)

51. Mathematics of angular momentum conservation (conservation of angular momentum)

52. In particular, angular momentum of planets moving around sun are conserved

53. b. Plausible (planetary) orbits with (polar coordinate)

54. Notice both closed (ellipse) and open (hyperbola) orbits exist!

55. (circle) (ellipse) (parabola) (hyperbola)

56. Kepler’s Law for closed orbits Law of Orbits: Planets move in elliptical orbits with the sun at one focus Law of Areas: A line joining any planet to the sun sweeps out equal areas in equal intervals of time Law of Periods: The square of the period of revolution of any planet is proportional to the cube of the semimajor axis a of the orbit.

57. Kepler’s Law for closed orbits Law of Orbits: consequence of gravitational force (shall not prove) Law of Areas: consequence of conservation of angular momentum Law of Periods: consequence of (1) + (2) + conservation of energy.

58. Kepler’s 2 nd Law Consider the area  A swap out by the orbit at a small time interval. r  r r

59. Kepler’s 2 nd Law L is conserved   A   t! (A line joining any planet to the sun sweeps out equal areas in equal intervals of time) r  r r

60. Kepler’s 3 rd Law Consider the rate of change of area  A/  t at the two extreme points A, B c a A B (Semi-major axis)

61. Kepler’s 3 rd Law Consider the rate of change of area  A/  t at the two extreme points A, B c a A B

62. Kepler’s 3 rd Law Consider the rate of change of area  A/  t at the two extreme points A, B c a A B

63. Kepler’s 3 rd Law Next we consider the total energies at points A and B Conservation of energy

64. Kepler’s 3 rd Law Eq.(1)+Eq.(2)+Eq.(3)  Substitute in Eq.(2) 

65. Kepler’s 3 rd Law The rate of change of area  A/  t at the extreme points B is c a A B

66. Kepler’s 3 rd Law Using property of ellipse a 2 =b 2 +c 2 c a A B b

67. Kepler’s 3 rd Law The area of an ellipse is A=  ab. Therefore, the period of the orbit is Kepler’s Third Law

68. Application of Kepler’s Law + conservation Laws Notice the importance of knowing the geometrical properties of ellipse

69. End of lecture III