1. Gravitation Chapter 7

2. Planetary Motion and Gravitation Kepler discovered the laws that describe the motions of every planet and satellite. Kepler’s first law states that the paths of the planets are ellipses, with the Sun at one focus. Kepler’s Laws Section 7.1

3. Planetary Motion and Gravitation Kepler found that the planets move faster when they are closer to the Sun and slower when they are farther away from the Sun. Kepler’s second law states that an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. Kepler’s Laws Section 7.1

4. Planetary Motion and Gravitation Kepler also found that there is a mathematical relationship between periods of planets and their mean distances away from the Sun. Kepler’s third law states that the square of the ratio of the periods of any two planets revolving about the Sun is equal to the cube of the ratio of their average distances from the Sun. Kepler’s Laws Section 7.1

5. Thus, if the periods of the planets are TA and TB, and their average distances from the Sun are rA and rB, Kepler’s third law can be expressed as follows: Planetary Motion and Gravitation Section 7.1 Kepler’s Laws

6. Planetary Motion and Gravitation According to his own third law, the force Earth exerts on the apple is exactly the same as the force the apple exerts on Earth. The force of attraction between two objects must be proportional to the objects’ masses, and is known as the gravitational force. Newton’s Law of Universal Gravitation Section 7.1

7. The law of universal gravitation states that objects attract other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. Planetary Motion and Gravitation Newton’s Law of Universal Gravitation Section 7.1

8. Newton stated his law of universal gravitation in terms that applied to the motion of planets about the Sun. This agreed with Kepler’s third law and confirmed that Newton’s law fit the best observations of the day. Planetary Motion and Gravitation Universal Gravitation and Kepler’s Third Law Section 7.1

9. Section Check Which of the following is true according to Kepler’s first law? Question 1 Section 7.1 A. Paths of planets are ellipses with Sun at one focus. B. Any object with mass has a field around it. C. There is a force of attraction between two objects. D. Force between two objects is proportional to their masses.

10. Section Check Answer: A Answer 1 Section 7.1 Reason: According to Kepler’s first law, the paths of planets are ellipses, with the Sun at one focus.

11. Section Check An imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. This is a statement of: Question 2 Section 7.1 A. Kepler’s first law B. Kepler’s second law C. Kepler’s third law D. Cavendish’s experiment

12. Section Check Answer: B Answer 2 Section 7.1 Reason: According to Kepler’s second law, an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals.

13. Using the Law of Universal Gravitation Newton used a drawing similar to the one shown below to illustrate a thought experiment on the motion of satellites. Orbits of Planets and Satellites Section 7.2

14. A satellite’s orbit around Earth is similar to a planet’s orbit about the Sun. The period of a planet orbiting the Sun is expressed by the following equation: Using the Law of Universal Gravitation A Satellite’s Orbital Period Section 7.2

15. Thus, the period for a satellite orbiting Earth is given by the following equation: Using the Law of Universal Gravitation A Satellite’s Orbital Period Section 7.2 The equations for speed and period of a satellite can be used for any object in orbit about another.

16. Using the Law of Universal Gravitation Orbital Speed and Period Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is 5.97×10 24 kg and the radius of Earth is 6.38×10 6 m, what are the satellite’s orbital speed and period? Section 7.2

17. Orbital Speed and Period Sketch the situation showing the height of the satellite’s orbit. Using the Law of Universal Gravitation Section 7.2

18. Using the Law of Universal Gravitation Orbital Speed and Period Identify the known and unknown variables. Section 7.2 Known: h = 2.25×10 5 m rE = 6.38×10 6 m mE = 5.97×10 24 kg G = 6.67×10 −11 N·m 2 /kg 2 Unknown: v = ? T = ?

19. Orbital Speed and Period Determine the orbital radius by adding the height of the satellite’s orbit to Earth’s radius. Using the Law of Universal Gravitation Section 7.2

20. Orbital Speed and Period Substitute h = 2.25×10 5 m, rE = 6.38×10 6 m. Using the Law of Universal Gravitation Section 7.2

21. Orbital Speed and Period Solve for the speed. Using the Law of Universal Gravitation Section 7.2

22. Orbital Speed and Period Substitute G = 6.67×10 -11 N.m 2 /kg 2, mE = 5.97×10 24 kg, r = 6.61×10 6 m. Using the Law of Universal Gravitation Section 7.2

23. Orbital Speed and Period Solve for the period. Using the Law of Universal Gravitation Section 7.2

24. Orbital Speed and Period Substitute r = 6.61×10 6 m, G = 6.67×10 -11 N.m 2 /kg 2, mE = 5.97×10 24 kg. Using the Law of Universal Gravitation Section 7.2

25. The acceleration of objects due to Earth’s gravity can be found by using Newton’s law of universal gravitation and his second law of motion. It is given as: Using the Law of Universal Gravitation Acceleration Due to Gravity Section 7.2 This shows that as you move farther away from Earth’s center, that is, as r becomes larger, the acceleration due to gravity is reduced according to this inverse square relationship.

26. Astronauts in a space shuttle are in an environment often called “zero- g” or ”weightlessness.” The shuttle orbits about 400 km above Earth’s surface. At that distance, g = 8.7 m/s 2, only slightly less than on Earth’s surface. Thus, Earth’s gravitational force is certainly not zero in the shuttle. Using the Law of Universal Gravitation Weight and Weightlessness Section 7.2

27. Using the Law of Universal Gravitation Any object with mass is surrounded by a gravitational field in which another object experiences a force due to the interaction between its mass and the gravitational field, g, at its location. The Gravitational Field Section 7.2

28. On Earth’s surface, the strength of the gravitational field is 9.80 N/kg, and its direction is toward Earth’s center. The field can be represented by a vector of length g pointing toward the center of the object producing the field. You can picture the gravitational field of Earth as a collection of vectors surrounding Earth and pointing toward it, as shown in the figure. The Gravitational Field Section 7.2 Using the Law of Universal Gravitation

29. Gravity is not a force, but an effect of space itself. Mass changes the space around it. Mass causes space to be curved, and other bodies are accelerated because of the way they follow this curved space. Einstein’s Theory of Gravity Section 7.2 Using the Law of Universal Gravitation

30. Einstein’s theory predicts the deflection or bending of light by massive objects. Light follows the curvature of space around the massive object and is deflected. Deflection of Light Section 7.2

31. Another result of general relativity is the effect on light from very massive objects. If an object is massive and dense enough, the light leaving it will be totally bent back to the object. No light ever escapes the object. Objects such as these, called black holes, have been identified as a result of their effect on nearby stars. The image on the right shows Chandra X-ray of two black holes (blue) in NGC 6240. Deflection of Light Section 7.2 Using the Law of Universal Gravitation

32. Section Check The period of a satellite orbiting Earth depends upon __________. Question 1 Section 7.2 A. the mass of the satellite B. the speed at which it is launched C. the value of the acceleration due to gravity D. the mass of Earth

33. Section Check Answer: D Answer 1 Section 7.2 Reason: The period of a satellite orbiting Earth depends upon the mass of Earth. It also depends on the radius of the orbit.

34. Section Check Your apparent weight __________ as you move away from Earth’s center. Question 2 Section 7.2 A. decreases B. increases C. becomes zero D. does not change

35. Section Check Answer: A Answer 2 Section 7.2 Reason: As you move farther from Earth’s center, the acceleration due to gravity reduces, hence decreasing your apparent weight.

36. End of Chapter 7 Chapter 7 Gravitation

37. Impulse and Momentum Section 9.1 Impulse and Momentum

38. The right side of the equation FΔt = mΔv involves the change in velocity: Δv = v f − v i. Therefore, mΔv = mv f − mv i An object’s momentum, also known as linear momentum, is represented by the following equation. Momentump = mv Momentum is measured in kg·m/s. Impulse and Momentum Section 9.1

39. Impulse and Momentum Because mv f = p f and mv i = p i, you get: FΔt = mΔv = p f − p i Thus, the impulse on an object is equal to the change in its momentum, which is called the impulse-momentum theorem. Impulse-Momentum Theorem FΔt = p f − p i Impulse and Momentum Section 9.1

40. Impulse and Momentum Because velocity is a vector, momentum also is a vector. Similarly, impulse is a vector because force is a vector. Impulse and Momentum Section 9.1

41. Impulse and Momentum What happens to the driver when a crash suddenly stops a car? An impulse force is needed to bring the driver’s momentum to zero. A large impulse can result either from a large force acting over a short period of time or from a smaller force acting over a long period of time. Using the Impulse-Momentum Theorem to Save Lives Section 9.1

42. Impulse and Momentum According to the impulse-momentum equation, FΔt = p f − p i. The final momentum, p f, is zero. The initial momentum, p i, is the same with or without an air bag. Thus, the impulse, FΔt, also is the same. Using the Impulse-Momentum Theorem to Save Lives Section 9.1

43. Impulse and Momentum Average Force A 2200-kg vehicle traveling at 94 km/h (26 m/s) can be stopped in 21 s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22 s if it hits a concrete wall. What average force is exerted on the vehicle in each of these stops? Section 9.1

44. Average Force Impulse and Momentum Section 9.1 Sketch the system. Include a coordinate axis and select the positive direction to be the direction of the velocity of the car.

45. Average Force Impulse and Momentum Section 9.1 Draw a vector diagram for momentum and impulse.

46. Impulse and Momentum Average Force Section 9.1 Known: m = 2200 kg Δt gentle braking = 21 s v i = +26 m/s Δt hard braking = 3.8 s v i = +0.0 m/s Δt hitting a wall = 0.22 s Unknown: F gentle braking = ? F hard braking = ? F hitting a wall = ? Identify the known and unknown variables.

47. Impulse and Momentum Section 9.1 p i = mv i Determine the initial momentum, p i, before the crash. Average Force

48. Impulse and Momentum Section 9.1 p i = (2200 kg) (+26 m/s) = +5.7×10 4 kg·m/s Substitute m = 2200 kg, v i = +26 m/s Average Force

49. Impulse and Momentum Average Force Section 9.1 p f = mv f Determine the initial momentum, p i, after the crash.

50. Impulse and Momentum Average Force Section 9.1 p f = (2200 kg) (+0.0 m/s) = +0.0×10 4 kg·m/s Substitute m = 2200 kg, v f = +0.0 m/s

51. Impulse and Momentum Average Force Section 9.1 FΔt = p f − p i Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle.

52. Impulse and Momentum Average Force Section 9.1 Substitute p f = 0.0 kg·m/s, v i = 5.7×104 kg·m/s FΔt = (+0.0×10 4 kg·m/s) − ( − 5.7×10 4 kg·m/s) = − 5.7×10 4 kg·m/s

53. Impulse and Momentum Average Force Section 9.1 Substitute Δt gentle braking = 21 s = − 2.7×10 3 N

54. Impulse and Momentum Average Force Section 9.1 Substitute Δt hard braking = 3.8 s = − 1.5×10 4 N

55. Impulse and Momentum Average Force Section 9.1 Substitute Δt hitting a wall = 0.22 s = − 2.6×10 5 N

56. Impulse and Momentum The angular velocity of a rotating object changes only if torque is applied to it. The angular momentum of an object is equal to the product of a rotating object’s moment of inertia and angular velocity. Angular Momentum Section 9.1

57. Impulse and Momentum Just as the linear momentum of an object changes when an impulse acts on it, the angular momentum of an object changes when an angular impulse acts on it. If there are no forces acting on an object, its linear momentum is constant. If there are no torques acting on an object, its angular momentum is also constant. Because an object’s mass cannot be changed, if its momentum is constant, then its velocity is also constant. Angular Momentum Section 9.1

58. Section Check Define momentum of an object. Question 1 Section 9.1 A. Momentum is the ratio of change in velocity of an object to the time over which the change happens. B. Momentum is the product of the average force on an object and the time interval over which it acts. C. Momentum of an object is equal to the mass of the object times the object’s velocity. D. Momentum of an object is equal to the mass of the object times the change in the object’s velocity.

59. Section Check Answer: C Answer 1 Section 9.1 Reason: Momentum of an object is equal to the mass of the object times the object’s velocity P = mv. Momentum is measured in kg·m/s.

60. Section Check Mark and Steve are playing cricket. Mark hits the ball with an average force of 6000 N and the ball snaps away from the bat in 0.2 ms. Steve hits the same ball with an average force of 3000 N and the ball snaps away in 0.4 ms. Which of the following statements about the impulse given to the ball in both the shots is true? Question 2 Section 9.1 A. Impulse given to the ball by Mark is twice the impulse given by Steve. B. Impulse given to the ball by Mark is four times the impulse given by Steve. C. Impulse given to the ball by Mark is the same as the impulse given by Steve. D. Impulse given to the ball by Mark is half the impulse given by Steve.

61. Section Check Answer: C Answer 2 Section 9.1 Reason: Impulse is the product of the average force on an object and the time interval over which it acts. Since the product of the average force on the ball and the time interval of the impact in both the shots is same, the impulse given to the ball by Mark is the same as the impulse given by Steve. Impulse given to the ball by Mark = (6000 N) (0.2×10 − 3 s) Impulse given to the ball by Steve = (3000 N) (0.4×10 − 3 s) = 1.2 N·s

62. Section Check In a baseball match, a pitcher throws a ball of mass 0.145 kg with a velocity of 40.0 m/s. The batter hits the ball with an impulse of 14.0 kg·m/s. Given that the positive direction is toward the pitcher, what is the final momentum of the ball? Question 3 Section 9.1 A. p f = (0.145 kg) (40.0 m/s)+14.0 kg·m/s B. p f = (0.145 kg) ( − 40.0 m/s) − 14.0 kg·m/s C. p f = (0.145 kg) (40.0 m/s) − 14.0 kg·m/s D. p f = (0.145 kg)( − 40.0 m/s)+14.0 kg·m/s

63. Section Check Answer: D Answer 3 Section 9.1 Reason: By the impulse-momentum theorem, P f = p i + F  t where, p i = mv i F  t = impulse P f = mv i + impulse Since the positive direction is toward the pitcher, v i is taken as negative as the ball is moving away from the pitcher before the batter hits the ball. The impulse is positive because direction of the force is toward the pitcher. Therefore, P f = mv i + impulse = (0.145 kg)( − 40 m/s) + 14 kg·m/s.

64. Two-Particle Collisions Section 9.2 Conservation of Momentum

65. Under what conditions is the momentum of the system of two balls conserved? The first and most obvious condition is that no balls are lost and no balls are gained. Such a system, which does not gain or lose mass, is said to be a closed system. Momentum in a Closed, Isolated System Section 9.2 When the net external force on a closed system is zero, the system is described as an isolated system. The second condition is that the forces involved are internal forces; that is, there are no forces acting on the system by objects outside of it.

66. Conservation of Momentum Systems can contain any number of objects, and the objects can stick together or come apart in a collision. Under these conditions, the law of conservation of momentum states that the momentum of any system does not change. The total momentum before the interaction is equal to the total momentum after. Momentum in a Closed, Isolated System Section 9.2

67. Conservation of Momentum Speed A 1875-kg car going 23 m/s rear-ends a 1025-kg compact car going 17 m/s on ice in the same direction. The two cars stick together. How fast do the two cars move together immediately after the collision? Section 9.2

68. Speed Conservation of Momentum Section 9.2 Define the system. Establish a coordinate system. Sketch the situation showing the “before” and “after” states. Sketch the system.

69. Speed Conservation of Momentum Section 9.2 Draw a vector diagram for the momentum.

70. Conservation of Momentum Speed Section 9.2 Known: m C = 2200 kg v Ci = +23 m/s m D = 1025 kg v Di = +17 m/s Unknown: v f = ? Identify the known and unknown variables.

71. Conservation of Momentum Section 9.2 p i = p i p Ci + p Di = p Cf + p Df m C v Ci + m D v Di = m C v Cf + m D v Df Momentum is conserved because the ice makes the total external force on the cars nearly zero. Speed

72. Conservation of Momentum Speed Section 9.2 v Cf = v Df = v f m C v Ci + m D v Di = (m C + m D ) v f Because the two cars stick together, their velocities after the collision, denoted as v f, are equal.

73. Conservation of Momentum Speed Section 9.2 Solve for v f.

74. Conservation of Momentum Speed Section 9.2 Substitute p f = 0.0 kg.m/s, v i = 5.7×104 kg.m/s

75. Conservation of Momentum Recoil Section 9.2 Assume that a girl and a boy are skating on a smooth surface with no external forces. They both start at rest, one behind the other. Skater C, the boy, gives skater D, the girl, a push. Find the final velocities of the two in-line skaters.

76. Conservation of Momentum After the collision, both skaters are moving, making this situation similar to that of an explosion. You can use the law of conservation of momentum to find the skaters’ relative velocities. The total momentum of the system was zero before the push. Therefore, it must be zero after the push. Recoil Section 9.2

77. Conservation of Momentum BeforeAfter p Ci + p Di = p Cf + p Df 0 = p Cf + p Df p Cf = − p Df m C v Cf = −m D v Df Recoil Section 9.2

78. Conservation of Momentum The coordinate system was chosen so that the positive direction is to the left. The momenta of the skaters after the push are equal in magnitude but opposite in direction. The backward motion of skater C is an example of recoil. Recoil Section 9.2

79. Conservation of Momentum How does a rocket in space change its velocity? The rocket carries both fuel and oxidizer. When the fuel and oxidizer combine in the rocket motor, the resulting hot gases leave the exhaust nozzle at high speed. Think Wall-E. Propulsion in Space Section 9.2

80. Conservation of Momentum If the rocket and chemicals are the system, then the system is a closed system. The forces that expel the gases are internal forces, so the system is also an isolated system. Thus, objects in space can accelerate using the law of conservation of momentum and Newton’s third law of motion. Propulsion in Space Section 9.2

81. Conservation of Momentum Until now, you have looked at momentum in only one dimension. The law of conservation of momentum holds for all closed systems with no external forces. It is valid regardless of the directions of the particles before or after they interact. But what happens in two or three dimensions? Two-Dimensional Collisions Section 9.2

82. Conservation of Momentum Consider the two billiard balls to be the system. Two-Dimensional Collisions Section 9.2 The original momentum of the moving ball is p Ci and the momentum of the stationary ball is zero. Therefore, the momentum of the system before the collision is equal to p Ci.

83. Conservation of Momentum After the collision, both billiard balls are moving and have momentum. As long as the friction with the tabletop can be ignored, the system is closed and isolated. Thus, the law of conservation of momentum can be used. The initial momentum equals the vector sum of the final momentum. So: Two-Dimensional Collisions Section 9.2 p Ci = p Cf + p Df

84. Conservation of Momentum Suppose the x-axis is defined to be in the direction of the initial momentum, then the y-component of the initial momentum is equal to zero. Therefore, the sum of the final y-components also must be zero. Two-Dimensional Collisions Section 9.2 p Cf, y + p Df, y = 0

85. Conservation of Momentum The y-components are equal in magnitude but are in the opposite direction and, thus, have opposite signs. The sum of the horizontal components also is equal. Two-Dimensional Collisions Section 9.2 p Ci = p Cf, x + p Df, x

86. Conservation of Momentum Earth spins on its axis with no external torques. Its angular momentum is constant. Thus, Earth’s angular momentum is conserved. As a result, the length of a day does not change. Conservation of Angular Momentum Section 9.2

87. The figure below shows an ice-skater spinning with his arms extended. Section Conservation of Momentum Two-Dimensional Collisions 9.2

88. When he pulls in his arms he begins spinning faster. Section Conservation of Momentum Two-Dimensional Collisions 9.2

89. End of Chapter Section 9 Momentum and Its Conservation