1. Nicholas J. Giordano www.cengage.com/physics/giordano Circular Motion and Gravitation

2. Chapter 5 Circular Motion and Gravitation

3. Introduction Circular motion Acceleration is not constant Cannot be reduced to a one-dimensional problem Examples Car traveling around a turn Parts of the motion of a roller coaster Centrifuge The Earth orbiting the Sun Gravitation Explore gravitational force in more detail Look at Kepler’s Laws of Motion Further details about g Introduction

4. Uniform Circular Motion Path of Motion: Circle The direction of the velocity is continually changing. The vector is always tangent to the circle Uniform circular motion (UCM) is circular motion at constant speed. Section 5.1

5. Circular Motion - Period v v v v r r: radius of circle Period T (s): time for a complete rotation If n rotations happen in time t: T=t/n. Frequency f (s-1, Hertz- Hz): number of rotations per unit of time - f=n/t. T=1/f, f=1/T, Tf=1

6. Tangential (Linear) Velocity C = 2πr = vT or

7. Exercises Set 1 1. A car’s tire rotates at 1200 RPM. a. What is the frequency?f = 1200/60s = 20 Hz b. What is the period?T= 1/f = 1/20 Hz = 0.05s c. What is the speed (R= 0.15 m) v = 2πRf = 2 x 3.14 x 0.15m x 20 Hz = 18.85 m/s

8. Ex 2 Question 1: An object is constrained to move in a circular orbit with radius 1 m. a) How far does the object travel after completing one orbit? b) If the orbital period (time to complete one orbit) is 0.5 s, what is its velocity? a) C = 2πR = 2 x 3.14 x 1 m = 6.28 m b) v = C/T = 6.28 m/0.5s = 12. 56 m/s

9. Ex3 Problem: The earth is 1.50 x10 11 m (93 million miles) from the sun. What is its speed in m/s (neglecting the motion of the sun through the galaxy)? sec

10. Angular Measure The position of an object can be described using polar coordinates— r and θ— rather than x and y. The figure at left gives the conversion between the two descriptions.

11. Angular Measure It is most convenient to measure the angle θ in radians:

12. Angular Speed and Velocity In analogy to the linear case, we define the angular speed: where θ is the angle swept by the radius in the time t. r θ [θ] = rad/s

13. Angular Speed and Velocity The angular velocity is (+) positive when the rotation is ccw and (-)negative if the rotation is cw.

14. Angular Speed and Velocity Relationship between tangential and angular speeds: This means that parts of a rotating object farther from the axis of rotation move faster.

15. Angular Speed, Period, and Frequency From: we get the relation of the frequency to the angular speed: and

16. Uniform Circular Motion and Centripetal Acceleration A careful look at the change in the velocity vector of an object moving in a circle at constant speed shows that the acceleration is toward the center of the circle.

17. Uniform Circular Motion and Centripetal Acceleration The change in the direction of the velocity vector is provided by the centripetal acceleration:

18. Exercises Set 4 2) A car is traveling along a circular path of 50 m radius at 22 m/s. a. What is the car’s acceleration? b. How much time to complete a circuit?

19. Example Set 5 3) Satellite, radius 1.3 x 10 7 m, g = 2.5 m/s 2. a) What is the speed? b) Period?

20. Ex. 6 n = 0.56 x 60 = 33.44 rev/min

21. Uniform Circular Motion and Centripetal Force The centripetal force is the mass multiplied by the centripetal acceleration. This force is the net force on the object. As the force is always perpendicular to the velocity, it does no work.

22. Centripetal Force Features: A force causing a centripetal acceleration acting toward the center of the circle. It causes a change in the direction of the velocity vector If the force vanishes, the object would move in a straight-line path tangent to the circle (“Centrifugal” effects which come from inertia) ! This is not a new force, it is a new application of a force …

23. Centripetal force The centripetal force can be produced in any number of ways (for example): Due to the tension in a string (pendulum) Due to friction between tires and the road Due to gravity

24. Conical Pendulum A small object of mass m is suspended from a string of length L. The object revolves in a horizontal circle of radius r with constant speed v, as shown. –Find the speed of the object. –Find the tension, T, in the string.

25. Conical Pendulum Need a free body diagram: –T is the force exerted by the string x y

26. Conical Pendulum The body does not move in the vertical direction …therefore no acceleration here! –Rearranging gives x y

27. Conical Pendulum In the x direction, there is centripetal acceleration! –Rearranging gives x y

28. Conical Pendulum Consider the x and y equations: We have two equations in two unknowns (v, T) –T can be eliminated by dividing 2 by 1 to give –and 1 2 x y

29. Conical Pendulum And solving for v

30. Conical Pendulum To find tension, just go back to the y- component equation: And solve for T

31. Problem Solving Strategy – Circular Motion Recognize the principle If the object moves in a circle, then there is a centripetal force acting on it Sketch the problem Show the path the object travels Identify the circular part of the path Include the radius of the circle Show the center of the circle Selecting a coordinate system that assigns the positive direction toward the center of the circle is often convenient Section 5.1

32. Problem Solving Strategy, cont. Identify the principles Find all the forces acting on the object A free body diagram is generally useful Find the components of the forces that are directed toward the center of the circle Find the components of the forces perpendicular to the center Apply Newton’s Second Law for both directions The acceleration directed toward the center of the circle is a centripetal acceleration Section 5.1 Solve for the quantities of interest Check your answer Consider what the answer means Does the answer make sense

33. Horizontal (Flat) Curve A 1500 kg car moving on a flat, horizontal road negotiates a curve with a radius of 35.0 m. If the static coefficient of friction between the tires and the dry pavement is 0.523, what is the maximum speed with which the car can negotiate the turn?

34. Centripetal Acceleration Example: Car A car rounding a curve travels in an approximate circle The radius of this circle is called the radius of curvature Forces in the y-direction Gravity and the normal force Forces in the x-direction Friction is directed toward the center of the circle Section 5.1

35. Horizontal (Flat) Curve What is the maximum force of friction F Friction ? We can get that from the vertical force balance and the static friction coefficient: So And,

36. Car Example, cont. Since friction is the only force acting in the x- direction, it supplies the centripetal force Solving for the maximum velocity at which the car can safely round the curve gives Section 5.1

37. Finish up, –Does it look okay? Increase μ s, v max increases √ Increase r, v max increases √ –Put in numbers Horizontal (Flat) Curve

38. Reality check You might want a bit more of a margin of safety here …

39. Angular Acceleration The average angular acceleration is the rate at which the angular speed changes: In analogy to constant linear acceleration:

40. Angular Acceleration If the angular speed is changing, the linear speed must be changing as well. The tangential acceleration is related to the angular acceleration:

41. Angular Acceleration

42. Circular Motion Example: Vertical Circle The speed of the rock varies with time At the bottom of the circle: Tension and gravity are in opposite directions The tension supports the rock (mg) and supplies the centripetal force Section 5.2

43. Vertical Circle Example, cont. At the top of the circle: Tension and gravity are in the same direction Pointing toward the center of the circle Section 5.2

44. Vertical Circle Example, Final There is a minimum value of v needed to keep the string taut at the top Let T top = 0 If the speed is smaller than this, the string will become slack and circular motion is no longer possible

45. Circular Motion Example: Roller Coaster The roller coaster’s path is nearly circular at the minimum or maximum points on the track There is a maximum speed at which the coaster will not leave the top of the track: If the speed is greater than this, N would have to be negative This is impossible, so the coaster would leave the track Section 5.2

46. Newton’s Law of Gravitation, Equation Law states: There is a gravitational attraction between any two objects. If the objects are point masses m 1 and m 2, separated by a distance r the magnitude of the force is Section 5.3

47. Law of Gravitation, cont. Note that r is the distance between the objects G is the Universal Gravitational Constant G = 6.67 x 10 -11 N. m 2 / kg 2 The gravitational force is always attractive Every mass attracts every other mass The gravitational force is symmetric The magnitude of the gravitational force exerted by mass 1 on mass 2 is equal in magnitude to the force exerted by mass 2 on mass 1 The two forces form an action-reaction pair Section 5.3

48. Gravitation and the Moon’s Orbit The Moon follows an approximately circular orbit around the Earth There is a force required for this motion Gravity supplies the force Section 5.3

49. Notes on the Moon’s Motion We assumed the Moon orbits a “fixed” Earth It is a good approximation It ignores the Earth’s motion around the Sun The Earth and Moon actually both orbit their center of mass We can think of the Earth as orbiting the Moon The circle of the Earth’s motion is very small compared to the Moon’s orbit Section 5.3

50. Gravitation and g Assuming a spherical Earth, we can consider all the mass of the Earth to be concentrated at its center The value of r in the Law of Gravitation is just the radius of the Earth Section 5.3

51. Gravitation Force From The Earth – Assumptions We assumed A spherical Earth That the gravitational force could be calculated as if all the mass of the Earth was located at its center Assumptions are true as long as the density of the object is spherically symmetric The object has a constant density The object’s density varies with depth as long as the density depends only on the distance from the center Section 5.3

52. Measuring G Henry Cavendish measured the force of gravity between two large lead spheres By an experimental set- up similar to the picture, he was able to determine the value of G Section 5.3

53. More About Gravity – Newton’s Apple G is a constant of nature Gravity is a weak force Much smaller than typical normal or tension forces Gravity is considered the weakest of the fundamental forces of nature Newton showed the motion of celestial bodies and the motion of terrestrial motion are caused by the same force and governed by the same laws of motion Section 5.3a

54. An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle  in the time t, through what angle did it rotate in the time 1/2 t? 1) 1/2  2) 1/4  3) 3/4  4) 2  5) 4  ConcepTest 8.3aAngular Displacement I

55. An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle  in the time t, through what angle did it rotate in the time 1/2 t? 1) 1/2  2) 1/4  3) 3/4  4) 2  5) 4  half the timeone-quarter the angle The angular displacement is  = 1/2  t 2 (starting from rest), and there is a quadratic dependence on time. Therefore, in half the time, the object has rotated through one-quarter the angle. ConcepTest 8.3aAngular Displacement I

56. An object at rest begins to rotate with a constant angular acceleration. If this object has angular velocity  at time t, what was its angular velocity at the time 1/2 t? 1) 1/2  2) 1/4  3) 3/4  4) 2  5) 4  ConcepTest 8.3bAngular Displacement II

57. An object at rest begins to rotate with a constant angular acceleration. If this object has angular velocity  at time t, what was its angular velocity at the time 1/2t? 1) 1/2  2) 1/4  3) 3/4  4) 2  5) 4  half the time half the speed The angular velocity is  =  t (starting from rest), and there is a linear dependence on time. Therefore, in half the time, the object has accelerated up to only half the speed. ConcepTest 8.3bAngular Displacement II

58. Example: Wheel And Rope l A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians)a R

59. Wheel And Rope... Use a =  R to find  :  = a / R = 4 m/s 2 / 0.4 m = 10 rad/s 2 l Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. = 0 + 0(10) + (10)(10) 2 = 500 rad a R 

60. Newton’s Law of Gravitation Newton’s law of universal gravitation describes the force between any two point masses m 1, m 2 separated by the distance r: G is called the universal gravitational constant: m1m1 m2m2 r FgFg -F g

61. Newton’s Law of Gravitation Gravity provides the centripetal force that keeps planets, moons, and satellites in their orbits. We can relate the universal gravitational force to the local acceleration of gravity:

62. Newton’s Law of Gravitation The gravitational potential energy is given by the general expression:

63. Kepler’s Laws and Earth Satellites (No!) Kepler’s laws were the result of his many years of observations. They were later found to be consequences of Newton’s laws. Kepler’s first law: Planets move in elliptical orbits, with the Sun at one of the focal points.

64. Kepler’s Laws and Earth Satellites Kepler’s second law: A line from the Sun to a planet sweeps out equal areas in equal lengths of time.

65. Kepler’s Laws and Earth Satellites Kepler’s third law: The square of the orbital period of a planet is directly proportional to the cube of the average distance of the planet from the Sun; that is,. This can be derived from Newton’s law of gravitation, using a circular orbit.

66. Kepler’s Laws and Earth Satellites If a projectile is given enough speed to just reach the top of the Earth’s gravitational well, its potential energy at the top will be zero. At the minimum, its kinetic energy will be zero there as well.

67. Kepler’s Laws and Earth Satellites This minimum initial speed is called the escape speed.

68. Kepler’s Laws and Earth Satellites Any satellite in orbit around the Earth has a speed given by

69. Kepler’s Laws and Earth Satellites

70. Astronauts in Earth orbit report the sensation of weightlessness. The gravitational force on them is not zero; what’s happening?

71. Kepler’s Laws and Earth Satellites What’s missing is not the weight, but the normal force. We call this apparent weightlessness. “Artificial” gravity could be produced in orbit by rotating the satellite; the centripetal force would mimic the effects of gravity.

72. Summary of Chapter 5 Angles may be measured in radians; the angle is the arc length divided by the radius. Angular kinematic equations for constant acceleration:

73. Summary of Chapter 5 Tangential speed is proportional to angular speed. Frequency is inversely proportional to period. Angular speed: Centripetal acceleration:

74. Summary of Chapter 5 Centripetal force: Angular acceleration is the rate at which the angular speed changes. It is related to the tangential acceleration. Newton’s law of gravitation:

75. Summary of Chapter 5 Gravitational potential energy: Kepler’s laws: 1. Planetary orbits are ellipses with Sun at one focus 2. Equal areas are swept out in equal times. 3. The square of the period is proportional to the cube of the radius.

76. Summary of Chapter 5 Escape speed from Earth: Energy of a satellite orbiting Earth:

77. ConcepTest 7.1 Tetherball Toward the top of the pole 1) Toward the top of the pole Toward the ground 2) Toward the ground Along the horizontal component of the tension force 3) Along the horizontal component of the tension force Along the vertical component of the tension force 4) Along the vertical component of the tension force Tangential to the circle 5) Tangential to the circle In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? W T

78. vertical component of the tensionweight horizontal component of tension centripetal force The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that points toward the center of the circle. Toward the top of the pole 1) Toward the top of the pole Toward the ground 2) Toward the ground Along the horizontal component of the tension force 3) Along the horizontal component of the tension force Along the vertical component of the tension force 4) Along the vertical component of the tension force Tangential to the circle 5) Tangential to the circle In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? W T W T ConcepTest 7.1 Tetherball

79. You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? 1) You are thrown to the right 2) You feel no particular change 3) You are thrown to the left 4) You are thrown to the ceiling 5) You are thrown to the floor ConcepTest 7.2a Around the Curve I

80. You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? 1) You are thrown to the right 2) You feel no particular change 3) You are thrown to the left 4) You are thrown to the ceiling 5) You are thrown to the floor ConcepTest 7.2a Around the Curve I The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.

81. 1) centrifugal force is pushing you into the door 2) the door is exerting a leftward force on you 3) both of the above 4) neither of the above During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? ConcepTest 7.2b Around the Curve II

82. 1) centrifugal force is pushing you into the door 2) the door is exerting a leftward force on you 3) both of the above 4) neither of the above During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path. ConcepTest 7.2b Around the Curve II

83. 1) car’s engine is not strong enough to keep the car from being pushed out 2) friction between tires and road is not strong enough to keep car in a circle 3) car is too heavy to make the turn 4) a deer caused you to skid 5) none of the above You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? ConcepTest 7.2c Around the Curve III

84. The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line! 1) car’s engine is not strong enough to keep the car from being pushed out 2) friction between tires and road is not strong enough to keep car in a circle 3) car is too heavy to make the turn 4) a deer caused you to skid 5) none of the above You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? ConcepTest 7.2c Around the Curve III Follow-up: What could be done to the road or car to prevent skidding?

85. ConcepTest 7.3 Missing Link A ping-pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?

86. ConcepTest 7.3 Missing Link l Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires! A ping-pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow? Follow-up: What physical force provides the centripetal force?

87. ConcepTest 7.4 Ball and String 1) T 2 = 1/4 T 1 2) T 2 = 1/2 T 1 3) T 2 = T 1 4) T 2 = 2 T 1 5) T 2 = 4 T 1 Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1?

88. T = mv 2 /r v 2 = 2v 1 r 2 = 2r 1 T 2 = 2T 1 The centripetal force in this case is given by the tension, so T = mv 2 /r. For the same period, we find that v 2 = 2v 1 (and this term is squared). However, for the denominator, we see that r 2 = 2r 1 which gives us the relation T 2 = 2T 1. ConcepTest 7.4 Ball and String Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? 1) T 2 = 1/4 T 1 2) T 2 = 1/2 T 1 3) T 2 = T 1 4) T 2 = 2 T 1 5) T 2 = 4 T 1

89. ConcepTest 7.5 Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5

90. normal force centripetal force downward force of gravity is balanced by the upward frictional force The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. ConcepTest 7.5 Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5 Follow-up: What happens if the rotation of the ride slows down?

91. ConcepTest 7.6a Going in Circles I 1) N remains equal to mg 2) N is smaller than mg 3) N is larger than mg 4) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

92. ConcepTest 7.6a Going in Circles I 1) N remains equal to mg 2) N is smaller than mg 3) N is larger than mg 4) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion? inward mg (down)N (up) N must be smaller than mg You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg?

93. R v 1) F c = N + mg 2) F c = mg – N 3) F c = T + N – mg 4) F c = N 5) F c = mg A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? ConcepTest 7.6b Going in Circles II

94. R v F c points toward the center of the circle, i.e., downward in this case.weight vector downnormal force up F c = mg – N F c points toward the center of the circle, i.e., downward in this case. The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is: F c = mg – N gmggmg N A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is F c of the skier equal to? ConcepTest 7.6b Going in Circles II Follow-up: What happens when the skier goes into a small dip? 1) F c = N + mg 2) F c = mg – N 3) F c = T + N – mg 4) F c = N 5) F c = mg

95. R v top 1) F c = T – mg 2) F c = T + N – mg 3) F c = T + mg 4) F c = T 5) F c = mg You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is F c equal to? ConcepTest 7.6c Going in Circles III

96. R v T gmggmg You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is F c equal to? F c points toward the center of the circle, i.e. downward in this case. weight vectordown tension down F c = T + mg F c points toward the center of the circle, i.e. downward in this case. The weight vector points down and the tension (exerted by the string) also points down. The magnitude of the net force, therefore, is: F c = T + mg ConcepTest 7.6c Going in Circles III Follow-up: What is F c at the bottom of the ball’s path? 1) F c = T – mg 2) F c = T + N – mg 3) F c = T + mg 4) F c = T 5) F c = mg

97. ConcepTest 7.7a Earth and Moon I 1) the Earth pulls harder on the Moon 2) the Moon pulls harder on the Earth 3) they pull on each other equally 4) there is no force between the Earth and the Moon 5) it depends upon where the Moon is in its orbit at that time Which is stronger, Earth’s pull on the Moon, or the Moon’s pull on Earth?

98. By Newton’s 3 rd Law, the forces are equal and opposite. ConcepTest 7.7a Earth and Moon I 1) the Earth pulls harder on the Moon 2) the Moon pulls harder on the Earth 3) they pull on each other equally 4) there is no force between the Earth and the Moon 5) it depends upon where the Moon is in its orbit at that time Which is stronger, Earth’s pull on the Moon, or the Moon’s pull on Earth?

99. ConcepTest 7.7b Earth and Moon II 1) one quarter 2) one half 3) the same 4) two times 5) four times If the distance to the Moon were doubled, then the force of attraction between Earth and the Moon would be:

100. increase distance2forcedecrease 4 The gravitational force depends inversely on the distance squared. So if you increase the distance by a factor of 2, the force will decrease by a factor of 4. ConcepTest 7.7b Earth and Moon II 1) one quarter 2) one half 3) the same 4) two times 5) four times If the distance to the Moon were doubled, then the force of attraction between Earth and the Moon would be: Follow-up: What distance would increase the force by a factor of 2?

101. You weigh yourself on a scale inside an airplane that is flying with constant speed at an altitude of 20,000 feet. How does your measured weight in the airplane compare with your weight as measured on the surface of the Earth? 1) greater than 2) less than 3) same ConcepTest 7.8 Fly Me Away

102. You weigh yourself on a scale inside an airplane that is flying with constant speed at an altitude of 20,000 feet. How does your measured weight in the airplane compare with your weight as measured on the surface of the Earth? 1) greater than 2) less than 3) same At a high altitude, you are farther away from the center of Earth. Therefore, the gravitational force in the airplane will be less than the force that you would experience on the surface of the Earth. ConcepTest 7.8 Fly Me Away

103. ConcepTest 7.9 Two Satellites 1) 1/8 2) 1/4 3) 1/2 4) it’s the same 5) 2 Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the centripetal force acting on B compared to that acting on A?

104. Using the Law of Gravitation: we find that the ratio is 1/4. ConcepTest 7.9 Two Satellites 1) 1/8 2) 1/4 3) 1/2 4) it’s the same 5) 2 Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the centripetal force acting on B compared to that acting on A? Note the 1/r 2 factor

105. ConcepTest 7.10 Averting Disaster 1) It’s in Earth’s gravitational field 2) the net force on it is zero 3) it is beyond the main pull of Earth’s gravity 4) it’s being pulled by the Sun as well as by Earth 5) none of the above The Moon does not crash into Earth because:

106. The Moon does not crash into Earth because of its high speed. If it stopped moving, it would, of course, fall directly into Earth. With its high speed, the Moon would fly off into space if it weren’t for gravity providing the centripetal force. ConcepTest 7.10 Averting Disaster The Moon does not crash into Earth because: Follow-up: What happens to a satellite orbiting Earth as it slows? 1) It’s in Earth’s gravitational field 2) the net force on it is zero 3) it is beyond the main pull of Earth’s gravity 4) it’s being pulled by the Sun as well as by Earth 5) none of the above

107. ConcepTest 7.11 In the Space Shuttle Astronauts in the space shuttle float because: 1) they are so far from Earth that Earth’s gravity doesn’t act any more 2) gravity’s force pulling them inward is cancelled by the centripetal force pushing them outward 3) while gravity is trying to pull them inward, they are trying to continue on a straight-line path 4) their weight is reduced in space so the force of gravity is much weaker

108. Astronauts in the space shuttle float because they are in “free fall” around Earth, just like a satellite or the Moon. Again, it is gravity that provides the centripetal force that keeps them in circular motion. ConcepTest 7.11 In the Space Shuttle Astronauts in the space shuttle float because: Follow-up: How weak is the value of g at an altitude of 300 km? 1) they are so far from Earth that Earth’s gravity doesn’t act any more 2) gravity’s force pulling them inward is cancelled by the centripetal force pushing them outward 3) while gravity is trying to pull them inward, they are trying to continue on a straight-line path 4) their weight is reduced in space so the force of gravity is much weaker

109. If you weigh yourself at the equator of Earth, would you get a bigger, smaller or similar value than if you weigh yourself at one of the poles? 1) bigger value 2) smaller value 3) same value ConcepTest 7.12 Guess My Weight

110. If you weigh yourself at the equator of Earth, would you get a bigger, smaller or similar value than if you weigh yourself at one of the poles? 1) bigger value 2) smaller value 3) same value normal force you are in circular motion net inward force normal force must be slightly less than mg The weight that a scale reads is the normal force exerted by the floor (or the scale). At the equator, you are in circular motion, so there must be a net inward force toward Earth’s center. This means that the normal force must be slightly less than mg. So the scale would register something less than your actual weight. ConcepTest 7.12 Guess My Weight

111. ConcepTest 7.13 Force Vectors A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth? 1 2 3 4 5 2d2d d 2m2m m Earth

112. 1 2 3 4 5 2d2d d 2m2m m mgreater 2m The force of gravity on the Earth due to m is greater than the force due to 2m, which means that the force component pointing down in the figure is greater than the component pointing to the right. 2m2d1/2 GMm / d 2 F 2m = GM E (2m) / (2d) 2 = 1/2 GMm / d 2 mdGMm / d 2 F m = GM E m / d 2 = GMm / d 2 A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth? ConcepTest 7.13 Force Vectors