CIRCULAR MOTION AND UNIVERSAL GRAVITATION

Transcription

1 CIRCULAR MOTION AND UNIVERSAL GRAVITATION

2 Uniform Circular Motion What holds an object in a circular path? A force. String Friction Gravity

3 What happens when the force is diminished? Object flies off in a straight line. The force must be maintained to keep the object moving in a circular path.

4 F C = Centripetal force Center seeking force. Acts at right angles to the velocity.

5 F C does not change the speed of the object. It changes the velocity. How? Changes the direction of the motion. Since the velocity changes, the object must be.

6 Centripetal Acceleration Newton s 2nd Law says the acceleration must be in the same direction as the force.

7 Centripetal acceleration = V = speed of object r = radius

8 Velocity (speed) = V = d in a circle t t T = = Time for 1 revolution in a circle:

9 Substituting in to: A C = V 2 r A C ( ) = 2 πr T 2 r

10 Since F Net = ma F C = ma C F C = m 4π 2 r T 2 F C = m V 2 r

11 In a lab, metals are tested for tensile strength (force needed to stretch break a metal). A 1.5 kg mass is attached to the end of a 2.0 m long wire, and swung in, a horizontal circle. What force is exerted on the wire if the mass is swinging at 2.0 revs/second?

12 2 m 1.5 kg F C = m 4π 2 r T 2 T = Time for 1 revolution T = seconds rev

13 F C = m 4π 2 r T 2 F C = 1.5 kg (4π 2 ) 2 m ( s ) 2

14 A 75 N weight is attached to a 25 cm string that has an angular speed of 3.00 rev/s. What is the centripetal acceleration of the weight? A C = 4 π 2 r T 2

15 A C = 4 π 2 (0.25 m) ( s) 2

16 A 500 gram object is placed on a turntable that is rotating at 30 RPM, at a distance of 15 cm from the center of the turntable. It begins to slide. A. What is the coefficient kinetic of friction between the object and the turntable?

17 µ K =F f F N F f F N 500 g 15 cm F w Σ F y = O N = F N + F w F N F N = F w = 4.9 N g kg N

18 F f = m 4π 2 r T 2 F f = (.5 kg ) 4π 2 (. ) ( ) 2 Watch F f = units

19 µ K =F f F N µ K =.74 N 4.9 N µ K =.151

20 B. What is the coefficient of friction between the object and the turntable if the mass of the object is increased to 1000 grams?

21 µ K =F f F N F f F N 1000 g 15 cm F w Σ F y = O N = F N + F w F N F N = F w = 9.8 N g kg N

22 F f = m 4π 2 r T 2 30 revs X 1 min 1 = 2 s min 60 sec x rev F f = ( 1 kg ) 4π 2 (.15 m) ( 2 s ) 2 Watch F f = 1.48 N units

23 µ K =F f F N µ K = 1.48 N 9.8 N µ K =.151 Contacting surfaces did not change. µ does not change.

24 C. Where should the 500 gram object be placed if the table turns at 78 RPM'S? F N F f 500 g?? cm F W

25 F f = m 4π 2 r T 2 r = F f T 2 m 4π 2

26 F f =.74 N This is the amount of friction needed to keep the block on the turn table. (Part A)

27 r =.74 N (.769 s ) 2 (.5 kg ) 4π 2 r = r = The faster the object spins, the closer to the center it must be to keep from sliding off.

28 A cylindrical spinning ride has a diameter of 10.0 m. If the ride rotates at a frequency of 0.25 rev/s, what is the minimum coefficient of static friction that would keep a rider pressed against the walls? Side view Top view F f F c F w

29 F c is holding the rider to the inside surface of the ride. F C = m 4π 2 r T 2 µ =F f F N Σ F y = O N =

30 µ =F w F c = m g m 4π 2 r T 2 T = time per revolution 0.25 rev/s 1 S 0.25 revs 4.0 s

31 What is the relationship between the magnitude of centripetal acceleration (a c ) and an object s speed (v)? A. B. C. a c α v a α c 1 2 v a α v c 2 D. a α c 1 v

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33 What is the direction of the velocity vector of an accelerating object? A. Toward the center of the circle. B. Away from the center of the circle. C. Along the circular path. D. Tangent to the circular path.

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35 2 MOTION 4 1 Which arrow represents the centripetal acceleration? Which arrow represents the tangential velocity vector? Which arrow represents the centripetal force vector? 3

36 Which horse has the greatest speed, the horse near the inside of a merry-go-round or the horse on the outside edge of the merry-go-round?

37 Which horse has the greatest period (frequency), a horse near the inside rail on a merry-go-round or the horse near the outside rail?

38 Kepler s Laws 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.

39 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.

40 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.

41 Thus, if the periods of the planets are T A and T B, and their average distances from the Sun are r A and r B, Kepler s third law can be expressed as follows: 2 3 T A r A = T r B B The squared quantity of the period of object A divided by the period of object B, is equal to the cubed quantity of object A s average distance from the Sun divided by Object B s average distance from the Sun.

42 The first two laws apply to each planet, moon, and satellite individually. The third law, however, relates the motion of several objects about a single body.

43 Interpretation The new planet takes times longer to orbit the sun than the earth does. Astronomers have recently discovered a new planet orbiting outside Pluto. If the planet is 50 times farther from the sun than earth, how much longer is its period of orbit than earth? 2 3 T A r A = T r B B

44 Callisto s Distance from Jupiter Galileo measured the orbital sizes of Jupiter s moons using the diameter of Jupiter as a unit of measure. He found that lo, the closest moon to Jupiter, had a period of 1.8 days and was 4.2 units from the center of Jupiter. Callisto, the fourth moon from Jupiter, had a period of 16.7 days.

45 Using the same units that Galileo used, predict Callisto s distance from Jupiter. Sketch the orbits of Io and Callisto.

46 Label the radii. Known: T C = 16.7 days T I = 1.8 days r I = 4.2 units Unknown: r C =?

47 Solve Kepler s third law for r C. C C I I T r T r = C C I I T r r T = 2 3 C 3 C I I T r r T =

48 Substitute r I = 4.2 units, T C = 16.7 days, T I = 1.8 days in: r C T = 3 r 3 C I TI days units = ( ) 1.8 days 2 = units 3 = 19 units

49 Which of the following is true according to Kepler s first law? 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.

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

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53 Newton s Law of Universal Gravitation Newton found that the magnitude of the force, F, on a planet due to the Sun varies inversely with the square of the distance, r, between the centers of the planet and the Sun. That is, F is proportional to 1/r 2. The force, F, acts in the direction of the line connecting the centers of the two objects.

54 The sight of a falling apple made Newton wonder if the force that caused the apple to fall might extend to the Moon, or even beyond. He found that both the apple s and the Moon s accelerations agreed with the 1/r 2 relationship.

55 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.

56 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. mm F = G r G = N m 2 /kg 2

57 The gravitational force is equal to the universal gravitational constant, times the mass of object 1, times the mass of object 2, divided by the square of the distance between the centers of the objects. mm F = G r 1 2 2

58 According to Newton s equation, F is inversely related to the square of the distance. F 1 r 2

59 Gravity Gravity, the weakest of the four forces, is about times the strength of the strong force. This weakness is easily seen - on a dry day, rub a comb across your shirt to give it a static charge, then hold it over a piece of paper on a desk. The piece of paper should lift off the desk. It takes an entire planet to keep the paper on the desk, but this force is easily overcome with everyday materials employing the electromagnetic force.

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61 What gravitational force exists between the Sun and Earth? mm F = G r m s = m E = r E = G = N m 2 /kg 2 F =

62 A 90 kg student is sitting 2.00 m from a 70 kg student in a classroom. What is the gravitational force that exists between them? mm F G r = G = N m 2 /kg 2 F =

63 Satellite Motion A satellite in an orbit that is always the same height above Earth moves in a uniform circular motion. Combining the equations for centripetal acceleration and Newton s second law, you can derive at the equation for the speed of a satellite orbiting Earth, v.

64 Thus, the period for a satellite orbiting Earth is given by the following equation: 3 r T = 2π Gm The period for a satellite orbiting Earth is equal to 2π times the square root of the radius of the orbit cubed, divided by the product of the universal gravitational constant and the mass of Earth. E

65 The equations for speed and period of a satellite can be used for any object in orbit about another. Central body mass will be replaced by m E, and r will be the distance between the centers of the orbiting body and the central body. If the mass of the central body is much greater than the mass of the orbiting body, then r is equal to the distance between the centers of the orbiting body and the central body. Orbital speed, v, and period, T, are independent of the mass of the satellite.

66 Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is kg and the radius of Earth is m, what are the satellite s orbital speed and period? Sketch the situation showing the height of the satellite s orbit.

67 Identify the known and unknown variables. Known: Unknown: h r E = m = m v =? T =? m E = kg G = N m 2 /kg 2

68 Determine the orbital radius by adding the height of the satellite s orbit to Earth s radius. r = h+ r E = m m = m

69 Solve for the speed. v = Gm E r = ( N.m 2 /kg 2 ) ( kg) ( m)

70 Solve for the period. r 3 T =2π Gm E = 2π ( m) ( )( N.m /kg kg) 3

71 Gravitation is expressed by the following equation: g = GM r 2 The gravitational field is equal to the universal gravitational constant times the object s mass, divided by the square of the distance from the object s center. The direction is toward the mass s center.

72 To find the gravitational field caused by more than one object, you would calculate both gravitational fields and add them as vectors. The gravitational field can be measured by placing an object with a small mass, m, in the gravitational field and measuring the force, F, on it. The gravitational field can be calculated using g = F/m. The gravitational field is measured in N/kg, which is also equal to m/s 2.

73 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.

74 The strength of the field varies inversely with the square of the distance from the center of Earth. The gravitational field depends on Earth s mass, but not on the mass of the object experiencing it.

75 A comet has a mass 8.00 X kg and an average radius of km. What is the acceleration due to gravity g on the comet? GM G = N m 2 /kg 2 g = r 2

76 A spacecraft was placed in orbit around a planet. The average radius from the planets center 1,896,000 m. If the mass of the planet is 6.42 x kg, what is the tangential velocity of the spacecraft? v = Gm E r G = N m 2 /kg 2

77 Newton figured that the moon is attracted to the Earth by the force of gravity. is a projectile falling around the Earth. has a tangential velocity that prevents it from falling into the Earth.

78 Einstein s Theory of Gravity 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.

79 Deflection of Light 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.

80 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.

81 The image on the right shows Chandra X-ray of two black holes (blue) in NGC 6240.

82 The period of a satellite orbiting Earth depends upon. 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

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84 The inertial mass of an object is measured by exerting a force on the object and measuring the object s using an inertial balance. A. gravitational force B. acceleration C. mass D. force

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86 Your apparent weight as you move away from Earth s center. A. decreases B. increases C. becomes zero D. does not change

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88 The End