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Chapter 7: Gravitation. Click the mouse or press the spacebar to continue. Splash Screen. In this chapter you will:. Learn the nature of gravitational force. Relate Kepler’s laws of planetary motion to Newton's laws of motion.

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  1. Chapter 7: Gravitation Click the mouse or press the spacebar to continue. Splash Screen

  2. In this chapter you will: • Learn the nature of gravitational force. • Relate Kepler’s laws of planetary motion to Newton's laws of motion. • Describe the orbits of planets and satellites using the law of universal gravitation. Chapter Intro

  3. Chapter 7: Gravitation Section 7.1:Planetary Motion and Gravitation Section 7.2:Using the Law of Universal Gravitation Chapter Menu

  4. In this section you will: • Relate Kepler’s laws to the law of universal gravitation. • Calculate orbital speeds and periods. • Describe the importance of Cavendish’s experiment. Section 1-1

  5. 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. Click image to view the movie. Section 1-2

  6. Kepler’s Laws 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. Click image to view the movie. Section 1-3

  7. Kepler’s Laws Kepler also found that there is a mathematical relationship between periods of planets and their mean distances away from the Sun. Section 1-4

  8. Kepler’s Laws 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. Click image to view the movie. Section 1-4

  9. Kepler’s Laws 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: The squared quantity of the period of planet A divided by the period of planet B, is equal to the cubed quantity of planet A’s average distance from the Sun divided by planet B’s average distance from the Sun. Section 1-5

  10. Kepler’s Laws 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. Section 1-6

  11. 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. Using the same units that Galileo used, predict Callisto’s distance from Jupiter. Section 1-7

  12. Callisto’s Distance from Jupiter Step 1: Analyze and Sketch the Problem Section 1-8

  13. Callisto’s Distance from Jupiter Sketch the orbits of Io and Callisto. Section 1-9

  14. Callisto’s Distance from Jupiter Label the radii. Known: TC = 16.7 days TI = 1.8 days rI = 4.2 units Unknown: rC = ? Section 1-10

  15. Callisto’s Distance from Jupiter Step 2: Solve for the Unknown Section 1-11

  16. Callisto’s Distance from Jupiter Solve Kepler’s third law for rC. Section 1-12

  17. Callisto’s Distance from Jupiter Substitute rI = 4.2 units, TC = 16.7 days, TI = 1.8 days in: Section 1-13

  18. Callisto’s Distance from Jupiter Step 3: Evaluate the Answer Section 1-14

  19. Callisto’s Distance from Jupiter Are the units correct? rC should be in Galileo’s units, like rI. Is the magnitude realistic? The period is large, so the radius should be large. Section 1-15

  20. Callisto’s Distance from Jupiter Step 1: Analyze and Sketch the Problem Sketch the orbits of Io and Callisto. Label the radii. Step 2: Solve for the Unknown Solve Kepler’s third law for rC. Step 3: Evaluate the Answer The steps covered were: Section 1-16

  21. 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/r2. The force, F, acts in the direction of the line connecting the centers of the two objects. Section 1-17

  22. Newton’s Law of Universal Gravitation 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/r2 relationship. Section 1-18

  23. Newton’s Law of Universal 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. Section 1-19

  24. Newton’s Law of Universal Gravitation 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. 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. Section 1-20

  25. Inverse Square Law According to Newton’s equation, F is inversely related to the square of the distance. Section 1-21

  26. Universal Gravitation and Kepler’s Third Law Newton stated his law of universal gravitation in terms that applied to the motion of planets around the Sun. This agreed with Kepler’s third law and confirmed that Newton’s law fit the best observations of the day. Section 1-22

  27. Universal Gravitation and Kepler’s Third Law In the equation below, squaring both sides makes it apparent that this equation is Kepler’s third law of planetary motion: the square of the period is proportional to the cube of the distance that separates the masses. Section 1-23

  28. Universal Gravitation and Kepler’s Third Law The factor 4π2/Gms depends on the mass of the Sun and the universal gravitational constant. Newton found that this derivative applied to elliptical orbits as well. Section 1-24

  29. Measuring the Universal Gravitational Constant Click image to view the movie. Section 1-25

  30. Importance of G Cavendish’s experiment often is called “weighing Earth,” because his experiment helped determine Earth’s mass. Once the value of G is known, not only the mass of Earth, but also the mass of the Sun can be determined. In addition, the gravitational force between any two objects can be calculated using Newton’s law of universal gravitation. Section 1-26

  31. Cavendish’s Experiment Determined the value of G. Confirmed Newton’s prediction that a gravitational force exists between two objects. Helped calculate the mass of Earth. Section 1-27

  32. Question 1 Which of the following helped calculate Earth’s mass? A. Inverse square law B. Cavendish’s experiment C. Kepler’s first law D. Kepler’s third law Section 1-28

  33. Answer 1 Reason:Cavendish's experiment helped calculate the mass of Earth. It also determined the value of G and confirmed Newton’s prediction that a gravitational force exists between two objects. Section 1-29

  34. Question 2 Which of the following is true according to Kepler’s first law? A. Paths of planets are ellipses with the 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. The force between two objects is proportional to their masses. Section 1-30

  35. Answer 2 Reason:According to Kepler’s first law, the paths of planets are ellipses, with the Sun at one focus. Section 1-31

  36. Question 3 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 Section 1-32

  37. Answer 3 Reason:According to Kepler’s second law, an imaginary line from the Sun to a planet sweeps out equal areas in equal time intervals. Section 1-33

  38. Section 1-33

  39. In this section you will: • Solve orbital motion problems. • Relate weightlessness to objects in free fall. • Describe gravitational fields. • Compare views on gravitation. Section 2-1

  40. Orbits of Planets and Satellites Newton used a drawing similar to the one shown below to illustrate a thought experiment on the motion of satellites. Click image to view the movie. Section 2-2

  41. Orbits of Planets and Satellites 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 the equation for the speed, v, of a satellite orbiting Earth. Section 2-3

  42. A Satellite’s Orbital Period A satellite’s orbit around Earth is similar to a planet’s orbit about the Sun. Recall that the period of a planet orbiting the Sun is expressed by the following equation: Section 2-4

  43. A Satellite’s Orbital Period Thus, the period for a satellite orbiting Earth is given by the following equation: 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. Section 2-5

  44. A Satellite’s Orbital Period The equations for speed and period of a satellite can be used for any object in orbit about another. Central body mass will replace mE, 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. Section 2-6

  45. A Satellite’s Orbital Period Satellites such as Landsat 7 are accelerated by large rockets such as shuttle-booster rockets to the speeds necessary for them to achieve orbit. Because the acceleration of any mass must follow Newton’s second law of motion, Fnet = ma, more force is required to launch a more massive satellite into orbit. Thus, the mass of a satellite is limited by the capability of the rocket used to launch it. Section 2-7

  46. Orbital Speed and Period Assume that a satellite orbits Earth 225 km above its surface. Given that the mass of Earth is 5.97×1024 kg and the radius of Earth is 6.38×106 m, what are the satellite’s orbital speed and period? Section 2-8

  47. Orbital Speed and Period Step 1: Analyze and Sketch the Problem Section 2-9

  48. Orbital Speed and Period Sketch the situation showing the height of the satellite’s orbit. Section 2-10

  49. Orbital Speed and Period Identify the known and unknown variables. Known: h = 2.25×105 m rE = 6.38×106 m mE = 5.97×1024 kg G = 6.67×10−11 N·m2/kg2 Unknown: v = ? T = ? Section 2-11

  50. Orbital Speed and Period Step 2: Solve for the Unknown Section 2-12

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