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CHAPTER 12 BLM ANSWER KEY BLM 12-2: Newton’s Law of Universal Gravitation/Reinforcement BLM 12-4: The Orbital Speed of Planets/Reinforcement Answers Answers m1m2 , where Fg is the force of gravity, G is r2 the universal gravitational constant, m1 is the mass of the Sun, m2 is the mass of the planet, and r is the orbital radius of the planet, gives the following answers. Equating the centripetal force equation, Fc Using Fg G Sun-Mercury: 1.30 1022 N Sun-Saturn: 3.68 1022 N Sun-Venus: 5.50 1022 N Sun-Uranus: 1.42 1021 N Sun-Earth: 3.58 1022 N Sun-Neptune: 6.75 1020 N Sun-Mars: 1.64 1021 N Sun-Pluto: Sun-Jupiter: 4.17 1023 N mv 2 , and r mm Newton’s law of universal gravitation, Fg G 1 2 2 , r where m1 is the mass of the Sun, m2 is the mass of the planet, and r is the orbital radius of the planet, gives m v G 1 . Solving this expression for each planet gives r the orbital speed of that planet. 4.96 1016 N Mercury: 4.79 104 m/s Saturn: 9.64 103 m/s Venus: 3.51 104 m/s Uranus: 6.80 103 m/s BLM12-3: The Acceleration Due to Gravity/Reinforcement Earth: 2.99 104 m/s Neptune: 5.43 103 m/s Mars: 2.41 104 m/s Pluto: 4.7 103 m/s Answers Jupiter: 1.31 104 m/s m1m2 , where m1 is the mass of the planet, r2 m m2 is the mass of an object, and Fg mg , gives g G 21 . r Solving this equation for each planet gives the acceleration due to gravity close to the surface of that planet. Equating Fg G Mercury: 3.31 m/s2 Saturn: 10.4 m/s2 Venus: 8.09 m/s2 Uranus: 8.24 m/s2 Earth: 9.80 m/s2* Neptune: 11.2 m/s2 Mars: Jupiter: 3.75 m/s2 24.6 m/s2 Pluto: 0.72 BLM 12-5: Chapter 12 Test/Assessment Answers 1. (a) 7. Brahe believed in an Earth-centred universe in which all planets other than Earth orbited the Sun. In formulating his laws of planetary motion, Kepler applied Brahe’s data to the motions of planets in a Sun-centred system. 8. During the period between July and December, Earth is moving closer to the Sun, and as any planet moves closer to the Sun, its orbital speed increases. m/s2 *The value of 9.81 m/s2 that is traditionally used for the acceleration due to gravity on Earth is obtained when the value is calculated from data that have more significant figures than those given in the Table of Planetary Data in BLM 12-1. 2. (b) 3. (b) 4. (a) 5. (b) 6. (b) 9. (a) If the mass of one object doubled, the gravitational force between the two bodies would be doubled. (b) If the distance between two bodies was doubled, the gravitational force between them would be one fourth as great. 10. To find a planet’s period using Kepler’s third law, you need to know the period of another planet and the orbital radii of both planets. 11. The surface gravity of each planet varies inversely with the square of the radius of the planet. Jupiter’s radius is much greater (almost 11 times) than Earth’s radius. CHAPTER 12 BLM ANSWER KEY 12. The satellite moves with uniform circular motion, so the curvature of Earth’s surface exactly matches the curvature of the trajectory of the satellite. Earth’s surface “falls away” from the satellite at the same rate that the satellite falls toward Earth. 13. The force of attraction will be greatest between the spheres in (c) set A and C. 16. The spheres would have to be 11.3 m apart to main the gravitational force between them. Gm1m2 r Fg (6.67 1011 Nkgm2 )(40.0 kg)(20.0 kg) 2 B mm Fg 1 2 2 r (32 kg)(25 kg) Fg (20 cm) 2 mm Fg 1 2 2 r (80 kg)(20 kg) (40 cm) 2 2.0 kg 2 cm 2 4.17 1010 N r 11.3 m A Fg r 17. The distance between the Sun and Mars is 1.52 AU. 2 TMars rMars TEarth rEarth 2 T rMars rEarth 3 Mars TEarth 1.0 kg 2 cm 2 C mm Fg 1 2 2 r (90 kg)(10 kg) (15 cm) 2 4.0 kg 2 cm 2 14. rMars 18. Fg 1.67 10 9 N The satellite is 1.84 108 m from Earth. 2 Tsatellite rsatellite TMoon rMoon The gravitational force between spheres is 4.17 1010 N. mm Fg G 1 2 2 r N m 2 (20 kg)(20 kg) Fg 6.67 1011 kg 2 (8.00 m) 2 The force between the spheres will be 1.67 109 N. mm Fg G 1 2 2 r N m 2 (20 kg)(20 kg) Fg 6.67 1011 kg 2 (4.00 m) 2 686 d 2 1 AU 3 365 d rMars 1.52 AU rsatellite 3 2 T satellite rMoon 3 TMoon 9.1d 2 rsatellite 3.84 108 m 3 27.3d 8 rsatellite 1.84 10 m Fg 4.17 1010 N 15. 3 19. The mass of the larger sphere is 50 kg. Fg r 2 m2 Gm1 m2 2.50 10 6.67 10 8 N 1.05 m 11 N m 2 kg 2 m2 5.0 101 kg 2 8.2 kg CHAPTER 12 BLM ANSWER KEY 20. (a) The period of the object is 32 years. 2 3 Tobject robject TEarth rEarth robject rEarth 3 10 AU 1 a 1 AU 32 a 3 Tobject TEarth Tobject Tobject (b) The speed of the object is 9.41103 m/s. GmSun v r v (6.67 1011 N m2 kg 2 )(1.99 1030 kg) 1.49 1012 m m v 9.44 103 s