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Appendix: Solutions to Problems Appendix: Solutions to Problems Chapter 2 1. 30 m/s2. 13,005 m. 2. 4,410 m. 3. 4,591.8 m. 4. PE is 9,800 J; (a) KE is 9,800 J; v is 140 m/s; (b) 14.286 s 5. The distances would be reduced proportionally by 0.3786. For the given times, distance is proportional to acceleration. 6. The distances would be increased proportionally by 1.138, by the same reasoning. 7. (a). 8. By virtue of the independence of forces, both hit the ground at the same time. 9. (c). 10. Since the mass is 1 kg, the force will be equal to the acceleration. Acceleration is constant at 9.8 m/s2. Velocities each second in meters per second: 9.8, 1.96, 2.93, 3.92; velocity at bottom is 4.427 m/s, which continues. Chapter 3 1. 0.00365 m. 2. 2.512 m. 3. 0.3759 m; period is 2 s. 4. 1.855 m. 5. 0.19 m/s2. 6. 0.0195 m. 7. 1.9; same for density. 8. Same reason that unequal masses hit the ground at the same time, as explained in Chap. 1. D.W. MacDougal, Newton’s Gravity: An Introductory Guide 419 to the Mechanics of the Universe, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-1-4614-5444-1, # Springer Science+Business Media New York 2012 420 Appendix: Solutions to Problems pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 9. P ¼ 2p lr2=GM; P ¼ 2pr32= = GM; P ¼ kr32= . 10. About 27.3 days, due to the greatly reduced value of the Earth’s surface gravity at the lunar distance. The actual lunar period is about 27.3 days. Chapter 4 1. e is 0.9962; a is 141.214 AU; b is 12.31 AU. 2. 5.458 years. 3. 541.436 AU. 4. 3.295 years. 5. q is 15.202 radii; Q is 15.269 radii. 6. (c). 7. P is1.76 years; b is 1.42 AU; e is 0.22253. 8. 1.0549 to 1; 0.46126 to 1. 9. [Graph] 10. [Graph] Chapter 5 1. a. 2. d. 3. a and c. 4. The bisector is perpendicular to the chord, and points toward the center of force; the arc and chord become unity as the two points of the arc merge. 5. (a) Newton used the term centripetal acceleration before he was prepared to call its cause gravity. (b) Centripetal acceleration acting on a mass requires for its existence, force, which opposed, by Newton’s Third Law, is sometimes described as centrifugal force. (c) Centrifugal force is an apparent force, not a real one, that appears to arise as a result of a moving object’s inertial tendency to continue in a straight line. (d) It takes a force to move a mass, by Newton’s Second Law, owing to the inertial quality inherent in mass. 6. Their accelerations are proportional to their distances from the center. 7. 1/8 to 1; 1/16 to 1. 8. 628 m/s; straight-line, tangent to the radius at point of release; forever until acted upon by an external force. 9. 2.943 m/s2; 5.86 Â 1017 N. 10. 1.65 to 1; 0.083 to 1. Chapter 6 1. (c). 2. 9.66 days. Appendix: Solutions to Problems 421 3. 1.91 times Earth’s. 4. 10.56 and 0.28, respectively. 5. 1 to 2.3. 6. 1 to 0.79. 7. 125 m or 12.5 cm. 8. 10.8 m; 0.00007 m/s squared; about the same. 9. About 1 to 2.3, or about the same as the answer to Problem 5, since the Moon’s velocity around the Sun also about 30 km/s, and its distance from the Sun is about the same as Earth’s. Calculating by masses, the ratio is 1 to 2.2. 10. 0.00000343 N. Chapter 7 1. 1.234 Â 10À13. 2. 1 to 6.263. 3. 1.582. 4. Phobos: 1.847 Â 105 km/day; 2.14 km/s. Deimos: 1.1675 Â 105 km/day; 1.35 km/s. 5. Phobos: 0.488 m/s2. Deimos: 0.0778 m/s2. Ratio: 6.264 to 1. 6. Eight times longer. 7. 1/16; yes. 8. 1/2 as fast. See also the proof of Corollary VI: velocities vary as the inverse square roots of the radii. 9. 3.71 m/s2. 10. The same. The source of the accelerations are the same. Chapter 8 1. 18.05 km/s. 2. It is the same. 3. 4.49 years. 4. 4.49 years. 5. 7.99 Â 10À4 m/s2. 6. It is the same. 7. 4.64 Â 1015 N. 8. It is the same. 9. 4.762 days. 10. 5.79 Â 1018; kg; 0.02%. Chapter 9 1. At the equator: 1.957 Â 106 kg. At the poles: 1.964 Â 106 kg. 2. The equation is Kepler’s Third Law. 422 Appendix: Solutions to Problems 3. 84.8 min; 7.88 km/s. 4. An increase of 0.98 m/s2. 5. 3,343 kg/m3. 6. 3.71 m/s2; 3.696 m/s2. 7. Density increases 49.3 times; g increases 13.45 times. 8. (a) 1/5; (b) 1/8; (c) square root of two times faster; (d) one over the square root of 2 less. 9. 11.274 m/s2; 0.0543 m/s2. 10. (a) 0.01214 m/s2; 8.884 days; (b) 318 to 1. Chapter 10 1. 1 to 0.1165 or 8.5837 to 1. 2. Pluto: 1.305 Â 1022 kg; Charon: 1.5203 Â 1021 kg. 3. Pluto: 2,042.2 km; Charon: 17,529.2 km; 0.1165 to 1. 4. a. 1.77; b. 16.97. 5. 8.5837 to 1. 6. Pluto: 23.26 km/s; Charon: 199.64 km/s. Ratio is 0.1165 to 1. 7. 222.9 km/s. 8. 1.36166 Â 1021 kg. 9. Yes. Each is 3.0351 Â 1023. 10. On Pluto: 2.64892 Â 10À4; on Charon: 2.27375 Â 10À3. Their periods are 6.38522 days, by all methods. Chapter 11 1. Each is approximately the same (about 7.5 Â 10À6), and approximately inde- pendent of the individual masses, since the masses of the planets are dwarfed by the mass of the Sun. 2. About 12% for Uranus and 3.4% for Neptune. 3. 1/22,894. 4. 1/19,411. 5. It should be close or equal to the previous answer. 6. Uranus: 8.6825 Â 1025 kg; 1/22,901. Neptune: 1.0245 Â 1026 kg; 1/19,414. 7. 14.5; 17.15. 8. 1.242 g/cm3; 1.1611 g/cm3. It would not be affected. 9. The mass should be close to or equal to the answer in Problem 6. 10. 3.506 Â 1026 kg; 3.177 days. Appendix: Solutions to Problems 423 Chapter 12 1. 5.753 km/s; 2.266 km/s. 2. 205,029 days; 561.34 years. 3. 3.61 km/s; results are the same as for Problem 1. 4. 5,753.4 m/s; 2,266.6 m/s. 5. 3,610.6 m/s; 1.28 Â 10À6 m/s2; perihelion, 3.25 Â 10À6 m/s2; aphelion, 5.05 Â 10À7 m/s2. 6. It is the same as in the previous problem. 7. Perigee: potential energy is À3.8287 Â 1029 J; kinetic energy is 2.7466 Â 1029 J; Apogee: potential energy is À1.5084 Â 1029 J; kinetic energy is 4.263 Â 1028 J. Their sum is the same due to the law of conservation of energy. 8. 5.206 km/s; 3.75 km/s; 111,867.2 days; 306.726 years. 9. 5.55 km/s; 3.72 km/s; 102,937 days; 281.826 years. 10. 4.02 Â 1021 kg; perihelion, 77.2 m/s; aphelion, 69.67 m/s. Chapter 13 1. Circular velocity: 377.587 km/s; perihelion velocity: 533.987 km/s; escape velocity: 533.989 km/s. The comet at perihelion is extremely close to solar system escape velocity. 2. 1.42571 Â 1011 J; 7.1286 Â 1010 J; 2 to 1; the energy of the comet at perihelion is twice the energy of its orbit in the circle. 3. Hyperbolic; p is 0.96489 AU; 41.338 km/s. 4. [graph] 5. 41.77 km/s. 6. 0.585975 AU; velocities at the respective orbits: Jupiter, 17.07 km/s; Mars, 33.43 km/s; Earth, 41.53 km/s; Venus, 49.03 km/s; Mercury, 67.34 km/s; velocity at perihelion, 54.57 km/s. 7. 7.464 Â 1024 J. 8. Simplify the equation after substituting a(e +1) for r at Q, and a(e À1) for r at q. 9. e is 0.43287761; p is 2.9422 AU; q is 2.053 AU; Q is 5.188 AU; perihelion velocity, 18.74 km/s; aphelion velocity, 11.79 km/s. 10. It is the ratio of the reciprocal of their semi-major axes; that is, their relative total orbital energies will be 1/a to 1/a0; Holmes, by 4.93 times. Chapter 14 1. 1.035 Â 106 J. 2. 1,618 m/s. 3. q, 9,376 km; Q, 23,458 km; a, 16,417 km; e, 0.4289; periapsis kick, 417.53 m/s; apoapsis kick, 330 m/s. 424 Appendix: Solutions to Problems 4.
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