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Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity 63 Orbital Motions under Gravity (b) F Note that circular orbits are special cases of elliptical orbits. (c) F If the time is not expressed in years we should not use T2 = a3. However, Kepler’s third law is still valid. (a) F The line only fits the data of the celestial bodies revolving around the Sun. (b) T Note that T2 is directly proportional to a3. (c) F The planets on the right of the Earth have larger orbits but are not necessarily more massive. Checkpoint Checkpoint 1 (p. 59) 1. 2. (a) F The Sun is at the focus of the elliptical orbit. (b) T (c) T While sweeping out equal areas in equal time intervals, a planet nearer the Sun has to travel a longer distance than it does when it is farther away. In other words, the planet travels a longer distance for a fixed time interval when it is nearer the Sun (i.e. higher speed). 3. | (a) p.1 Checkpoint 3 (p. 69) 1. The orbital period in seconds is The orbital semi-major axis in metres is (b) Perihelion distance = 3.05/2 − 0.145 = 1.38 AU Since , we have Aphelion distance = 3.05/2 + 0.145 = 1.67 AU (c) 3. (a) Perihelion. It is nearer the Sun than the aphelion (and in fact it is the nearest point to the Sun). 2. T Given that the areas shown are equal, the six time intervals should be equal. Therefore, the orbital period is six times any one of the time interval. (b) T (c) F In fact, the speed of the planet is the lowest at aphelion. (a) Y Note that Kepler’s third law is always valid. (b) Y (c) N Phobos and Deimos are not planets revolving around the Sun and this relation cannot be applied. Checkpoint 4 (p. 72) 1. (a) Checkpoint 2 (p. 63) 1. Applying (a) , T The law states that . When either mass is halved, the gravitational PE becomes halved. (b) Y (c) N When the distance is halved, the gravitational PE is doubled. (a) The gravitational PE is always negative. (b) The gravitational PE is always negative. This implies a = 2.8 AU. 2. Y Recall that 2. . Also, it is indirectly proportional to r. Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity (a) The escape speed becomes higher. Checkpoint 5 (p. 75) (b) The escape speed remains unchanged. 1. (c) The escape speed becomes higher. (a) C Its distance from the star is the longest. Gravitational PE is always negative and is zero at infinity. The farther away the two bodies the larger the gravitational PE between them. (b) (c) 2. Since mechanical energy is conserved, 1. Option B is incorrect. The gravitational force varies with the distance. Since the objects are in an elliptical orbit, the force must have changed in magnitude (and also direction) when they travel. 2. C 2. (a) 3. 0.5U 3. Reduced by 0.5U Since mechanical energy is conserved, the change of KE is equal to the change in PE i.e. U – 0.5U = 0.5U. F It is the gravitational force that keeps a spacecraft in the orbit. (b) F Note that the spacecraft and the objects inside are exerted the same gravitational force from the massive object. (c) F The astronaut is falling with the same acceleration as the spacecraft. He feels weightless because the spacecraft exert no reaction on him. Time intervals II and III Neglecting air resistance, gravitational force is the only force that acts on the aeroplane and the astronaut when the engine is shut down. Recall that , When the distance is doubled, the gravitational PE becomes halved. (b) (a) Moving at the same speed cannot guarantee weightlessness. For example, a man in a lift travelling downwards at a uniform speed can still feel his weight. Option A is incorrect because as the KE of the spacecraft has been reduced to decelerate during the change in orbit. Option B is incorrect. The PE does not change while the spacecraft is at the point it decelerates. The PE depends on the position of the spacecraft. C Option A is incorrect. It is the gravitational force from the Moon that keeps the astronauts in the orbit. Checkpoint 6 (p. 82) 1. p.2 Checkpoint 7 (p. 86) A The speed of the planet is the highest when it is closest to the star. The planet has the same mechanical energy at all positions as the mechanical energy is conserved. | You can imagine that the aeroplane is undergoing projectile motion during the time interval. Exercise Exercise 3.1 (p. 63) 1. The escape speed on the Moon is A By T2 = a3, the period should be The above is valid only when the time is expressed in years and the length is expressed in AU. 2. 4. Recall that , the escape speed is related to the mass and the radius of the planet but independent of the mass of the object itself. A The orbital period T and the semi-major axis a must satisfy the relation T2 = a3. See the table below. T/y T2 / y2 a / AU a3 / AU3 Active Physics Full Solutions to Textbook Exercises 3. 2.83 8.01 2 8 2.83 8.01 4 64 4 16 2 8 4 16 4 64 (a) (b) 4. 63 Orbital Motions under Gravity 4. By 5. By 6. (a) By (b) (i) , the orbital speed is , the period is Therefore, the ratio is (a) The semi-major axis is (b) By T2 = a3, the period is (a) The aphelion distance is (ii) (2)(2.217)−0.33 ≈ 4.10 AU (c) (i) (ii) , the mass is It has a shorter period. The shorter the orbital radius, the shorter the orbital period ( ). By T2 = a3, the semi-major axis is a = T2/3 = 3.32/3 = 2.217 ≈ 2.22 AU (b) p.3 The orbital period is not proportional to the orbital radius. The correct relation should be T = a3/2 = 1.5253/2 ≈ 1.88 yr 5. | Y The aphelion distance of Comet Encke is shorter than the semi-major axis of Jupiter. By , OR: By , the period is The orbital period of Jupiter is longer. By T2 = a3, the longer the semi-major axis, the longer the orbital period. Note that 1 d = 24 × 60 × 60 s. Exercise 3.2 (p. 70) 1. Recall 2. 7 A , we have . B From , when the mass of the star is doubled, the relation becomes where T is expressed in years and a is expressed in AU. The relation is independent of the mass of the planet. 3. By B By , we have Or: 8. (a) When the comet travels along its orbit, it experiences the gravitational pull from the Sun. As it moves away from the Sun, the pull always has a component opposite to the comet’s velocity in direction. Therefore the comet slows down when it moves away. Note that the component of the pull perpendicular to the velocity does no work on the comet and will not change the speed of the comet. Active Physics Full Solutions to Textbook Exercises (b) 63 Orbital Motions under Gravity If the mass M of the star is not large enough, the centre of the circular orbit will not be the star. The orbital radius does not equal the distance between the star and the planet. As a result, we cannot derive from 7. . | p.4 (a) The acceleration is (b) (i) The gravitational PE is (ii) Since the spacecraft is moving in a circular orbit, the mechanical energy is Exercise 3.3 (p. 82) 1. A The PE is Since the speed is unknown, we cannot find the KE. Options B and C cannot be correct. Option D is incorrect (due to conservation of energy). 2. (iii) The KE is E – U = −1.299 × 1010 – (−2.599 × 1010) ≈ 1.30 × 1010 J B The mechanical energy is 8. The KE of the probe just before it hits the surface of the Moon is (in joules) where r is the distance of the satellite from the centre of the Earth. 3. A Imagine that a spacecraft is initially in the orbit X. It has to accelerate (e.g. by firing a rocket) at the leftmost point in order to enter orbit Y. In other words, the spacecraft has gained KE during the orbital transition. In contrast, a spacecraft initially in orbit Z has to decelerate at the rightmost point in order to enter orbit Y. Therefore, a body should have the lowest mechanical energy in orbit X and the highest in orbit Z. 4. 6. 9. (a) Applying , the escape speed is (b) By the conservation of energy, the speed should be also 60.2 km s−1. C Recall that 5. The velocity is . (a) They have the same acceleration of . (b) The object in orbit 2 has less mechanical energy. (c) They have the same PE of (d) The object in orbit 2 has less KE. (e) The object in orbit 2 is slower. Substitute: (f) [Deleted] (a) The gravitational PE is v1 = 4538 m s−1, G = 6.67 × 10−11N m2 kg−2, M = 6.42 × 1023 kg, r1 = 1.71 × 105 + 3390 × 103 = 3.561 × 106 m, and r2 = 1.78 × 107 + 3390 × 103 = 2.119 × 107 m. 10. By the conservation of energy, . Now we have (b) The PE is smaller (more negative). Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity 1. Statement (3) is incorrect. According to Kepler’s third law, T2 ∝ a3.Therefore, it should be the period squared that is proportional to AC cubed. 2. B C 3. The contents are in weightless condition and appear to remain at the original position. 3. (a) 4. 5. (b) Recall 4. The astronaut and the spacecraft are free falling together and hence there is no reaction from the spacecraft on the astronaut. As a result, the astronaut feels weightless. (i) (ii) 5. Statement (1) is incorrect. The planet moves with the lowest speed at the aphelion. aphelion / AU a / AU 1.24 5.68 3.46 19P/Borelly 1.35 5.83 3.59 9P/Tempel 1.51 4.74 3.125 81P/Wild 1.59 5.31 3.45 C By , the ratio should be 13/2 : 23/2 : 33/2 = 1 : 23/2 : 33/2 6. C Since , the nearer to the Earth, the faster the satellite travels. Therefore, Y should have more KE. Since , the farther away from the Earth, the more the PE the satellite have. (Note the minus sign!) Therefore, X should have more PE. 7. C Since the satellites are moving in circular orbits, we have E = U/2. Options A and B are incorrect as E must be changing with r similar to U. Also, E must be more ‘positive’ than U and option C should be the best representation. (Note the minus sign!) Chapter Exercise: Multiple-choice Questions (p. 91) D perihelion / AU 67P/C-G The longer the semi-major axis, the longer the orbital period. Therefore, 19P/Borelly should have the longest orbital period. The magnitude of acceleration gradually decreases as the gravitational force acting on the spacecraft decreases with the distance between it and the Earth. The astronaut feels weightless throughout the whole journey when he is moving in the orbit. B comet Chapter Exercise 1. . Therefore, the period ratio is The semi-major axes of the comets a are as shown. Since no net force acts on the marble after the throw, by Newton’s first law of motion, it moves in a straight line at a uniform velocity. The normal reaction from the balance on the block is zero when they fall freely at the same rate. As a result, the balance reads zero. (a) D Yes. Since the spacecraft is travelling in space at a uniform velocity, the only possible scenario for her to experience weightlessness is that the spacecraft is so far away from all celestial bodies so that the total gravitational force acting on the spacecraft is too small to be influential. (b) D The centripetal force is provided by the gravitational force: The acceleration is 2. p.5 Statement (2) is incorrect. The time is not proportional to the angular separation. According to Kepler’s second law, the imaginary line joining the Sun and the planet sweeps out equal areas in equal time intervals. The spacecraft has a speed is 765 m s−1 when it is farthest away from Mars. Exercise 3.4 (p. 87) | 8. A Note that 9. A . Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity Statement (2) is correct only when both the astronaut and the spacecraft remain still relatively to each other. 10. (b) | p.6 Applying T2 = a3, the period is T = 1.4553/2 = 1.755 ≈ 1.76 y (1M+1A) Note that the distance must be in AU and the period is in years. D Option A is incorrect. The Moon is orbiting around the Earth due to the gravitational pull from the Earth. It is not logical to deduce that the gravitational field produced by the Earth is zero at a position nearer than the Moon. (c) (i) The change in PE is Option B is incorrect because an astronaut orbiting the Earth is not always between the two bodies. Option C is incorrect. Note that centripetal force is not a real force. 11. (1M+1A) A (ii) Since , the closer the comet to the Sun, the smaller the PE (the more negative the PE). 12. Therefore, the KE change is −1.91 × 1024 J. (1A) A When the engine is on, it produces a constant thrust. At the same time, the gravitational force acting on the astronaut gradually decreases (with the distance between the rocket and the Earth). Hence the g-force rises with an increasing trend during the first 10 s. By conservation of energy, the KE decrease should be equal to the PE increase in value. (1M) 17. (d) The Sun is not at the centre of the orbit. (1A) (a) The semi-major axis is 205 + 6370 × 2 + 50 930 = 31937.5 ≈ 3.19 × 104 km. (1A) (b) Applying , the period is When the engine is switched off, the rocket and the astronaut inside are only subjected to the gravitational force. Therefore, the astronaut will feel weightless and the g-force is hence zero. 13. (1M+1A) B Also accept 56 830 / (60 × 60) ≈ 15.8 h. The change in PE is −5.8 × 109 – (−3 × 109) = −2.8 × 109 J. (c) By conservation of energy, Options C and D must be incorrect because the PE has to be decreasing when the distance becomes smaller. 14. Substitute: A The original PE is is r1 = (205 + 6370) × 103 = 6.575 × 106 m, and r2 = (50 930 + 6370) × 103 =5.73 × 107 m. and the original KE . When the spacecraft Now we have is very far away, the PE becomes zero and the KE is . The speed is thus 15. B (1M+1A) Note that the orbit of the Earth is not completely circular. (d) (i) The radius of the orbit is (1740+200) × 103 = 1.94 × 106 m Applying Chapter Exercise: Structured Questions (p. 93) 16. (a) The semi-major axis is (1.13+1.78)/2 = 1.455 AU. (1A) Moon is , the mass of the Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity | p.7 (1A) The mechanical energy E is the sum of PE and KE, i.e. (1M+1A) (ii) Applying . (1A) (b) , the speed is (i) The mechanical energy E decreases. (1A) As and r decreases, E also decreases (more ‘negative’). (1A) OR: Part of the mechanical energy has been converted to heat and internal energy when the satellite moves across air. (1M+1A) (e) (f) If the total mechanical energy is conserved, Chang’e 1 would have travelled in a fixed orbit but would not change its orbit. (1A) In the circular orbit (1A) At the point of transition, Chang’e 1 decelerates to transit from the elliptical orbit to the less elliptical one and finally to the circular one. In other words, the PE of Chang’e 1 remains unchanged at that point but the KE decrease. Hence we can deduce that Chang’e 1 had the least mechanical energy in the circular orbit. (1A) 18. (a) (ii) The speed increases. (1A) As and E takes a more negative value, we can deduce that v increases. (1A) 20. (a) By , the orbital speed is (1M+1A) (b) The acceleration is The gravitational PE is (1M+1A) (b) (c) W = mg = (100)(9.8049 × ≈ 0.980 N (1M+1A) 10−3) (1M+1A) The mechanical energy is E = U/2 = −7.612 × 1010 / 2 = −3.806 × 1010 ≈ −3.81 × 1010 J (1M+1A) The escape speed is (c) The minimum extra KE needed is 3.806 × 1010 J. The speed is (2A) Therefore, the ball can escape because its speed is higher than the escape speed. (1A) (d) For the spacecraft to just escape from the Earth forever, its mechanical energy after the turning off the thrusters has to be zero. (1M) The speed should be (1A) (d) No. (1A) The spacecraft is not on the Earth’s surface initially. The KE needed and thus the value of v are smaller. (1A) (1M+1A) 19. (a) The PE is 21. (a) (i) At infinity, all KE has changed to PE. Since the PE at infinity is zero, KEmin is +GMm/r. (1A) (ii) This is only the minimum because there may be work done against other resistive forces. (1A) . (1A) The centripetal force is provided by the gravitational force. Therefore, the KE can be deduced by Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity (iii) The initial KE is . (1A) Rearranging p.8 The satellite actually keeps ‘falling’ but do not fall to the surface of the Earth. , we have (1A) (b) (i) The centripetal force need to keep the satellite in orbit is (iv) Substitute | . (1A) This is provided by the gravitational force, i.e. into the above. (1A) (1A) (1A) (v) (1A) Since v is indirectly proportional to √r, v decreases as r increases. (1A) (b) (i) The KE is . The velocity is thus (ii) Subsitute v = 2πr/T into the above. (1A) (1M+1A) 22. (a) (ii) Its speed is far lower than the escape speed required. (1A) (i) The length of the orbit is 2 π (1.8 × 108) = 1.13 × 109 m. (1A) r3/2, Since T ∝ T increases as r increases. From (i) we can deduce that T increases as v decreases. (1A) (c) The period is The speed is thus 1.13 × 109/(21 × 60 × 60) ≈ 1.5 × 104 m s−1. (1A) (ii) The centripetal force is . (1A) (1M+1A) The value of the force is (d) The gravitational PE has been converted to internal energy (heat) and KE. (2A) (a) A satellite in a geostationary orbit appears to be stationary as viewed from the ground because its orbital period is exactly the same as the self-rotation period of the Earth. (1A) . (1A) (iii) The gravitational force is 24. . (1A) Therefore, the mass is The transmitter and receiver on the ground do not need to track such satellite as time passes by. (1A) (1A) (b) (i) By , the PE is (b) The centripetal force is provided by the gravitational force. (1M+1A) (ii) The KE is . (1M+1A) (1A) (iii) The total mechanical energy (PE+KE) is negative. The rock does not have sufficient KE to escape. (1A+1C) 23. (a) (c) (i) . The KE is . (1A) The phrase ‘gravity is diminishing’ means that the gravitational force is getting weaker as the distance increases ( The PE is The total mechanical energy is ). (1A) There is no such a force pointing away from the Earth acting on a satellite orbiting around the Earth. (1A) Also, the forces on the satellite are actually unbalanced. (1A) (1A) (ii) The change in energy is Active Physics Full Solutions to Textbook Exercises 63 Orbital Motions under Gravity (1M+1A) (iii) Since T2 ∝ a 3, we have (1M) The time is 10.57/2 ≈ 5.29 h. (1A) You can also find T using . | p.9