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(cs39963) schroeder Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 1 This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points An ant of mass m clings to the rim of a ywheel of radius r , as shown. The ywheel rotates clockwise on a horizontal shaft S with constant angular velocity . As the wheel rotates, the ant revolves past the stationary points I , II , III , and IV . The ant can adhere to the wheel with a force much greater than its own weight. I Ant r 002 (part 2 of 2) 10.0 points What is the magnitude of the minimum adhesion force necessary for the ant to stay on the ywheel at point III ? 1. F = m 2 r + m g correct 2. F = m 2 r m g 3. F = m 2 r 2 4. F = m 2 r 2 + m g 5. F = m g Explanation: According to the explanation in the previous part, the minimum adhesion force is IV II G + Fc = m 2 r + m g . S III It will be most dicult for the ant to adhere to the wheel as it revolves past which of the four points? 1. II 2. It will be equally dicult for the ant to adhere to the wheel at all points. 3. I 4. III correct 5. IV Explanation: The sum of the ants gravity and its adhesion force is the centripetal force Fc with magnitude m 2 r . Thus Fc = m g + Fad Fad = Fc m g . The maximum Fad is when Fc and m g are in opposite directions; i.e., at III . 003 (part 1 of 2) 10.0 points An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the oor drops away (see gure). The coecient of static friction between the person and the wall is and the radius of the cylinder is R. R What is the minimum tangential velocity needed to keep the person from slipping downward? 1. v = 1 gR 2. v = g R schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 2 3. v = 4. v = 5. v = 6. v = 2 gR 2gR gR correct gR 004 (part 2 of 2) 10.0 points Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall. 1. F = m g correct 2. F = m v2 R m v2 R 7. v = 2 g R 8. v = 9. v = 10. v = 2gR 2gR 3. F = m g + 4. F = m g 5. F = 6. F 7. F 8. F 9. F 10. F m v2 r max Frictional force: fs N = fs Solution: The maximum frictional force due to friction is fmax = N , where N is the inward directed normal force of the wall of the cylinder on the person. To support the person vertically, this maximal friction force max fs must be larger than the force of gravity m g so that the actual force, which is less than N , can take on the value m g in the positive vertical direction. Now, the normal v2 force supplies the centripetal acceleration R on the person, so from Newtons second law, Centripetal force: F = N= Since max fs = N = 1 gR 2 Explanation: Basic Concepts: m v2 mg R mg = m v2 = R m v2 = mg R m v2 = mg + R m v2 = R m v2 . R m v2 mg, R Explanation: The vertical friction force must equal m g , in order to balance the force of gravity and not have any acceleration in the vertical direction. 005 10.0 points A ball rolls around a circular wall, as shown in the gure below. The wall ends at point X . the minimum speed required to keep the person supported is at the limit of this inequality, which is 2 m vmin = m g, R or 1 2 vmin = gR . schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 3 D C B A E X Ferris wheel moves in a circle. 007 (part 1 of 2) 10.0 points A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2 . When the ball gets to X , which path does the ball follow? 1. Path E 2. Path D 3. Path B 4. Path A 5. Path C correct Explanation: As soon as the ball reaches point X the centripetal force is removed, so the ball moves in a straight line (tangent to the circle at oint X). Path C is the correct answer. 006 10.0 points A passenger on a Ferris wheel moves in a vertical circle at constant speed. Are the forces on her balanced? 1. Yes; it moves in a vertical circle. 2. Yes; the speed is constant. 3. No; there is an inward net force. correct 4. No; the direction of net force does not change. Explanation: Even though she moves at constant speed, she does not move in a straight line, so, by Newtons rst law, the forces on her are not balanced. The net force is inward since the 21 9.8 m/s2 6 kg What is the speed of the ball when it is in circular motion? Correct answer: 1.55777 m/s. Explanation: Let : = 1.8 m , = 21 , g = 9.8 m/s2 , m = 6 kg . Use the free body diagram below. 1. 8 m v and T mg The tension on the string can be decomposed into a vertical component which balances the weight of the ball and a horizontal component which causes the centripetal acceleration, acentrip that keeps the ball on its horizontal circular path at radius r = sin . If T is the magnitude of the tension in the string, then Tvertical = T cos = m g (1) schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 4 and Thoriz = m acentrip or T sin = Solving (1) for T yields T= mg cos (3) 2 m vball . sin Tperiod = 2 (2) = 2 cos g (1.8 m) cos 21 9.8 m/s2 = 2.60183 s . as sin cancels. 009 (part 1 of 2) 10.0 points Calculate the period of a ball tied to a string of length 1.9 m making 6.4 revolutions every second. Correct answer: 0.15625 s. Explanation: and substituting (3) into (2) gives m g tan = Solving for v yields v= = g tan sin ( 9. 8 m/s2 ) (1.8 m) tan 21 sin 21 2 m vball . sin = 1.55777 m/s . 008 (part 2 of 2) 10.0 points How long does it take Tperiod for the ball to rotate once around the axis? Correct answer: 2.60183 s. Explanation: Basic Concept: d = vt. Solution: Because the tangential speed of the ball around the circle is constant, we have vball = s Tperiod . Let : f = 6.4 rev/s , R = 1. 9 m . and T= 1 1 = = 0.15625 s . f 6.4 rev/s 010 (part 2 of 2) 10.0 points Calculate the speed of the ball. Correct answer: 76.4035 m/s. Explanation: D 2R = T T 2 (1.9 m) = 0.15625 s = 76.4035 m/s . v= s is the distance the ball travels in one revolution, which is the perimeter of the circle of radius sin , therefore, we have s = 2 sin . Equating both expressions for vball , we have 2 sin = vball = Tperiod 2 sin = Tperiod g tan sin g sin2 cos 011 10.0 points Why is the linear speed greater for a horse on the outside of a merry-go-round than for a horse closer to the center? 1. None of these 2. The horse on the outside has longer legs. schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 5 3. The outside horse moves easier. 4. The horse on the outside feels less force from the merry-go-round. 5. The horse on the outside is larger. 6. The tangential speed of the horse is directly proportional to the distance from the center. correct Explanation: Every horse on a merry-go-round has the same rotational speed; the tangential speed of the horse is directly proportional to the distance from the center. 012 10.0 points A string under a tension of 48 N is used to whirl a rock in a horizontal circle of radius 2.2 m at a speed of 20.05 m/s. The string is pulled in, and the speed of the rock increases. When the string is 0.548 m long and the speed of the rock is 57 m/s, the string breaks. What is the breaking strength of the string? Correct answer: 1557.41 N. Explanation: At the beginning, neglecting gravity, we have 2 mv0 T= , r where T is the tension, v0 is the initial velocity and r is the radius of the circle. So the mass of the rock is Tr m= 2 v0 (48 N) (2.2 m) = 20.05 m/s2 = 0.262685 kg . At the breaking point, if the breaking strength is Tb then we have m v2 Tb = (0.262685 kg) (57 m/s2 ) = 0.548 m = 1557.41 N . 013 10.0 points Mass m1 moves in a circular path of radius r on a frictionless horizontal table. It is attached to a string that passes through a frictionless hole in the center of the table. A second mass m2 is attached to the other end of the string. v m1 r m2 Determine an expression for r in terms of m1 , m2 , and the period T (the time for one revolution). 1. r = 2. r = 3. r = 4. r = 5. r = m2 g T 2 2 2 m1 m2 g T 2 2 m1 m2 g T 2 correct 4 2 m1 m1 g T 2 4 2 m2 m1 g T 2 2 2 m2 Explanation: The linear velocity v can be expressed in terms of the distance it travels each revolution: v= 2r . T Consider the forces acting on each mass: T N m1 g T m2 g schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 6 The tension T in the string provides the centripetal force required to keep m1 moving in a circular path. Applying Fx = m ax to m1 , we obtain T = m1 ac = m1 v2 r and 015 (part 1 of 2) 10.0 points A car rounds a slippery curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is and the coecient of friction is . Fy = m ay = 0 to m2 , so m2 g T = 0 m2 g = m1 M v2 r 4 2 r m2 g = m1 T2 m2 g T 2 . r= 4 2 m1 014 10.0 points A race car travels 77 m/s around a circular track of radius 183 m. g 2. 2 m 0. = 14 18 What should be the cars speed in order that there is no frictional force between the car and the road? 1. v = 2. v = 3. v = 4. v = 5. v = g R tan correct gR g R tan gR cos g R sin g sin R g R cos What is the magnitude of the resultant force on the 2200 kg driver and his car if the car does not slip? Correct answer: 71.2776 kN. Explanation: Let : m = 2200 kg , v = 77 m/s , r = 183 m , = 18 , and = 0.14 . v2 r 1 kN (77 m/s)2 = (2200 kg) (183 m) 1000 N 6. v = 7. v = Explanation: Basic Concepts: Consider the free body diagram for the car. The forces acting on the car are the normal force, the force due to gravity, and possibly friction. Fnet = m ar = m N N sin mg N cos y x N = 71.2776 kN . schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 7 To keep an object moving in a circle requires a force directed toward the center of the circle; the magnitude of the force is Fc = m ac = m Also remember, F= i 1. vmin = 2. vmin = 3. vmin = v2 . r g R (sin + cos ) cos cos g R (cos + cos ) sin cos g R (sin + cos ) sin + cos g R (sin cos ) correct sin + cos g R (sin cos ) sin cos gR sin + cos g R (sin + cos ) Fi . 4. vmin = 5. vmin = v2 (1) r (2) (3) (4) 7. vmin = 6. vmin = Using the free-body diagram, we have Fx i N sin N cos = m Fy i N cos + N sin = m g (m g ) = m g sin v2 cos r and, if = 0, we have v2 tan = gr ma = m (5) Solution: If the car is not slipping, then its acceleration is just the centripetal acceleration that keeps it in a circular motion. This acceleration is directed inward and parallel to the horizontal. When there is no frictional force, the horizontal component of the normal force is what generates the centripetal acceleration. Since there is no acceleration in the vertical direction, we will have N cos = mg , where m is mg the mass of the car. Thus, N = . Newcos tons equation in the horizontal direction will v2 be N sin = m , and then R v2 = R N sin = g R tan m g R tan . Explanation: Since we want to calculate the minimum speed in order for the car to not slip, we need to consider the case in which the frictional force is Ff = N and is directed up the incline. Once again, there is no acceleration in the vertical direction and thus N cos +Ff sin = N cos + N sin = m g . Solving for the normal force, we get N= mg . sin + cos In the horizontal direction we have N sin Ff cos = N (sin cos ) v2 = m min , R and then using the expression we obtained for N, g R(sin cos ) 2 vmin = sin + cos vmin = g R (sin cos ) . sin + cos v= 016 (part 2 of 2) 10.0 points What is the minimum speed required in order for the car not to slip? 017 10.0 points A race car starts from rest on a circular track schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 8 of radius 220 m. Its speed increases at the constant rate of 0.6 m/s2 . At the point where the magnitudes of the radial and tangential accelerations are equal, determine the speed of the race car. Correct answer: 11.4891 m/s. Explanation: Basic Concepts: Fr = m ar = m v2 r 019 (part 1 of 2) 10.0 points Compare the gravitational force on a 5.5 kg mass at the surface of the Earth (with radius 6.4 106 m and mass 6 1024 kg) with that on the surface of the Moon with mass 1 ME and radius 0.27 RE . 81.3 What is the force on the Earth? Correct answer: 53.7378 N. Explanation: Let : m = 5.5 kg , ME = 6 1024 kg , RE = 6.4 106 m , and G = 6.67 1011 N m2 /kg2 . G m ME 2 RE (6.67 1011 N m2 /kg2 ) (5.5 kg) (6.4 106 m)2 (6 1024 kg) The tangential acceleration of the car is a, v2 . If we and the centripetal acceleration is r v2 equate these two, a = , we obtain r v = ar = (0.6 m/s2 )(220 m) = 11.4891 m/s . 018 10.0 points A 1220 kg car rounds a circular turn of radius 22 m. The road is at and the coecient of friction between tires and road is 0.52. The acceleration of gravity is 9.8 m/s2 . How fast can the car go without skidding? Correct answer: 10.5883 m/s. Explanation: Given : m = 1220 kg , r = 22 m , and k = 0.52 . Friction must supply the required centripetal force, so fk = Fc v2 k m g = m r v = k r g = (0.52) (22 m) (9.8 m/s2 ) FE = = = 53.7378 N . 020 (part 2 of 2) 10.0 points What is it on the Moon? Correct answer: 9.06696 N. Explanation: Let : MM = 81.3 ME and RM = 0.27 RE . F= G m MM Gm ME = 2 (0.27RE )2 81.3 RM G m ME 1 = 2 (0.27)2(81.3) RE FE = (0.27)2(81.3) 53.7378 N = (0.27)2(81.3) = 9.06696 N . = 10.5883 m/s . schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 9 021 10.0 points of Which the objects a. a book b. the nearest star c. the Sun d. a distant galaxy exert(s) a gravitational force on you? 1. a, b, c and d correct 2. c and d 3. Another combination 4. a, b and c 5. b, c and d 6. b 7. a and d 8. a and b 9. a 10. c Explanation: All objects exert a gravitational force on you, no matter how far away they are. 022 10.0 points An apparatus like the one Cavendish used to nd G has large lead balls that are 5.9 kg in mass and small ones that are 0.075 kg. The center of a large ball is separated by 0.046 m from the center of a small ball. G. As the small spheres of mass m are attracted to the large spheres of mass M , the rod between the two small spheres rotates through a small angle. Find the magnitude of the gravitational force between the masses if the value of the universal gravitational constant is 6.67259 1011 N m2 /kg2 . Correct answer: 1.39538 108 N. Explanation: Let : m1 = 5.9 kg , m2 = 0.075 kg , r = 0.046 m , and G = 6.67259 1011 N m2 /kg2 . By Newtons universal law of gravitation, F =G m1 m2 r2 = (6.67259 1011 N m2 /kg2 ) (5.9 kg)(0.075 kg) (0.046 m)2 = 1.39538 108 N . 023 10.0 points The earth has about 80 times the mass of the earths moon. The gravitational force exerted on the moon by the earth has what relation to the gravitational force exerted on the earth by the moon? 1. the same as correct 2. 1 80 Mirror Light source 3. zero; the earth exerts a force on the moon, but the moon is too small and too far away to exert a force on the earth. 4. 80 r M m The Cavendish apparatus for measuring Explanation: By Newtons third law, the gravitational force exerted on the moon by the earth must schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 10 be the same the gravitational force exerted on the earth by the moon. 024 10.0 points How does the force of gravity between two bodies change when the distance between them doubles? 1. quadruples 2. drops to one quarter of its original value correct 3. doubles 4. Unable to determine; needed. 5. remains the same 6. halves Explanation: The force of gravity between two bodies is inversely proportional to the square of the 1 distance separating them F 2 , so the r force of gravity will be F 1 11 1 = F, 2 2 (2 r ) 4r 4 the mass is Fm,s = G mm mS d2 m,s (7.36 1022 kg) (1.99 1030 kg) (1.49216 1011 m)2 dE,m = 3.84 108 m , mm = 7.36 1022 kg , mS = 1.99 1030 kg , and G = 6.673 1011 N m2 /kg2 . The distance between the moon and the Sun is dm,s = dE,S dE,m = 1.496 1011 m 3.84 108 m = 1.49216 1011 m , so = 6.673 1011 N m2 /kg2 = 4.38956 1020 N . 026 (part 2 of 3) 10.0 points What gravitational force is exerted on the moon by Earth? Correct answer: 1.99176 1020 N. Explanation: Let : mE = 5.98 1024 kg . Fm,E = G mm mE d2 E,m (7.36 1022 kg)(5.98 1024 kg) (3.84 108 m)2 i.e., one fourth of its original value. 025 (part 1 of 3) 10.0 points During a solar eclipse, the moon (of mass 7.36 1022 kg), Earth (of mass 5.98 1024 kg), and Sun (of mass 1.99 1030 kg) lie on the same line, with the moon between Earth and the Sun. What gravitational force is exerted on the moon by the Sun? The universal gravitational constant is 6.673 1011 N m2 /kg2 , the Earth-moon distance is 3.84 108 m, and the Earth-Sun distance is 1.496 1011 m. Correct answer: 4.38956 1020 N. Explanation: = 6.673 1011 N m2 /kg2 = 1.99176 1020 N . 027 (part 3 of 3) 10.0 points What gravitational force is exerted on Earth by the Sun? Correct answer: 3.54823 1022 N. Let : dE,S = 1.496 1011 m , schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 11 Explanation: FE,S mE mS =G 2 dE,S = 6.673 1011 N m2 /kg2 (5.98 1024 kg) (1.99 1030 kg) (1.496 1011 m)2 On the surface of the planet, GM m . R2 When the object is moved to a distance 4 R from the center of the planet, the gravitational force on it will be GM m F= (4 R)2 GM m = 16 R2 1 GM m = 16 R2 1 = W. 16 W= 030 10.0 points Earths gravitational eld is 7.98 N/kg at the altitude of the space shuttle. What is the size of the force of attraction between a student of mass 47.3 kg and Earth? Correct answer: 377.454 N. Explanation: Let : g = 7.98 N/kg m = 47.3 kg . and = 3.54823 1022 N . 028 10.0 points Suppose the gravitational force between the moon and the earth were equal to S. If the moons mass were quadrupled, by what factor of S would the gravitational force between the earth and moon be? Correct answer: 4. Explanation: S m = S m The mass is to be quadrupled, so 4m S = S m S = 4S 029 10.0 points An object has a weight W when it is on the surface of a planet of radius R. What will be the gravitational force on the object after it has been moved to a distance of 4 R from the center of the planet? 1. F = 16 W 2. F = 4 W 3. F = 1 W 4 The force of attraction between the student and the Earth is his weight: W = mg = (47.3 kg) (7.98 N/kg) = 377.454 N . 031 10.0 points The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. If the radius of the Moon is about 0.29 RE , nd the ratio M oon : Earth of their densities. Correct answer: 0.574713. Explanation: Let : gM = rM 1 gE and 6 = 0.29 RE . 4. F = W 1 W correct 16 Explanation: 5. F = schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 12 Density is mass per unit volume, so V= M 4 = R3 3 M 4 = R, R2 3 If the planetoid has mass M and radius R which expression gives the speed of the rock in its surface-level orbit? 1. v = 2. v = 3. v = 4. v = G M RM gM = gE G E RE gM RE M = E gE RM RE 1 = 6 0.29 RE = 0.574713 . 032 10.0 points Near the surface of the Earth, the acceleration due to gravity is 9.8 m/s/s. After falling 6 s from rest, an object would have a velocity of 1. 3.8 m/s. 2. 9.8 m/s. 3. 58.8 m/s. correct 4. 15.8 m/s. Explanation: v = vo + a t = 0 + (9.8 m/s2) (6 s) = 58.8 m/s . 033 10.0 points A small planetoid is almost perfectly spherical and has no trace of atmosphere. A spacesuited astronaut on the surface picks up a rock and launches it into orbit by tossing it horizontally. GM R2 gR GM correct R GM R and the gravitational acceleration is g=G Thus 4 M = GR G . 2 R 3 Explanation: v2 The acceleration a = is supplied by the R gravitational force: v2 mM =G R R2 M v2 = G R GM . v= R m 034 10.0 points An astronaut weighs 131 N on the moons surface. When she is in a circular orbit about the moon at an altitude above the moons surface equal to 0.8 Rmoon , what gravitational force does the moon exert on her? Correct answer: 40.4321 N. Explanation: Let : W = 131 N and h = 0.8 Rmoon . The astronauts distance from the center of the moon is R = 0.8 Rmoon + Rmoon = 1.8 Rmoon . GM m , the gravitational R2 moon force that the moon exerts on her is Since Wmoon = a= GM m GM m = 2 R (1.8 Rmoon )2 schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 13 = GM m Wmoon = 2 R2 (1.8) moon (1.8)2 131 N = 40.4321 N . = (1.8)2 X U Sun 035 10.0 points If Spacecraft X has twice the mass of Spacecraft Y , then true statements about X and Y include which of the following? I) On Earth, X experiences twice the gravitational force that Y experiences. II) On the Moon, X has twice the weight of Y. III) When both are in the same circular orbit, X has twice the centripetal acceleration of Y . 1. I, II, and III 2. III only 3. I only 4. I and II only correct 5. II and III only Explanation: I) gravitational force mass II) weight mass III) The centripetal acceleration is determined by ac = G M , r2 a Z b S P Using Keplers second law of planetary motion, rank those positions in order of their relative speeds, with the position for the fastest speed rst. 1. 2. 3. 4. 5. 6. U S S Z P Z X P U P S X Z Z X X U U P X P U X S S correct U Z S Z P Explanation: The closer the comet is to the Sun, the faster it is traveling, so (U , X , Z , P , S ) is correct. 037 10.0 points Halleys comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.652 AU and its greatest distance being 29.9 AU (1 AU=the Earth-Sun distance). If the comets speed at closest approach is 44.2 km/s, what is its speed when it is farthest from the Sun? You may assume that its angular momentum about the Sun is conserved. Correct answer: 0.963826 km/s. Explanation: Using conservation of angular momentum, we have Lapogee = Lperihelion , or where G is the gravitational constant, M is the planets mass and r is the radius of the spacecrafts orbit. Thus X and Y should have the same centripetal acceleration when they are in the same circular orbit. 036 10.0 points Consider the orbit of a typical comet around the sun, which is marked at ve dierent positions, P , X , S , U , and Z . schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 14 (m ra 2 ) a = (m rp 2 ) p , thus vp va m ra 2 = m rp 2 , giving ra rp ra va = rp vp , or rp va = vp ra (0.652 AU) (44.2 km/s) = (29.9 AU) = 0.963826 km/s . 038 10.0 points Two planets are never seen at midnight. Which two? 1. Neptune and Mercury 2. Saturn and Jupiter 3. Venus and Mercury correct 4. Jupiter and Mars 5. Neptune and Pluto Explanation: At midnight you face away from the sun, and cannot see the planets (Mercury and Venus) which lie inside the Earths orbit. 039 10.0 points If you stopped an Earth satellite dead in its tracks, it would simply crash into the Earth. Why, then, dont the communications satellites that hover motionless above the same spot on Earth crash into the Earth? 1. The satellites are not attracted by the Earth. 2. The satellites orbital period coincides with the daily rotation of the Earth. correct 3. There is no power on the satellites. 4. The moon attracts the satellites at the same time. Explanation: If the orbital period and direction of motion Figure: Eliptical orbit of planet. How far from the Sun will the comet travel before it starts its return journey? Correct answer: 24.2521 AU. Explanation: Let : = 11 AU , T = 74 years , KS = 1 year2 /AU3 and 2a = + x. By Keplers third law in the case of negligible mass of the comet T 2 = K S a3 , where a is the semi-major axis of the comets orbit. For any object orbiting the Sun, with the period T in years and the semi-major axis a in AU, where 2 a = + x (from the gure above). 3 +x 2 T = KS , 2 vA A 11 AU Sun vB x B of an Earth satellite coincide with the daily rotation of the Earth, it is motionless relative to a point on the Earths surface. Thus the communications satelites actually move at the same pace as the Earth. 040 10.0 points A comet approaches the Sun to within 11 AU, and its orbital period is 74 years. Note: AU is the abbreviation for astronomical unit, where 1 AU = 1.50 108 km is the mean Earth-Sun distance. Thus, KS 1 year2 /AU3 . schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 15 so the farthest distance x (in AU) is x=2 =2 3 T2 KS Correct answer: 6.22857 1030 kg. Explanation: From Keplers third law, we have 4 2 r3 G T2 (74 years)2 (11 AU) (1 year2 /AU3 ) = 24.2521 AU . 3 M= = 041 10.0 points The distance of Mars to the sun is 1.5 that of Earth. How many Earth years does it take for Mars to orbit the sun? 1. 2.5 yr 2. 1 yr 3. 1.5 yr 4. 2 yr 5. 1.8 yr correct Explanation: Let : dM = 1.5 dE TE = 1 yr . and 4 2 (6.67259 1011 N m2 /kg2 ) (2.83 1011 m)3 (4.64 107 s)2 = 6.22857 1030 kg . 043 (part 1 of 3) 10.0 points Assume: For this problem, which consists of 3 parts, we assume that we are on Planet-I. The satellites considered below are moving along circular orbits which are concentric to the center of Planet-I. The gravitational constant G = 6.67259 1011 N m2 /kg2 . The radius of Planet-I is R =2440 km, the gravitational acceleration at the surface is gI = 7.98 m/s2 . #3 r3 #1 r1 R r2 #2 From Keplers third law, the square of the orbital period is proportional to the cube of the semimajor axis, so TM = = d3 M TE d3 E (1.5 dE )3 yr d3 E Satellite #1, r1 R Satellite #2, r2 = nR Satellite #3, r3 = 10R (The sketch is not to scale) = 1.83712 yr . 042 10.0 points Given: G = 6.67259 1011 N m2 /kg2 A planet is orbiting a star. Calculate the mass of the star using the fact that the period of the planet is 4.64 107 s and its distance from the star is 2.83 1011 m. Consider the rst satellite which is moving very close to the surface of Planet-I (i.e., r1 R), where the gravitational acceleration may be approximated by gI = 7.98 m/s2 . Determine its period (which in the remainder of this problem will be referred to as T0 ). Correct answer: 0.965097 hr. Explanation: schroeder (cs39963) Resource 6: Circular Motion and Gravitation balasubramanya (1401310) 16 The gravitational acceleration is the centripetal acceleration of the satellite; i.e., v2 gI = , or v = gI R. So the period R 2R T0 = v = 2 R gI (2440 km)(1000 m/km) 7.98 m/s2 Note: You may work on this part even if you dont get previous results. What is the kinetic energy of a third satellite which has a radius r3 = 10 R = 24400 km, assuming that the satellite has a mass of m =5530 kg? Correct answer: 5.38379 109 J. Explanation: From Part 2, we have m v 2 = G the surface of Planet-I, m gI = G M or 044 (part 2 of 3) 10.0 points Note: Even if you did not get the value of T0 in part 1, you may still be able to work on this part. Consider a second satellite whose orbit has a radius r2 = n R = 5.36 R = 13078.4 km and a period T. T Find the ratio . T0 Correct answer: 12.4093. Explanation: Equating the gravitational attraction force and the centripetal force, we have m v2 m 2r = r r T GM m , = r2 or T2 = 4 2 GM r3 . 2 = 2 (0.000277778 h/s) Mm . At r = 0.965097 hr . m , R2 G M = gI R 2 . So at r3 = 10 R, K= 1 m v2 2 GM m = 2 r3 m gI R 2 = 20 R 1 (5530 kg)(7.98 m/s2 ) = 20 (2440 km)(1000 m/km) = 5.38379 109 J . This implies T 2 r 3 , which is Keplers Third Law. So the ratio T r = T0 R 3 2 3 2 13078.4 km = 2440 km = 12.4093 . 045 (part 3 of 3) 10.0 points ... View Full Document