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4/24/09 3:24 AMMasteringPhysics Page 1 of 12http://session.masteringphysics.com/myct [ Assignment View ] Course PHY121(TTH) HW #12 Due at 11:59pm on Sunday, April 26, 2009 View Grading Details Problem 11.74 If you put a uniform block at the edge of a table, the center of the block must be over the table for the block not to fall off. Part A If you stack two identical blocks at the table edge, the center of the top block must be over the bottom block, and the center of gravity of the two blocks together must be over the table. In terms of the length of each block, what is the maximum overhang possible? ANSWER: Part B Repeat part (A) for three identical blocks. ANSWER: Part C Repeat part (A) for four identical blocks. ANSWER: Part D Is it possible to make a stack of blocks such that the uppermost block is not directly over the table at all? ANSWER: yes no Part E How many blocks would it take to do this? (Try this with your friends using copies of this book.) ANSWER: 4 Problem 11.83: Pyramid Builders Ancient pyramid builders are balancing a uniform rectangular slab of stone tipped at an angle above the horizontal using a rope (the figure ). The rope is held by five workers who share the force equally. 4/24/09 3:24 AMMasteringPhysics Page 2 of 12http://session.masteringphysics.com/myct Part A If 21.6, what force does each worker exert on the rope? Express your answer in terms of (the weight of the slab). ANSWER: = 8.13×10 ?3 Part B As increases, does each worker have to exert more or less force than in part A, assuming they do not change the angle of the rope? Why? ANSWER: Answer not displayed Part C At what angle do the workers need to exert no force to balance the slab? ANSWER: = 25.0 Part D What happens if exceeds this value? ANSWER: Answer not displayed Problem 11.91 You hang a floodlamp from the end of a vertical steel wire. The floodlamp stretches the wire a distance of 0.20 and the stress is proportional to the strain. Part A How much would it have stretched if the wire were twice as long? Express your answer using two significant figures. ANSWER: = 0.40 Part B How much would it have stretched if the wire had the same length but twice the diameter? Express your answer using two significant figures. ANSWER: = 5.0×10 ?2 Part C How much would it have stretched for a copper wire of the original length and diameter? Express your answer using two significant figures. ANSWER: = 0.36 A Matter of Some Gravity Learning Goal: To understand Newton's law of gravitation and the distinction between inertial and gravitational masses. In this problem, you will practice using Newton's law of gravitation. According to that law, the magnitude of the gravitational force between two small particles of masses and , separated by a distance , is given by , 4/24/09 3:24 AMMasteringPhysics Page 3 of 12http://session.masteringphysics.com/myct where is the universal gravitational constant, whose numerical value (in SI units) is . This formula applies not only to small particles, but also to spherical objects. In fact, the gravitational force between two uniform spheres is the same as if we concentrated all the mass of each sphere at its center. Thus, by modeling the Earth and the Moon as uniform spheres, you can use the particle approximation when calculating the force of gravity between them. Be careful in using Newton's law to choose the correct value for . To calculate the force of gravitational attraction between two uniform spheres, the distance in the equation for Newton's law of gravitation is the distance between the centers of the spheres. For instance, if a small object such as an elephant is located on the surface of the Earth, the radius of the Earth would be used in the equation. Note that the force of gravity acting on an object located near the surface of a planet is often called weight. Also note that in situations involving satellites, you are often given the altitude of the satellite, that is, the distance from the satellite to the surface of the planet; this is not the distance to be used in the formula for the law of gravitation. There is a potentially confusing issue involving mass. Mass is defined as a measure of an object's inertia, that is, its ability to resist acceleration. Newton's second law demonstrates the relationship between mass, acceleration, and the net force acting on an object: . We can now refer to this measure of inertia more precisely as the inertial mass. On the other hand, the masses of the particles that appear in the expression for the law of gravity seem to have nothing to do with inertia: Rather, they serve as a measure of the strength of gravitational interactions. It would be reasonable to call such a property gravitational mass. Does this mean that every object has two different masses? Generally speaking, yes. However, the good news is that according to the latest, highly precise, measurements, the inertial and the gravitational mass of an object are, in fact, equal to each other; it is an established consensus among physicists that there is only one mass after all, which is a measure of both the object's inertia and its ability to engage in gravitational interactions. Note that this consensus, like everything else in science, is open to possible amendments in the future. In this problem, you will answer several questions that require the use of Newton's law of gravitation. Part A Two particles are separated by a certain distance. The force of gravitational interaction between them is . Now the separation between the particles is tripled. Find the new force of gravitational interaction . Express your answer in terms of . ANSWER: = Note that the gravitational force between two objects is inversely proportional to the square of the distance between them. If the distance is tripled, the force of gravitational attraction is nine times weaker. Part B A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite moves to a different orbit, so that its altitude is tripled. Find the new force of gravitational interaction . Hint B.1Altitude versus distance Hint not displayed Express your answer in terms of . ANSWER: = Part C A satellite revolves around a planet at an altitude equal to the radius of the planet. The force of gravitational interaction between the satellite and the planet is . Then the satellite is brought back to the surface of the planet. Find the new force of gravitational interaction . Express your answer in terms of . ANSWER: = Part D Two satellites revolve around the Earth. Satellite A has mass and has an orbit of radius . Satellite B has mass and an orbit of unknown radius . The forces of gravitational attraction between each satellite and the Earth is the same. Find . Express your answer in terms of . ANSWER: = 4/24/09 3:24 AMMasteringPhysics Page 4 of 12http://session.masteringphysics.com/myct Part E An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far from the center of the Earth should an elephant be placed so that its weight equals that of the elephant shrew on the surface of the Earth? The radius of the Earth is 6400 . ( .) Express your answer in kilometers. ANSWER: = 2.00×10 6 The table below gives the masses of the Earth, the Moon, and the Sun. NameMass (kg) Earth Moon Sun The average distance between the Earth and the Moon is . The average distance between the Earth and the Sun is . Use this information to answer the following questions. Part F Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the new moon (when the moon is located directly between the Earth and the Sun). Hint F.1 How to apprach the problem Hint not displayed Hint F.2 Find the force of attraction between the Earth and the Moon Hint not displayed Hint F.3 Find the force of attraction between the Earth and the Sun Hint not displayed Express your answer in newtons to three significant figures. ANSWER: = 3.54×10 22 Part G Find the net gravitational force acting on the Earth in the Sun-Earth-Moon system during the full moon (when the Earth is located directly between the moon and the sun). Express your answer in newtons to three significant figures. ANSWER: = 3.50×10 22 Escape Velocity Learning Goal: To introduce you to the concept of escape velocity for a rocket. The escape velocity is defined to be the minimum speed with which an object of mass must move to escape from the gravitational attraction of a much larger body, such as a planet of total mass . The escape velocity is a function of the distance of the object from the center of the planet , but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?" Part A The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances. Hint A.1Consider various cases Hint not displayed ANSWER: = 0 Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether 4/24/09 3:24 AMMasteringPhysics Page 5 of 12http://session.masteringphysics.com/myct the following statements are true or false. Part B Angular momentum about the center of the planet is conserved. ANSWER: true false Part C Total mechanical energy is conserved. ANSWER: true false Part D Kinetic energy is conserved. ANSWER: true false Part E The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the object moves from small to large (like a rocket being launched) or from large to small (like a comet approaching the earth). ANSWER: true false What if the object is not moving directly away from or toward the planet but instead is moving at an angle from the normal? In this case, it will have a tangential velocity and angular momentum . Since angular momentum is conserved, for any , so will go to 0 as goes to infinity. This means that angular momentum can be conserved without adding any kinetic energy at . The important aspect for determining the escape velocity will therefore be the conservation of total mechanical energy. Part F Find the escape velocity for an object of mass that is initially at a distance from the center of a planet of mass . Assume that , the radius of the planet, and ignore air resistance. Hint F.1 Determine the gravitational potential energy Hint not displayed Hint F.2 Determine the kinetic energy Hint not displayed Hint F.3 Putting it all together Hint not displayed Express the escape velocity in terms of , , , and , the universal gravitational constant. ANSWER: = Does it surprise you that the escape velocity does not depend on the mass of the object? Even more surprising is that it does not depend on the direction (as long as the trajectory misses the planet). Any angular momentum given at radius can be conserved with a tangential velocity that vanishes as goes to infinity, so the angle at which the object is launched does not have a significant effect on the energy at large . Properties of Circular Orbits Learning Goal: To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth. The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass . 4/24/09 3:24 AMMasteringPhysics Page 6 of 12http://session.masteringphysics.com/myct For all parts of this problem, where appropriate, use for the universal gravitational constant. Part A Find the orbital speed for a satellite in a circular orbit of radius . Hint A.1Find the force Hint not displayed Hint A.2A basic kinematic relation Hint not displayed Hint A.3Newton's 2nd law Hint not displayed Express the orbital speed in terms of , , and . ANSWER: = Part B Find the kinetic energy of a satellite with mass in a circular orbit with radius . Express your answer in terms of , , , and . ANSWER: = Part C Express the kinetic energy in terms of the potential energy . Hint C.1Potential energy Hint not displayed ANSWER: = This is an example of a powerful theorem, known as the Virial Theorem. For any system whose motion is periodic or remains forever bounded, and whose potential behaves as , Rudolf Clausius proved that , where the brackets denote the temporal (time) average. Part D Find the orbital period . Hint D.1How to get started Hint not displayed Express your answer in terms of , , , and . ANSWER: = Part E Find an expression for the square of the orbital period. Express your answer in terms of , , , and . ANSWER: = This shows that the square of the period is proportional to the cube of the semi-major axis. This is Kepler's Third Law, in the case of a circular orbit where the semi-major axis is equal to the radius, . 4/24/09 3:24 AMMasteringPhysics Page 7 of 12http://session.masteringphysics.com/myct Part F Find , the magnitude of the angular momentum of the satellite with respect to the center of the planet. Hint F.1 Definition of angular momentum Hint not displayed Express your answer in terms of , , , and . ANSWER: = Part G The quantities , , , and all represent physical quantities characterizing the orbit that depend on radius . Indicate the exponent (power) of the radial dependence of the absolute value of each. Hint G.1Example of a power law Hint not displayed Express your answer as a comma-separated list of exponents corresponding to , , , and , in that order. For example, -1,-1/2,-0.5,-3/2 would mean , , and so forth. ANSWER: -0.500,-1,-1,0.500 Understanding Mass and Weight Learning Goal: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton's law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word "weight" often replaces "mass," as in "My weight is seventy-five kilograms" or "I need to lose some weight." Of course, mass and weight are related; however, they are also very different. Mass, as you recall, is a measure of an object's inertia (ability to resist acceleration). Newton's 2nd law demonstrates the relationship among an object's mass, its acceleration, and the net force acting on it: . Mass is an intrinsic property of an object and is independent of the object's location. Weight, in contrast, is defined as the force due to gravity acting on the object. That force depends on the strength of the gravitational field of the planet: , where is the weight of an object, is the mass of that object, and is the local acceleration due to gravity (in other words, the strength of the gravitational field at the location of the object). Weight, unlike mass, is not an intrinsic property of the object; it is determined by both the object and its location. Part A Which of the following quantities represent mass? Check all that apply. ANSWER: 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN Part B Which of the following quantities would be acceptable representations of weight? Check all that apply. ANSWER: 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN Weight is a force and is measured in newtons (or kilonewtons, meganewtons, etc.) or in pounds (or tons, megatons, etc.). Using the universal law of gravity, we can find the weight of an object feeling the gravitational pull of a nearby planet. We 4/24/09 3:24 AMMasteringPhysics Page 8 of 12http://session.masteringphysics.com/myct can write an expression , where is the weight of the object, is the gravitational constant, is the mass of that object, is mass of the planet, and is the distance from the center of the planet to the object. If the object is on the surface of the planet, is simply the radius of the planet. Part C The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported from the moon to the earth, which properties of the rock change? ANSWER: mass only weight only both mass and weight neither mass nor weight Part D An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet. As a result, which of the following changes in the properties of the object take place? ANSWER: mass increases; weight decreases mass decreases; weight decreases mass increases; weight increases mass increases; weight remains the same mass remains the same; weight decreases mass remains the same; weight increases mass remains the same; weight remains the same Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system: Name Mass ( ) Radius ( ) Tehar 2.1 0.80 Loput 5.6 1.7 Cremury0.36 0.30 Suven 12 2.8 Pentune8.3 4.1 Rams 9.3 4.0 In this table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii. Part E If acceleration due to gravity on the earth is , which formula gives the acceleration due to gravity on Loput? Hint E.1What equations to use Hint not displayed ANSWER: Part F If the acceleration due to gravity on the earth is 9.8 , what is the acceleration due to gravity on Rams? Express your answer in meters per second squared and use two significant figures. ANSWER: 5.7 4/24/09 3:24 AMMasteringPhysics Page 9 of 12http://session.masteringphysics.com/myct Part G Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer. Hint G.1Determine the percentage difference in weight To make the scale read 118 lb, the 236-lb Punch has to travel to a planet where the gravitational field is what percentage of that on the earth? ANSWER: 25% 50% 200% 400% ANSWER: Tehar Loput Cremury Suven Pentune Rams Part H As Punch Taut travels to Pentune, what actually happens to his mass and his weight? ANSWER: mass increases; weight decreases mass decreases; weight decreases mass increases; weight increases mass increases; weight remains the same mass remains the same; weight decreases mass remains the same; weight increases mass remains the same; weight remains the same Answer not displayed Understanding Newton's Law of Universal Gravitation Learning Goal: To understand Newton's law of universal gravitation and be able to apply it in two-object situations and (collinear) three-object situations; to distinguish between the use of and . In the late 1600s, Isaac Newton proposed a rule to quantify the attractive force known as gravity between objects that have mass, such as those shown in the figure. Newton's law of universal gravitation describes the magnitude of the attractive gravitational force between two objects with masses and as , where is the distance between the centers of the two objects and is the gravitational constant. The gravitational force is attractive, so in the figure it pulls to the right on (toward ) and toward the left on (toward ). The gravitational force acting on is equal in size to, but exactly opposite in direction from, the gravitational force acting on , as required by Newton's third law. The magnitude of both forces is calculated with the equation given above. The gravitational constant has the value and should not be confused with the magnitude of the gravitational free-fall acceleration constant, denoted by , which equals 9.80 near the surface of the earth. The size of in SI units is tiny. This means that gravitational forces are sizeable only in the vicinity of very massive objects, such as the earth. You are in fact gravitationally attracted toward all the objects around you, such as the computer you are using, but the size of that force is too small to be noticed without extremely sensitive equipment. Consider the earth following its nearly circular orbit (dashed curve) about the sun. The earth has mass and the sun has mass . They are separated, center to center, by . 4/24/09 3:24 AMMasteringPhysics Page 10 of 12http://session.masteringphysics.com/myct Part A What is the size of the gravitational force acting on the earth due to the sun? Hint A.1What units to select Hint not displayed Express your answer in newtons. ANSWER: 3.53×10 22 This force causes the earth to orbit the sun. Part B At the moment shown in the figure of the earth and sun , what is the direction of the gravitational force acting on the earth? The possible directions are displayed in this figure . ANSWER: A As the earth proceeds around its orbit, the direction of the gravitational force acting on it changes so that the force always points directly toward the sun. Part C What is the size of the gravitational force acting on the sun due to the earth? Hint C.1Newton's third law Hint not displayed ANSWER: The earth does not exert any gravitational force on the sun. The earth exerts some force on the sun, but less than because the earth, which is exerting the force, is so much less massive than the sun. The earth exerts of force on the sun, exactly the same amount of force the sun exerts on the earth found in Part A. The earth exerts more than of force on the sun because the sun, which is experiencing the force, is so much more massive than the earth. Also note that the force exerted on the sun by the earth is directed from the sun toward the earth, downward at the moment shown in the figure. This is exactly the opposite direction of the force exerted on the earth by the sun found in Part B, in accordance with Newton's third law. Recall that Newton's second law states that . The sun does not accelerate as much as the earth due to this gravitational force since the sun is so much more massive. This force causes the earth to follow a nearly circular path as it orbits the sun, but the same amount of force only causes the sun to wobble back and forth very slightly. Part D Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the value found in Part A? Check all that apply. ANSWER: Reduce the mass of the earth to one-fourth its normal value. 4/24/09 3:24 AMMasteringPhysics Page 11 of 12http://session.masteringphysics.com/myct Reduce the mass of the sun to one-fourth its normal value. Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value. Increase the separation between the earth and the sun to four times its normal value. Answer not displayed Part E Part not displayed Transition will be visible after you complete previous item(s). Part F Part not displayed A Satellite in a Circular Orbit Consider a satellite of mass that orbits a planet of mass in a circle a distance from the center of the planet. The satellite's mass is negligible compared with that of the planet. Indicate whether each of the statements in this problem is true or false. Part A The information given is sufficient to uniquely specify the speed, potential energy, and angular momentum of the satellite. Hint A.1What constitutes sufficient initial conditions? Hint not displayed ANSWER: true false Part B The total mechanical energy of the satellite is conserved. Hint B.1When is mechanical energy conserved? Hint not displayed ANSWER: true false Part C The linear momentum vector of the satellite is conserved. Hint C.1When is linear momentum conserved? Hint not displayed ANSWER: true false Part D The angular momentum of the satellite about the center of the planet is conserved. Hint D.1When is angular momentum conserved? Hint not displayed ANSWER: true false Part E The equations that express the conservation laws of total mechanical energy and linear momentum are sufficient to solve for the speed necessary to maintain a circular orbit at without using . Hint E.1How are conservation laws used? Hint not displayed ANSWER: true false 4/24/09 3:24 AMMasteringPhysics Page 12 of 12http://session.masteringphysics.com/myct Energy of a Spacecraft Very far from earth (at ), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of the earth were to act on it (i.e., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into the earth. The mass of the earth is and its radius is . Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space. Part A Find the speed of the spacecraft when it crashes into the earth. Hint A.1How to approach the problem Hint not displayed Hint A.2Total energy Hint not displayed Hint A.3Potential energy Hint not displayed Express the speed in terms of , , and the universal gravitational constant . ANSWER: = Part B Now find the spacecraft's speed when its distance from the center of the earth is , where . Hint B.1General approach Hint not displayed Hint B.2First step in finding the speed Hint not displayed Express the speed in terms of and . ANSWER: = Summary8 of 10 items complete (79.53% avg. score) 8.97 of 11 points clockwork MasteringPhysics

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Author: Anonymous

Textbook: University Physics (12th Edition)

University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2009-04-24

Updated: 2009-04-24

File Size: 12 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 6179

Textbook: University Physics (12th Edition)

University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2009-04-24

Updated: 2009-04-24

File Size: 12 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 6179

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