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`MasteringPhysics: Assignment Print ViewA Matter of Some GravityLearning 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 , 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 would be such as an elephant is located on the surface of the Earth, the radius of the Earth 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: 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 . We can now refer toMasteringPhysics: Assignment Print Viewgood 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 Part B.1 Altitude versus distance Part not displayed Express your answer in terms of ANSWER: = Part C . .MasteringPhysics: Assignment Print ViewA 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: = 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: = 2.00×106 The table below gives the masses of the Earth, the Moon, and the Sun. Name Mass (kg) Earth Moon SunThe average distance between the Earth and the Moon is. The average distanceMasteringPhysics: Assignment Print Viewbetween the Earth and the Sun is questions. Part F. Use this information to answer the followingFind 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 Hint not displayed Part F.2 Part not displayed Part F.3 Part not displayed Express your answer in newtons to three significant figures. ANSWER: = 3.54×1022 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: = 3.50×1022Understanding Mass and WeightMasteringPhysics: Assignment Print ViewLearning 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 &quot;weight&quot; often replaces &quot;mass,&quot; as in &quot;My weight is seventy-five kilograms&quot; or &quot;I need to lose some weight.&quot; 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: 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? A. B. C. D. E. F. G. 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MN . Mass is an intrinsic property of an object and is independent of the object'sEnter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think that C and D correspond to the correct answers, enter CD. ANSWER: BC Part BMasteringPhysics: Assignment Print ViewWhich of the following quantities would be acceptable representations of weight? A. B. C. D. E. F. G. 12.0 lbs 0.34 g 120 kg 1600 kN 0.34 m 411 cm 899 MNEnter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if you think that C and D correspond to the correct answers, enter CD. ANSWER: ADG 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 , where is the weight of the of a nearby planet. We can write an expression is the mass of that object, is mass of the planet, and object, is the gravitational constant, 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 weightPart DMasteringPhysics: Assignment Print ViewAn 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 samePunch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and &quot;weighs in&quot; 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: Mass ( 2.1 5.6 0.36 12 8.3 9.3 Radius ) ( ) 0.80 1.7 0.30 2.8 4.1 4.0Name Tehar Loput Cremury Suven Pentune RamsIn 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MasteringPhysics: Assignment Print ViewIf acceleration due to gravity on the earth is , which formula gives the acceleration due to gravity on Loput? Hint E.1 What equations to use Hint not displayed ANSWER:Part F If the acceleration due to gravity on the earth is 9.8 on Rams? Express your answer in meters per second squared and use two significant figures. ANSWER: 5.70 Part G , what is the acceleration due to gravityMasteringPhysics: Assignment Print ViewWhich 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. Part G.1 Part not displayed ANSWER: Tehar Loput Cremury Suven Pentune RamsPart 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 sameOf course, the &quot;weight classes&quot; in boxing are really &quot;mass classes&quot;: It is the relative mass of the boxers that matters. The masses and the weights of the athletes are directly proportional--as long as everyone is on the same planet!Escape VelocityMasteringPhysics: Assignment Print ViewLearning 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 &quot;How fast does my rocket have to go to escape from the surface of the planet?&quot; Part A The key to making a concise mathematical definition of escape velocity is to consider the energy. of the If an object is launched at its escape velocity, what is the total mechanical energy 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.1 Consider various cases Hint not displayed ANSWER: = 0Consider the motion of an object between a point close to the planet and a point very very far from the planet. Indicate whether the following statements are true or false. Part B Angular momentum about the center of the planet is conserved. ANSWER: Part C Total mechanical energy is conserved. ANSWER: Part D Kinetic energy is conserved. ANSWER: Part E true false true false true falseMasteringPhysics: Assignment Print ViewThe 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 ANSWER: true to small false (like a comet approaching the earth).What if the object is not moving directly away from or toward the planet but instead is moving at and an angle from the normal? In this case, it will have a tangential velocity 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. Part F.1 Part not displayed Part F.2 Part not displayed Hint F.3 Hint not displayed Express the escape velocity in terms of 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 . , , , and , the universal gravitational constant.Properties of Circular OrbitsMasteringPhysics: Assignment Print ViewLearning 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 . For all parts of this problem, where appropriate, use Part A Find the orbital speed Part A.1 Part not displayed Part A.2 Part not displayed Hint A.3 Hint not displayed Express the orbital speed in terms of ANSWER: = Part B Find the kinetic energy of a satellite with mass , , , and . in a circular orbit with radius . , , and . for a satellite in a circular orbit of radius . for the universal gravitational constant.Express your answer in terms of ANSWER: = Part CMasteringPhysics: Assignment Print ViewExpress the kinetic energy Part C.1 Potential energyin terms of the potential energy.Part 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.1 Hint not displayed Express your answer in terms of ANSWER: = , , , and . .Part EMasteringPhysics: Assignment Print ViewFind an expression for the square of the orbital period. Express your answer in terms of 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, . Part F Find , the magnitude of the angular momentum of the satellite with respect to the center of the planet. Hint F.1 Hint not displayed Express your answer in terms of ANSWER: = Part G The quantities , depend on radius of each. Hint G.1 , , and all represent physical quantities characterizing the orbit that . Indicate the exponent (power) of the radial dependence of the absolute value , , , and . , , , and .Hint not displayed Express your answer as a comma-separated list of exponents corresponding to , that order. For example, -1,-1/2,-0.5,-3/2 would mean ANSWER: -0.500,-1.00, -1.00,0.500 , , , and , in, and so forth.Geosynchronous SatelliteMasteringPhysics: Assignment Print ViewA satellite that goes around the earth once every 24 hours is called a geosynchronous satellite. If a geosynchronous satellite is in an equatorial orbit, its position appears stationary with respect to a ground station, and it is known as a geostationary satellite.Part A Find the radius of the orbit of a geosynchronous satellite that circles the earth. (Note that is measured from the center of the earth, not the surface.) You may use the following constants:qThe universal gravitational constant The mass of the earth is The radius of the earth isis . ..qqPart A.1 Part not displayed Part A.2 Part not displayed Part A.3 Part not displayed Give the orbital radius in meters to three significant digits. ANSWER: = mSatellites and Kepler's LawsMasteringPhysics: Assignment Print ViewThis applet shows satellites orbiting a central body from various starting distances. The mass of the central body is much greater than the masses of the satellites. You should think of this applet as data obtained in an experiment (e.g., a set of videos made by compiling data of observations of satellites). Part A of the purple (innermost) satellite when it has an initial speed of Determine the orbital period 2.0. (For this problem, the units are chosen for convenience and may be ignored for the first few parts of the problem.) Note that simply watching one orbit and seeing how long it takes will not give you three significant figures. Try to think of a technique that will. Hint A.1 How to approach the problem Hint not displayed Express your answer to three significant figures. ANSWER: = 3.160 Part B Determine the period Hint B.1 Hint not displayed Express your answer to three significant figures. ANSWER: = 10.70 Part C Find the length of the major axis the length of the major axis of the orbit of the purple satellite with initial speed 2.0 and for the major of the orbit of the red satellite with initial speed 1.5. Recall that of revolution for the red satellite if its initial speed is 1.5.the semi-major axis of an ellipse is usually denoted ; hence we use the notation axis of the purple satellite's orbit. Hint C.1 Hint not displayedGive the length of the major axis for the purple satellite followed by the length for that of the red satellite separated by a comma. Express your answers to two significant figures. ANSWER: = 2.00,4.60 Part DMasteringPhysics: Assignment Print ViewNow, use the periods (obtained in Parts A and B) and Kepler's laws to determine the ratio the semi-major axes. Hint D.1 Hint not displayed Express your answer to three significant figures. Note that if you use your results from part C, your answer will have two significant figures instead of three. ANSWER: = 2.250 , you will get 2.3, Notice that if you take your answer from Part C and calculate the ratio which is the same as your answer here to two significant figures. Now, assume that the speed is measured in kilometers per second and the distance is measured in hundreds of thousands of kilometers. (e.g., A distance measured as 5 would be .) Part E Which of the following techniques could be used to determine the mass using only information available in the applet? of the central object,ofA. Determine the initial potential energy of one of the satellites and use it to find the central mass. B. Determine the escape speed of one of the satellites from its initial position, and use it to find the central mass. C. Determine the speed for a circular orbit of one of the satellites and use it to find the central mass. Enter the letters of all of the options that you think are correct in alphabetical order, without spaces or commas between them. ANSWER: BC Part FMasteringPhysics: Assignment Print ViewUse one of these techniques to determine the mass of the central object. Be sure to think about which satellite will give the most accurate information for the method you choose. Experiment with the applet to help you make this choice. Part F.1 Part not displayed Hint F.2 Hint not displayed Part F.3 Part not displayed Hint F.4 Hint not displayed Express your answer in kilograms to two significant figures and remember that speed is measured in kilometers per second and distance is measured in hundreds of thousands of kilometers. ANSWER: = 5.87×1024 Part G Using only information available in the applet to determine the initial potential energy of the red satellite, can you find the red satellite's mass? ANSWER: Part H Using only information available in the applet to determine the escape speed of the red satellite from its initial position, can you find the red satellite's mass? ANSWER: Part I Using only information available in the applet to determine the speed for a circular orbit of the red satellite, can you find the red satellite's mass? ANSWER: yes no yes no yes noNone of the techniques will work and hence the mass of an orbiting satellite cannot be found using information in the applet. Interestingly, the mass of the satellite cancels out of the expressions for the speed in a circular orbit and the escape speed. The potential energy could be used to determine the satellite's mass, but the potential energy cannot be determined directly from the applet.MasteringPhysics: Assignment Print ViewWeight on a Neutron StarNeutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. Part A If you weigh 670 on the earth, what would you weigh on the surface of a neutron star that has the same mass as our sun and a diameter of 22.0 ? Take the mass of the sun to be-11= 1.99×1030, the gravitational constant to be= 6.67×10 ., and the acceleration due to gravity at the earth's surface to be= 9.810Hint A.1 How to approach the problem Hint not displayed Part A.2 Part not displayed Part A.3 Part not displayed Express your answer in newtons. ANSWER: = 7.49×1013 This is over times your weight on earth!`

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