Homework 7-1

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 1/21 Homework 7-1 Due: 11:57pm on Tuesday, February 27, 2024 To understand how points are awarded, read the Grading Policy for this assignment. ± 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 magnitude 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 magnitude 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 magnitude 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 .

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 2/21 Part A What is the magnitude of the gravitational force acting on the earth due to the sun? Express your answer in newtons. Hint 1. What units to select For the force to come out in newtons, all masses and distances used must be expressed in SI units. Note that neither ( ) nor ( ) is in the SI unit of length, which is meters. ANSWER: Correct 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 . 3.53×10 22

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 3/21 ANSWER: Correct 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 magnitude of the gravitational force acting on the sun due to the earth? Hint 1. Newton's third law Newton's third law states that if object A exerts a force on object B, then object B must exert a force of the same type on object A. These two forces are equal in magnitude and exactly opposite in direction. Since they act on different objects, these two forces never cancel one another out. ANSWER: A B C D E

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 4/21 Correct 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: Correct The actions in choices A, B, and C would reduce the force to one-fourth its original value by modifying the product of the masses. Increasing the separation distance by a factor of 4 would reduce the force by a factor of . To reduce the force by only a factor of 4, increase the separation distance by a factor of 2 instead because . Part E With the sun and the earth back in their regular positions, consider a space probe with mass launched from the earth toward the sun. When the probe is exactly halfway between the earth and the sun along the line connecting them, what is the direction of the net gravitational force acting on the probe? Ignore the effects of other massive objects in the solar system, such as the moon and other planets. 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. Reduce the mass of the earth to one-fourth its normal value. 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.

3/1/24, 11:42 PM Homework 7-1 5/21 Hint 1. How to approach the problem The sun and the earth each exert a gravitational force on the probe. Use Newton's law of universal gravitation to find the magnitude and direction of each of these forces. Then sum them (as vectors!) to find the net force acting on the probe. Hint 2. Direction of the gravitational force due to the earth What is the direction of the gravitational force acting on the probe due to the earth (refer to the figure)? ANSWER: Hint 3. Direction of the gravitational force due to the sun https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 A B C D E

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 6/21 What is the direction of the gravitational force acting on the probe due to the sun (refer to the figure)? ANSWER: Hint 4. Comparing the gravitational forces acting on the probe Which exerts the stronger gravitational force on the probe? ANSWER: ANSWER: A B C D E The sun exerts a stronger force. The earth exerts a stronger force the forces exerted on the probe by the sun and by the earth are the same magnitude The force is toward the sun. The force is toward the earth. There is no net force because neither the sun nor the earth attracts the probe gravitationally at the midpoint. There is no net force because the gravitational attractions on the probe due to the sun and the earth are equal in magnitude but point in opposite directions, so they cancel each other out.

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 7/21 Correct In calculating these forces, the mass of the probe is the same, as is the distance involved. However, the sun is over 300,000 times as massive as the earth, so at this location the probe is attracted much more strongly toward the sun than toward the earth. Perhaps the most-common gravitational calculation is to determine the "weight" of an object of mass sitting on the surface of the earth (i.e., the magnitude of the gravitational attraction between the earth and the object). According to Newton's law of universal gravitation, the gravitational force on the object is , where is the mass of the earth and is the distance between the object and the center of the earth (i.e., = radius of the earth = ). Part F What is the value of the composite constant , to be multiplied by the mass of the object in the equation above? Express your answer numerically in meters per second per second. ANSWER: Correct This particular combination of constants occurs so often that it is given its own name, , and called the "gravitational free-fall acceleration near the surface of the earth." Using leads to the more familiar formula for calculating the gravitational weight of an object of mass , namely . This equation is limited to calculating the weight of objects near the surface of the earth. When calculating gravitational forces between objects under other conditions, such as in the earth/probe calculation done earlier, use Newton's law of universal gravitation instead. 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. Express the speed in terms of , , and the universal gravitational constant . 9.80

3/1/24, 11:42 PM Homework 7-1 8/21 Hint 1. How to approach the problem Use a conservation-law approach. Specifically, consider the mechanical energy of the spacecraft when it is (a) very far from the earth and (b) at the surface of the earth. Hint 2. Total energy What is the total mechanical energy of the spacecraft when it is far from earth, at a distance ? ANSWER: Hint 3. Potential energy If the spacecraft has mass , what is its potential energy at the surface of the earth? Note that, at the surface of the earth, the spacecraft is a distance from the center of the earth. Express your answer in terms of , , , and the universal gravitational constant . Hint 1. Formula for the potential energy The gravitational potential energy of a system of 2 masses and separated by a distance is . ANSWER: ANSWER: Correct Part B Now find the spacecraft's speed when its distance from the center of the earth is , where the coefficient . Express the speed in terms of and . Hint 1. General approach This problem is very similar to the problem that you've just done. Note that the potential energy of the spacecraft at a distance is different from its potential energy at the earth's surface. Hint 2. First step in finding the speed https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 = 0 = =

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 9/21 Find the spacecraft's speed at in terms of , , , and . ANSWER: ANSWER: Correct ± Gravitational Force of Three Identical Masses Three identical very dense masses of 7200 each are placed on the x axis. One mass is at = -140 , one is at the origin, and one is at = 330 . Part A What is the magnitude of the net gravitational force on the mass at the origin due to the other two masses? Take the gravitational constant to be = 6.67×10 −11 . Express your answer in newtons to three significant figures. Hint 1. How to approach the problem Calculate the force on the mass at the origin, including magnitude and direction, exerted by each of the other masses, then use vector addition to find the total force on it. Hint 2. Calculate the gravitational force from the first mass Calculate the gravitational force exerted on the mass at the origin by the mass at = -140 . Express your answer in newtons to three significant figures. Hint 1. Law of gravitation Recall that Newton's law of gravitation states that the force exerted on a mass by a second mass is , where is the distance between the two masses and 6.67×10 −11 is the gravitational constant. ANSWER: = =

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 10/21 Hint 3. Determine the direction of the gravitational force from the first mass What is the direction of the gravitational force due to the mass at = -140 ? ANSWER: Hint 4. Calculate the gravitational force from the second mass Calculate the gravitational force exerted on the mass at the origin by the mass at = 330 . Express your answer in newtons to three significant figures. Hint 1. Law of gravitation Recall that Newton's law of gravitation states that the force exerted on a mass by a second mass is , where is the distance between the two masses and 6.67×10 −11 is the gravitational constant. ANSWER: Hint 5. Determine the direction of the gravitational force from the second mass What is the direction of the gravitational force due to the mass at = 330 ? ANSWER: ANSWER: Correct = 1.76×10 −3 + x direction - x direction = 3.18×10 −4 + x direction - x direction = 1.45×10 −3

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 11/21 Part B What is the direction of the net gravitational force on the mass at the origin due to the other two masses? ANSWER: Correct The closer together two masses are, the stronger is the gravitational attraction between them. Thus, the mass at the origin is more strongly attracted to the mass at = -140 than it is to the mass at = 330 . Thus, the net force on the mass at the origin is in the - x direction. Gravitational Acceleration inside a Planet Consider a spherical planet of uniform density . The distance from the planet's center to its surface (i.e., the planet's radius) is . An object is located a distance from the center of the planet, where . (The object is located inside of the planet.) Part A Find an expression for the magnitude of the acceleration due to gravity, , inside the planet. Express the acceleration due to gravity in terms of , , , and , the universal gravitational constant. Hint 1. Force due to planet's mass outside radius From Newton's Principia , Proposition LXX, Theorem XXX: If to every point of a spherical surface there tend equal centripetal forces decreasing as the square of the distances from those points, I say, that a corpuscle placed within that surface will not be attracted by those forces any way. In other words, you don't have to worry about the portion of the planet's mass that is located outside of the radius . The net gravitational force from this "outer shell" of mass will equal zero. You only have to worry about that portion of the planet's mass that is located within a radius . Hint 2. Find the force on an object at distance Suppose the object has a mass . Find the magnitude of the gravitational force acting on this object when it is located a distance from the center of the planet. Express the force in terms of , , , , and , the universal gravitational constant. Hint 1. Find the mass within a radius Find the net mass of the planet located within the radius . Remember, the volume of a sphere is . Express your answer in terms of , , and . ANSWER: + x direction - x direction

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 12/21 Hint 2. Magnitude of gravitational force The general equation for the magnitude of the gravitational force is . ANSWER: Hint 3. Finding from According to Newton's 2nd law, the net force acting on an object is given by . In this problem, and since the only force acting on the object is the gravitational force. Therefore, , where is the force you found in the previous hint. Note that in this usage, both and are magnitudes and hence are positive. By convention, (or in this case) is the magnitude of the gravitational field. This gravitational field is a vector, with direction downward. ANSWER: Correct Part B Rewrite your result for in terms of , the gravitational acceleration at the surface of the planet, times a function of R. Express your answer in terms of , , and . Hint 1. Acceleration at the surface Note that the acceleration at the surface should be equal to the value of the function from Part A evaluated at the radius of the planet: . ANSWER: Correct Notice that increases linearly with , rather than being proportional to . This assures that it is zero at the center of the planet, as required by symmetry. = = = =

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 13/21 Part C Find a numerical value for , the average density of the earth in kilograms per cubic meter. Use for the radius of the earth, , and a value of at the surface of . Express your answer to three significant figures. Hint 1. How to approach the problem You already derived the relation needed to solve this problem in Part A: . At what distance is known so that you could use this relation to find ? ANSWER: Correct ± 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 1. Consider various cases If the object is given some speed less than its escape velocity, it will eventually return to the planet. If it is given some speed greater than its escape velocity, then even at very large distances, it will still have some velocity directed away from the planet. If the object is given its escape velocity exactly , its velocity at a very large distance from the planet will tend to zero (so, its kinetic energy will go to zero). ANSWER: = 5500 = 0

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 14/21 Correct Consider 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: Correct Part C Total mechanical energy is conserved. ANSWER: Correct Part D Kinetic energy is conserved. ANSWER: Correct 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). true false true false true false

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 15/21 ANSWER: Correct 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. Express the escape velocity in terms of , , , and , the universal gravitational constant. Hint 1. Determine the gravitational potential energy Find , the gravitational potential energy of the object at a distance from the center of the planet, with the gravitational potential energy taken to be zero when the object is infinitely far away from the planet. Express your answer in terms of , , , and , the universal gravitational constant. ANSWER: Hint 2. Determine the kinetic energy Suppose that the object is given its escape velocity . Find the initial kinetic energy of the object when it is at its initial distance from the center of the planet. Express your answer in terms of and other given quantities. ANSWER: Hint 3. Putting it all together You can assume that no nonconservative forces act on the object as it moves through space. Hence, using conservation of mechanical energy, you can equate the object's initial mechanical energy (when it is at a distance from the planet) to the object's mechanical energy at very large distances from the planet. This should allow you to solve for . ANSWER: true false = =

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 16/21 Correct 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 . 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 . Express the orbital speed in terms of , , and . Hint 1. Find the force Find the radial force on the satellite of mass . (Note that will cancel out of your final answer for .) Express your answer in terms of , , , and . Indicate outward radial direction with a positive sign and inward radial direction with a negative sign. ANSWER: Hint 2. A basic kinematic relation Find an expression for the radial acceleration for the satellite in its circular orbit. Express your answer in terms of and . Indicate outward radial direction with a positive sign and inward radial direction with a negative sign. ANSWER: = = =

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 17/21 Hint 3. Newton's 2nd law Apply to the radial coordinate. ANSWER: Correct 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: Correct Part C Express the kinetic energy in terms of the potential energy . Hint 1. Potential energy What is the potential energy of the satellite in this orbit? Express your answer in terms of , , , and . ANSWER: ANSWER: = = = =

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 18/21 Correct 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 . Express your answer in terms of , , , and . Hint 1. How to get started Use the fact that the period is the time to make one orbit. Then . ANSWER: Correct Part E Find an expression for the square of the orbital period. Express your answer in terms of , , , and . ANSWER: Correct 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. Express your answer in terms of , , , and . Hint 1. Definition of angular momentum = =

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3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 19/21 Recall that , where is the momentum of the object and is the vector from the rotation point. Here the rotation point is the center of the planet, and since the object is moving in a circular orbit, is perpendicular to . ANSWER: Correct 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. 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. Hint 1. Example of a power law The potential energy behaves as , so depends inversely on . Therefore, the appropriate power for this is (i.e., ). ANSWER: Correct Problem 13.31 Mars rotates on its axis once every 24.8 hours. Part A What is the speed of a geosynchronous satellite orbiting Mars? Express your answer with the appropriate units. ANSWER: Correct = -0.500,-1,-1,0.500 1.44

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 20/21 Part B What is the altitude of a geosynchronous satellite orbiting Mars? Express your answer with the appropriate units. ANSWER: Correct Problem 13.57 The figure shows two planets of mass orbiting a star of mass . The planets are in the same orbit, with radius , but are always at opposite ends of a diameter. Part A Find an exact expression for the orbital period . Hint: Each planet feels two forces. Express your answer in terms of , , , and . ANSWER: Correct Problem 13.34 1.72×10 7 =

3/1/24, 11:42 PM Homework 7-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 21/21 Pluto moves in a fairly elliptical orbit around the sun. Pluto's speed at its closest approach of is 6.12 . Part A What is Pluto's speed at the most distant point in its orbit, where it is from the sun? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 100%. You received 9 out of a possible total of 9 points. = 3.71

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