course Phy 121
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14:17:52 q001. Note that this assignment contains 8 questions. Masses attract each other. The forces of attraction are equal and opposite: The force exerted by one small concentrated mass on another is equal in magnitude but in the opposite direction from the force exerted on it by the other. Greater masses exert greater attractions on one another. If two such objects remain separated by the same distance while one object increases to 10 times its original mass while the other remains the same, there will be 10 times the original force. If both objects increase to 10 times their original masses, there will be 100 times the original force. The force of attraction is inversely proportional to the square of the distance between the objects. That means that if the objects move twice as far apart, the force becomes 1 / 2^2 = 1/4 as great; if they move 10 times as far apart, the force becomes 1 / 10^2 = 1/100 as great. The same statements hold for spherical objects which have mass distributions which are symmetric about their centers, provided we regard the distance between the objects as the distance between their centers. Suppose a planet exerts a force of 10,000 Newtons on a certain object (perhaps a satellite) when that object is 8000 kilometers from the center of the planet. How much force does the satellite exert on the planet?
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RESPONSE --> The satellite is exerting a force of 10,000 Newtons on the planet because the force is equal and opposite.
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14:17:57 The gravitational forces exerted by the planet and the object are equal and opposite, and are both forces of attraction, so that the object must be exerting a force of 10,000 Newtons on the planet. The object is pulled toward the planet, and the planet is pulled toward the object.
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RESPONSE --> ok
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14:19:13 `q002. If the object and the planet are both being pulled by the same force, why is it that the object accelerates toward the planet rather than the planet accelerating toward the object?
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RESPONSE --> The planet is much more massive than the satellite, and there may be some acceleration of the planet but it is not enough to notice.
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14:19:16 Presumably the planet is much more massive than the object. Since the acceleration of any object is equal to the net force acting on it divided by its mass, the planet with its much greater mass will experience much less acceleration. The minuscule acceleration of the planet toward a small satellite will not be noticed by the inhabitants of the planet.
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RESPONSE --> ok
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14:20:10 `q003. If the mass of the object in the preceding exercise is suddenly cut in half, as say by a satellite burning fuel, while the distance remains at 8000 km, then what will be the gravitational force exerted on it by the planet?
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RESPONSE --> 5,000 Newtons
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14:20:13 Halving the mass of the object, while implicitly keeping the mass of the planet and the distance of the object the same, will halve the force of mutual attraction from 10,000 Newtons to 5,000 Newtons.
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RESPONSE --> ok
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14:21:05 `q004. How much force would be experienced by a satellite with 6 times the mass of this object at 8000 km from the center of a planet with half the mass of the original planet?
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RESPONSE --> 1/2 * 6 = 3 3 * 10,000 N = 30,000 N
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14:21:09 The distance is the same as in the previous examples, so increasing the mass by a factor of 6 would to result in 6 times the force, provided everything else remained the same; but halving the mass of the planet would result in halving this force so the resulting force would be only 1/2 * 6 = 3 times is great as the original, or 3 * 10,000 N = 30,000 N.
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RESPONSE --> ok
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14:22:14 `q005. How much force would be experienced by the original object at a distance of 40,000 km from the center of the original planet?
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RESPONSE --> 40,000 km / (8000 km) = 5 1 / 5^2 = 1/25 1/25 * 10,000 N = 400 N
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14:22:18 The object is 40,000 km / (8000 km) = 5 times as far from the planet as originally. Since the force is proportional to the inverse of the square of the distance, the object will at this new distance experience a force of 1 / 5^2 = 1/25 times the original, or 1/25 * 10,000 N = 400 N.
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RESPONSE --> ok
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14:24:25 `q006. The relationship between the force of attraction and the masses and separation can be expressed by a proportionality. If the masses of two small, uniformly spherical objects are m1 and m2, and if the distance between these masses is r, then the force of attraction between the two objects is given by F = G * m1 * m2 / r^2. G is a constant of proportionality equal to 6.67 * 10^-11 N m^2 / kg^2. Find the force of attraction between a 100 kg uniform lead sphere and a 200 kg uniform lead sphere separated by a center-to-center distance of .5 meter.
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RESPONSE --> F = G * m1 * m2 / r^2 F = 6.67 * 10^-11 N m^2 / kg^2 * 100 kg * 200 kg / (.5 m)^2 = 5.3 * 10^-6 Newton
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14:24:33 We are given the two masses m1 = 100 kg and m2 = 200 kg and the separation r = .5 meter between their centers. We can use the relationship F = G * m1 * m2 / r^2 directly by simply substituting the masses and the separation. We find that the force is F = 6.67 * 10^-11 N m^2 / kg^2 * 100 kg * 200 kg / (.5 m)^2 = 5.3 * 10^-6 Newton. Note that the m^2 unit in G will be divided by the square of the m unit in the denominator, and that the kg^2 in the denominator of G will be multiplied by the kg^2 we get from multiplying the two masses, so that the m^2 and the kg^2 units disappear from our calculation.
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RESPONSE --> ok
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14:27:10 `q007. If these two objects were somehow suspended so that the net force on them was just their mutual gravitational attraction, at what rate would the first object accelerate toward the second, and if both objects were originally are rest approximately how long would it take it to move the first centimeter?
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RESPONSE --> a = 5.3 * 10^-6 N / (100 kg) = 5.3 * 10^-8 m/s^2 `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * .01 m / (5.3 * 10^-8 m/s^2) ) = `sqrt( 3.8 * 10^5 m / (m/s^2) ) = 6.2 * 10^2 sec
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14:27:14 A mass of 100 kg subject a net force of 5.3 * 10^-6 N will have acceleration of a = 5.3 * 10^-6 N / (100 kg) = 5.3 * 10^-8 m/s^2. At this rate to move from rest (v0 = 0) thru the displacement of one centimeter (`ds = .01 m) would require time `dt such that `ds = v0 `dt + .5 a `dt^2; since v0 = 0 this relationship is just `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * .01 m / (5.3 * 10^-8 m/s^2) ) = `sqrt( 3.8 * 10^5 m / (m/s^2) ) = 6.2 * 10^2 sec, or about 10 minutes. Of course the time would be a bit shorter than this because the object, while moving somewhat closer (and while the other object in turn moved closer to the center of gravity of the system), would experience a slightly increasing force and therefore a slightly increasing acceleration.
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RESPONSE --> ok
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14:28:06 `q008. At what rate would the second object accelerate toward the first?
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RESPONSE --> a = 5.3 * 10^-6 N / (200 kg) = 2.7 * 10^-8 m/s^2
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14:28:10 The second object, with its 200 kg mass, would also a subject to a net force of 5.3 * 10^-6 N and would therefore experience and acceleration of a = 5.3 * 10^-6 N / (200 kg) = 2.7 * 10^-8 m/s^2. This is half the rate at which the first object changes its velocity; this is due to the equal and opposite nature of the forces and to the fact that the second object has twice the mass of the first.
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RESPONSE --> ok assignment #027 [뺗܃ܝ Physics I 04-04-2006
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14:29:52 Query intro probs set 7, 1-7 Knowing the 9.8 m/s^2 gravitational field strength of the Earth's field at the surface of the Earth, and knowing the radius of the Earth, how do we find the gravitational field strength at a given distance 'above' the surface of the Earth?
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RESPONSE --> Field strength=(Re/r)^2*9.8m/s^2
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14:29:58 ** You have an inverse square force. Square the ratio of Earth radius to orbital radius and multiply by 9.8 m/s^2: Field strength=(Re/r)^2*9.8m/s^2 **
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RESPONSE --> ok
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14:30:56 If we double our distance from the center of the Earth, what happens to the gravitational field strength we experience?
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RESPONSE --> The strength will reduce to 1/4.
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14:31:04 ** We have an inverse square force so if r2 = 2 * r1 the ratio of the gravitational field will be g2 / g1 = (1 / r2^2) / (1 / r1^2) = r1^2 / r2^2 = (r1 / r2)^2 = (r1 / (2 * r1))^2 = r1^2 / 4 r1^2 = 1/4. In a nutshell double the radius gives us 1 / 2^2 = 1/4 the gravitational field. **
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RESPONSE --> ok
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14:33:06 How do we approximate the energy required to move a given mass from the surface of the Earth to a given height 'above' the Earth, where the field strength at the given height differ significantly from that at the surface?
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RESPONSE --> Average the force at the surface and the force at the maximum altitude. Then multiply by the distance.
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14:33:09 STUDENT SOLUTION AND INSTRUCTOR RESPONSE: mass*[(Re + distance)/Re]^2=force Force*distance=KE INSTRUCTOR RESPONSE: The first approximation would be to average the force at the surface and the force at the maximum altitude, then multiply by the distance. The result would give you the work necessary to 'raise' the object against a conservative force, which would be equal to the change in PE. ADDENDUM FOR UNIVERSITY PHYSICS STUDENTS ONLY:The exact work is obtained by integrating the force with respect to position. You can integrate either G M m / r^2 or g * (RE / r)^2 from r = RE to rMax. **
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RESPONSE --> ok
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14:33:27 Query class notes #24
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RESPONSE --> ok
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14:38:38 Describe the paths of various particles 'shot' parallel to the surface of the Earth from the top of a very high tower, starting with a very small velocity and gradually increasing to a velocity sufficient to completely escape the gravitational field of the Earth.
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RESPONSE --> Each particle releases to follow an orbit around the center of the earth. But for particles shot at low velocity, this path is interupted by the surface of the eath and stops there. The faster the velocity, the further x distance becomes before the particle lands. However, if it given a large enough velocity, it will fall around the earth.
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14:38:42 GOOD STUDENT ANSWER: Each particle sets out to follow an orbit around the center of mass of the earth. But for particles shot at slower speeds, this path is interupted by the surface of the eath and simply stops there. The faster it is shot, the further x distance becomes before the particle lands. However, if it given a great enough velocity, it will fall around the curviture of the earth. If is shot even faster than that, it will follow an eliptical oribit with varying speeds and distances from center of earth. GOOD STUDENT ANSWER: With a very low velocity the projectile will not travlel as far. It will fall to earth in a nearly parabolic fashion since it gains vertical velocity as it travels horizontally at a steady pace. If the projectile is fired at a very strong velocity it will leave the earths vacinity but will still be pulled by the forces acting on it from the earths center. This will cause it to go only so far at which point it has slowed down considerabley, since it has lost most of its kinetic energy. It turns and begins to gain energy as it approaches the earths area, using the potential energy it gained on the trip out. (Causing it to speed up). The path that this projectile will take will be eliptical, and it will continue to loop around the earth. If the projectile is fired at the correct velocity to form a circular orbit, it will also fall at a parabolic fashion, although the earth's surface will also be descending at the same rate so that the object will appear to be 'not falling'. It is falling but at the same rate the earth is 'falling' under it. It will circle the earth until something causes it to stop. INSTRUCTOR RESPONSE: The path of the projectile will always be an ellipse with the center of the Earth at one focus. For low velocities and low altitude this path is very nearly parabolic before being interrupted by the surface of the Earth. One of these ellipses is a perfect circle and gives us the circular orbit we use frequently in this section. **
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RESPONSE --> ok
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14:39:34 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> Only one.
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14:39:40 ** There is only one velocity for a given orbital radius; the orbital radius is determined by the height of the 'tower', so there is only one velocity for a given tower. **
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RESPONSE --> ok
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14:40:13 Is it necessary in order to achieve a circular orbit to start the object out in a direction parallel to the surface of the Earth?
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RESPONSE --> yes
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14:40:18 12:40:19
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RESPONSE --> ok
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14:40:29 ** If you have just one 'shot' then you must start out parallel to the surface of the Earth. The reason is that any circle about the center must be perpendicular at every point to a radial line--a line drawn from the center to the circle. Any radial line will intercept the surface of the Earth and must be perpendicular to it, and the circular orbit must also be perpendicular to this line. Therefore the orbit and the surface are perpendicular to the same line and are therefore parallel. **
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RESPONSE --> ok
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