course PHY 121 ???????????????assignment #024
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14:22:58 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> because of the gravitational pull?
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14:23:20 ** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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14:24:09 Why do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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RESPONSE --> Because it doesn't go vertical until it starts going back around the circle.
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14:24:22 ** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity being perpendicular to this vertical must be wholly in the horizontal direction. **
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14:25:52 What is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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RESPONSE --> The centripetal acceleration would be v^2/r
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14:26:47 ** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **
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RESPONSE --> Is v^2/r the same as the centripetal acceleration equal to acceleration of gravity, or is there another equation for this?
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????????X?w??assignment #025 X??w??????p??Physics I 07-15-2006 ?j?V?????E?????? assignment #026 X??w??????p??Physics I 07-15-2006 ?S????^?z????assignment #027 X??w??????p??Physics I 07-15-2006
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14:37:24 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 --> (Re/R)^2 *g RE stands for the Earth's radius.
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14:37:44 ** 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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14:41:30 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 gravitational field strength will double.
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14:44:13 ** 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 --> doubling the radius gives us 1/4 the gravitational field.
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14:48:24 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 --> F = G m1 (m2/r^2) I don't think this is right because you want energy and this is force? Would I find the force and then divide by the distance to get the energy?
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14:50:30 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 --> m * [(Re + distance)/Re]^2 Force * distance = Ke
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14:55:56 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 --> As an object travels away from the Earth, it is pulled toward the center, it does not become circular however and the force pulling decreases. If the velocity is great enough it escapes the pull of the Earth.
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14:57:31 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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14:59:12 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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14:59:57 ** 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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15:01:29 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 --> No a component perpendicular to motion bends towards the Earth.
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15:01:37 12:40:19
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15:02:10 ** 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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????????????? assignment #028 X??w??????p??Physics I 07-15-2006
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15:35:05 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
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15:35:42 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
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RESPONSE --> What does the k stand for?
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