course Phy 201 ???????????assignment #027??????????????
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14:48:57 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 --> g=9.8m/s^2 field strength at a given distance= (R/r1)^2 *g where r1 is the given distance and R is the radius
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14:49:53 ** 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 --> Field strength = (Re/r)^2*9.8m/s^2
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14:55:34 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 of the gravitational field is 1/4 as strong.
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14:56:14 ** 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 --> Double the radius gives 1/2^2=1/4 the graviational field
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15:08:39 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 --> Graviational field at distance r1= g(R/r1)^2 F(r1)=m*gravitational field at distance r1= mg(R/r1)^2 work=force*distance=F(r1)*`dr=mg(R/r1)^2*`dr Should break the distance between into segments and calculate the energy required for each segment and add them up
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15:10:21 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 --> mass*{(Re +distance)/Re}^2=force Force*distance=KE
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15:10:30 Query class notes #24
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RESPONSE -->
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15:17:31 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 --> At a very small velocity, the object would fall back to the earth. As the ball is shot with a greater velocity, it will travel further before coming back to earth. If the corrrect velocity is used, the ball will not return to the earth, but will orbit it. If above this velocity is used, the ball or object may escape the gravitational field of the earth. The paths curve less toward the earth as their velocities increase.
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15:21:07 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 --> The ball will fall to earth in a parabolic path until it is propeled with enough velocity to orbit the earth, or with even more velocity, escape into outer space. As is orbits the earth, it will take an elliptical orbit.
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15:21:34 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> None becuase the Earth is not perfectly circular
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15:22:57 ** For a given distance from the center of the Earth, there is only one velocity for which centripetal acceleration is equal to gravitational acceleration, so there is only one possible velocity for a circular orbit of given orbital radius. The orbital radius is determined by the height of the 'tower', so for a given tower there is only one velocity which will achieve a circular orbit. **
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RESPONSE --> There is one velocity for which centripetal acceleration is equal to gravitational acceleration, so only one possible velocity fo a circular orbit at a given orbital radius. The orbital radius is determined by the height of the tower.
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15:23:55 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
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15:26:10 ** 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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RESPONSE --> You must start out parallel to the surface of the Earth because any circle about the center must be perpendicular at every point to the radial line. Any radial line will intercept the surface of the Earth and must be perpendicular to it, and the circular obit must also be perpendicular to this line. The orbit and the surface are perpendicular to the same line and are parallel.
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15:31:09 Principles of Physics and General College Physics Problem 5.2: A jet traveling at 525 m/s moves in an arc of radius 6.00 km. What is the acceleration of the jet?
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RESPONSE --> a=v^2/r v^2=(525m/s)^2=275625m^2/s^2 r=6000m a=45.94m/s^2
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15:31:54 The jet will have centripetal acceleration a_cent = v^2 / r, where v is its speed and r the radius of the circle on which it is traveling. In this case we have v = 525 m/s and r = 6.00 km = 6000 meters. The centripetal acceleration is therefore a_cent = v^2 / r = (525 m/s)^2 / (6000 m) = 45 m/s^2, approx.. One 'g' is 9.8 m/s^2, so this is about (45 m/s^2) / (9.8 m/s^2) = 4.6 'g's'.
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RESPONSE --> 45m/s^2 TO get the answer in gs 45m/s^2/9.8m/s^2=4.6g
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15:32:01 Univ. Why is it that the center of mass doesn't move?
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15:32:08 ** There is no net force on the system as a whole so its center of mass can't accelerate. From the frame of reference of the system, then, the center of mass remains stationary. **
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