course Phy 231 7/13 9:30 027. `query 27*********************************************
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Given Solution: `a** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qIf we double our distance from the center of the Earth, what happens to the gravitational field strength we experience? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The strength of the gravitational field will be a fourth of what it was before. The ratio before was 1:1; the new ratio is 1:2, which you square and you get 1/4. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow do we approximate the energy required to move a given mass from the surface of the Earth to a given height 'above' the Earth, where the field strength at the given height differ significantly from that at the surface? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You use the average of the force on the mass at the surface of Earth and the force on the mass at the new height. Then you can multiply that by the increase in the radius and you get the energy required to move the mass. To be more exact, you integrate the function for force with respect to height from the least force to the greatest. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: mass*[(Re + distance)/Re]^2=force Force*distance=KE 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 m g * (RE / r)^2 from r = RE to rMax. Integrating G M m / r^2 from r = r1 to r = r2 you get G M m / r1 - G M m / r2, which is the change between r = r1 and r = r2 of the potential energy function -G M m / r (note that this function is an antiderivative with respect to r of G M m / r^2).** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery class notes #24 Describe the paths of various particles 'shot' parallel to the surface of the Earth from the top of a very high tower, starting with a very small velocity and gradually increasing to a velocity sufficient to completely escape the gravitational field of the Earth. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A very small velocity will have a relatively small horizontal range and will go to the ground very quickly. A greater velocity will allow the object to move farther by it will gain eventually hit the ground. The minimum velocity to cause an orbit the object will travel fast enough that the force of gravity does not cause it to come back to the ground, but curves the path so that it continues around the planet. The greatest velocity will cause the object to go really far away from the planet; it can sometimes make a really big elliptical orbit, or the object can leave the gravitational field altogether. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. With a very low velocity the projectile will not traveled 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Only the exact velocity will allow the x range for the object to be exactly right and for the orbit to be a perfect circle. Too fast and too slow velocities will either cause an ellipse with the earth at one focus, or the object will hit the surface of the earth because the ellipse was not big enough to go around the earth. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qIs it necessary in order to achieve a circular orbit to start the object out in a direction parallel to the surface of the Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, this is the only way to get the orbit and the surface of the earth to be parallel. You want all the lines going from the center of the earth to the orbit to be perpendicular to the orbit and the center of the earth. The only way you get this is by starting parallel. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** The main reason it has to be very tall is because of the atmosphere. If you shoot the projectile with sufficient speed while it's in the atmosphere, it will quickly lose most of its kinetic energy to air resistance; the energy goes mostly into heating the object, which as a result proceeds to melt as its orbit decays. Whether it melts before hitting the ground or not depends on how quickly the orbit decays and how high it was in the first place. If there was no atmosphere, you still would need to be careful about the flattening of the Earth at the equator (the radius at the equator is about 20 km greater than the radius at the poles, which means that if you wanted an orbit that took you over the equator, then even in the absence of atmosphere a tower at the pole would have to be at least 20 km high. You want to get out of the atmosphere as quickly as possible, to minimize the work you need to do against air resistance. A vertical launch position is much more stable than one at an angle away from vertical. Same reason we build towers vertical rather than leaning. The launch starts out vertical, and gradually curves toward the tangential direction. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qPrinciples of Physics and General College Physics Problem 5.2: A jet traveling at 525 m/s moves in an arc of radius 6.00 km. What is the acceleration of the jet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Centripetal acceleration = 525m/s ^2 / 6000 m = 45.9 m/s^2 1 g is 9.8m/s^2, so the acceleration is 4.68 g’s, or 4.68 times the acceleration due to gravity confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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'. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK Question: `qUniv. Why is it that the center of mass doesn't move? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The external net force is almost 0, so the center of mass does not accelerate. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was a little bit confused by this question. I looked in the textbook and it was talking about how the center of mass moves as though all of the forces were working at that one point. ?????In the question, does it mean that since the external forces cause a net force of 0, the center of mass does not move?????