course phy 121 }\~Ijwassignment #027
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21:25:26 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 above the surface of Earth equals g (R/r1)^2 where g = gravitational field strength at surface = 9.8m/s^2 and R = Earth's radius = 6400 and r1 = distance above surface of earth.
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21:25:47 ** 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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21:26:11 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 --> It becomes 1/4th as strong
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21:26:24 ** 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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21:32:08 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 --> I had a hard time understanding this one... First find the field strength at the given distance: mass * gravity at surface * (R/r1)^2 and then multiply it by the distance the object is to move (dr) mg(R/r1)^2 * dr
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21:35:14 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 --> Did I just find the force at the maximum altitude and multiply by distance? Would it be: [ (mgR + mg r1)/2 ] * dr That doesn't seem right
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21:37:10 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. The objects would all arc down to the Earth, but at longer distances. The object that finally escapes the grav. field of the Earth would continue to arc, but would then be in orbit around the Earth
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21:38:41 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 --> That's kind of what I meant.
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21:39:07 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> 1
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21:39:32 ** 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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21:48:17 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, I think it needs the horizontal velocity. It is the horiz velocity along with the vertical velocity (the pull of gravity) that cause it to arc. If it just has vertical velocity I think it would just go up and then come back down. ... But then the Earth is rotating and causes the object to have a horizontal velocity anyway... but I think the horizontal velocity would be in the opposite direction of the movement of the object. If the Earth is rotating counterclockwise (I'm too tired to figure out which way it does rotate) and the object goes up and comes back down sufficiently long enough time wise then it will have gone in a clockwise manner. The object goes up at 12:00oclock position, the Earth rotates counterclockwise and it comes down and hits at the 1:00 position.... so I am back to thinking it does need to start out with a horizontal velocity.... But then rockets go straight up (don't they) and then release satellites that orbit... I don't know!!!
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21:49:03 ** 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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21:52:13 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 --> v = 525m/s r = 6.00km what is a a = v^2 /r 525^2 / 6000m 275625/6000m = 45.9375 = 45.94 m/s^2 what is accel in g's acceleration due to gravity = 9.8m/s^2 45.95/9.8 = 4.69g
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21:52:57 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 --> I didn't think that was right - cool!
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21:53:07 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> Not applicable
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21:53:10 ** 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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