course PHY 201 xy̓Oassignment #027
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17:58:31 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 --> The field strength would propotional to the inverse square of the ratio of the earth's radius to the radius + height.
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17:58:43 ** 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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17:59:33 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 --> we get gravity / 2^2 = 1/4 of the original gravity
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17:59:43 ** 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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18:12:05 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 --> By calculating the gravity at the first radius and the second, and averaging the two to find an average acceleration. Then use this this acceleration and the mass to calculate the number of Newtowns required to lift the object and multiply by meteres to yield the total number of Joules required.
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18:12:39 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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18:12:43 Query class notes #24
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16:24:18 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 --> The lower the velocity, the more the path of a particle resembles a ball rolling of a table. As the velocity increases eventually the particles will go into a sustainable orbit. As velocity increases further, the orbit will become more eliptical, and eventually the initial velocity will be so high they will just fly off into space, never to be seen again.
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16:25:51 ** 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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17:15:01 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 --> I don't think so, no. If you start an object off on a trajectory upward such that its momentum is cancelled by gravity at the precise moment where it reaches a circular orbit, it should maintain that circular orbit.
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17:15:50 ** 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 --> Hmm, okay, well I missed that one :-)
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17:16:10 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 --> 525 m/s^2 / 6000 m = 275625 m^2/s^2 / 6000 m = 45.9375 m/s/s
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17:16:52 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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17:16:56 Univ. Why is it that the center of mass doesn't move?
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17:16:58 ** 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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