course Phy 121
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16:48:00 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 --> By using the formula : field at distance r1 = g (R / r1) ^ 2; Where r1 is the distance, g is the gravity, R is the raduius of the earth
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16:48:08 ** 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 --> ok
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16:52:28 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 closer the object is to the center of the earth the smaller the spread of gravitational effect resulting in a larger gravitational field. The farther the distance from the center fo the earth the larger the spread of gravitational effect resulting in a less gravitational field
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16:54:26 ** 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 --> sorry... I get I could have given you back the distance^2 proportion ratio: Thus a double of distance creates 1/2^2 or 1/4 the gravitational field
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16:55:48 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 would assume that we would take the average of the g from the two heights ... and plug into our normal equations
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16:57:00 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 --> ok
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16:57:20 Query class notes #24
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RESPONSE --> ok
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17:20:28 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 the object is fired at a low velocity the object will act as a projectile and follow a parabolic path as an object falling off a table would. The second object was fired at a greater velocity than the first and still acted like a projectile but traversed a noticable portion of the earth. The last was fired with a very very fast velocity and its path curved slightly toward the earth in which the video showed a larger picture of the path... It actually travels in a circular path all the way around the earth (assuming that the earth is perfectly spherical)
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17:22:40 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 --> ok
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17:27:08 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> The last two velocities.... both of these velocitys are just so that the curvature of the earch The second to last particle tends to follow a parabolic path, but as it falls the Earth curves away from it at the same rate that it falls and it maintains the same distance from the Earth. This object will then travel in a circular path all the way around the Earth (assuming that the Earth is perfectly spherical, which is almost but not exactly so). The last path starting with the red streak actually becomes an orbit, though not a circular one. As the object travels away from the Earth, it is pulled back toward the center of the Earth, though with a decreasing force. An object fired at a great enough velocity could in fact completely escape the pull of the Earth and travel away forever.
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17:29:41 ** 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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RESPONSE --> ...I was think that an orbit was to ""keep itself in space"" ... I guess this is a misconception since we want the orbit of the object to travel back around to the tower. The last velocity that I mentioned would have too much of a velocity to maintain a patch back to the tower. After all of this I understand much more of an orbit of an object.
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17:32:54 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 would think not as long as the object was on a path of that which is in the direction of the surface of the earth... In other words once the object obtained velocity and then adusted it direction in the direction parallel to the surface of the earth at the right altitude
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17:33:05 12:40:19
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RESPONSE --> ok
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17:34:43 ** 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 --> applying this to the previous questions in having only a single shot to place the object in motion...this shot would have to be parallel to the surface of the earth
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17:58:02 Query gen phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.
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RESPONSE --> g = G* m/r^2 g = 6.67*10^-11 * 7.35*10^22 / 1.74 * 10^6 ^2 = g = 1.619 m/s^2
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17:58:09 ** The acceleration due to gravity on the Moon is found using the equation g' = G (Mass of Moon)/ radius of moon ^2 g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m) = 1.619 m/s^2 **
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RESPONSE --> ok
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17:58:15 Univ. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m. When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change?
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RESPONSE --> Not Required
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17:58:19 The force would be F = (6.67 * 10^-11 * 25 * 100) / 20^2 F = 4.17 * 10^-10 a1 = 4.17 * 10^-10 / 25 a1 = 1.67 * 10^-11 m/s/s a2 = 4.17 * 10^-10 / 100 a2 = 4.17 * 10^-12 m/s/s The position of center of mass doesn't change because the two spheres are the same size. ** At separation r the force is F = G m1 m2 / r^2. For any small increment `dr of change in separation the approximate work done by the gravitational force is F `dr = G m1 m2 / r^2 * `dr. We take the sum of such contributions, between the given separations, to form an approximation to the total work done by the gravitational force. We then take the limit as `dr -> 0 and obtain the integral of G m1 m2 / r^2 with respect to r from separation r1 to separation r2. An antiderivative is - G m1 m2 / r; evaluating between the two separations we get - G m1 m2 / r1 - (-G m1 m2 / r2) = G m1 m2 ( 1/r2 - 1 / r1). This expression is evaluated at r1 = 40 m and r2 = 20 m to get the change G m1 m2 ( 1/(20 m) - 1 / (40 m) ) in KE. I get around 1.49 * 10^-9 Joules but it isn't guaranteed so you should verify that carefully. Assuming a reference frame initially at rest with respect to the masses the intial momentum is zero. If the velocities at the 20 m separation are v1 and v2 we know that m1 v1 + m2 v2 = 0, so that v2 = -(m1 / m2) * v1. The total KE, which we found above, is .5 m1 v1^2 + .5 m2 v2^2. Substituting v2 = - (m1 / m2) v1 and setting equal to the KE we can find v1; from this we easily find v2. You might get something like 4.1 * 10^-6 m/s for the velocity of the 100 kg mass; this number is again not guaranteed so verify it yourself. The position of the center of mass does not change because there is no external force acting on the 2-mass system. The center of mass is at position r with respect to m1 (take m1 to be the 25 kg object) such that m1 r - m2 (40 meters -r) = 0; substituting m1 and m2 you get 25 r - 100 (40 meters - r ) = 0. I believe you get r = 4 / 5 * 40 meters = 32 m, approx., from the 25 kg mass, which would be 8 meters from the 100 kg mass.
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RESPONSE --> ok
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17:58:23 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> ok
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17:58:24 Univ. Why is it that the center of mass doesn't move?
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RESPONSE -->
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17:58:25 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> ok
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17:58:28 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> ok
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17:58:31 ** 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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RESPONSE --> ok
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