course Phy 201 Ù¿éÓ~úDó‘ü©ožœ‡Ÿœ£³ÔÅŒïÄúassignment #028 P«£Ïð·xðÀ‹ÂÄ•²x¨vàÃc«ÙÜÓi‹ÜïÚ Physics I 08-06-2008
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12:21:44 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
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12:36:02 ** 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 --> I understand, I was misreading the formula
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12:52:56 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 --> N/A
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12:52:59 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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13:14:48 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This query was a lot harder than the rest. I don't think I answered half of them.
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