course Phy 231 7/16 4 028. `query 28*********************************************
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Given Solution: `a** 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 ( rE / r ) ^2. If we set the acceleration equal to v^2 / r, we obtain v^2 / r = g ( rE / r)^2 so that v^2 = g ( rE / r) and v = sqrt( 2 g rE / r). Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth. If we do know G and the mass of the Earty, we can proceed as follows: The gravitational force on mass m at distance r from the center of the Earth is F = G m M / r^2, Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have m v^2 / r = G m M / r^2, which we solve for v to get v = sqrt( G M / r). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qPrinciples of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g = G (mass) / r^2 g = G(7.35*10^22) / (1.74*10^6)^2 g = 1.62 m/s^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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)^2 = 1.619 m/s^2 ** Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km). What is the total force on Earth due to the planets, assuming perfect alignment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Venus pulls toward the sun, Jupiter and Saturn pull in the other direction Fv = G(.815 * 4.87*10^24)(5.97*10^24) / 1.50*10^11 - 1.08*10^11 ^2 = 1.1*10^18 N Fj = G(318*1.9*10^27)(5.97*10^24) / 7.78*10^11 1.50*10^11 ^2 = 1.9*10^18 N Fs = G(95.1*5.68*10^26)(5.97*10^24) / 1.43*10^12 1.50*10^11 ^2 = 1.4*10^17 N Net force = 1.1*10^18 1.9*10^18 1.4*10^17 = -9.4*10^17 N confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Using F = G m1 m2 / r^2 we get Force due to Venus: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (.815 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.08 * 10^11 m)^2 = 1.1 * 10^18 N, approx. Force due to Jupiter: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (318 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 7.78 * 10^11 m)^2 = 1.9 * 10^18 N, approx. Force due to Saturn: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (95.7 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.43 * 10^11 m)^2 = 1.4 * 10^17 N, approx. Venus being 'inside' the Earth's orbit pulls in the direction of the Sun while Jupiter and Saturn pull in the opposite direction so the net force is -1.1 * 10^18 N + 1.9 * 10^18 N + 1.4 * 10^17 N = .9 * 10^18 N = 9 * 10^17 N, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qUniv. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m. When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Force1 = G(25kg)(100kg) / 20m ^2 = 4.17*10^-10 N Force2 = G(25kg)(100kg) / 40m ^2 = 1.04*10^-10 N 40m: a of first mass: 1.04*10^-10 / 25kg = 4.16*10^-12 A of the second mass: 1.04*10^-10 / 100kg = 1.04*10^-12 20m: a of first mass: 4.17*10^-10 / 25kg = 1.67*10^-11 (16.7*10^-12) A of second mass: 4.17*10^-10 / 100kg = 4.17*10^-12 Change in KE = G(25kg)(100kg)(1/20 1/40) = 4.17*10^-9 Joules Conservation of momentum: v2 = -(m1/m2)(v1) KE = .5m1 v1^2 + .5 m2 v2^2 4.17*10^-9 = .5(100)(v1^2) + .5(100)[(-100/25)(v1)] ^2 V1 = 2.21*10^-6 m/s V2 = 4.2*10^-6 m/s The center of mass does not changes because there are no other forces acting on the masses. Center of mass = m1(r) m2(40-r) = 0 25 r 40(100) + (100)r = 0 12.5 r = 400 R = 32m => the point of collision relative to the 25kg; this is 8m from the 100 kg confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The question as phrased here didn't ask you to find the force or the acceleration at any point, but to put the problem in context we will do so below: The forces at the r1 = 40 m and r2 = 20 m separations would be F1 = (6.67 * 10^-11 N m^2 / kg^2 * 25 kg * 100 kg) / (20 m)^2 = 1 * 10^-10 N F2 = (6.67 * 10^-11 N m^2 / kg^2 * 25 kg * 100 kg) / (20 m)^2 = 4 * 10^-10 N Their accelerations are therefore accel of first mass at 40 m separation: a = F1 / m1 = 1 * 10^-10 N / (25 kg) = 4 * 10^-12 m/s^2 accel of first mass at 20 m separation: a = F2 / m1 = 4 * 10^-10 N / (25 kg) = 16 * 10^-12 m/s^2 (written in this form for easy comparison with the first value, but more properly written 1.6 * 10^-11 m/s^2) accel of 2d mass at 40 m separation: a = F1 / m1 = 1 * 10^-10 N / (100 kg) = 1 * 10^-12 m/s^2 accel of 2d mass at 20 m separation: a = F2 / m1 = 4 * 10^-10 N / (100 kg) = 4 * 10^-12 m/s^2 ** 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. The change in the antiderivative between the separations r1 and r2 is - 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. We get KE of about 4 * 10^-9 Joules but you should verify that carefully. We use conservation of momentum and kinetic energy to determine the final velocities: Within a reference frame at rest with respect to initial state of the masses the initial momentum is zero. If the velocities at the 20 m separation are v1 and v2 the from conservation of momentum, we have 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, then 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 (taking 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. The solution for r is r = 4 / 5 * 40 meters = 32 m, approx. This is the position of the collision relative to the initial position of the 25 kg mass. This position is 8 meters from the 100 kg mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery gen phy problem 5.50 24 m diam wheel, rot period 12.5 s, fractional change in apparent weight at top and at bottom. What is the fractional change in apparent weight at the top and that the bottom of the Ferris wheel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12m radius Period = 12.5 seconds C = 24 pi velocity = 24 pi m/12.5s = 1.92 pi m/s Acceleration = 1.92 pi ^2 / 12m = 3.03 m/s^2 Gravity + apparent weight = centripetal force M(-9.8m/s^2) + app weight = m(-3.03m/s^2) the force is going down at the top of the wheel App weight = 6.77 m/s^2 * m M(-9.8m/s^2) + app weight = m(3.03m/s^2) force is going up this time App weight = 12.8 m/s^2 * m Ratio of the weights: 12.8 m/s^2 m / 6.77 m/s^2 *m = 1.91 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Centripetal acceleration is a = v^2 / r. For a point on the rim of the wheel, v = dist in 1 rev / time for 1 rev = `pi * 24 m / (12.5 sec) = 1.9 m/s, approx. Thus v^2 / r = (`pi * 1.9 m/s)^2 / 12 m = 3 m/s^2, approx. At the top the only accel is the centripetal, and it is acting toward the center, therefore downward. The forces acting on any mass at the top are the gravitational force and the force exerted by the wheel on the mass. At the top of the wheel the latter force is the apparent weight. Thus grav force + apparent weight = centripetal force - m * 9.8 m/s^2 + wtApparent = m * (-3 m/s^2 ) wtApparent = m (-3 m/s^2) + m ( 9.8 m/s^2) = m (6.8 m/s^2). A similar analysis at the bottom, where the centripetal force will be toward the center, therefore upward, gives us - m * 9.8 m/s^2 + wtApparent = m * (+3 m/s^2 ) wtApparent = m (+3 m/s^2) + m ( 9.8 m/s^2) = m (12.8 m/s^2). The ratio of weights is thus 12.8 / 6.8, approx. ** A more elegant solution obtains the centripetal force for this situation symbolically: Centripetal accel is v^2 / r. Since for a point on the rim we have v = `pi * diam / period = `pi * 2 * r / period, we obtain aCent = v^2 / r = [ 4 `pi^2 r^2 / period^2 ] / r = 4 `pi^2 r / period^2. For the present case r = 12 meters and period is 12.5 sec so aCent = 4 `pi^2 * 12 m / (12.5 sec)^2 = 3 m/s^2, approx. This gives the same results as before. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. So if you have 2 spheres in space and they both start at rest, they will start moving toward each other because of the gravitational attraction. Is the reason that doesnt happen on Earth because of friction and the fact that the attraction is so tiny compared to the pull of the earth? #$&*