#$&* course PHY 201 4/29 about 5:45 pm 028. `query 28
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Given Solution: `a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k / rE^2. Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as a = g rE^2 / r^2. If we set the acceleration equal to v^2 / r, we obtain v^2 / r = g ( rE / r)^2 so that v^2 = g ( rE^2 / r) and v = sqrt( g rE^2 / r) = rE sqrt( g / r) Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth. If we do know G and the mass of the Earth, we can proceed as follows: The gravitational force on mass m at distance r from the center of the Earth is F = G m M / r^2, Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have m v^2 / r = G m M / r^2, which we solve for v to get • v = sqrt( G M / r). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qPrinciples of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A_g = (G*m_moon)/(r^2) A_g = ((6.67*10^-11 N*m^2/kg^2)*(7.35*10^22 kg))/(1.74*10^6 m)^2 A_g = (4.90*10^12 N*m^2/kg)/(3.03*10^12 m^2) = 1.62 N/kg = 1.62 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ** STUDENT COMMENT The problem is set up correct but was not solved for. The answer for the acceleration from gravity is 1.619 m/s^2 INSTRUCTOR RESPONSE The numbers work out to 6.67 * 7.35 / (1.74)^2 * 10^-11 * 10^22 / (10^12), with 10^12 being the square of 10^6. The powers of 10 therefore work out to 10^(-11 + 22 - 12) = 10^(-1). You shouldn't require a calculator to get this must. You can then use the calculator to find 6.67 * 7.35 / (1.74)^2. Your result is 16 * 10^-1 = 1.6. The units work out as indicated. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: G=6.67*10^-11 N*m^2/kg^2 M_e = 5.97*10^24 kg M_v = 0.815*m_e = 4.84*10^24 kg M_j = 318*m_e = 1.90*10^27 kg M_s = 95.1*m_e = 5.68*10^26 kg R_v = 1.50*10^11 m - 1.08*10^11 m = 4.20*10^10 m R_j = 7.78*10^11 m - 1.50*10^11 m = 6.28*10^11 m R_s = 1.43*10^12 m - 1.50*10^11 m = 1.28*10^12 m F_g = (G*m_e*m)/(r^2) F_v = (G*m_e*m_v)/(r_v)^2 = 1.10*10^18 N F_j = (G*m_e*m_j)/(r_j)^2 = 1.92*10^18 N F_s = (G*m_e*m_s)/(r_s)^2 = 1.38*10^17 N F_net = -f_v+f_j+f_s = 9.57*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.. ** STUDENT QUESTION Where do we get 1.5 * 10^11 m as part of r? I’m slightly confused as to where that came from? INSTRUCTOR RESPONSE Each calculation will be based on the masses of Earth and the planet in question, and on the distance between them when they are perfectly aligned. The distance between Earth and another planet at perfect alignment is the difference of their distances from the Sun. For example Venus is 1.08 * 10^11 m from the Sun, while Earth is 1.5 * 10^11 m from the Sun. The distance between Earth and Venus, when they line on a straight line with the Sun (and on the same side of the Sun) is therefore 1.5 * 10^11 m - 1.08 * 10^11 m. Thus the quantity (1.5 * 10^11 m - 1.08 * 10^11 m)^2 in the denominator of the calculation for Venus. Calculations for Jupiter and Saturn are similar, but each uses the appropriate distance of that planet from the Sun, and of course the mass of the planet. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: C=2*’pi*r = 2*’pi*12m = 75.4 m V=75.4 m / 12.5 s = 6 m/s A=v^2/r = (6 m/s)^2 / 12 m = 3 m/s^2 Since the motion is circular, the forces will add up to the centripetal force. F_g + F_N = F_c, where F_N is the force the car of the ferris wheel exerts on the person, or the apparent weight. F_N = F_c - F_g At the top F_g and F_c both act downward downward. F_N = (-(3 m/s^2 * mass))-(-(9.8 m/s^2 * mass)) = 6.8 m/s^2 * mass At the bottom, F_g is acting down but F_c is acting up to the center of the wheel. F_N = (3 m/s^2 * mass)-(-(9.8 m/s^2 * mass)) = 12.8 m/s^2 * mass Ratio of weight at the top vs. the bottom is 6.8 / 12.8 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 pi m/s, approx. (about 6 m/s) 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. STUDENT QUESTION 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 doesn’t happen on Earth because of friction and the fact that the attraction is so tiny compared to the pull of the earth? INSTRUCTOR RESPONSE Suppose you have two 1 kg spheres, with their centers separated by 10 cm. The gravitational attraction between the spheres would be about 10^-10 as great as the gravitational attraction of each to the Earth. So for example if one of the masses was suspended by a thread 1 meter long, then if the other mass is brought to within .1 meter, the position of the suspended mass would change by about 10^-10 meters, the approximate diameter of a single atom. We wouldn't be able to measure that. The gravitational forces between objects on the Earth are just too insignificant to be measured in this way. Now we could put the system into a vacuum and allow the suspended mass to oscillate back and forth past the other mass. The gravitational attraction would affect the period of the pendulum, and over a large number of oscillations we might be able to detect the effect. If we modify this idea slightly and use a torsion pendulum, then we have the basic idea of the Cavendish balance (you can search that term, which I believe you will encounter fairly soon in your text). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. STUDENT QUESTION 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 doesn’t happen on Earth because of friction and the fact that the attraction is so tiny compared to the pull of the earth? INSTRUCTOR RESPONSE Suppose you have two 1 kg spheres, with their centers separated by 10 cm. The gravitational attraction between the spheres would be about 10^-10 as great as the gravitational attraction of each to the Earth. So for example if one of the masses was suspended by a thread 1 meter long, then if the other mass is brought to within .1 meter, the position of the suspended mass would change by about 10^-10 meters, the approximate diameter of a single atom. We wouldn't be able to measure that. The gravitational forces between objects on the Earth are just too insignificant to be measured in this way. Now we could put the system into a vacuum and allow the suspended mass to oscillate back and forth past the other mass. The gravitational attraction would affect the period of the pendulum, and over a large number of oscillations we might be able to detect the effect. If we modify this idea slightly and use a torsion pendulum, then we have the basic idea of the Cavendish balance (you can search that term, which I believe you will encounter fairly soon in your text). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!