course PHY 201 Hello Professor Smith, thought you might enjoy this cartoon which seems somewhat related to the coursework...http://xkcd.com/123/
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08:42:31 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
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RESPONSE --> WE know that gravitational acceleration is 9.8 m/s/s at the surface of earth, so we could multiply this value by an inverse square of the ratio of the two distances to give us gravity at some point above the surface. accel@radiusR = 9.8 m/s/s * 1/(r/rE)^2 9.8 m/s/s * 1/(r/Re)^2 = v^2 / r v = sqrt( r * 9.8 m/s/s * (1/(r/rE)^2)
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10:17:47 ** 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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RESPONSE --> I don't really understand this, I need to move on. I have found that I waste a lot of time ""roadblocking"" on certain things, but they tend to become clear over time if I just keep moving.
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14:42:33 Principles 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.
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RESPONSE --> Pick some value for mass 2: 1 kg Fgrav = G*m1* m2/r^2 = 6.67 x 10^-11 N m^2/kg^2 * 7.35 * 10^22 kg * 1 kg/ 1.74 * 10^6 ^2 Fgrav = 4.905 x 10^12 N m^2 / 3.03 * 10^12 m^2 Fgrav = 1.619 N a = F/m = 1.619 N / 1 kg a = 1.619 m/s^2
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14:45:19 ** 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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14:49:32 Query 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?
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RESPONSE --> Distances: Venus = 150 - 108 = 4.2 x 10^7 km Jupiter = 778 - 150 = 6.28 x 10^8 km Saturn = 1430 - 150 = 1.28 x 10^9 km Masses: Venus = .815 * 5.97 x 10^24 = 4.866 x 10^24 kg Jupiter = 318 * 5.97 x 10^24 = 1.90 x 10^27 kg Saturn = 95.1 * 5.97 x 10^24 = 5.68 x 10^26 kg FVenus = G * 4.866 x 10^24 kg * 5.97 x 10^24 / (4.2 x 10^7 km)^2 = 1.099 x 10^24 N FJupiter = G * 1.90 x 10^27 kg * 5.97 x 10^24 / (6.28 x 10^8 km)^2 = 1.919 x 10^24 N FSaturn = G * 5.68 x 10^26 kg * 5.97 x 10^24 / (1.28 x 10^9 km)^2 = 1.381 x 10^23 N 1.919x10^24N + 1.381x10^23N - 1.099x10^24 N = 9.581 x 10^23 N
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15:24:57 ** 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.. **
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RESPONSE --> I ended up with values that were significantly different than the given answer. I think that the difference lies in the use of 10^11 consistently for all values of the distances in the given answer. I interpreted the problem in the book differently and accordingly used a range of exponents between 10^7 and 10^9. I am not entirely sure which is right.
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15:25:10 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 univ phy.
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15:25:13 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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15:33:53 Query 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?
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RESPONSE --> circumference = 24 m * pi = 75.398 m velocity = 75.938 m / 12.5 s = 6.03 m/s a = v^2/r = 6.03 m/s^2 / 12 m = 3.03 m/s^2 aBottom = 9.8 m/s^2 + 3.03 m/s^2 = 12.83 m/s^2 aTop = 9.8 m/s^2 - 3.03 m/s^2 = 6.77 m/s^2 12.83 / 9.8 = 31% increase in apparent weight 6.77/9.8 = 31% decrease in apparent weight
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15:35:04 ** 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. **
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15:35:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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