course Phy 231 a·|åÞ¶ï®×êÜvʇ’Þzšó^ƒÚassignment #027
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21:17:56 `q001. Note that this assignment contains 8 questions. Masses attract each other. The forces of attraction are equal and opposite: The force exerted by one small concentrated mass on another is equal in magnitude but in the opposite direction from the force exerted on it by the other. Greater masses exert greater attractions on one another. If two such objects remain separated by the same distance while one object increases to 10 times its original mass while the other remains the same, there will be 10 times the original force. If both objects increase to 10 times their original masses, there will be 100 times the original force. The force of attraction is inversely proportional to the square of the distance between the objects. That means that if the objects move twice as far apart, the force becomes 1 / 2^2 = 1/4 as great; if they move 10 times as far apart, the force becomes 1 / 10^2 = 1/100 as great. The same statements hold for spherical objects which have mass distributions which are symmetric about their centers, provided we regard the distance between the objects as the distance between their centers. Suppose a planet exerts a force of 10,000 Newtons on a certain object (perhaps a satellite) when that object is 8000 kilometers from the center of the planet. How much force does the satellite exert on the planet?
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RESPONSE --> The force that the satelite exerts on the planet will be equal and opposite of the force that the planet exerts on the satelite. At a distance of 8000km, the force exerted will be 10000N * 1/8000km^2 = 1.56 * 10 ^-10N confidence assessment: 2
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21:29:28 The gravitational forces exerted by the planet and the object are equal and opposite, and are both forces of attraction, so that the object must be exerting a force of 10,000 Newtons on the planet. The object is pulled toward the planet, and the planet is pulled toward the object.
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RESPONSE --> Well I was right about the equal and opposite, but was confused by the rest of the information and therefore bungled it. It would be more appropriate to say that F is inversely proportional to r^2 where r is the distance between the two bodies, not what I had at the end. self critique assessment: 2
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21:34:27 `q002. If the object and the planet are both being pulled by the same force, why is it that the object accelerates toward the planet rather than the planet accelerating toward the object?
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RESPONSE --> The smaller object has less mass. Say two objects have 10N exerted against each other and one object has 1kg and another has 5kg. The one with more mass has much less acceleration at 10N/5kg = 2m/s^2 and the other has 10N/1kg = 10m/s^2 so the smaller object will accelerate more than the other. confidence assessment: 3
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21:34:47 Presumably the planet is much more massive than the object. Since the acceleration of any object is equal to the net force acting on it divided by its mass, the planet with its much greater mass will experience much less acceleration. The minuscule acceleration of the planet toward a small satellite will not be noticed by the inhabitants of the planet.
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RESPONSE --> self critique assessment: 3
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21:45:08 `q003. If the mass of the object in the preceding exercise is suddenly cut in half, as say by a satellite burning fuel, while the distance remains at 8000 km, then what will be the gravitational force exerted on it by the planet?
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RESPONSE --> The force will be cut in half. confidence assessment: 2
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21:45:21 Halving the mass of the object, while implicitly keeping the mass of the planet and the distance of the object the same, will halve the force of mutual attraction from 10,000 Newtons to 5,000 Newtons.
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RESPONSE --> self critique assessment: 3
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21:47:28 `q004. How much force would be experienced by a satellite with 6 times the mass of this object at 8000 km from the center of a planet with half the mass of the original planet?
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RESPONSE --> It would be 3 times as much force. 6 times as much for the satellite mass and half as much for the planet, so overal it would be 3 times as much force. confidence assessment: 3
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21:47:38 The distance is the same as in the previous examples, so increasing the mass by a factor of 6 would to result in 6 times the force, provided everything else remained the same; but halving the mass of the planet would result in halving this force so the resulting force would be only 1/2 * 6 = 3 times is great as the original, or 3 * 10,000 N = 30,000 N.
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RESPONSE --> self critique assessment: 3
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22:16:50 `q005. How much force would be experienced by the original object at a distance of 40,000 km from the center of the original planet?
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RESPONSE --> Since the mass is constant, we can say: F = k * 1/r^2 where F is 10,000N and r = 8000km, then k = 6.4 * 10^10 kg * km^2 * m/s^2 If the distance is increased to 40,000,000 m then the force would be 6.4 * 10^10 kg * km^2 * m/s^2 / (4.0 * 10^4 km)^2 = 40N confidence assessment: 3
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22:21:43 The object is 40,000 km / (8000 km) = 5 times as far from the planet as originally. Since the force is proportional to the inverse of the square of the distance, the object will at this new distance experience a force of 1 / 5^2 = 1/25 times the original, or 1/25 * 10,000 N = 400 N.
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RESPONSE --> My calculation was off because of an error in 64 * 10^10 to 6.4 * 10^10 when it should have been 6.4 * 10^11. This should have been apparent to me to look at the proportionality and would have been much less error prone. self critique assessment: 2
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22:26:16 `q006. The relationship between the force of attraction and the masses and separation can be expressed by a proportionality. If the masses of two small, uniformly spherical objects are m1 and m2, and if the distance between these masses is r, then the force of attraction between the two objects is given by F = G * m1 * m2 / r^2. G is a constant of proportionality equal to 6.67 * 10^-11 N m^2 / kg^2. Find the force of attraction between a 100 kg uniform lead sphere and a 200 kg uniform lead sphere separated by a center-to-center distance of .5 meter.
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RESPONSE --> F = 6.67 * 10^-11 N m^2/kg^2 * 100kg * 200kg/(.5m)^2 F = 5.3* 10^-6 N confidence assessment: 3
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22:26:30 We are given the two masses m1 = 100 kg and m2 = 200 kg and the separation r = .5 meter between their centers. We can use the relationship F = G * m1 * m2 / r^2 directly by simply substituting the masses and the separation. We find that the force is F = 6.67 * 10^-11 N m^2 / kg^2 * 100 kg * 200 kg / (.5 m)^2 = 5.3 * 10^-6 Newton. Note that the m^2 unit in G will be divided by the square of the m unit in the denominator, and that the kg^2 in the denominator of G will be multiplied by the kg^2 we get from multiplying the two masses, so that the m^2 and the kg^2 units disappear from our calculation.
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RESPONSE --> self critique assessment: 3
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22:45:52 `q007. If these two objects were somehow suspended so that the net force on them was just their mutual gravitational attraction, at what rate would the first object accelerate toward the second, and if both objects were originally are rest approximately how long would it take it to move the first centimeter?
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RESPONSE --> The acceleration of the 100kg mass would be a = F/m, so 5.3 * 10^-6 N / 100kg = 5.3 * 10^-8 m/s^2 The acceleration for the 200kg mass would be: 5.3 * 10^-6N / 200kg = 2.65 * 10^-8 m/s^2 Since the accelerations are moving both objects toward each other, it makes sense to add them together to reach the acceleration of the system. This gives us 7.95 * 10^-6 cm/s^2 Using `ds = v0`dt + .5a`dt^2 and using `ds = 1cm and a = 7.95 * 10^-6 cm/s^2, `dt = sqrt(1cm/ 7.95 * 10^-6 cm/s^2) = 355s confidence assessment: 2
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22:50:24 A mass of 100 kg subject a net force of 5.3 * 10^-6 N will have acceleration of a = 5.3 * 10^-6 N / (100 kg) = 5.3 * 10^-8 m/s^2. At this rate to move from rest (v0 = 0) thru the displacement of one centimeter (`ds = .01 m) would require time `dt such that `ds = v0 `dt + .5 a `dt^2; since v0 = 0 this relationship is just `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * .01 m / (5.3 * 10^-8 m/s^2) ) = `sqrt( 3.8 * 10^5 m / (m/s^2) ) = 6.2 * 10^2 sec, or about 10 minutes. Of course the time would be a bit shorter than this because the object, while moving somewhat closer (and while the other object in turn moved closer to the center of gravity of the system), would experience a slightly increasing force and therefore a slightly increasing acceleration.
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RESPONSE --> While I had the formula right, I missed the 2 * `ds. Even still, does it not make sense that the other object is accelerating toward the smaller object as well? I also understand that actually the acceleration would be increasing as the objects moved closer together. self critique assessment: 2
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22:51:32 `q008. At what rate would the second object accelerate toward the first?
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RESPONSE --> I think I answered that before. It would be accelerating at 5.3 * 10^-6N / 200kg = 2.65 * 10^-8 m/s^2 confidence assessment: 3
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22:51:50 The second object, with its 200 kg mass, would also a subject to a net force of 5.3 * 10^-6 N and would therefore experience and acceleration of a = 5.3 * 10^-6 N / (200 kg) = 2.7 * 10^-8 m/s^2. This is half the rate at which the first object changes its velocity; this is due to the equal and opposite nature of the forces and to the fact that the second object has twice the mass of the first.
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RESPONSE --> self critique assessment: 3
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