#$&* course Phy 121 4.20.11 at 9:45pm
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Would it be negative, as I wrote? Or is it safer just to say 10,000? ------------------------------------------------ Self-critique rating:
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Given Solution: 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. COMMON ERROR: The planet is bigger so it attracts the satellite, but the satellite doesn't attract the planet. INSTRUCTOR RESPONSE: It's not correct to say that the planet, being much bigger, attracts the satellite while the satellite doesn't attract the planet. They attract one another, exerting equal and opposite forces on one another and therefore accelerate one another, but the due to its greater mass the planet's acceleration is much less than that of the satellite.< &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Gravity may or may not play a role, right? It depends on whether the planet has a gravitational force? I've read that all things with mass experience gravity although, I think I've also heard otherwise. ------------------------------------------------ Self-critique rating:
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It will experience 6x the force. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/5^2 = 0.04 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION What does it mean when the force is proportional to the inverse of the square of the distance?? INSTRUCTOR RESPONSE That means the if the distance changes by factor r, the force changes by factor 1 / r^2. So for example if the distance becomes half as great the force becomes 1 / (1/2)^2 = 1 / 4 = 4 times as great if the distance doubles the force becomes 1 / 2^2 = 1/4 as great if the distance increases by a factor of 3 the force would become 1 / 3^2 = 1/9 as great This goes back to the image of the gravitational effect of the mass being spread, as distance increases, over increasingly large concentric spheres. The areas of the spheres increase, spreading the gravitational effect over their increasing areas and reducing the gravitational field accordingly. Since the area of a sphere increases in proportion to the square of its radius, the gravitational field decreases in proportion to the square of the distance. So we say that the gravitational field, and hence the force felt by a small object at different distances, is inversely proportional to the distance. This explanation is intended just to give you the general idea of the proportionality and the reasons for it. A full explanation requires the concept of flux and the techniques of multivariable calculus. STUDENT QUESTION 40,000/8000= 5 so it is five times less 1/r^2 = 1/5^2 = 1/25 * 10,000= 400N I guess you divide what you have by what the original was always? INSTRUCTOR COMMENT There are more formal ways, but when it comes down to intuitive calculations you can do as follows: Find the ratio of the distances and square that ratio. Depending on which ratio you find, to get the new force you will either multiply or divide by the squared ratio, whichever makes sense. The one that makes sense will increase the force if the distance decreases, decrease the force if the distance increases. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = 6.67*10^-11 * 100 * 200 / 0.5^2 F = 5.3*10^-6 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = 5.3*10^-6 / 100kg a = 5.336*10^-8m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A mass of 100 kg subject a net force of 5.3 * 10^-6 N will have acceleration of a = 5.3 * 10^-6 N / (100 kg) = 5.3 * 10^-8 m/s^2. At this rate to move from rest (v0 = 0) thru the displacement of one centimeter (`ds = .01 m) would require time `dt such that `ds = v0 `dt + .5 a `dt^2; since v0 = 0 this relationship is just `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * .01 m / (5.3 * 10^-8 m/s^2) ) = `sqrt( 3.8 * 10^5 m / (m/s^2) ) = 6.2 * 10^2 sec, or about 10 minutes. Of course the time would be a bit shorter than this because the object, while moving somewhat closer (and while the other object in turn moved closer to the center of gravity of the system), would experience a slightly increasing force and therefore a slightly increasing acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. At what rate would the second object accelerate toward the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = 5.3 * 10^-6 / 200 kg a = 2.65 * 10^-8 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. At what rate would the second object accelerate toward the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = 5.3 * 10^-6 / 200 kg a = 2.65 * 10^-8 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!