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course phy 201
11/26 1:00 am
`q001. Using the fact that the gravitational force between two given particles is inversely proportional to the square of the distance between them, and that the gravitational force exerted on an object the Earth is inversely proportional to its distance from the center of the Earth:What would be the ratio of the force on a given object located two Earth radii from the center, to the force exerted on the same object at a distance of one Earth radii from the center?
Difference: 1 radian; 1/1^2 = 1; ratio of one;
What would be the ratio of the force on a given object located three Earth radii from the center, to the force exerted on the same object at a distance of one Earth radii from the center?
Difference: 2 radians; ¼ ratio;
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The correct word would be 'radii', not 'radians'. Distance can be measured in the same units as radius. A radian is not a measure of distance but of angle.
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What would be the ratio of the force on a given object located two Earth radii from the center, to the force exerted on the same object at a distance of three Earth radii from the center?
Difference: 1 radian; 1 ratio;
Sketch a circle representing the Earth, and sketch points representing a certain object at distances of 1, 2 and 3 Earth radii from the center. For each position sketch a vector representing the force exerted on the object at that position. The relative lengths of your vectors should be approximately in proportion to the forces.
According to your sketch what is the approximate length of each of the other vectors, as percents of the length of the longest?
1 radii: 100%
2 radii: 66%
3 radii: 30%
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You calculated that at 2 Earth radii the ratio of forces would be 1/4, which is 25%. This is not consistent with your 66% estimate.
What would the ratio of forces be at 3 Earth radii, and what should be the percent?
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`q002. A certain object has mass 800 kilograms.
How much force does gravity exert on it at the surface of the Earth?
9.8 m/s^2 * 800 kg = 7840 newtons of force
How much force does gravity exert on it at a distance of two Earth radii from the center of the Earth?
One radian = radius; two radii = twice radius; magnitude of force: 1; 7840 newtons of force;
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2 radii is 2 times as great as 1 radius, so the ratio would be 1 / 2^2. The force at a distance of 2 radii wouldn't be nearly as great as at the surface.
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How much force does gravity exert on it at a distance of three Earth radii from the center of the Earth?
Magnitude: ¼; 7840 * ¼ = 1960 newtons;
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RIght idea, However at 3 Earth radii the object is 3 times as far from the center of the Earth, so the ratio would be 1/ 3^2.
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`q003. Reasoning in terms of the inverse square proportionality, how much force would the Earth exert on the 800 kg object of the previous problem if it was located 1.4 Earth radii from the center?
0.4 difference; 4/10; 2/5; inverse of square: 25/4; 6.25 magnitude; 7840 * 6.25 = 49000 newtons;
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The ratio to use is the ratio of distances from the center of the Earth. You appear to be using the ratio of the distance of the object from the surface of the Earth to the distance from the center to the surface. By this reasoning an object at the surface, being 0 Earth radii from the surface, would experience a force of 1 / 0^2 times as great as itself, which makes no sense.
This is easily correct.
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Assume Earth to have radius 6400 km (this is in fact about half a percent greater than Earth's average radius).
How far from the center of the Earth would an object be if it was at 1.4 Earth radii from the center?
Distance from center: 0.4 (radius) + radius = 8960 km;
Using F = G M m / r^2, where G = 6.67 * 10^-11 N m^2 / kg^2, calculate the force exerted on the object at this position. The mass of the Earth is about 6 * 10^24 kg.
6.67e^-11 * 8e^2 * 6e^24 = 320.16e^15; 3.2e^17/(6400 km)^2;
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The mass is at 1.4 Earth radii from the center, or 8960 km. You denominator should be the square of this distance.
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3.2e^17/4.1e^10; 0.78e^7; 7.8e^6; 7,800,000 newtons;
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You didn't use units in your calculation. With the numbers you did use the units of the result would not be Newtons.
I suspect that you did think about the fact that a km is 1000 meters, but neglected to notice that the distance is squared.
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You need to straighten a couple of things out here. You're on the right track and it shouldn't take you more than a few minutes to do so.
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