Herron.Darland.Aaronson.PHY211.Lab8Worksheet

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University of Kentucky *

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211

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Physics

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Dec 6, 2023

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PI: Rachel Herron DA: Nicole Darland Researcher: Holden Aaronson Lab 8 Worksheet: Gravity DA-1: Sun Earth Mass 1000 kg 100 kg X 0 122.123 Y 0 21.342 Center of mass X center of mass = 11.1020909 Y center of mass= 1.94018182 Researcher-1: The center of mass formula is as follows: x centerof mass = ( i m i x i ) i m i . This equation tells us that the sum of the mass multiplied by the distance of the objects (sun and earth) over the sum of both the masses of each object in the system gives us the center of mass for the two masses. Overall, the center of mass formula refers to the position of the objects or the average location of the mass in the system.
PI-1: We found that the center of mass was much closer to the more massive object (the sun), so the Earth is revolving the sun, but the sun is still slightly going around the Earth; it’s just less obvious. We calculated the center of mass (x) to be approximately 11.1 and (y) to be about 1.94, as indicated by the pink dot on the photo above. DA-2: TRIAL 1 G= 1.160827 7 Initial r1 Final r2 Mass Speed 500 Mass Speed 32.90784 1000 0 1000 0.874836 100 0 100 7.6324 TRIAL 2 G= 1.228429 9 Initial r1 Final r2 Mass Speed 500 Mass Speed 25.4981 1000 0 1000 0.91175 100 0 100 9.117457 TRIAL 3 G= 1.646569 3 Initial r1 Final r2 Mass Speed 1000 Mass Speed 33.34256 1000 0 1000 0.748044 100 0 100 9.4804 Researcher-2: To calculate G, we must use the Law of Conservation of Energy. The Law of Conservation of Energy says that energy cannot be created or destroyed. This means that we must find the energy for the whole system. This equation is as follows: Total Energy = KE initial + GPE initial + W external = KE final + GPE final . In this equation KE represents kinetic energy, GPE represents the gravitational potential energy, and W refers to the work done
by the person who is operating the simulation; however, we can conclude that this value is equal to 0. Because we’re dealing with gravitational potential energy our equation for that is as follows: GPE = G M 1 m 2 r . With these equations in mind, we can then go forward in calculating G. PI-2: Trial 1 calculated G to be 1.16, Trial 2: 1.22, and Trial 3: 1.65. They unfortunately did not measure perfectly equivalent G values, but they are decently close to each other and the goal value of 1. Any errors in this area are likely due to tiny inconsistencies in our values. DA-3: In order to calculate the areas, we needed the x and y coordinates of the initial of both the Sun and Earth, as well as the final for the Earth when orbiting in a elliptical around the sun. Area 1: x1=0 y1=0 area= 2117 x2= 79.5 y2= 57.7 time= 2.16 s x3=116.5 y3=31.3 5 A/T= 980 Area 2: x1=0 y1=0 area=2017. 3 x2=118.5 y2=15.1 9 time=2.03 s x3=90.1 y3=-22.5 A/T= 992.7 Researcher-3: The formulas used to calculate the two areas are Gauss’s Magic Shoelace Formula and Kepler’s 2 nd Law. We used Kepler’s 2 nd Law to determine that the planet traces equal areas on its orbit about the Sun in equal intervals of time. These areas form triangles which we then can calculate by using Gauss’s Magic Shoelace Formula. The formula is as follows: 1 2 x 1 y 2 + x 2 y 3 + x 3 y 1 x 2 y 1 x 3 y 2 x 1 y 3 ¿ .
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PI-3: The areas A 1 and A 2 are not exactly equal, but they are close. It was difficult to make the times precisely the same, so some of the coordinates were hard to catch and that may be why they aren’t exactly the same, but they still are very close. They agree with a 95% ratio.