Lab 8

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Researcher: Sahil Shah Data Analysist: Olivia Swisher Principal Investigator: Molly Tierney Lab 8 Gravity Worksheet Part 1: The Sun Moves?! DA1: Take the data needed to find the exact position of the center of mass of the star planet system. In addition, include a screenshot of the simulation with the orbits traced out. Mass(kg) Velocity x (m) y (m) M_sun 1000 0.043453 92.38284 38.19445 M_earth 100 2.623339 101.9885 -105.338 Table 1. Data to find position of the center of mass of the star planet system

R1: explain how the Center of Mass formula was used to find the center of mass. The Center of Mass is the place on the axis or the coordinate where the free system is balenced due to gravity, and due to the mass of the system at the certain coordinate. The Center of Mass formula was used to find the center of mass by finding the torque. To find this, add the torque of the sun to the torque of the planet, this will provide the torque of the combined system. PI1: Make the explicit connection between the numerical value calculated for center of mass and its location on the screenshot of simulation provided by the DA and the question “What is it that the Sun is going around?”. You can use the tape measure option on the right-hand side of the screen to elaborate. Use the center of mass formula to explain what happens to the center of mass position when the mass of the star is extremely large when compared to the mass of the planet (like the sun earth system)? The value we calculated for center of mass for the x value was 93.26 and for y it was 25.15. This value is closer to the coordinates of the Sun at this point in time. The Sun had a x value of 92.38 and a y value of 38.19. Since the center of mass is closer to the sun, it looks to us as though the earth is orbiting the sun. The sun is going around the center of mass which causes it not to be as evident of a motion compared to the Earth since the center of mass is almost in the same position as the sun. Using the center of mass formula, if the mass of the star is extremely large compared to the mass of the planet, then the center of mass position will end up being closer to the star since it has the larger mass. The center of mass is a position that indicates where the average location of the mass is in the system, so when an object has a larger mass than another, the center of mass will be closer to the object containing the larger mass since it has more of an effect of where the average mass of the system is located. Part 2: Calculating G R2: explain the formulas used to solve for G. The formula that was used to solve is below, We used 9.81 m/s^2 because this is the acceleration of gravity on the surface of the earth at sea level. Therefore, the acceleration due to gravity is given by = GM/r^2. DA2: Record and present the data needed to find the solution.

Table 2. Data to calculate G for trial 1 Table 3. Data to calculate G for trial 2 Table 4. Data to calculate G for trial 3 PI2: G is one of the main results of this lab (but this value is only good for this computer simulation). Report on your three trials and comment on whether the three trials measured equivalent G values. Hint: Is the total energy constant for all three trials? The value of G for trial 1 was 1.16. The value of G for trial 2 was 0.99. The value of G for trial 3 was 1.0. The G values are very similar between the three trials which tells us that G is constant for all three trials. Task 3: Testing Kepler’s 2nd law R3: Explain in detail, the formulas used to calculate the two areas. The shoelace method was used to calculate the two areas. The purpose was a test trying to find the sum of the Area with the formula and the Area without the formula. By incorporating the assumption that

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coordinates for the initial position of the planet is (×1, y1), the final position of the sun is (x2, y2), and the final position of the planet is (×3, y3). The formula that was used was... 0.5 • ((X1 • Y2 - X2 • Y1) + (X2 • Y3 - X3 • Y2) This equation can be used for both of the triangles that were on either side of the orbit. The shoelace formula, is a derivative of the area formula for a triangle. (A=.S*b*h) DA3: Report and present all the data required for calculating the areas. Planet x initial y initial time(s) 192.6432 0 1.4 x final y final 192.6168 -4.19993 Sun x initial y initial Area 1 0 0 -809.089 Table 5. Data for area 1 Planet x initial y initial Time(s) -1238.25 -0.22175 1.4 x final y final

-1238.24 0.511315 Sun x initial y initial Area 2 0 0 -907.717 Table 6. Data for area 2 PI3: Based on the data provided by DA3, are the areas A1 and A2 equal? To what extent do these two areas agree? Hint: calculate the ratio of the areas (smaller area/ larger area) and report your result in percentage The areas are almost exactly equal. The A1 area we calculated was 809.089 while the A2 area was 907.717. The ratio of the areas given in a percentage is 89.1% which is a very close similarity which agrees with Kiepler’s second law.

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