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C A S I O w w w. C A S I O e d u C At I O n. C O m 1. POLYNOMIAL Investigation 6.5: Earth s Revolution Before the 1530s, many people believed the Earth was the center of the universe and that every celestial body revolved around it, including the sun! Copernicus, a Polish astronomer, presented a cosmological theory stating that the Earth and other heavenly bodies actually revolved around the sun. Although today his heliocentric system of the universe is widely accepted, in his day Copernicus s theories were considered extremely controversial. Reference: http://info-poland.buffalo.edu/classroom/kopernik/copernicus.shtml Kepler s First Law of Planetary Motion states that the orbit of a planet is elliptical, with the Sun at one of the foci. A complete revolution of the Earth about the Sun takes approximately 365.26 days. During its revolution around the sun, the Earth winter solstice and summer solstice occur at the ends of the major axis. On January 3, the Earth is at its perihelion, approximately 147 million km from the sun. On July 4, it is at its aphelion (its farthest point), approximately 152 million km from the sun. (We re actually closer to the sun in the Northern Hemisphere s winter than we are in summer.) These two values are at the endpoints of the major axis of its elliptical path. a. Set up an axis system that can describe the Earth s revolution about the Sun using the center of the major axis of the ellipse as the origin. What are the coordinates of the Earth at its perihelion and aphelion? b. What are the coordinates of the Sun? c. What are the coordinates of the Earth when it is an equal distance from both foci? d. Graph this ellipse and select an appropriate viewing window. Which view window best represents the eccentricity of the ellipse? e. Use the calculator to verify the location of the foci determined in part a. Why might these answers vary slightly? f. What is the eccentricity of the ellipse? Discuss what that this means in general terms. Reference: www.physicalgeography.net/fundamentals/6h.html AND LOGARITHMIC

1. POLYNOMIAL C A S I O p u t va l u e b A C k i n t h e e q u at I O n Sample Solution: Earth s Revolution AND a. Set up an axis system that can describe the Earth s revolution about the Sun using the center of the major axis of the ellipse as the origin. What are the coordinates of the Earth at its perihelion and aphelion? The total distance along the major axis is 2A. We ll find the sum of the distance from the earth to the Sun at both the perihelion and aphelion. This will give us the length of the entire major axis. By dividing by 2, we ll have the value of A. In RUN: Enter the values, using the c key as shown. This can help us avoid potential mistakes with inputting the correct number of 0 s. LOGARITHMIC Consequently, A is 1.495 x 10 8 km. At the perihelion, the Earth would be at -1.495 x 10 8 km, and at the aphelion, it would be at 1.495 x 10 8 km. b. What are the coordinates of the Sun? We will assume we are orienting the ellipse with the sun at the left focus. The Sun is located 1.47 x 10 8 km to the right of the perihelion and 1.52 x 10 8 km to the left of the aphelion. Though we need do only one of the calculations, for confirmation, we do both calculations as shown. From this, we find that the sun is located at (-2.5 x 10 6, 0). c. What are the coordinates of the Earth when it is an equal distance from both foci?

C A S I O w w w. C A S I O e d u C At I O n. C O m 1. POLYNOMIAL Earth s revolution The minor vertices are equal distances from each focus. Letting (0, B) and (0, -B) represent these vertices; the Earth will be at these points when it is equidistant from the two foci. We note that either of these points, we ll use (0, B), the Sun, and the origin form a right triangle. The hypotenuse of the triangle is the distance from (0, B) to the Sun, which, as we found earlier, is at (-2.5 x 10 6, 0). We know, however, that the sum of the distances from the Sun to (0, B) and from (0, B) to the other focus is 2A, so the distance from the Sun to (0, B) is precisely A, which we already know is 1.495 x 10 8 km. The leg from the origin to our point of interest has length B. The leg from the origin to the sun has length 2.5 x 10 6 km. Using the Pythagorean Theorem, we find that (2.5 x 10 6 ) 2 + B 2 = (1.495 x 10 8 ) 2. To find B, we used the RUN mode on the calculator. AND LOGARITHMIC Thus we find that B is approximately 1.495 x 10 8 km, which we note is almost identical to A. d. Graph this ellipse and select an appropriate viewing window. Which view window best represents the eccentricity of the ellipse? From the Conic Graphs mode: Scroll down until the form of the ellipse you want is highlighted and press l. Enter the values we have determined. Press u(draw) to draw the graph. At this point, you will likely want to adjust the viewing window. Press Le(V-Window) and make the following changes: Xmin: -2.0c8l; Xmax: 2.0c8l; Ymin: -2.0c8l; Ymax: 2.0c8l; and scales of 2.0c7l on both axes.

1. POLYNOMIAL C A S I O p u t va l u e b A C k i n t h e e q u at I O n Earth s revolution Press du(draw) to view the graph with the new view window settings. AND LOGARITHMIC The window is not yet squared, however, so we cannot tell how far removed from a circle this elliptical path really is. Here the ellipse appears elongated. Press Le(V-Window). Select y(square) for a square view window, followed by q(y-base) to use the y-base. The calculator will then select the appropriate x-range for the square view. The screen shot shows the calculator s adjusted values for the Xmin and Xmax. Redraw the graph. Note: how circular the path seems to be now. This is a much better visualization of how close the Earth s orbit is to a circle. e. Use the calculator to verify the location of the foci determined in part a. Why might these answers vary slightly? With the graph displayed, press Ly(G-Solv)q(FOCUS) to find the focus. Use! to find the other focus.

C A S I O w w w. C A S I O e d u C At I O n. C O m 1. POLYNOMIAL Earth s revolution Though we note slight rounding error, relative to the distances we are discussing, these are minor. What is noteworthy here is how close to the center the two foci appear to be. f. What is the eccentricity of the ellipse? Discuss what that means in general terms. With the graph displayed, press Ly(G-Solv)uq(e) to find the eccentricity value. AND The eccentricity of an ellipse is a number between 0 and 1. The closer the eccentricity is to 0, the closer the ellipse is to a circle. The closer it is to 1, the more the ellipse collapses on itself, making a very narrow ellipse. Here we see the eccentricity is approximately 0.0167; the Earth s trip around the sun is very close to circular. LOGARITHMIC