Elliptical Orbits in the Solar SystemDate: 05/22/2005 at 16:29:42 From: Rick Subject: Contructing ellipses with specific sizes I teach middle school math and I want to have my students draw a scale model of the solar system that shows the orbits of the planets. Assuming I have the apogee and perigee of each planet's orbit about the sun they need to construct 9 ellipses with some degree of accuracy. I am hoping there is some variation on the tack and string method that wouldn't involve nine different tack settings and string lengths. Date: 05/22/2005 at 22:59:51 From: Doctor Peterson Subject: Re: Contructing ellipses with specific sizes Hi, Rick. Let me start with a quick side comment. The words "apoGEE" and "periGEE" refer to the greatest and least distances of a satellite from the EARTH ("ge" in Greek); for planets, you should use "apHELION" and "periHELION", which refer to the SUN. See the following page, for a term I didn't know! http://mathworld.wolfram.com/Apoapsis.html If I recall correctly, most of the planets have a small enough eccentricity that you can get a pretty accurate approximation to the ellipse by just marking the aphelion and perihelion, then finding the midpoint and drawing a circle with that center. You could then check the minor semiaxis and see whether it is noticeably different from the radius of the circle you drew. Here are the basics on ellipses: The semiaxes are a and b, in the x- and y- directions respectively; that is, these are half the "diameters" in each direction. The distance from the center to the focus is c, and a^2 = b^2 + c^2 | b* * | \ * * | \ * --*--+-------+-------+--*-- -a*-c | c *a * | * -b* | <-----------------> r1 <--> r2 If the apoapsis is r1 and the periapsis is r2, then r1 = a+c r2 = a-c Therefore, a = (r1+r2)/2 c = (r1-r2)/2 b = sqrt(a^2-c^2) = sqrt[(a+c)(a-c)] = sqrt(r1*r2) Note that the major semiaxis is the arithmetic mean of the two distances, and the minor semiaxis is the geometric mean! If your focus (sun) is at the origin, and you mark the perihelion on the left and the aphelion on the right, like this: center <-----+----------+-o------------+----> -r2 0 r1 then the center of the circle would be at c=(r1-r2)/2, and its radius would be a=(r1+r2)/2. Thus the radius of the circle is the correct value of a, the major semiaxis, but the correct value of b would be smaller than that. By how much? I looked up the data for earth, and find aphelion: r1 = 152.6 million km perihelion: r2 = 147.5 million km a = 150.05 million km c = 2.55 million km b = 150.03 million km You can see that in this case, the minor semiaxis is so close to the major semiaxis that there is no need to draw an exact ellipse; it looks practically identical to a circle, even though its center is offset noticeably from the sun. At least Pluto, and possibly others, will require you to draw the ellipse with more care. I looked up the data for Pluto and find aphelion: r1 = 7,000 million km perihelion: r2 = 4,500 million km a = 5750 million km c = 1250 million km b = 5612.5 million km In this case the minor semiaxis is significantly smaller than the major semiaxis, so you would want to use some method to accurately draw the ellipse. If you don't like pencil-and-string, try this one: Accurate Drawing of an Ellipse http://mathforum.org/library/drmath/view/55085.html That uses the circle we've already drawn, and another with the same center, to draw accurate points on the ellipse, which you can connect with a smooth curve. You can find more on the string method at the bottom of this page, and in the links: Defining an Ellipse http://mathforum.org/library/drmath/view/63463.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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