MOONPASS Moonlight in Vermont - Don't let the moon phase you - at least without plotting It!! BRIEF ABSTRACT Here's a simple question to ask, but a hard one to answer! How much of the lit moon is visible? One solution is made easier with spherical coordinates. GENERAL INFORMATION FileName: MOONPASS Full title: Moonlight in Vermont - Don't let the moon phase you - at least without plotting It!! Last Update: 6/1/96 Developer: Lynn Kiaer, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering. STATEMENT OF PROBLEM How much of the lit moon is visible? At any given time, half of the moon is visible (that is, facing the earth), and half of the moon is lit. If we model the visible moon as a circle, what proportion of the circle is lit? KEYWORDS Spherical coordinates, integration. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem seems to be most appropriate late in a calculus course, but I have also proposed it as a modeling and discussion problem in the first week. In that case, I expect the students to come up with an approximation that involves a trigonometric function (usually cosine), and often ask for tracings, which can then be transferred to poster board, cut out, and weighed on a simple balance scale. If time is tight, the poster board shapes can be made in advance. Prerequisites This problem has a strong geometric component, and can be solved using a variety of coordinate systems. Time allotment - time management Expectations Future payoffs Extensions References and Sources POSSIBLE SOLUTION(S) It is clear that the lit proportion of the visible moon is a function of a single variable, which I will call d, the number of days past the new (the entire visible moon is unlit) moon. Model the moon as a unit sphere. Place the origin at the center of the moon, and let the great circle that defines the line between the visible and non-visible portions of the moon lie in the y-z plane. Direct the positive x-axis toward the earth. (This is a standard right-hand system.) Since we are only interested in the visible moon, we are only interested in the hemisphere whose circular base lies in the y-z plane, and in which all points have a non-negative x-coordinate. When the moon is new, none of the visible moon is lit. If we take 29 days as the period of the moon, then 14.5 days after a new moon, all of the visible moon is lit. At 7.25 and 21.75 days after a new moon, exactly half of the visible moon is lit. We will use these benchmarks to validate our model. Since the sun, earth and moon are all in approximately the same plane, we will model them as being in exactly the same plane. Thus, if we view the x-y plane as defining the moon's equator, the point on the surface of the moon that is closest to the sun at time d will always lie on the moon's equator. The point on the surface of the moon that is closest to the sun is a useful point. In spherical coordinates, we call that point SunPoint. (1, 2 Pi d/29 - Pi, pi/2). In[2]:= rho = 1; phi = Pi/2; theta = 2 Pi d / 29 - Pi; In[3]:= SunPoint = {rho, phi, theta}; We need a way of referring to the points that lie on the great circle that separates the lit and unlit halves of the moon. One way of identifying those points is to note that they are all points on the surface of the moon such that the vectors connecting these points to the origin are at a right angle to the vector to SunPoint. This is easier (at least for some of us) to set up in cartesian coordinates. In[4]:= CartSunPoint = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}; So we are seeking the set of all points (x, y, z) that satisfy In[5]:= eq1 = {x,y,z}.CartSunPoint == 0; and also lie on the surface of the moon. In[6]:= eq2 = x^2 + y^2 + z^2 == 1; So we have two equations in four unknowns (x, y, z and d). Since we expect our final answer to be in terms of d, we will have an infinite number of solutions to this pair of equations - which is good because we have an infinite number of points on this circle. We can solve this pair of equations for any two of the variables. It makes no sense to solve for d, but we can see how well we have progressed by solving In[7]:= sol = Solve[{eq1,eq2},{y,z}] Out[7]= 2 2 2 d Pi 2 {{z -> -Sqrt[1 - x - x Cot[------] ], 29 2 d Pi y -> -(x Cot[------])}, 29 2 2 2 d Pi 2 {z -> Sqrt[1 - x - x Cot[------] ], 29 2 d Pi y -> -(x Cot[------])}} 29 Since the upper portion of the moon has the same area as the lower portion, we can just use the upper part. In[8]:= ztop[x_] = sol[[2,1,2]]; y[x_] = sol[[2,2,2]]; A quick check on our accuracy so far: When x = 0, z should be 1 (or -1, but we are using the top part) and y should be 0. In[10]:= ztop[0] Out[10]= 1 In[11]:= y[0] Out[11]= 0 But this is not the most useful way to solve the equations. Now, if we could find y as a function of z, we could integrate with respect to z as z goes from 0 to 1. This would give us the area between the vertical axis and the curve. So why not solve for x and y? In[12]:= sol = Solve[{eq1,eq2},{x,y}] Out[12]= 2 2 d Pi Sqrt[1 - z ] Sin[------] 29 {{x -> -(---------------------------------), 2 d Pi 2 2 d Pi 2 Sqrt[Cos[------] + Sin[------] ] 29 29 2 2 d Pi Sqrt[1 - z ] Cos[------] 29 y -> ---------------------------------}, 2 d Pi 2 2 d Pi 2 Sqrt[Cos[------] + Sin[------] ] 29 29 2 2 d Pi Sqrt[1 - z ] Sin[------] 29 {x -> ---------------------------------, 2 d Pi 2 2 d Pi 2 Sqrt[Cos[------] + Sin[------] ] 29 29 2 2 d Pi Sqrt[1 - z ] Cos[------] 29 y -> -(---------------------------------)}} 2 d Pi 2 2 d Pi 2 Sqrt[Cos[------] + Sin[------] ] 29 29 It may not be immediately clear which solution we want, but since they are the same anyway except for sign, we can pick one and experiment. In[13]:= yofz[z_] = sol[[2,2,2]]; The lit proportion of the visible moon, LitAndVisible[d], will be one half plus or minus the area between the vertical axis and the curve. Conveniently the sign of the area changes as needed. (Convince yourself.) In[14]:= LitAndVisible[d_] = 1/2 + 2 Integrate[yofz[z],{z,0,1}]/Pi Out[14]= 2 d Pi Cos[------] 1 29 - - ----------- 2 2 We see that our benchmarks now work. In[15]:= LitAndVisible[0] LitAndVisible[7.25] //N LitAndVisible[14.5] //N LitAndVisible[21.75] //N Out[15]= 0 Out[16]= 0.5 Out[17]= 1. Out[18]= 0.5 In[21]:= Plot[LitAndVisible[d], {d, 0, 29}, AxesLabel -> {"day", "fraction lit"}] Out[21]= -Graphics- If students are asked to guess the function, and are convinced that a straight-line function will not work (perhaps by experimentation), a reasonable first guess regarding the answer would be: In[22]:= Plot[1/2 + 1/2 Sin[2 Pi d/29 - Pi/2], {d,0,29}, AxesLabel -> {"day", "fraction lit"}] Out[22]= -Graphics- It is a straightforward exercise in trig identities to show that these functions are equivalent! ISSUES IN SOLUTION This is a very complex problem. For most students an introductory step, how much is visible on a particular day after the new moon, is necessary. Having the students predict the lit proportion of the visible moon for that evening is a good start. If students do this problem as part of an introduction to trig functions, the more complicated part is a nice confirmation of the initial guess.