Distance to the SunDate: 04/16/99 at 12:12:32 From: Larissa Buchholz Subject: Calculus My teachers and I can't figure this one out. We would appreciate it if you could help! Thank you. According to Kepler's laws, the planets in our solar system move in elliptical orbits around the sun. If a planet's closest approach to the sun occurs at time t=0, then the distance r from the center of the planet to the center of the sun at some later time t can be determined from the equation r=a(1-ecos(phi)) where a is the average distance between centers, e is the positive constant that measures the "flatness" of the elliptical orbit, and phi is the solution of Kepler's equation: (2*pi*t)/T=phi-esin(phi) in which T is the time it takes for one complete orbit of the planet. Find the distance from Earth to the sun when t=90 days. {First find phi from Kepler's equation, and use this value of phi to find the distance. Use a = 150x10^6 km, e = 0.0167 and T = 365 days.} The problem we had was that we couldn't find a way to isolate phi to solve for phi in Kepler's equation. Date: 04/19/99 at 13:13:20 From: Doctor Fwg Subject: Re: Calculus Dear Larissa, Thank you for sending such an interesting question. This kind of problem is a little uncommon, except in more advanced mathematical and scientific situations, so it is not surprising that you and your teachers were stumped. However, there are at least a few different ways of dealing with it successfully. I am assuming that your own references show all the correct relations and definitions between the variables and constants that are needed here, along with appropriate diagrams, so I will not go into a lot of excessive detail about all that and will instead concentrate on one method of solution. The method I will describe here for you involves solution by successive approximation (SA) but you might also be able to use what is sometimes called Newton's Method (or the Newton-Raphson Method), but, in this case, I think Newton's Method will be more difficult to apply. When you have time you should look into Newton's Method though; it is very powerful, not too difficult to apply in most situations, and you will probably enjoy knowing more about it if you really like mathematics. Now the method of SA: Important Constants and Variables Phi = Angle (in radians) between center of elliptical orbit and planet Pi = 3.14159... e = Eccentricity (dimensionless) of elliptical orbit = approx. 0.0167 for earth T = Orbital period = approx. 365 for earth t = Time (in days) after Perihelion a = Semi major axis of elliptical orbit = approx. 150 x 10^6 km for earth R = Distance between earth and Sun (will come out with same dimensions as "a") Equations Phi = [(2 Pi t)/T] + [e Sin(Phi)], rearranged slightly R = a [1.0 - e Cos(Phi)] Notice that I have written the first equation for Phi with Phi on both sides. This technique allows us to guess at a value for Phi, place the guessed value in the RHS of the equation only (along with values of Pi, t, T, and e) and then calculate a new value for Phi. In this case, the new value for Phi is on the lefthand side (LHS) of the equation. If we happen to guess exactly right, the new value for Phi will equal the guessed value (this seldom occurs). If we are lucky or clever in our initial guesses and in the way we have chosen to re- write the original equation, the new value of Phi will be a little closer (at least) to the true value we are after. This process is repeated by putting every new value of Phi only into the righthand side (RHS) of the equation again and repeating the process until there is no difference between the calculated and guessed value. In some cases, the process of convergence occurs very rapidly, i.e., only a few iterations are necessary before the differences between the guessed and calculated values agree. As a side comment, note: agreement of almost any reasonable number of significant figures can usually be obtained using this technique over and over again, but usually it is not necessary to go beyond the number of significant figures known to exist within the least well known quantity employed in the calculation. In this case, three significant figures are probably all that are needed. Do you see why? In some applications of this method, what may be referred to as divergence occurs, and each calculated answer moves farther and farther away from the "correct" solution. If this happens, it will be obvious. Sometimes this situation can be corrected by re-expressing the original equation in another way. In general, it is best to try to put the term or expression that is most sensitive to change on the LHS of the equation before attempting the method of SA. A hand calculator is okay for this kind of thing but if you know how to use a computer spreadsheet you can more easily follow how each iteration affects your answer. Now, a few more comments: the elliptical orbit of the earth is almost circular, so in 90 days, Phi will be about 90 degrees or Pi/2 (i.e., 1.57080) radians because 90 days is very near the time interval needed for the earth to complete 1/4 of its orbit around the Sun. So, a good 1st approximation (i.e., guess) for Phi is 1.57080 radians. Placing that value of Phi into the RHS of the equation (above), you should get a second (new) approximation for Phi of about 1.56598. Note: for the time being I am deliberately ignoring my earlier comments about significant figures. Using the second approximation for Phi of 1.56598, and repeating the calculation, you should get a third (new) approximation of about 1.56598. You can stop here because the new answer is equal (within six significant figures) to the previous estimate. If you convert this last value of Phi (1.56598 radians) to degrees, you should get about 89.724 deg. So, you can see that our first approximation (i.e., guess) of about 90 deg in 90 days wasn't too bad. We were a little lucky here; if the orbit had been "highly elliptical," our first guess might not have been so good. But even then it would have probably only taken a few more iterations for our solutions to converge. You might want to try this kind of calculation with Pluto - it has a highly elliptical orbit but you need to remember to use the values of a, e, t, and T that apply to Pluto. In any case, you should check over my results since I cannot guarantee that I haven't made some error myself, and you also need to complete the calculation to get the distance between the earth and the Sun 90 days after Perihelion. You may also want to try this calculation for Earth at t = 180 days. What would be a good initial guess for Phi in this case? I hope this helps out. With Best Wishes, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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