HW5_solution

HW5_solution - H45 Harm/org #f grgblgm ,2;1. Since the...

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Unformatted text preview: H45 Harm/org #f grgblgm ,2;1. Since the problem is to calculate the position of Halley’s comet at your current date, the actual solution depends on the date you are doing the problem. a date of October 29, 1988. The semimajor axis of the orbit is 17.9564 A.U. = 11 In this solution, we will assume 2.68625x109 km. 2 The value of u for the sun is he = 1.32715 x 10 km3/sec . The period of the orbit is then given by Kepler’s third law (2.37) as T = 2 " a”: = 2.40127 x 109 sec VuG = 27792.5 days = 76 years 33.5 days at 86,400 sec/day (exact) and 365.25 days/year (approximate). The date of the next perihelion passage is then approximately March 13, 2062. To calculate the position of Halley's comet on October 29, 1988, we need to find the time interval since perihelion passage, t - To. This is, for the given date, 993 days, or 8.579 x 107 seconds. Tb? mean motion of Halley’s comet is a _ u _ -9 . n - ——§ — 2.6166 x 10 radians/sec a The mean anomaly is then M = n( t — To) = 0.22447 radians. It is necessary to iteratively solve Kepler's equation to find the value oF the eccentric anomaly E which duplicates this value of M. Begin with the approximate solution (2.65) E = M + e sin M = .4397805 radians The Newton — Raphson method to solving Kepler's equation begins with the calculation of the error AMk (2.66) AMk = Ek — e Sln Ek — M = —O.196507 rad W5 14K. However? #f fob/Ibo . and the slope of Kepler’s equation (2.68) dM 3E = l - e cos B = 0.124745 Then, the correction to the current value ER is found from AEk = - AMk / dM/dE = +1.57527 rad This correction is added to the first value of E to find the new valuef Ez. Obviously, the correction was rather large, Raphson method cannot be said to have converged. next six iterations of the method. so the Newton —§ The table shows the§ Convergence is slow at first, buté then becomes extremely rapid once the method has found the vicinity of the true answer (about 4). this happens, the; convergence is quadratic, as can be seen by the approximate doubling; of the number of leading zeros in the third and last columns of the; iteration Once 2/5 table. k AE 1 .4397805 -0.196507 0.124745 1.575270 2 2.015054 0.917182 1.415733 “0.647849 3 1.367204 0.195414 0.804423 -0.242924 4 1.124279 0.027348 0.582295 -0.0469671 5 1.077312 0.000954 ' 10.541794 ‘0.0017624 6 1.075550 . 0.0000013 0.540293 -0.0000024 7 1.075547 0.0 The value of E must then be converted into a true anomaly using _ (2.70) This gives v tan ( v/z ) = 2.715791 radians. section equation (2.71) r a a( 1 - e2) 1 + e cos v u“ a /—%§§— tan ( 3/2) Inserting this into the conic 9.70167 A.U. or, about the distance to the planet Saturn. MM w- Hamworfl 45‘)” swan. The case of Halley's comet is extreme, since theeccentricity of the elliptical orbit is quite high. The initial approximation (2.65} is of relatively little value, and relatively many iterations of the Newton — Raphson method are needed. For a much smaller eccentricity, this is not true. Also, for a small eccentricity, the three anomalies 3, E and t» are much closer together than they are in the case of fialley's comet. Ragim Z : [ago/sch. Mamag‘ Ergblgm 2—6. The approximate solution procedure began by assumingé e=0, so E a M. Substituting this into Kepler’s equation in the form Ef = M + e sinE led to equation (2.65) E a M + e sin M To continue this process, substitute the above approximation for E into the right hand side of Kepler's equation, E = 14 + e sinE, to obtain" E a M + e sin { M + e sin M } This form is rather messy, and includes the unfamiliar sine of a sine function. If we apply the trigonometric identity for thesine of a sum, the above takes the form E a M + e{ sinM cos( e sinM } + cosM sin{ e sinM }} How, in this form we can profitably apply the assumption that the orbital eccentricity e is small. If e << 1, then the cosine and sine above can be approximated as cos( e sinM } a l a sin( e sinM } a e SinM éyg WE Mo/ _ Hana/w! #fegéofiéw. 4/23 These are just the usual small angle approximations of sine and cosine, where the eccentricity e ensures that the 'angle’ remains small. Inserting these into Kepler’s equation, we have E w M + e sinM + e2 cosM sinM = M + e sinM + «3— e2 sin 2M 2 This is the desired result, involving powers of the eccentricity and sines of multiples of the mean anomaly M. 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HW5_solution - H45 Harm/org #f grgblgm ,2;1. Since the...

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