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Some Graphical Solutions of the Kepler Problem

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Some Graphical Solutions of the Kepler Problem Some Graphical Solutions of the Kepler Problem Marc Frantz 1. INTRODUCTION. When a particle P moves in an inverse square acceleration field, its orbit is a conic section. If the acceleration is repulsive?that is, of the form /x||FP ||~3Fr for a positive constant ?jland a fixed point F?then the orbit is a branch of a hyperbola with F at the focus on the convex side of the branch. If the acceleration is attractive (?/x||TP\\~3FP), then the orbit is a circle with F at the center, an ellipse or parabola with F at a focus, or a branch of a hyperbola with F at the focus on the concave side of the branch. We refer to such orbits as Keplerian orbits after Johannes Kepler, who proposed them as models of planetary motion. It takes rather special conditions to set up circular or parabolic orbits, so it is proba bly safe to say that inNature most closed orbits are ellipses, and most nonclosed orbits are hyperbolas. Ironically, these are precisely the cases in which it is impossible to formulate the position of P as an elementary function of time t?the very thing we most want to know! Nevertheless, for any given time t it is possible to locate P using graphs of elemen tary functions, thereby achieving a partial victory. This endeavor has an interesting history in the case of elliptical orbits, where locating P is called the "Kepler problem." It involves a famous equation known as "Kepler's equation," which we discuss later. In his book Solving Kepler's Equation over Three Centuries Peter Colwell says [5, p. ix]: In virtually every decade from 1650 to the present there have been papers devoted to the Kepler a problem and its solution. We can see from a list of them that the problem has enticed wide variety of scientists to comment on or involve themselves in its solution. [T]he Kepler problem has acquired an undeniable luster and allure for the modern practi tioner. Any new technique for the treatment of transcendental equations should be applied to this illustrious test case; any new insight, however slight, lets its conceiver join an eminent list of contributors. Colwell's exhaustive survey of the subject includes power series solutions, iterative procedures, and graphical solutions. Names from the history of graphical solutions include those of Kepler himself, John-Dominique Cassini (namesake of the "Cassini division" in the rings of Saturn) and his son Jacques, Johann Franz Encke (namesake of Comet Encke and the "Encke gap" in Saturn's rings), Christopher Wren, John Wallis, and Isaac Newton [5]. This impressive history suggests that new graphical solutions should have some thing extra to recommend them. In this article we present an approach that gives double solutions for both elliptic and hyperbolic orbits, includes both attractive and repulsive accelerations, and leads to interesting animations linking Keplerian motion with other physical processes. Throughout the article we encourage the reader to look at and en joy these animations, which appear on the website [9]. 2. THE COSINE SOLUTIONS. We begin with two particularly simple solutions and derive forms of Kepler's equation in their proofs. January 2006] graphical solutions of the kepler problem 47 Proposition 1. Let S be the ellipse with equation 2 2 y x a1 b2 = = ? eccentricity ?, and foci F\ (0, as) and F2 (0, a?). Fix a positive parameter ? ?i, and let yt signify the cosine curve moving to the right at constant speed v (b/?)^/??L/a3 whose equation at time t is = ? y a cos((?/b)(x vt)). (1) = Then, at any time t, ? (~)yt {P\(t), P2(t)}, where the points P\ and P2 have the following properties: ? (i) Pi orbits anticlockwise relative to F{, with acceleration ?jl\\F\P\ \\~3F\Pi; (ii) P2 orbits clockwise relative to F2, with acceleration ??jl\\F2P2\\~3 F2P2. Figure 1. Illustration of Proposition 1. Proposition 2. Let T? be the upper branch of the hyperbola with equation 2 2 ___ Z_ = i a2 b2 = = ? eccentricity ?, and foci F\ (0, a?) and F2 (0, a?). Fix a positive parameter \i, and let Vt signify the hyperbolic cosine curve moving to the right at constant speed v = (b/?)y/?jl/o3 whose equation at time t is ? ? y a cosh ((?/b)(x vt)). (2) = Then, at any time t, TiC\Yt {P\(t), P2(t)}, where the points P\ and P2 have the following properties: ? (ir) Pi orbits anticlockwise relative to F\, with acceleration ?jl\\F\P\\\~3F\P\', (ii7) P2 orbits clockwise relative to F2, with (repulsive) acceleration /x||F2P21|~3P^n>. The proofs of Propositions 1 and 2 depend on facts from basic orbital mechanics, including the following. Consider a fixed point F and an attractive acceleration field -/x||Fa ||~3Pa for X ^ F. Any ellipse with F as a focus?and any branch of a hy perbola with F as a focus on the concave side?is a possible orbit for a point P subject 48 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113 Figure 2. Illustration of Proposition 2. ~3 to the given acceleration. Similarly, given a repulsive acceleration field ?jl \\Fa \\ Fa, any branch of a hyperbola with F as a focus on the convex side is a possible orbit for a point P. In either case, the motion of P is completely determined if we specify the field, the orbit curve, the direction of motion along the curve, and the initial position P(0). For example, the field and the orbit curve determine the energy of the orbit, which in turn determines the speed of P as a function of its position. Proof of Proposition 1. By the previous remarks there exist position functions P\(t) = and P2W satisfying (i) and (ii), respectively. To specify the two we want, put Pi (0) = = P2(0) (0, a). It remains to show that 8 n yt {Pi, P2} for all times t. By our choice of Pi and P2, they can be parameterized as = = Pj (?bsinEi, a cos Ei), P2 (Z?sin?2, a cos ?2), (3) = = where Ex and E2 are strictly increasing functions of time t with ?i(0) ?"2(0) 0. = = = For / 1, 2 define r? F? P? and r? dr?/dt. The positions and velocities are then = ? = 1*1 (?b sin Ei, a cos Ei as), r2 (b sin E2, a cos ?2 + as), = = ri (?bEi cos?i, ?aE\ sin?i), r2 (??2cos E2, -a?2 sin ?2) It is well known that = - ||r/ x r?|| y/?a(l s2) = ? (for example, see [6, p. 122]). Since b a\J\ s2, we have x = ||r/ Vi\\/(ab) y/?/a3. Computing cross products, we obtain x . = "|ri rill ^ ? d ? yff?/?3 e?i cos ?1 (Ei esin?i) ab dt and x = |r2 r2|| = = ^fi/a3 ?2 + s?2cos E2 ?(?2 + ?sin?2). ab January 2006] graphical solutions of the kepler problem 49 = = Using the initial values E{(0) E2(0) 0 we find that ? ? = V/x/a31 Ei ? sin Ei, y/fi/a31 E2 + ?sin?2. (4) := as mean can The quantity M >Jii/a31 is known the anomaly. We write the equa tions in (4) in terms of M as = - M Ei ? sin Ei (5) and = M E2 + ??nE2. (6) Equation (5) is Kepler's equation for elliptical orbits, and Ex is known as the ec centric anomaly. The quantity E2 is not the eccentric anomaly for P2, but we use it here to parallel our approach to the hyperbolic case. Notice that dM/dE? > 0, hence for given M the solutions of (5) and (6) for E{ and E2 are unique. now t = = = We fix in R and show that ? H yt {Pu P2}. Since vt (b/s)y/'?i/a31 (b/s)M, we can rewrite (1) as = - y a cos ((?/b)x M). (7) = ? Now define 0 (s/b)x M, so that = x (b/?)(M + (P). (8) Then by (7) and (8) a point (x, y) on y, has the form ((b/?)(M + 0), a cos 0). Con sequently, each point of ? Hyt has the form (b sin a, a cos a) for some real number a satisfying = (bsina, acosa) ((b/e)(M + 0), acose/)). (9) = = Equating y-components in (9) we have cosa cos0, so either sin a sin0 or = ? sin a sin0. = = If sin a sin0, the intersection point has the form Qx (??sin0, acos0), and = comparing gi with the right side of (9) gives (b/e)(M + 0) b sin0. Hence = = ? Q\ (b sin 0, a cos 0) =>>M 0 + s sin0. = The latter equation is just (5) with E\ replaced by ?0, so the unique solution is 0 ? = = E\. In this instance, gi (-?>sin?i, a cos E\) P\. = ? = If sin a sin 0, the intersection point has the form Q2 (?b sin 0, a cos 0), and = ? comparing Q2 with the right side of (9) gives (b/?)(M + 0) ?>sin0. Accordingly, = = ? 02 (-?>sin0, acos0) =^>M -0 ?sin0. = The latter equation is (6) with E2 replaced by ?0, yielding the unique solution 0 ? = ? E2. As a result, g2 (& sin ?2> a cos E2) P2. Remark. Although we could have argued the case for P2 by symmetry, we structured the proof as we did in order to abbreviate the proof in the hyperbolic case. 50 ? THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 113 Proof of Proposition 2 We sketch the proof, which exactly parallels that of Proposi tion 1. By a similar argument there exist position functions P\(t) and P2(t) satisfy = = ing (i') and (iir), respectively, with Pi(0) P2(0) (0,a).
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