THE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI

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1 THE 2-BODY PROBLEM ROBERT J. VANDERBEI ABSTRACT. In his shor noe, we show ha a pair of ellipses wih a common focus is a soluion o he 2-body problem. INTRODUCTION. Solving he 2-body problem from scrach is doable bu difficul. Bu, wha if we simply wan o verify ha here are ellipical orbis where he wo ellipses share a common focus, which is also he cener of mass of he sysem? Wih hese supposiions, maybe his problem isn so bad. In oher words, le s ry o use guess-n-check mehod. I should be easier. (We all believe ha P NP, righ!) Here we go... Firs, choose he coordinae sysem so ha he foci lie on he horizonal x-axis and so ha he shared focus is a he origin (see Figure). One body s second focus is on he posiive x-axis and he oher body s second focus is on he negaive x-axis. Call he righ body body one and he lef body will be body wo. The orbi of Key words and phrases. Celesial mechanics, n-body problem, ellipse. (c,b) c c+a FIGURE. A pair of ellipses sharing a common focus.

2 2 ROBERT J. VANDERBEI body one can be given paramerically as x c + a cos θ y b sin θ. Here, a, b, and c are consans whereas x, y, and θ are funcions of ime. The consan c is he x-coordinae of he cener of he ellipse and he consans a and b are he semi-major and semiminor axes, respecively. Clearly all hree consans are posiive numbers and a > b. Furhermore, an imporan propery of ellipses is ha he disance from he cener of he ellipse o a focus is a2 b 2. Since (c, 0) is he cener of he ellipse and (0, 0) is a focus, i follows ha c a 2 b 2. The assumpion ha he cener of mass of he sysem coincides wih he focus a he origin implies ha ] ] x2 x. y 2 y The disance r beween he wo bodies plays and imporan role in Newon s law of graviy, so we sar by compuing i: r (x 2 x ) 2 + (y 2 y ) 2 (2x ) 2 + (2y ) 2 2 x 2 + y 2 2 (c + a cos θ) 2 + (b sin θ) 2 2 c 2 + 2ac cos θ + a 2 cos 2 θ + b 2 sin 2 θ 2 c 2 + 2ac cos θ + a 2 cos 2 θ + (a 2 c 2 ) sin 2 θ 2 c 2 cos 2 θ + 2ac cos θ + a 2 2(a + c cos θ). Newon s laws involve acceleraions and so we differeniae once and hen a second ime ẋ a sin θ θ ẏ b cos θ θ. ẍ a sin θ θ a cos θ θ 2 ÿ b cos θ θ b sin θ θ 2. For simpliciy, and wihou loss of generaliy, suppose ha unis are chosen in such a way ha he graviaional consan G equals one. Also for simpliciy, bu wih some loss of generaliy, suppose

3 THE 2-BODY PROBLEM 3 ha he masses of he wo bodies are boh equal o one. Wih hese assumpions, Newon s law of graviy is ẍ x 2 x x r 3 ÿ y 2 y r 3 y. Subsiuing he formulas for x, y, and heir second derivaives derived above, we ge a sin θ θ a cos θ θ 2 b cos θ θ b sin θ θ 2 (c + a cos θ) (b sin θ). Negaing boh sides and wriing in marix form, we see ha ] ] c/a + cos θ sin θ cos θ θ cos θ sin θ θ 2 sin θ Muliplying by he marix inverse, we ge θ θ 2 ] sin θ cos θ cos θ sin θ ] c/a + cos θ sin θ ] sin θ(c/a + cos θ) cos θ sin θ cos θ(c/a + cos θ) + sin θ sin θ ] (c/a) sin θ (c/a) cos θ + 4a(a + c cos θ) 3 4a(a + c cos θ) 2 In oher words, for a soluion o exis, he funcion θ mus simulaneously be a soluion o boh of he following, possibly inconsisen, equaions: θ 4a(a + c cos θ) 3 θ 2 a(a + c cos θ)

4 4 ROBERT J. VANDERBEI To verify consisency, we check ha differeniaing he equaion for θ brings us o he equaion derived above for θ: θ 2 θ a (a + c cos θ) 2 2 a (a + c cos θ) 2 2 a(a + c cos θ) 4a(a + c cos θ). 3 Hence, we see ha he formulas for he firs and second derivaive of θ are indeed consisen wih each oher. Picking θ(0) can be given any value as i merely deermines where he bodies are in heir orbi a ime 0. The semi-major and semi-minor axes a and b deermine c (as already menioned) and θ(0): θ(0) (0) 4a(a + c cos θ(0)) 3. Suppose, for simpliciy, ha θ(0) 0. The differenial equaion for θ() can be rewrien as (a + c cos θ) dθ d 2 a. Since ε c/a is he well-known eccenriciy of he ellipse, i is common o divide boh sides of his equaion by a hereby replacing c wih he eccenriciy ε: Inegraing from 0 o, we ge ( + ε cos θ) dθ d θ() 0 ( + ε cos θ) dθ The inegrals can be compued explicily. The resul is θ() + ε sin θ() 0 d Unforunaely, his is a ranscendenal equaion for θ() and so i does no have a simple closed form soluion. However, in he case where he eccenriciy is zero, we have an exac soluion: θ()

5 THE 2-BODY PROBLEM 5 And, when he eccenriciy is no zero, he following recursion quickly converges on he correc answer: θ (0) () 0 θ () () 2a ε sin 3/2 θ(0) () θ (k+) ().. 2a 3/2 ε sin θ(k) () DEPARTMENT OF OPERATIONS RESEARCH AND FINANCIAL ENGINEERING, PRINCETON UNIVERSITY, PRINCE- TON, NJ address: rvdb@princeon.edu

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