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The Two-Body Problem Computational Astrophysics I: Introduction and basic concepts Helge Todt Astrophysics Institute of Physics and Astronomy University of Potsdam SoSe 2021, 3.6.2021 H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 1 / 32 The two-body problem H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 2 / 32 Equations of motionI We remember (?): The Kepler’s laws of planetary motion (1619) 1 Each planet moves in an elliptical orbit where the Sun is at one of the foci of the ellipse. 2 The velocity of a planet increases with decreasing distance to the Sun such, that the planet sweeps out equal areas in equal times. 3 The ratio P2=a3 is the same for all planets orbiting the Sun, where P is the orbital period and a is the semimajor axis of the ellipse. The 1. and 3. Kepler’s law describe the shape of the orbit (Copernicus: circles), but not the time dependence ~r(t). This can in general not be expressed by elementary mathematical functions (see below). Therefore we will try to find a numerical solution. H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 3 / 32 Equations of motionII Earth-Sun system ! two-body problem ! one-body problem via reduced mass of lighter body (partition of motion): M m m µ = = m (1) m + M M + 1 as mE M is µ ≈ m, i.e. motion is relative to the center of mass ≡ only motion of m. Set point of origin (0; 0) to the source of the force field of M. Moreover: Newton’s 2. law: d 2~r m = F~ (2) dt2 and force field according to Newton’s law of gravitation : GMm F~ = − ~r (3) r 3 H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 4 / 32 Equations of motion III Kepler’s laws, as well as the assumption of a central force imply ! conservation of angular momentum ! motion is only in a plane (! Kepler’s 1st law). So, we use Cartesian coordinates in the xy-plane: GMm F = − x (4) x r 3 GMm F = − y (5) y r 3 The equations of motion are then: d 2x GM = − x (6) dt2 r 3 d 2y GM = − y (7) dt2 r 3 p where r = x2 + y 2 (8) H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 5 / 32 Excursus: Analytic solution of the Kepler problemI To derive the analytic solution for equation of motion ~r(t) ! use polar coordinates: φ, r 1 use conservation of angular momentum `: µr 2φ_ = ` = const. (9) ` φ_ = (10) µr 2 2 use conservation of total energy: 1 `2 GMµ E = µr_2 + − (11) 2 2µr r 2E `2 2GM r_2 = − + (12) µ µ2r 2 r ! two coupled equations for r and φ H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 6 / 32 Excursus: Analytic solution of the Kepler problemII 3 α decouple Eq. (10), use the orbit equation r = 1+e cos φ with numeric eccentricity e 2 ` _ (= f1 O=a) and α = r(π=2) ≡ GMµ2 gives separable equation for φ dφ G 2M2µ3 φ_ = = (1 + e cos φ)2 (13) dt `3 Z t Z φ 0 0 dφ t = dt = k 0 2 = f (φ) (14) t0 φ0 (1 + e cos φ ) right-hand side integral can be looked up in, e.g., Bronstein: e sin φ 1 Z dφ t=k = − (15) (e2 − 1)(1 + e cos φ) e2 − 1 1 + e cos φ ! e 6= 1: parabola excluded; the integral can be further simplified for the elllipse: Z dφ 2 (1 − e) tan φ 0 ≤ e < 1 : = p arctan p 2 (16) 1 + e cos φ 1 − e2 1 − e2 H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 7 / 32 Excursus: Analytic solution of the Kepler problem III and for the hyperbola: p Z dφ 1 (e − 1) tan φ + e2 − 1 e > 1 : = p ln 2 p (17) 1 + e cos φ 2 φ 2 e − 1 (e − 1) tan 2 − e − 1 ! Eqn. (16) & (17) t(φ) must be inverted to get φ(t) ! (e.g., by numeric root finding) ! only easy for e = 0 ! circular orbit Z G 2M2µ3 t = k dφ0 = kφ ! φ(t) = k−1t = t (18) `3 `2 and from orbit equation r = α = GMµ2 = const. For the general case, it is much easier to solve the equations of motion numerically. H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 8 / 32 Excursus: The Kepler equationI Alternative formulation for time dependency in case of an ellipse (0 ≤ e < 1): Q P Orbit, circumscribed by auxillary circle with radius a (= semi-major axis); true anomaly φ, eccentric anomaly . Sun at S, planet at P, b circle center at O. Perapsis (perhelion) Π and ψ φ apapsis (aphelion) A: A Π consider a line normal to AΠ through P on a OR S the ellipse, intersecting circle at Q and AΠ at R. consider an angle (or E, eccentric anomaly) defined by \ΠOQ H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 9 / 32 Excursus: The Kepler equationII Then: position (r; φ) of the body P can be described in terms of : xP = r cos φ = a cos − ae (19) p 2 yP = r sin φ = a sin 1 − e (20) p (with PR=QR = b=a = 1 − e2), square both equations and add them up: r = a(1 − e cos ) (21) Now, to find = (t), need relationship between dφ and d , so combine Eqn. (20) & (21) b sin sin φ = jd=dx0 (22) a(1 − e cos ) b (cos (1 − e cos )d − e sin2 d ) cos φdφ = (23) a (1 − e cos )2 b dφ = d (24) a(1 − e cos ) H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 10 / 32 Excursus: The Kepler equation III ` together with the angular momentum dφ = µr 2 dt: ` (1 − e cos )d = dt (25) µab = set t = 0 ! (0) = 0; integration: (26) `t − e sin = (27) µab πab ` use Kepler’s 2nd law P = 2µ with πab the area of the ellipse, we get `=(µab) = 2π=P ≡ ! (orbital angular frequency), so: Kepler’s equation − e sin = !t (28) E − e sin E = M (astronomer’s version) (29) M: mean anomaly = angle for constant angular velocity H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 11 / 32 Excursus: The Kepler equationIV Kepler’s equation E(t) − e sin E(t) = M(t) is a transcendental equation can be solved by, e.g., Newton’s method because of E = M + e sin E, also (Banach) fixed-point iteration possible (slow, but stable), already used by Kepler (1621): E = M ; for (int i = 0 ; i < n ; ++i) E = M + e * sin(E) ; can be solved, e.g., by Fourier series ! Bessel (1784-1846): 1 X 2 E = M + J (ne) sin(nM) (30) n n n=1 1 Z π Jn(ne) = cos(nx − ne sin x)dx (31) π 0 H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 12 / 32 Circular orbits A special case as a solution of the equations of motion (6) & (7) is the circular orbit. Then: v 2 ¨r = (32) r mv 2 GMm = (equilibrium of forces) (33) r r 2 r GM ) v = (34) r The relation (34) is therefore the condition for a circular orbit. Moreover, Eq. (34) yields together with 2πr P = (35) v 4π2 ) P2 = r 3 (36) GM H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 13 / 32 Astronomical units For our solar system it is useful to use astronomical units (AU): 1 AU = 1:496 × 1011 m and the unit of time is the (Earth-) year 1 a = 3:156 × 107 s (≈ π × 107 s), so, for the Earth P = 1 a and r = 1 AU Therefore it follows from Eq. (36): 4π2r 3 GM = = 4π2 AU3 a−2 (37) P2 I.e. we set GM ≡ 4π2 in our calculations. Advantage: handy numbers! Thus, e.g. r = 2 is approx. 3 × 1011 m and t = 0:1 corresponds to 3:16 × 106 s, and v = 6:28 is roughly 30 km/s. cf.: our rcalc program with “solar units” for R, T , L; natural units in particle physics ~ = c = kB = 0 = 1 ! unit of m, p, T is eV (also for E) H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 14 / 32 The Euler methodI The equations of motion (6) & (7): d 2~r GM = − ~r (38) dt2 r 3 are a system of differential equations of 2nd order, that we shall solve now. Formally: integration of the equations of motion to obtain the trajectory ~r(t). Step 1: reduction Rewrite Newton’s equations of motion as a system of differential equations of 1st order (here: 1d): dx(t) dv(t) F (x; v; t) v(t) = & a(t) = = (39) dt dt m H. Todt (UP) Computational Astrophysics SoSe 2021, 3.6.2021 15 / 32 The Euler methodII Step 2: Solving the differential equation Differential equations of the form (initial value problem) dx = f (x; t); x(t ) = x (40) dt 0 0 can be solved numerically (discretization1) by as simple method: Explicte Euler method (“Euler’s polygonal chain method”) 1 choose step size ∆t > 0, so that tn = t0 + n∆t; n = 0; 1; 2;::: 2 calculate the values (iteration): xn+1 = xn + f (xn; tn)∆t Obvious: The smaller the step size ∆t, the more steps are necessary, but also the more accurate is the result. 1I.e. we change from calculus to algebra, which can be solved by computers.
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