A645/A445: Exercise #1 The Kepler Problem Due: 2017 Sep 26 1 Numerical solution to the one-degree-of-freedom implicit equation The one-dimensional implicit solution is: t = t o + x x o 2(E U(x)). (1) The turning points are defined as the coordinates for which ẋ = 0: {x : 2(E U(x)) = 0}. There will be two turning points in general: x,x +. It immediately follows that the orbital period found a bound orbit is: P = 2 x+ x 2(E U(x)). (2) To evaluate the integral in equation (2), make the following trigonometric change variables: where θ [ π/2, π/2] and x = a + + a sinθ (3) a ± x + ± x. 2 Substituting equation (3) into equation (2) gives: π/2 dθcosθ P = 2. (4) π/2 2(E U(x)) 1
Near a turning point, x to be explicit, the quantity 2(E U(x)) goes to zero linearly, that is: lim 2(E U(x)) = K(x x ) = Ka (1 + cosθ) = 2Ka cos 2 (θ/2) x x where K 2 du. x From this, it is easy to convince yourself from this that the square root divergence is removed from the integrand using this transformation. This works for square root divergences of all types. Similarly, one can solve for the one-dimensional implicit solution: θ(x) dθcosθ t = t o +. (5) θ(x o ) 2(E U(x)) where x and θ are related by equation (3). (a) Use the transformation in equations (3) and (4) to solve for the period of a harmonic oscillator numerically. To perform the integral, use the trapezoidal rule: b a f (x) h N i=1 where h = (b a)/n and x i = a + ih. f (x i 1 ) + f (x i ) 2 (b) Compare with the exact, analytic solution as a function of N. E.g. graph the numerical error as a function of N on a log log plot. Interpret this. 1. Pick a point 2/3 of the way from the lower to upper turning point for your harmonic oscillator and solve equation (??) using equation (6). Repeat your log log error plot for this problem. 2 The full two-body problem In Problem Set 1, you showed that the two-body problem for reduced to a single equation for the radial degreeof-freedom in polar coordinates r,θ is where L = r 2 θ is the angular momentum. (6) r = du dr + L2 r 3 (7) 2017 Sep 10/MDW 2
(a) Write down the implicit solution for t = t(r). (b) Using L = const, write down an analogous implicit solution for theta(t) = θ(r(t)). 3 The analytic solution of the gravitational n-body problem The gravitational potential for the gravitational two-body problem is most often written as: U(r) = GM (8) r where G is the gravitational constant and M is the total mass. We will now show that equation (7) can be solved analytically! (a) Change variables equations (7) and (8) to derive: d 2 u dθ 2 + u = GM L 2. This is a linear inhomogeneous 2nd-order differential equation. Solve this to yield: where e and ϖ are constants. u = 1 r = GM [1 + ecos(θ ϖ)] (9) L2 (b) Replace the angular momentum L by another constant of integration a defined by L 2 = GMa(1 e 2 ) to show that the shape of the orbit is given by: r = a(1 e2 ) 1 + ecos f (10) where f = θ ϖ is known as the true anomaly in celestial mechanics. (c) The closest approach of the two bodies occurs at f = 0 or θ = ϖ and ϖ is known, therefore, as the azimuth of periapsis. Using these relations, show that the periapsis value is r p = a(1 e) and that the total energy is E = GM 2a. 2017 Sep 10/MDW 3
(d) Equation (10) is a parametric equation for an ellipse (Kepler s first law)! Show that that major axis (twice the full extent of the long axis) has the value a and that the minor axis (twice the full extent of the short axis) is related to a and e as e 2 = 1 b 2 /a 2. (e) Since θ = L/r 2 we have t = t2 t 1 dt = 1 L f2 f 1 d f r 2. (11) This equation, then implicitly determines θ = ϖ + f (t) and r = r(t) follows using equation (9). Geometrically, (r r d f /2) is the infinitesimal triangle subtended by the arc d f to the origin and therefore the integral above is twice the area of the ellipse if we integrate from f 1 = 0 to f = 2π. Use this to show that P = 2π(a 3 /GM) 1/2, which is Kepler s third law! There are more fancy things we can do with these equations but this is enough for now. 4 Numerical solution of the gravitational n-body problem Now that we know what to expect, let s solve the differential equations numerically. This is also good practice for problems whose characteristics are unknown. (a) To do this, let us return to the general equation of motion for the reduced two-body problem: d 2 x dt 2 = U(r) = GM x x 3. Choose the units G = M = 1. With no loss of generality, we can consider motion on a single plane (why?). Choose the initial conditions x = (0.140,0.078,0.0) and ẋ = ( 1.56,2.78,0.0), t 1 = 0 and solve Newton s equations Cartesian coordinates with an ODE solver to get the trajectory between t 1 and t 2 = 2. Decrease the step size until the solution converges; that is, it gives nearly the same trajectory. (b) Repeat the ODE solution using polar coordinates for the same initial conditions. (c) Finally, compute the solution using the methods describes in the previous problem, and compare with the ODE solutions. 2017 Sep 10/MDW 4
(d) If any of these results are substantially different from each other, try to figure out why... [Hint: This is not a trick question. All methods should give the same trajectory!] 5 Making sense of the Poisson equation For a single point mass, the Poisson equation tells us that: 2 Φ(r) = ( GM r ) r 3 = 4πGMδ 3 (r). On the other hand, away for the origin, we know that ( r ) r 3 = 0. This tension can be explained by the singularity at r = 0 and motivates the definition of the Dirac delta function. This whole thing is a bit easier to swallow if we deal with non-singular potentials. We can remove the singularity by softening the point mass: (a) Show explicitly that where S is a spherical volume. x F ε (x) = (x 2 + ε 2 ) 3/2. lim F ε = 4π ε 0 S (b) Similarly, show that lim F ε 0 ε 0 for x 0. This expression, which is a density, is a representation of the Dirac delta function. 2017 Sep 10/MDW 5