Problem Set 5 Solution

Transcription

1 Problem Set 5 Solution Friday, 14 October 011 Physics 111 Proble The Setup For a pair of point masses, and m, interacting via a conservative force directed along the line joining the masses, show that (a) the Lagrangian may be written L = µ (ṙ + r θ ) U(r) where the reduced mass µ is defined by m /( ), and (b) conservation of angular momentum implies Kepler s second law, that the radial vector from the Sun (well, really from the center of mass) to a planet sweeps out equal areas in equal times. (c) Is the center of mass of the Sun-Jupiter system inside the Sun? (d) In the derivation of Kepler s third law, I used without proof the fact that the area of an ellipse is πab. Prove it. [Not that I used the formula, but the formula itself!] Solution: (a) The center of mass is defined by r 1 = m r, where the two distances are measured from the center of mass. If the separation between the masses is r = r 1 + r, then r = r 1 + r 1 m = r 1 m r 1 = m r and r = We can now express the kinetic energy of the system in terms of r: T = (ṙ 1 + r 1 θ ) + m (ṙ + r θ ) m = [(ṙ + r θ ) ( ) ] + m [(ṙ + r θ ) ( ) ] = 1 (ṙ + r θ ) m + m m 1 ( ) = 1 (ṙ + r θ ) m = µ (ṙ + r θ ) r Therefore, L = T U = µ (ṙ + r θ ) U(r) Physics Peter N. Saeta

2 (b) An infinitesimal wedge of angular extent dθ is that of a triangle of base r dθ and height r, which is da = 1 r(r dθ). So, the rate that area is swept out is Similarly, the total angular momentum is Therefore, Ȧ = 1 L = r 1 θ + m r θ m da dt = 1 dθ r dt = [ ( ) + m ( ) ] r θ = µr θ L µ, which is constant. (c) Jupiter s mass is times the solar mass. The radius of its orbit is 5.04 AU, whereas the radius of the Sun is AU. So, the center of mass of the Sun-Jupiter system is at m r CM = a Jupiter = AU M + m which is greater than the radius of the Sun. So, the center of mass is about 0.1R solar outside the Sun. (d) The equation of an ellipse in Cartesian coordinates is so the area is A = x a + y b = 1 y dx = Let x = a cos ϕ, so dx = a sin ϕ dϕ. Then A = b π 0 a b a 1 ( x a ) dx 1 cos ϕ ( a sin ϕ) dϕ = ab Now use the trig identity sin ϕ = (1 cos ϕ)/ to get 0 π sin ϕ dϕ π A = ab (1 cos ϕ) dϕ = ab (ϕ 1 π 0 sin ϕ) = πab 0 Problem Cone Head A massive particle moves without friction under the acceleration of gravity, g, on the surface of a cone of half angle α. The axis of the cone is vertical and the point of the cone is at the bottom. Peter N. Saeta Physics 111

3 (a) Find the condition for a circular orbit at radius ρ. (b) Investigate the stability of the circular orbit. If it is stable, is the perturbed orbit closed or open? (A closed orbit retraces itself; an open orbit does not.) Solution: Noting that tan α = ρ/z, the Lagrangian is L = m (ṙ + r θ + ż ) + mgz = m [ṙ (1 + cot α) + r θ ] mgr cot α Since 1 + cot α = sin α+cos α sin α = csc α, L = 1 m[ṙ csc α + r θ ] mgr cot α. The Lagrangian is cyclic in θ, so L θ = mr θ = l = constant Remember that we cannot use this relation to substitute into L before finding the Euler- Lagrange equation for r, so d dt ( L ṙ ) L r = m r csc α (mr θ mg cot α) = 0 The condition for a circular orbit is r = 0 and ṙ = 0; therefore, a circular orbit at radius ρ satisfies ρ θ = l g m ρ 3 = g cot α θ = ρ tan α Now we may substitute with impunity to eliminate θ, getting r csc α l m r 3 + g cot α = 0 r + g sin α cos α l sin α m r 3 = 0 We can investigate the stability of the circular orbit by substituting r = ρ + x The radial equation of motion becomes where x ρ 1 0 = ẍ + g sin α cos α l sin α m ρ 3 (1 + x/ρ) 3 0 ẍ + g sin α cos α l sin α m ρ 3 (1 3 x ρ ) = ẍ + 3l sin α m ρ 4 x Physics Peter N. Saeta

4 where we have neglected terms of order (x/ϕ). This is the simple-harmonic-oscillator equation with frequency ω = 3 ρ l sin α m ρ 3 = 3 sin α g cot α = 3 g sin α cos α ρ ρ Since the expression for ω > 0, the circular orbit is stable for arbitrary values of ρ and α. To see whether the orbits are closed, let s compare the oscillation frequency to the orbital frequency: ω θ = 3gρ 1 sin α cos α gρ 1 = 3 sin α cot α ω θ = 3 sin α Therefore, the ratio of the frequencies is independent of ρ and depends only on the cone half-angle α. For certain cone half-angles, including sin 1 (1/ 3) 35 and 60, the orbits are closed, but in general they are open. Problem 3 Bertrand s Theorem Bertrand s theorem states that the only central potentials that produce stable closed orbits for all bound particles are U(r) = kr (Hooke s law) and U(r) = kr 1 (e.g., gravitation). Investigate Bertrand s theorem using Mathematica by assuming U(r) = kr n for various values of n. Describe your approach and results in sufficient detail that someone utterly unfamiliar with Mathematica can understand it. You may do this entirely within the Mathematica notebook. Suggestions: (a) Start from Eq. (6.46) of Helliwell & Sahakian, replacing GMmu with the potential you are planning to investigate, expressed as a function of u = 1/r. (b) Differentiate once with respect to θ to get rid of the energy. (c) For given angular momentum and n, find the value of u that produces a circular orbit. (Don t forget to choose the appropriate sign for k to yield a bound orbit.) (d) Use NDSolve to solve for u(θ) for a range of θ that includes a few tours around the center of force. (e) Prepare a plot using ParametricPlot of y(θ) vs. x(θ) for one or more suitable ranges of θ. Peter N. Saeta 4 Physics 111

5 Bertrand s Theorem HW05, Problem 3 Peter N. Saeta, 14 October 011 Bertrand's theorem states that the only central potentials that produce stable closed orbits for all bound particles are UHrL = k r (Hooke s law) and UHrL = -k r -1. We will investigate Bertrand s theorem for potentials for the form UHrL = k r n for various values of n to see whether orbits are closed. From Eq. (6.46) in the Helliwell & Sahakian, which is written in terms of u = 1, we begin r with Hu'L = m L JE - L m u - k u -n N (where L is the angular momentum) and then differentiate with respect to q, giving u' u'' = - u u' + m n k u -n u' L u or u'' + u = m n L UH1êuL u = m n k L u -n-1 Our approach will be to integrate this equation numerically for several revolutions, and plot the trajectory in the vicinity of the starting point on each successive pass. This has the advantage of expanding the orbit so we can look at differences on subsequent trajectories. Alternatively, we can do a Poincaré section, but we'll do a streaked version of Poincaré. To follow the order of the turns around, the successive revolutions will be displayed in a progressively lighter and greener hue, starting from dark blue. We will calculate nloops revolutions and plot 0 < q < 0.05 radians for each revolution. The initial condition will be that the radial velocity will be zero, but the initial value of u will be greater than the one yielding a circular orbit by 1%. For a closed orbit, only a single streak will appear; otherwise two or more will show up. In[1]:= nloops = 5; H* the number of revolutions to calculate and plot *L e = 0.01; H* the relative departure of the initial radius from circular *L In[3]:= BertrandLoops@L_, n_d := ModuleB8u0, s, k<, H* Set the value of k for an attractive potential *L k = L n ; u0 = 1; s = FirstBNDSolveB:u@0D ã u0 H1 + el, u'@0d ã 0, u''@qd + u@qd ã u@qd, 8q, 0, p nloops<, MaxSteps Ø 10 5 FF; ParametricPlotB EvaluateB TableB : Cos@qD u@qd, Sin@qD > ê. s ê. u@qd q Ø q + p j, 8j, nloops, 0, -1<FF, 8q, 0, 0.05<, n k L u@qd-n-1 >, Frame Ø True, AspectRatio Ø 1 ê, PlotLabel Ø StringJoin@"n = ", ToString@nDD, GridLines Ø 8None, None<, Axes Ø None, FrameTicks Ø None, j PlotStyle Ø TableB:Thick, HueB j, 1 -, 1F>, 8j, nloops, 0, -1<FFF nloops + 1

6 Bertrand.nb In[4]:= nd, 88n, <, -3, 4<D n n = 1.97 Out[4]= Now we prepare a grid of values exploring the range between n = 3 and n = 4. As you can see, only n = 1 and n = give closed orbits with a single streak.

7 Bertrand.nb 3 GraphicsGrid@ Table@ BertrandLoops@1, i + jd, 8i, -3, 3<, 8j, 0.5, 1, 0.5<DD n = -.75 n = -.5 n = -.5 n = -. n = n = -1.5 n = -1.5 n = -1. n = n = -0.5 n = -0.5 n = 0. n = 0.5 n = 0.5 n = 0.75 n = 1. n = 1.5 n = 1.5 n = 1.75 n =. n =.5 n =.5 n =.75 n = 3. n = 3.5 n = 3.5 n = 3.75 n = 4. Just for grins, we can explore more carefully the region around the two values that yield closed orbits. GraphicsGrid@ Table@ BertrandLoops@1, i + jd, 8i, -1,, 3<, 8j, -0.05, 0.05, 0.05<DD n = n = n = -1. n = n = n = 1.95 n = n =. n =.05 n =.05