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Quiz # 2 Answer ALL Questions 1. Find the Fourier series for the periodic extension of f (t) = ½ 0, −2 < t < 0 sin(πt/2), 0 < t < 2 2. Find an expression for the central force that will provide radial quadratic spiral orbits given by r = bθ2 . Here, r and θ are polar coordinates and b is a constant. 3. A particle P of unit mass is moving under a central force to the origin O, show that the equation of its path must satisfy d2 u F (u) +u= 2 2, 2 dθ hu where r = 1/u and θ are polar coordinates, F (u) is the strength of the force, and h is the angular momentum. Find the law of force to the pole under which a particle can describe the curve r3 = a3 {1 + cos(3θ)} , where a is a positive constant. If F (u) = µu2 + λu3 , where µ and λ are positive constants, and P is projected with speed V at right angles to OP from a point where OP = r0 and θ = 0, find the equation of the orbit on the assumption that λ < V 2 r02 < 2µr0 + λ. An “apse” is defined as a point on an orbit at a maximum or minimum distance from the center of force. Show that at an apse, du/dθ = 0, and that between consecutive apses, the radius vector turns through an angle π/(1 − λV −2 r0−2 )1/2 . 4. (a) A particle of unit mass is moving under a central force µ/r2 in an elliptic orbit of semi-major axisqa and eccentricity e. Show that its angular momentum about the center of force is µa(1 − e2 ). (b) The period of Neptune is 164.8 years. Show that it is about thirty times as far from the sun as the earth is. 5. A body of mass m is projected with speed u in a medium that exerts a resistance force of magnitude (i ) mk|v|, or (ii) mK|v|2 , where k and K are positive constants and v is the velocity of the body. Gravity can be ignored. Determine the subsequent motion in each case, i.e., calculate the position coordinate x as a function of time. Verify that the motion is bounded in case (i), but not in case (ii). 2 Solutions 1. The Fourier series is given by ∞ X 1 f (t) = a0 + (an cos(nt) + bn sin(nt)) , 2 n=1 where µ ¶ ∙ µ 1Z 2 1Z 2 πt 1 πt a0 = dt f (t) = dt sin = − cos 2 0 2 0 2 π 2 ¶¸2 = 0 2 π µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ 1Z 2 nπt 1Z 2 nπt πt an = dt f (t) cos = dt cos sin 2 0 2 2 0 2 2 1Z 2 1Z 2 nπt nπt πt bn = dt f (t) sin = dt sin sin 2 0 2 2 0 2 2 But, a1 = 1Z 2 dt sin(πt) = 0 4 0 and for n > 1, we have an ( à ! à 1Z 2 (n + 1)πt (n − 1)πt = dt sin − sin 4 0 2 2 " à ! à !) 1 1 1 (n + 1)πt (n − 1)πt = cos − cos 2π n + 1 2 n−1 2 2 1 , for n even = − 2 πn −1 = 0 , otherwise . !#0 2 (1) We also have for n > 1 bn µ ¶ µ ¶ nπt πt 1Z 2 = dt sin sin 2 0 2 2 ( à ! à !) 1Z 2 (n − 1)πt (n + 1)πt = dt cos − cos 4 0 2 2 = 0 and b1 = 1/2. 3 (2) 2. Since u = 1/bθ2 , we have d2 u 6 = θ−4 = 6bu2 2 dθ b This gives d2 u F (u) + u = u(1 + 6bu) = . dθ2 mh2 u2 Thus, the magnitude of the attractive force is à 6b mh2 F = 3 1+ 2 r r ! . 3. (b) From u= 1 1 a (1 + cos 3θ)1/3 we have du 1 sin 3θ = dθ a (1 + cos 3θ)4/3 and d2 u 1 3 cos 3θ − cos2 3θ + 4 = dθ2 a (1 + cos 3θ)7/3 So, we have d2 u + u = 5a3 u4 2 dθ But, this last equation is equal to F/mh2 u2 . Therefore, F = (5a3 mh2 ) u6 (b) Here 4 F = µu2 + λu3 , h = V r0 . So, d2 u µ + λu + u = dθ2 V 2 r02 which we rewrite as à ! d2 u λ µ + 1− 2 2 u= 2 2 2 dθ V r0 V r0 The coefficient of u on the left-hand side γ2 ≡ 1 − λ V 2 r02 is positive since we have λ < V 2 r02 . So, we have as solution µ u= V 2 r2 γ 2 0 + A cos γθ , where A is a constant. The total energy is conserved. The initial kinetic energy is 1 T = V2 . 2 The potential energy is U =− λ µ − 2 . r 2r So that the total (initial) energy is ´ µ λ 1 ³ 1 E = V2− − = 2 V 2 r02 − 2µr0 − λ 2 r0 2r0 2r0 which is negative since V 2 r02 < 2µr0 + λ. The coefficient A” is obtained by substituting the equation of the orbit into the energy equation à du dθ !2 + u2 = This is the same as 5 2 (E − U) . V 2 r02 à du dθ !2 + γ 2 u2 − 2µ 2E u= 2 2 . 2 2 V r0 V r0 After some algebra, we obtain 1 A= γ s µ2 2E + 2 2 . 4 2 4 V r0 γ V r0 4. (a) From our lecture notes, we have for the total energy E=− µm 2a and the eccentricity is given in terms of the energy by e= s 1+ 2Eh2 mµ2 and E = −µm/2a follows since l = a(1 − e2 ) and l = h2 /µ. Substituting the value of E from the first equation into the second, we obtain h= q µa(1 − e2 ) . (b) According to Kepler’s third law, we have τN2 = C a3N for Neptune, and τE2 = C a3E for earth. where C is a constant, τ is the periodic time and a is the semi-major axis. Therefore, from these two equations, we obtain µ aN τN = aE τE ¶2/3 6 = (164.8)2/3 ≈ 30.6 5. (i) Suppose that the body starts at the origin and moves along the positive x axis. Then, the equation of motion is m dv = −mkv , dt where v = ẋ. Separating the variables, rewrite this equation as Z Z dv = −k dt . v This leads to ln v = −kt + C , where C is a constant. But, v = u at t = 0. Therefore, C = ln u. So, the velocity at time t is given by v = u e−kt Integrating this equation and using the initial conditions, we obtain x= u ³ −kt ´ 1e k This motion is bounded e−kt → 0 as t → ∞bso that x → u/k. (ii) The equation of motion is m dv = −mKv 2 dt assuming the body starts at the origin and moves along the positive x axis. Separate the variables and integrate to obtain Z Z dv = −K dt , v2 which gives 1 − = −Kt + C v The constant C is obtained from the initial conditions and we have v= 7 u Ktu + 1 Integrating this equation with respect to time and applying the initial conditions, we obtain x= 1 ln(Kut + 1) , K which tends to infinity as t → ∞ and so the motion is unbounded. The acceleration in polar coordinates is a = ar r̂ + aθ θ̂ 1d 2 (r ω)θ̂ r dt = (r̈ − ω 2 r)r̂ + 2ṙω θ̂ . = (r̈ − ω 2 r)r̂ + (3) Applying d r̂ = ω θ̂ , dt d θ̂ = −ωr̂ , dt to obtain d d a = (ar r̂ + aθ θ̂) dt dt = (ȧr r̂ + ȧθ θ̂) + (ar ω θ̂ − aθ ωr̂) . Setting the coefficient of θ̂ in Eq. (4) equal to zero, we obtain ȧθ + ar ω = 2r̈ω + (r̈ − ω 2 r)ω = 0 . This equation gives 1 r̈ = ω 2 r . 3 Multiply this equation by ṙ and integrate over time. We have ṙ2 = ´ 1³ 2 r − a2 . 3 Rewrite as dr 1 √ = √ dt . 2 2 r −a 3 Integrating, we obtain 8 (4) cosh −1 µ ¶ x 1 =√ t a 3 or à t x = a cosh √ 3 9 ! .