Quiz # 2. Answer ALL Questions. 1. Find the Fourier series for the periodic extension of. 0, 2 <t<0 f(t) = sin(πt/2), 0 <t<2

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1 Quiz # Answer ALL Questions 1 Find the Fourier series for the periodic extension of ½, <t< f(t) = sin(πt/), <t< Find an expression for the central force that will provide radial quadratic spiral orbits given by r = bθ 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, showthat the equation of its path must satisfy d u dθ + u = F (u) h u, 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 r 3 = a 3 {1+cos(3θ)}, where a is a positive constant If F (u) =µu + λu 3,whereµ and λ are positive constants, and P is projected with speed V at right angles to OP from a point where OP = r and θ =,find the equation of the orbit on the assumption that λ <V r < µr + 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 =, and that between consecutive apses, the radius vector turns through an angle π/(1 λv r ) 1/

2 4 (a) A particle of unit mass is moving under a central force µ/r in an elliptic orbit of semi-major axis a and eccentricity e Show that its angular momentum about the center of force is q µa(1 e ) (b) The period of Neptune is 1648 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,wherek and K arepositiveconstants and v is the velocity of the body Gravity can be ignored Determine the subsequent motion in each case, ie, calculate the position coordinate x as a function of time Verify that the motion is bounded in case (i), but not in case (ii)

3 1 The Fourier series is given by Solutions where a = 1 Z f(t) = 1 a + dt f(t) = 1 X (a n cos(nt)+b n sin(nt)), n=1 Z dt sin µ πt = 1 µ πt cos = π π a n = 1 Z µ nπt dt f(t)cos = 1 Z µ nπt dt cos sin µ πt But, b n = 1 Z µ nπt dt f(t)sin = 1 Z µ nπt dt sin sin µ πt and for n>1, we have a 1 = 1 4 Z dt sin(πt) = a n = 1 Z ( Ã! Ã!) (n +1)πt (n 1)πt dt sin sin 4 = 1 " Ã! 1 (n +1)πt π n +1 cos 1 Ã!# (n 1)πt n 1 cos = 1 π n 1, for n even =, otherwise (1) We also have for n>1 and b 1 =1/ b n = 1 Z µ µ nπt πt dt sin sin = 1 Z ( Ã! Ã!) (n 1)πt (n +1)πt dt cos cos 4 = () 3

4 Since u =1/bθ,wehave This gives d u dθ = 6 b θ 4 =6bu d u F (u) + u = u(1 + 6bu) = dθ mh u Thus, the magnitude of the attractive force is F = mh r 3 Ã 1+ 6b! r 3 (b) From we have and u = 1 1 a (1 + cos 3θ) 1/3 du dθ = 1 sin 3θ a (1 + cos 3θ) 4/3 d u dθ = 1 3cos3θ cos 3θ +4 a (1 + cos 3θ) 7/3 So, we have d u dθ + u =5a3 u 4 But, this last equation is equal to F/mh u Therefore, F =(5a 3 mh ) u 6 (b) Here 4

5 F = µu + λu 3, h = Vr So, d u dθ + u = µ + λu V r which we rewrite as d à u dθ + 1 λ! u = µ V r V r The coefficient of u on the left-hand side γ 1 λ V r is positive since we have λ <V r So, we have as solution u = µ + A cos γθ, V rγ where A is a constant The total energy is conserved The initial kinetic energy is The potential energy is T = 1 V U = µ r λ r So that the total (initial) energy is E = 1 V µ λ = 1 ³ V r r r r µr λ A is obtained by substi- which is negative since V r < µr + λ Thecoefficient tuting the equation of the orbit into the energy equation Ã! du + u = (E U) dθ V r This is the same as 5

6 Ã! du + γ u µ u = dθ V r E V r After some algebra, we obtain A = 1 γ s µ V 4 r 4 γ + E V r 4 (a) From our lecture notes, we have for the total energy E = µm a and the eccentricity is given in terms of the energy by e = s 1+ Eh mµ and E = µm/a follows since l = a(1 e )andl = h /µ Substituting the value of E from the first equation into the second, we obtain h = q µa(1 e ) (b) According to Kepler s third law, we have for Neptune, and τ N = Ca3 N τ E = Ca3 E 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 a N τn /3 = =(1648) /3 36 a E τ E 6

7 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 dt = mkv, where v = ẋ Separating the variables, rewrite this equation as This leads to Z dv v = k Z dt ln v = kt + C, where C is a constant But, v = u at t = Therefore, C =lnu So,thevelocityat time t is given by v = ue kt Integrating this equation and using the initial conditions, we obtain x = u ³ 1e kt k This motion is bounded e kt ast bso that x u/k (ii) The equation of motion is m dv dt = mkv assuming the body starts at the origin and moves along the positive x axis Separate the variables and integrate to obtain which gives Z dv Z v = K dt, 1 v = Kt + C The constant C is obtained from the initial conditions and we have v = 7 u Ktu +1

8 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 = a rˆr + a θ ˆθ = ( r ω r)ˆr + 1 d r dt (r ω)ˆθ = ( r ω r)ˆr +ṙωˆθ (3) Applying to obtain d dt ˆr = ωˆθ, d dt ˆθ = ωˆr, d dt a = d dt (a rˆr + a θ ˆθ) = (ȧ rˆr + ȧ θ ˆθ)+(ar ωˆθ a θ ωˆr) (4) Setting the coefficient of ˆθ in Eq (4) equal to zero, we obtain This equation gives ȧ θ + a r ω = rω +( r ω r)ω = r = 1 3 ω r Multiply this equation by ṙ and integrate over time We have ṙ = 1 3 ³ r a Rewrite as dr r a = 1 3 dt Integrating, we obtain 8

9 µ x cosh 1 = 1 t a 3 or x = a cosh Ã! t 3 9