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Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid December 1, 2001 Chapter 3 Problem 3.1 A particle of mass m is constrained to move under gravity without friction on the inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional problem equivalent to its motion. What is the condition on the particle’s initial velocity to produce circular motion? Find the period of small oscillations about this circular motion. We’ll take the paraboloid to be deﬁned by the equation z = αr 2 . The kinetic and potential energies of the particle are m 2 ˙ T = ˙ ˙ (r + r2 θ2 + z 2 ) 2 m ˙ ˙ ˙ = (r2 + r2 θ2 + 4α2 r2 r2 ) 2 V = mgz = mgαr2 . Hence the Lagrangian is m ˙ L= ˙ (1 + 4α2 r2 )r2 + r2 θ2 − mgαr2 . 2 This is cyclic in θ, so the angular momentum is conserved: ˙ l = mr2 θ = constant. 1 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 2 For r we have the derivatives ∂L ˙ ˙ = 4α2 mrr2 + mrθ2 − 2mgαr ∂r ∂L ˙ = m(1 + 4α2 r2 )r ∂r˙ d ∂L ˙ = 8mα2 rr2 + m(1 + 4α2 r2 )¨. r ˙ dt ∂ r Hence the equation of motion for r is ˙ ˙ ˙ 8mα2 rr2 + m(1 + 4α2 r2 )¨ = 4α2 mrr2 + mrθ2 − 2mgαr r or ˙ ˙ m(1 + 4α2 r2 )¨ + 4mα2 rr2 − mrθ2 + 2mgαr = 0. r In terms of the constant angular momentum, we may rewrite this as l2 ˙ m(1 + 4α2 r2 )¨ + 4mα2 rr2 − r + 2mgαr = 0. mr3 So this is the diﬀerential equation that determines the time evolution of r. ˙ If initially r = 0, then we have l2 m(1 + 4α2 r2 )¨ + − r + 2mgαr = 0. mr3 ¨ ˙ Evidently, r will then vanish—and hence r will remain 0, giving circular motion— if l2 = 2mgαr mr3 or ˙ θ = 2gα. So if this condition is satisﬁed, the particle will execute circular motion (assum- ing its initial r velocity was zero). It’s interesting to note that the condition on ˙ θ for circular motion is independent of r. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 3 Problem 3.2 A particle moves in a central force ﬁeld given by the potential e−ar V = −k , r where k and a are positive constants. Using the method of the equivalent one- dimensional potential discuss the nature of the motion, stating the ranges of l and E appropriate to each type of motion. When are circular orbits possible? Find the period of small radial oscillations about the circular motion. The Lagrangian is m 2 ˙ e−ar L= ˙ r + r2 θ2 + k . 2 r As usual the angular momentum is conserved: ˙ l = mr2 θ = constant. We have ∂L ˙ e−ar = mrθ2 − k (1 + ar) 2 ∂r r ∂L ˙ = mr ˙ ∂r so the equation of motion for r is ˙ k e−ar r = rθ2 − (1 + ar) 2 ¨ m r l2 k e−ar = 2 3− (1 + ar) 2 . (1) m r m r The condition for circular motion is that this vanish, which yields ˙ k e−ar0 /2 θ= (1 + ar0 ) 3/2 . (2) m r 0 What this means is that that if the particle’s initial θ velocity is equal to the above function of the starting radius r0 , then the second derivative of r will remain zero for all time. (Note that, in contrast to the previous problem, in this case the condition for circular motion does depend on the starting radius.) To ﬁnd the frequency of small oscillations, let’s suppose the particle is exe- cuting a circular orbit with radius r0 (in which case the θ velocity is given by (2)), and suppose we nudge it slightly so that its radius becomes r = r0 + x, where x is small. Then (1) becomes k e−ar0 k e−a[r0 +x] ¨ x= 1 + ar0 − (1 + a[r0 + x]) (3) m 2 r0 m [r0 + x]2 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 4 Since x is small, we may write the second term approximately as k e−ar0 x ≈ m r02 (1 + ar0 + ax)(1 − ax) 1 − 2 r 0 k e −ar0 k e −ar0 (1 + ar0 ) ≈ (1 + ar0 ) 2 + 2 a − a(1 + ar0 ) − 2 x m r0 m r0 r0 k e −ar0 k e −ar0 2 ≈ (1 + ar0 ) 2 − 2 2a + + a2 r0 x. m r0 m r0 r0 The ﬁrst term here just cancels the ﬁrst term in (??), so we are left with k e−ar0 2 ¨ x= 2 2a + + a2 r0 x m r0 r0 The problem is that the RHS here has the wrong sign—this equation is satisﬁed by an x that grows (or decays) exponentially, rather than oscillates. Somehow I messed up the sign of the RHS, but I can’t ﬁnd where–can anybody help? Problem 3.3 Two particles move about each other in circular orbits under the inﬂuence of grav- itational forces, with a period τ . Their motion is suddenly stopped, and they are then released and allowed to fall into each other. Prove that they collide after a √ time τ /4 2. Since we are dealing with gravitational forces, the potential energy between the particles is k U (r) = − r and, after reduction to the equivalent one-body problem, the Lagrangian is µ 2 ˙ k L= ˙ [r + r2 θ2 ] + 2 r where µ is the reduced mass. The equation of motion for r is ˙ k µ¨ = µrθ2 − 2 . r (4) r If the particles are to move in circular orbits with radius r0 , (4) must vanish at r = r0 , which yields a relation between r0 and θ:˙ 1/3 k r0 = ˙ µθ2 1/3 kτ 2 = (5) 4π 2 µ Homer Reid’s Solutions to Goldstein Problems: Chapter 3 5 where we used the fact that the angular velocity in the circular orbit with period ˙ τ is θ = 2π/τ . When the particles are stopped, the angular velocity goes to zero, and the ﬁrst term in (4) vanishes, leaving only the second term: k ¨ r=− . (6) µr2 This diﬀerential equation governs the evolution of the particles after they are stopped. We now want to use this equation to ﬁnd r as a function of t, which we will then need to invert to ﬁnd the time required for the particle separation r to go from r0 to 0. ˙ The ﬁrst step is to multiply both sides of (6) by the integrating factor 2r: 2k ˙r 2r¨ = − ˙ r µr2 or d 2 d 2k ˙ r =+ dt dt µr from which we conclude 2k ˙ r2 = + C. (7) µr ˙ The constant C is determined from the boundary condition on r. This is simply ˙ that r = 0 when r = r0 , since initially the particles are not moving at all. With the appropriate choice of C in (7), we have 1/2 dr 2k 1 1 ˙ r= = − dt µ r r0 1/2 2k r0 − r = . (8) µ rr0 We could now proceed to solve this diﬀerential equation for r(t), but since in fact we’re interested in solving for the time diﬀerence corresponding to given boundary values of r, it’s easier to invert (8) and solve for t(r): 0 dt ∆t = dr r0 dr 0 −1 dr = dr r0 dt 1/2 0 1/2 µ rr0 = dr 2k r0 r0 − r Homer Reid’s Solutions to Goldstein Problems: Chapter 3 6 We change variables to u = r/r0 , du = dr/r0 : 1/2 0 1/2 µ 3/2 u = r0 du 2k 1 1−u Next we change variables to u = sin2 x, du = 2 sin x cos x dx : 0 µ 1/2 3/2 =2 r0 sin2 x dx 2k π/2 µ 1/2 3/2 π = r0 . 2k 4 Now plugging in (5), we obtain 1/2 µ 1/2 kτ 2 π ∆t = 2k 4π 2 µ 4 τ = √ 4 2 as advertised. Problem 3.6 (a) Show that if a particle describes a circular orbit under the inﬂuence of an attractive central force directed at a point on the circle, then the force varies as the inverse ﬁfth power of the distance. (b) Show that for the orbit described the total energy of the particle is zero. (c) Find the period of the motion. ˙ ˙ (d) Find x, y, and v as a function of angle around the circle and show that all three quantities are inﬁnite as the particle goes through the center of force. Let’s suppose the center of force is at the origin, and that the particle’s orbit is a circle of radius R centered at (x = R, y = 0) (so that the leftmost point of the particle’s origin is the center of force). The equation describing such an orbit is √ r(θ) = 2R(1 + cos 2θ)1/2 so 1 1 u(θ) = =√ . (9) r(θ) 2R(1 + cos 2θ)1/2 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 7 Diﬀerentiating, du sin 2θ = √ dθ 2R(1 + cos 2θ)3/2 du 1 2 cos 2θ sin2 2θ = √ +3 dθ 2R (1 + cos 2θ)3/2 (1 + cos 2θ)5/2 1 1 = √ 2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ . (10) 2 2R (1 + cos 2θ)5/2 Adding (9) and (10), d2 u 1 +u= √ (1 + cos 2θ)2 + 2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ dθ2 2R(1 + cos 2θ) 5/2 1 =√ [4 + 4 cos 2θ] 2R(1 + cos 2θ)5/2 4 =√ 2R(1 + cos 2θ)3/2 = 8R2 u3 . (11) The diﬀerential equation for the orbit is d2 u m d 1 +u=− 2 V (12) dθ2 l du u Plugging in (11), we have m d 1 8R2 u3 = − V l2 du u so 1 2l2R2 4 2l2 R2 V =− u −→ V (r) = − (13) u m mr4 so 8l2R2 f (r) = − (14) mr5 which is the advertised r dependence of the force. (b) The kinetic energy of the particle is m 2 ˙ T = [r + r2 θ2 ]. ˙ (15) 2 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 8 We have √ r= 2R(1 + cos 2θ)1/2 r2 = 2R2 (1 + cos 2θ) √ sin 2θ r = 2R ˙ ˙ θ (1 + cos 2θ)1/2 sin2 2θ ˙2 ˙ r2 = 2R2 θ 1 + cos 2θ Plugging into (15), ˙ sin2 2θ T = mR2 θ2 + 1 + cos 2θ 1 + cos 2θ ˙ sin2 2θ + 1 + 2 cos 2θ + cos2 2θ = mR2 θ2 1 + cos θ ˙ = 2mR2 θ2 ˙ In terms of l = mr2 θ, this is just 2R2 l2 = mr4 But this is just the negative of the potential energy, (13); hence the total particle energy T + V is zero. (c) Suppose the particle starts out at the furthest point from the center of force on its orbit, i.e the point x = 2R, y = 0, and that it moves counter-clockwise from this point to the origin. The time required to undergo this motion is half the period of the orbit, and the particle’s angle changes from θ = 0 to θ = π/2. Hence we can calculate the period as π/2 dt τ =2 dθ 0 dθ π/2 dθ =2 0 ˙ θ ˙ Using θ = l/mr2 , we have π/2 m =2 r2 (θ) dθ l 0 4R2 m π/2 = (1 + 2 cos 2θ + cos2 2θ) dθ l 0 4R2 m 3π = · l 4 3πR2 m = . l Homer Reid’s Solutions to Goldstein Problems: Chapter 3 9 Problem 3.8 (a) For circular and parabolic orbits in an attractive 1/r potential having the same angular momentum, show that the perihelion distance of the parabola is one half the radius of the circle. (b) Prove that in the same central force as in part (a) the speed of a particle at √ any point in a parabolic orbit is 2 times the speed in a circular orbit passing through the same point. (a) The equations describing the orbits are 2 l (circle) mk r= l2 1 (parabola.) mk 1 + cos θ Evidently, the perihelion of the parabola occurs when θ = 0, in which case r = l2 /2mk, or one-half the radius of the circle. (b) For the parabola, we have l2 sin θ ˙ ˙ r= θ (16) mk (1 + cos θ)2 ˙ sin θ = rθ 1 + cos θ so ˙ ˙ v 2 = r2 + r2 θ2 ˙ sin2 θ = r2 θ2 +1 (1 + cos θ)2 ˙ sin2 θ + 1 + 2 cos θ + cos2 θ = r2 θ2 (1 + cos θ)2 ˙ 1 = 2r2 θ2 1 + cos θ 2mkr3 θ2 ˙ = 2 l 2k = (17) mr ˙ in terms of the angular momentum l = mr 2 θ2 . On the other hand, for the circle ˙ r = 0, so ˙ l2 k v 2 = r2 θ2 = = (18) m2 r 2 mr Homer Reid’s Solutions to Goldstein Problems: Chapter 3 10 where we used that fact that, since this is a circular orbit, the condition k/r = l2 /mr2 is satisﬁed. Evidently (17) is twice (18) for the same particle at the √ same point, so the unsquared speed in the parabolic orbit is 2 times that in the circular orbit at the same point. Problem 3.12 At perigee of an elliptic gravitational orbit a particle experiences an impulse S (cf. Exercise 9, Chapter 2) in the radial direction, sending the particle into another elliptic orbit. Determine the new semimajor axis, eccentricity, and orientation of major axis in terms of the old. The orbit equation for elliptical motion is a(1 − 2 ) r(θ) = . (19) 1 + cos(θ − θ0 ) For simplicity we’ll take θ0 = 0 for the initial motion of the particle. Then perigee happens when θ = 0, which is to say the major axis of the orbit is on the x axis. Then at the point at which the impulse is delivered, the particle’s momentum is entirely in the y direction: pi = piˆ After receiving the impulse S in the radial j. (x) direction, the particle’s y momentum is unchanged, but its x momentum is now px = S. Hence the ﬁnal momentum of the particle is pf = Sˆ iˆ Since the i+p j. particle is in the same location before and after the impulse, its potential energy is unchanged, but its kinetic energy is increased due to the added momentum: S2 Ef = Ei + . (20) 2m Hence the semimajor axis length shrinks accordingly: k k ai af = − =− = . (21) 2Ef 2Ei + S 2 /m 1 + S 2 /(2mEi ) Next, since the impulse is in the same direction as the particle’s distance from the origin, we have ∆L = r × ∆p = 0, i.e. the impulse does not change the particle’s angular momentum: Lf = Li ≡ L. (22) With (20) and (22), we can compute the change in the particle’s eccentricity: 2Ef L2 f = 1+ mk 2 2Ei L2 L2 S 2 = 1+ 2 + 2 2. (23) mk m k Homer Reid’s Solutions to Goldstein Problems: Chapter 3 11 What remains is to compute the constant θ0 in (19) for the particle’s orbit after the collision. To do this we need merely observe that, since the location of the particle is unchanged immediately after the impulse is delivered, expression (19) must evaluate to the same radius at θ = 0 with both the “before” and “after” values of a and : ai (1 − 2 ) af (1 − 2 ) f i = 1+ i 1 + f cos θ0 or 1 af (1 − 2 ) f cos θ0 = −1 . f ai (1 − i ) Problem 3.13 A uniform distribution of dust in the solar system adds to the gravitational attrac- tion of the sun on a planet an additional force F = −mCr where m is the mass of the planet, C is a constant proportional to the gravitational constant and the density of the dust, and r is the radius vector from the sun to the planet (both considered as points). This additional force is very small compared to the direct sun-planet gravitational force. (a) Calculate the period for a circular orbit of radius r0 of the planet in this com- bined ﬁeld. (b) Calculate the period of radial oscillations for slight disturbances from this cir- cular orbit. (c) Show that nearly circular orbits can be approximated by a precessing ellipse and ﬁnd the precession frequency. Is the precession the same or opposite direction to the orbital angular velocity? (a) The equation of motion for r is l2 m¨ = r + f (r) mr3 l2 k = − 2 − mCr. (24) mr3 r For a circular orbit at radius r0 this must vanish: l2 k 0= 3 − r2 − mCr0 (25) mr0 0 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 12 −→ l = mkr0 + m2 Cr0 4 ˙ l 1 −→ θ = 2 = mr2 mkr0 + m2 Cr0 4 mr0 0 k 3 mCr0 = 3 1+ mr0 k 3 k mCr0 ≈ mr03 1+ 2k Then the period is 2π 3/2 m mCr03 τ= ≈ 2πr0 1− ˙ θ k 2k 2 Cτ0 = τ0 1 − 8π 2 3/2 where τ0 = 2πr0 m/k is the period of circular motion in the absence of the perturbing potential. (b) We return to (24) and put r = r0 + x with x r0 : l2 k m¨ = x − − mC(r0 + x) m(r0 + x)3 (r0 + x)2 l2 x k x ≈ 3 1−3 − 2 1−2 − mCr0 − mCx mr0 r0 r0 r0 Using (25), this reduces to 3l2 2k m¨ = − x 4 + r3 − mC x mr0 0 or ¨ x + ω2x = 0 with 1/2 3l2 2k ω= 2 r4 − 3 −C m 0 mr0 1/2 2l2 k = 4 − mr3 m2 r0 0 where in going to the last line we used (25) again. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 13 Problem 3.14 Show that the motion of a particle in the potential ﬁeld k h V (r) = − + 2 r r is the same as that of the motion under the Kepler potential alone when expressed in terms of a coordinate system rotating or precessing around the center of force. For negative total energy show that if the additional potential term is very small compared to the Kepler potential, then the angular speed of precession of the ellip- tical orbit is ˙ 2πmh Ω= 2 . l τ The perihelion of Mercury is observed to precess (after corrections for known plan- etary perturbations) at the rate of about 40 of arc per century. Show that this precession could be accounted for classically if the dimensionless quantity k η= ka (which is a measure of the perturbing inverse square potential relative to the grav- itational potential) were as small as 7 × 10−8 . (The eccentricity of Mercury’s orbit is 0.206, and its period is 0.24 year). The eﬀective one-dimensional equation of motion is L2 k 2h m¨ = r 3 − 2+ 3 mr r r L2 + 2mh k = + 2 mr3 r L2 + 2mh + (mh/L)2 − (mh/L)2 k = + 2 mr3 r [L + (mh/L)]2 − (mh/L)2 k = + 2 mr3 r If mh L2 , then we can neglect the term (mh/L)2 in comparison with L2 , and write [L + (mh/L)]2 k m¨ = r 3 + 2 (26) mr r which is just the normal equation of motion for the Kepler problem, but with the angular momentum L augmented by the additive term ∆L = mh/L. Such an augmentation of the angular momentum may be accounted for by Homer Reid’s Solutions to Goldstein Problems: Chapter 3 14 augmenting the angular velocity: ˙ mh ˙ mh L = mr2 θ −→ L 1+ = mr2 θ 1 + L2 L2 2˙ 2˙ = mr θ + mr Ω where ˙ mh 2πmh Ω= = ˙ L2 θ L2 τ is a precession frequency. If we were to go back and work the problem in the ˙ reference frame in which everything is precessing with angular velocity Ω, but there is no term h/r in the potential, then the equations of motion would come 2 ˙ out the same as in the stationary case, but with a term ∆L = mr 2 Ω added to the eﬀective angular momentum that shows up in the equation of motion for r, just as we found in (26). To put in the numbers, we observe that ˙ 2π m Ω= (h) τ L2 2π mka h = τ L2 ka 2π 1 h = τ 1 − e2 ka so h τΩ˙ = (1 − e2 ) ka 2π = (1 − e2 )τ fprec where in going to the third-to-last line we used Goldstein’s equation (3-62), and ˙ in the last line I put fprec = Ω/2π. Putting in the numbers, we ﬁnd h 1◦ 1 revolution 1 century−1 = (1 − .2062 ) · 0.24 yr · 40 yr−1 ka 3600 360◦ 100 yr−1 = 7.1 · 10−8 . Homer Reid’s Solutions to Goldstein Problems: Chapter 3 15 Problem 3.22 In hyperbolic motion in a 1/r potential the analogue of the eccentric anomaly is F deﬁned by r = a(e cosh F − 1), where a(1 − e) is the distance of closest approach. Find the analogue to Kepler’s equation giving t from the time of closest approach as a function of F . We start with Goldstein’s equation (3.65): r m dr t= 2 r0 k − l2 +E r 2mr 2 r m r dr = . (27) 2 r0 Er2 + kr − l2 2m With the suggested substitution, the thing under the radical in the denom- inator of the integrand is l2 l2 Er2 + kr − = Ea2 (e2 cosh2 F − 2e cosh F + 1) + ka(e cosh F − 1) − 2m 2m 2 l2 = Ea e cosh F + ae(k − 2Ea) cosh F + Ea − ka − 2 2 2 2m It follows from the orbit equation that, if a(e − 1) is the distance of closest approach, then a = k/2E. Thus k 2 e2 k 2 e2 l2 = cosh2 F − − 4E 4E 2m k2 2 2El2 = e cosh F − 1 + 2 4E mk 2 k 2 e2 k 2 e2 = cosh2 F − 1 = sinh2 F = a2 e2 E sinh2 F. 4E 4E Plugging into (27) and observing that dr = ae sinh F dF , we have F m ma2 t= a(e cosh F − 1) dF = [e(sinh F − sinh F0 ) − (F − F0 )] 2E F0 2E and I suppose this equation could be a jumping-oﬀ point for numerical or other investigations of the time of travel in hyperbolic orbit problems. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 16 Problem 3.26 Examine the scattering produced by a repulsive central force f = kr −3 . Show that the diﬀerential cross section is given by k (1 − x)dx σ(Θ)dΘ = 2E x2 (2 − x)2 sin πx where x is the ratio Θ/π and E is the energy. The potential energy is U = k/2r 2 = ku2 /2, and the diﬀerential equation for the orbit reads d2 u m dU mk +u=− 2 =− 2 u dθ2 l du l or d2 u mk 2 + 1+ 2 u=0 dθ l with solution u = A cos γθ + B sin γθ (28) where mk γ= 1+ . (29) l2 We’ll set up our coordinates in the way traditional for scattering experiments: initially the particle is at angle θ = π and a great distance from the force center, and ultimately the particle proceeds oﬀ to r = ∞ at some new angle θs . The ﬁrst of these observations gives us a relation between A and B in the orbit equation (28): u(θ = π) = 0 −→ A cos γπ + B sin γπ = 0 −→ A = −B tan γπ. (30) The condition that the particle head oﬀ to r = ∞ at angle θ = θs yields the condition A cos γθs + B sin γθs = 0. Using (30), this becomes − cos γθs tan γπ + sin γθs = 0 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 17 or − cos γθs sin γπ + sin γθs cos γπ = 0 −→ sin γ(θs − π) = 0 −→ γ(θs − π) = π or, in terms of Goldstein’s variable x = θ/π, 1 γ= . (31) x−1 Plugging in (29) and squaring both sides, we have mk 1 1+ = . l2 (x − 1)2 Now l = mv0 s = (2mE)1/2 s with s the impact parameter and E the particle energy. Thus the previous equation is k 1 1+ = 2Es2 (x − 1)2 or k (x − 1)2 s2 = − . 2E x(x − 2) Taking the diﬀerential of both sides, k 2(x − 1) (x − 1)2 (x − 1)2 2s ds = − − 2 − dx 2E x(x − 2) x (x − 2) x(x − 2)2 k 2x(x − 1)(x − 2) − (x − 1)2 (x − 2) − x(x − 1)2 =− 2E x2 (x − 2)2 k 2(1 − x) =− . (32) 2E x2 (x − 2)2 The diﬀerential cross section is given by | s ds | σ(θ)dΩ = . sin θ Plugging in (32), we have k (1 − x) σ(θ)dΩ = dx 2E x2 (x − 2)2 sin θ as advertised.