Reich and the Falling Blob Solution

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Reich and the Fallin Blob Solution Monday, 5 December 011 Physics 111 Problem 1 Chemist Ferdinand Reich conducted early experiments on the Coriolis acceleration in 1831, shortly before Coriolis made

(a) Calculate the expected eastward deflection, clearly statin any simplifyin assumptions you make. (b) Reich also found a small southerly deflection, as have subsequent investiators.

1 Reich and the Fallin Blob Solution Monday, 5 December 011 Physics 111 Problem 1 Chemist Ferdinand Reich conducted early experiments on the Coriolis acceleration in 1831, shortly before Coriolis made his investiation in Reich dropped pellets down a m-deep mine shaft in Freiber, Germany, and observed a mean eastward deflection of 8.3 mm. The latitude of Freiber is λ = 51. (a) Calculate the expected eastward deflection, clearly statin any simplifyin assumptions you make. (b) Reich also found a small southerly deflection, as have subsequent investiators. This is a second-order Coriolis effect. The eastward velocity arises from the dominant vertical (z) motion; the eastward velocity (ẏ) produces a Coriolis acceleration in the x direction, proportional to ω, where ω is the anular speed of the Earth. You also need to keep track of the variation of and the centrifual force with heiht, since they are also proportional to ω. In fact, all terms are proportional to h ω sin λ cos λ. Accordin to Marion and Thornton (10-13), the total southerly deflection is 4h ω sin λ cos λ But they re wron. The correct leadin coefficient is not 4. Find the correct coefficient analytically by keepin track of all terms throuh quadratic order in ω. Then check your work with an exact numerical calculation in Mathematica. Solution: Orientin z alon the radially outward direction at latitude λ, with x pointin south and y pointin east as shown in Fi. 1, the forces and pseudo forces actin on the fallin particle are F = m(z)ẑ + mω (R + z) cos λ(sin λˆx + cos λẑ) mω v (1) where R is the radius of the Earth. Before analyzin the particle, however, a word about the plumb line. The plumb bob is suspended from the top of the tower via a fine line, which holds it just about Physics Peter N. Saeta

2 ω z λ y x Fiure 1: Coordinate system for analyzin the effect of pseudo forces on the descent of a point mass from heiht h at latitude λ. the round. We assume the line has neliible mass, so the line will han consistent with the value of and the centripetal acceleration at the surface of the Earth. From Eq. (1), the net force on the bob (which is at rest in the rotatin frame) is F bob = mrω cos λ sin λ ˆx m( Rω cos λ) ẑ () Therefore, the slope of the plumb line is where dx dz = Rω cos λ sin λ sin λ cos λ Rω cos = γ λ 1 γ cos λ = γ sin λ cos λ(1 + γ cos λ) γ = Rω = ( m) π 9.8 m/s ( s ) = (3) and I have expanded throuh second order in the small quantity γ. For a tower of heiht h, therefore, the plumb bob will be displaced from the base of the tower by x plumb = γh sin λ cos λ(1 + γ cos λ) (4) To et a sense of scale, if the tower is 00-m tall, and we are at λ = π/4, then the leadin term in x plumb is m and the small correction term is 0.59 mm. Now we turn to the fallin particle. The acceleration is the force divided by the mass, so we will concern ourselves only with F/m. In the chosen coordinate system, the anular velocity vector of the Earth is ω = ω(sin λ ẑ cos λ ˆx) Peter N. Saeta Physics 111

3 Usin the zeroth-order solution that inores everythin but the ravitational acceleration, we computed the vertical velocity durin the descent, which ave rise to an eastward Coriolis acceleration to first order in ω. This y component of velocity, combined with the z component of ω, ives a second-order deflection to the south (the x direction). If we inore the centrifual pseudo force to determine the vertical velocity as a function of time, we have ż(t) = t z(t) = h t T 0 = h/ where T 0 is the time to reach the round when we inore everythin but uniform ravity. The Coriolis acceleration in the y direction is so υ y is ÿ = ω x ż = ω( cos λ)( t) ẏ = t ω cos λ The Coriolis acceleration in the x direction 1 is therefore ẍ = ω z ẏ = ω sin λ t ω cos λ Interatin twice with respect to time, we et the net x displacement produced by the second-order Coriolis effect: so C a = /3. x Cor = ω sin λ cos λ 6 ( h ) = 3 h ω sin λ cos λ As mentioned in the prompt, there are two additional second-order terms in the southward deflection, both of which arise by considerin the variations in the acceleration with elevation. The reater the value of z, the smaller the acceleration due to ravity and the reater the centrifual acceleration. From Eq. (1), the centrifual acceleration in the x direction is Interatin twice ives ẍ = ω (R + z) sin λ cos λ = ω sin λ cos λ (R + h t /) (6) x = ω sin λ cos λ ( R + h t 4 t4 ) 1 There is also a z component of acceleration proportional to ω, which we inore, since its modification to the vertical acceleration produces Coriolis and centrifual terms that are of order reater than in ω. They may therefore be nelected in a second-order computation. (5) Physics Peter N. Saeta

4 Evaluatin at t = T 0 ives x = ω (R + h)h sin λ cos λ [ 4h 4 ] = ω sin λ cos λ ( Rh + 5h 6 ) The first term is the first term in the deflection of the plumb line, Eq. (4). The second is the displacement due to the variation in the centrifual force with heiht, so C b = 5/6. Expandin the ravitational acceleration to first order in the elevation z, we et (z) = GM (R + z) = GM R (1 + z R ) (1 z R ) where we have assumed that the acceleration due to ravity is at the Earth s surface. The reduction in the value of with heiht means that we have overestimated the downward acceleration and velocity, independent of the value of ω. This means that it will take loner for the mass to fall, allowin the centrifual acceleration to operate loner than we oriinally estimated. Lookin back to Eq. (6), we will et a contribution from the R term when we interate twice to et x, since we will use a slihtly loner time T to reach the round. Secondly, because will be reduced with heiht, the final term in Eq. (6) will be reduced, further increasin the southerly acceleration. First we will calculate the chane in the fall time when we account for the variation in with elevation, throuh second order in h and ω. Then we will evaluate the consequences to the first and third terms in the parentheses of Eq. (6). Usin the first-order expansion of (z), we et the equation of motion in the vertical, z = (1 z/r) z z = (7) R The solution to the homoeneous equation is z = C cosh βt, where β = /R and we have chosen the solution that has ż(0) = 0. Since z(0) = h, we can solve for C to et From Eq. (6), we now have z(t) = R + (h R ) cosh βt (8) ẍ = ω sin λ cos λ[r + z(t)] (9) with a more complicated function for z(t). The additional contribution arisin from considerin the variation in the fall time with the drop in will arise from (a) a different value of T and a difference in the expression for z(t). We face a choice. We can either interate first, and then simplify or expand first and then interate. I think it miht be a bit simpler to expand Eq. (8) first, fiure out at what time T the mass hits the round, expand the cosh βt term in Eq. (8) Peter N. Saeta 4 Physics 111

5 to appropriate order, and then interate the whole lot. So, that s how we ll proceed, always keepin in mind that we need to keep terms to order h. To help us keep track of the quantities that are small, define so that we can rewrite Eq. (8) as є = h R (10) z(t) = R [1 (1 є) cosh βt] (11) So, we need to keep terms throuh є. However, the hyperbolic cosine term in Eq. (8) depends on t, not h. How far do we need to o in Taylor expandin it? The time it takes to fall is oin to be slihtly loner than the T = h/ value we found before, so T is of order h. Therefore, we will need to keep terms up to T 4. Taylor expandin the hyperbolic cosine in Eq. (8) throuh fourth order in t ives z(t) R [1 (1 є) (1 + β t where s β t /. + β4 t 4 )] = R 4! s [є (1 є) (s + )] (1) 6 When does the particle hit the round? When z = 0, which yields a quadratic equation for s: s 6 + s є 1 є = 0 s = 1 ± ( /3 є 1 є )1/ = 3 { 1 ± [1 + є 3 (1 + є)] 1/ } In the last step I have used the binomial expansion to express the fraction throuh quadratic order in the small parameter є. Since s must be positive, we have to take the positive sin. Usin the binomial expansion once more, we have to be careful to include all terms that are second order in є. Since, (1 + ξ) 1/ = ξ + 1 ( 1 ) ξ! + O(ξ3 ) = 1 + ξ ξ 8 + O(ξ3 ) the positive root for s throuh quadratic order in є is s = 3 { 1 + [1 + 1 So, to second order in h, T R = h R є (є ) є (є ) ]} = 3 { є + є є } = є + 5є 6 4h R T = h h R = T 0 + T 1 (13) Physics Peter N. Saeta

6 The first term is, of course, just what we had before. The correction for the fall time is the second term. Now we return to Eq. (9), substitutin the expansion in Eq. (1) for z(t), ettin ẍ = ω sin λ cos λ R {1 + 1 Interatin twice with respect to time ives [є (1 є) ( t R + t 4 6 )]} x = ω sin λ cos λ R [ t + є 4 t (1 є) ( t4 4R ) + O(t6 )] = ω sin λ cos λ [ Rt + ht t4 4 + O(T6 )] (14) All this fanciness in є, however, has yielded the same expression for x(t) throuh quartic order in t. We have already evaluated the terms for t = T 0. The new term is T 1 substituted into the leadin term of Eq. (14), which yields x = ω sin λ cos λ At lon last, therefore, RT 1 = 5 3 ω h sin λ cos λ C c = 5 3 C a = 3 C b = 5 6 C c = 5 3 and x = ω h sin λ cos λ (C a + C b + C c ) with C a + C b + C c = 19 6, which is a bit shy of the factor of 4 advertised by Marion. Confirmation: Peter N. Saeta 6 Physics 111

7 Fallin blob Peter N. Saeta 5 November 009, updated 4 December 011 We seek to evaluate the deviation in the southward direction from a plumb line of a particle fallin from heiht h at latitude l. We will first compute the x displacement of the plumb line from top to bottom, then turn to the displacement of the fallin particle. We assume that the line suspendin the plumb bob has neliible mass, so the direction will be set entirely by the local acceleration at the surface of the Earth. This combines two effects: and the centrifual acceleration. ü Plumb line In[1]:= az = - + w R Cos@lD ; ax = w R Sin@lD Cos@lD; x plumb = ax az H-hL; In[4]:= We will keep thins symbolic, but substitute numerical values when required. To facilitate this, here are the substitutions we will use (all quantities in SI): sub = :w Ø In[5]:= x plumb ê. sub Out[5]= In[6]:= Out[10]= p, Ø 9.8, R Ø , l Ø p ê 4, h Ø 88>; 4 µ 3600 So, the plumb line will extend 1.47 m from the base of the tower, when the tower is 88-m tall (the heiht of Burj Khalifa in Dubai, which I ot to see on the way to Kenya in the summer of 010). ü Fallin mass The fallin particle experiences the same forces as the plumb line, althouh they now depend on the elevation z, plus an additional Coriolis force. W = w 8- Cos@lD, 0, Sin@lD<; acoriolis = - Cross@W, 8x'@tD, y'@td, z'@td<d; acentrifual = - Cross@W, Cross@W, 8x@tD, y@td, R + z@td<dd; agrav = z@td - 80, 0, 1<; R a = acoriolis + acentrifual + agrav :R w sinhll coshll + w sin HlL xhtl + w sinhll y HtL + w sinhll coshll zhtl, - Hw sinhll x HtL + w coshll z HtLL + w sin HlL yhtl + w cos HlL yhtl, - + R w cos HlL + w sinhll coshll xhtl + w coshll y HtL + w cos HlL zhtl> + 1M I zhtl R We now seek to solve the nonlinear differential equation for the trajectory of the particle usin numerical methods.

8 FallinBlob.nb In[11]:= sol = FirstB NDSolveB 8 x''@td ã Ha ê. subl@@1dd, y''@td ã Ha ê. subl@@dd, z''@td ã Ha ê. subl@@3dd, x'@0d ã 0, y'@0d ã 0, z'@0d ã 0, x@0d ã 0, y@0d ã 0, z@0d ã h ê. sub<, 8x, y, z<, :t, 0, 1.1 h ê ê. sub>ff Out[11]= 8x Ø InterpolatinFunction@H L, <>D, y Ø InterpolatinFunction@H L, <>D, z Ø InterpolatinFunction@H L, <>D< In[1]:= tf = t ê. FindRoot@ Evaluate@ z@td ê. sold, 8t, 14<D Out[1]= In[13]:= x@tfd ê. sol Out[13]= In[14]:= IHx@tfD ê. soll - Ix plumb ê. submm 1000 Out[14]= So, the difference between the spot the particle lands and the plumb bob is 0.59 mm! What is h w sinhll coshll? In[15]:= 1000 h w Out[15]= Sin@lD Cos@lD ê. sub In[16]:= Or 10.8 µm. The ratio is Hx@tfD ê. soll - Ix plumb h w Out[16]= ê. subm Sin@lD Cos@lD ê. sub and the answer we et from series expansion is 19 6 êê N Note that Marion and Thornton, problem 10.13, lists the answer as 4, but they are clearly double countin the chane in centrifual acceleration with heiht. Kudos to Wynn, DK, and Cecily for uncoverin this. Reich s data Supplement 17 November 010 for Reich's data. For a mine shaft m-deep at l = 51, the eastward deflection should be y = 1 3 t3 w cos l, or = 9.8; h = 158.5; w = p 4 µ 3600 ;

9 FallinBlob.nb h ê w Cos@ 51 D In other words, we expect an eastward deflection of 7.5 mm, which is rather close to Reich's reported value of 8.3 mm. On the other hand, our estimate of the southward deflection is % H - 3 µ.5 ê 5.5L 6 w h Sin@51 D Cos@51 D or 13.4 µm deflection to the south. This is about 1/8 the diameter of a human hair. There is no way that Reich could have observed this small a deflection.

Homework # 2. SOLUTION - We start writing Newton s second law for x and y components: F x = 0, (1) F y = mg (2) x (t) = 0 v x (t) = v 0x (3)

Homework # 2. SOLUTION - We start writing Newton s second law for x and y components: F x = 0, (1) F y = mg (2) x (t) = 0 v x (t) = v 0x (3) Physics 411 Homework # Due:..18 Mechanics I 1. A projectile is fired from the oriin of a coordinate system, in the x-y plane (x is the horizontal displacement; y, the vertical with initial velocity v =