1. The Subterranean Brachistochrone

Transcription

1 1 1. The Subteanean Bachistochone A Bachistochone is a fictionless tack that connects two locations and along which an object can get fom the fist point to the second in minimum time unde only the action of gavity. Above gound shot-ange Bachistochones ae well undestood. Hee we conside a subteanean vesion in which the stat and end points ae widely sepaated on the suface of the Eath. This vesion of the poblem was fist consideed by seveal authos back in 1966 (see Coope (1966a); Kimse (1966); Venezian (1966); Mallett (1966); Laslett (1966); Coope (1966b)). A vey nice moden teatment can be found in Calkin (1999). Fo any path, the time T to tavese fom one end to the othe can be witten quite simply as ds T v whee ds dx + dy d + dθ denotes the incemental squaed ac length along the path and v denotes the speed at each point along the path. Instantaneous speed is detemined fom consevation of enegy as follows. The foce due to gavity above the suface of the Eath is F GMm/. Below the suface, the effective mass of the Eath is educed fom M to M(/R) 3, whee R denotes the adius of the Eath and denotes the distance the tack is fom the cente of the Eath. Hence, the gavitational foce below the Eath s suface is F GMm/R 3. The gavitational foce is the negative of the gadient of the potential enegy field. Hence, the kinetic plus potential enegy at adius and velocity v is given by KE + PE 1 mv + 1 GMm. R 3 At the stat, R and v 0. Hence, by consevation of enegy, 1 mv + 1 GMm 1 GMm R 3 R.

2 Solving fo v, we get v ( ) GM 1. R R Hence, ou integal fo the time to tavese the path is given by d + T dθ R ( ) 3 GM R 1 GM R + R dθ. Following standad notational conventions fom the calculus of vaiations, let L(, ) denote the integand in the ight-hand integal above. Because L depends only on and and not on θ, we can use the Beltami equation to descibe a minimize of this integal: L L C whee C is an abitay constant. Fo ou paticula poblem, we compute explicitly as follows: L L + R + R + R C. Solving fo, we get Intoducing a new notation fo the constant, ( ) C (R ). C (R ) 0 C 1 + C R, we can simplify the fomula fo the squae of the deivative: ( ) d R dθ 0 0 R. Taking the (positive) squae oot of both sides and isolating the θ dependant quantities fom the dependant ones, we get R R dθ 0 0 d.

3 3 So, to aive at ou fomula elating to θ, we integate: R R θ 0 0 d. All that emains is to compute this integal explicitly. To this end, we make the following change of vaiable: u 0 R. To compute d, it is helpful to solve fo as a function of u: u R u Fo d, we get d u R 0 (1 + u ) du. Putting this altogethe, we compute the integal as follows: R 0 d 1 u 1 + u u R 0 (1 + u ) du R 0 u R + 0 (1 + u ) du R 0 (u R + 0)(1 + u ) du R + u R u R 0 du (u R + 0)(1 + u ) R du du du u R u du R 1 Ru tan tan 1 u + C 0 0 R tan 1 R R tan 1 0 R + C Hence, ou equation elating θ to can now be witten as θ tan 1 R 0 0 R 0 R tan 1 0 R + C.

4 4 Without loss of geneality, we may assume that C 0. If we let 0, then θ 0. If we let R, then θ ( ) 1 0 π. R The fomula as deived has one endpoint at the suface of the Eath and the othe endpoint at the nade given by 0. The entie path stats at the suface and etuns to the suface. In othe wods, it is two cuves of the fom shown. Hence, the angula extent of the path, θ is simply ( θ 1 ) 0 R π. Example. Fo a tunnel that extends 45, we have R.. Solution Via Numeical Optimization 3. Compaison Between Numeical Computation and Exact Result

5 5 paam pi : 4*atan(1); paam pi : pi/; paam eps : 1e-15; paam n : 51; paam G : e-11; paam M : 5.97e+4; paam R : 6.371e+6; # mˆ3 / kg sˆ # kg # m paam theta {j in 0..n}; paam dtheta {j in 1..n} : (theta[j] - theta[j-1]); va {j in 0..n} > 0; va mid {j in 1..n} ([j]+[j-1])/; va d {j in 1..n} ([j] - [j-1]); va ds {j in 1..n} sqt( d[j]ˆ + (mid[j]*dtheta[j])ˆ ); va dt {j in 1..n} *ds[j] / ( sqt(eps+1-([j]/r)ˆ) + sqt(eps+1-([j-1]/r)ˆ) ); minimize time: sqt(r/(*g*m)) * sum {j in 1..n} dt[j]; fix [0] : R; fix [n] : R; let {j in 0..n} [j] : (j/n)*[n] + (1-j/n)*[0] -.0*R*(j/n)*(1-j/n); let theta[0] : pi/; let theta[n] : pi/4; #use a nonlinea intepolation that bunches nea the endpoints let {j in 0..n} theta[j] : sin(pi*j/n)ˆ*theta[n] + cos(pi*j/n)ˆ*theta[0]; option loqo_options "vebose itelim000 sigfig1 inftol1.3e-1"; solve;

6 deg 30 deg 45 deg 60 deg 90 deg 10 deg The Subteanean Bachistochone Fig. 1. Tunnel paths spanning 10, 30, 45, 60, 90, and 10 degees. The numeically obtained solution is shown as a solid line. The dashed line of the same colo is the exact solution. Note that the two solutions match with high pecision in all cases except the 10 degee example.

7 7 REFERENCES M.G. Calkin. Lagangian and Hamilonian Mechanics: Solution to the Execises. Wold Scientific, P.W. Coope. Though the Eath in Foty Minutes. Am. J. Phys., 34:68 70, 1966a. P.W. Coope. Futhe Commentay on Though the Eath in Foty Minutes. Am. J. Phys., 34: , 1966b. P.G. Kimse. An Example of the Need fo Adequate Refeences. Am. J. Phys., 34:701, L.J. Laslett. Tajectoy fo Minimum Tansit Time Though the Eath. Am. J. Phys., 34:70 703, R.L. Mallett. Comments on Though the Eath in Foty Minutes. Am. J. Phys., 34:70, G. Venezian. Teestial Bachistochone. Am. J. Phys., 34:701, This manuscipt was pepaed with the AAS L A TEX macos v5..