index_12.html

Numerical Solutions and Plots

Here rf=final radial distance from earth center, rmin=trajectory minimum radial distance from earth center, R = radius of earth's surface, G = Gravitational Constant, and M = earth's mass

In[29]:=

$tnum[rf_, rmin_, R_, G_, M_] := NIntegrate[-dxdr[r, rmin]/dxdt[r, R, G, M], {r, R, rf}]$

${6378140 Meter, (6.673*10^-11 Meter^2 Newton)/Kilogram^2, 5.9742*10^24 Kilogram}$

$Plot[tnum[rf, 0, 6378140, 6.673 * 10^(-11), 5.9742 * 10^24], {rf, 6378140, 0}, PlotLabel->"Time vs Radius on freefall to earth center", AspectRatio→1]$

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From the figure above, it appears that elapsed time for a frictionless dive is the same regardless of the distance of the trip!

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${{rmin→ -1/2 (-d^2 + 4 R^2)^(1/2)}, {rmin→1/2 (-d^2 + 4 R^2)^(1/2)}}$

In[30]:=

rmn[d_, R_] := 1/2 (4 R^2 - d^2)^(1/2) (* Minimum radius as a function of Trip Length d *)

$Table[{Convert[10^n Meter, Kilo Meter], Convert[N[6378140 - rmn[10^n, 6378140]] Meter, Meter], 180./Pi * (Pi/2 - ArcSin[rmn[10^n, 6378140]/6378140]) }, {n, 3, 6, .5}]//MatrixForm$

The above table lists Trip_Length, Maximum_Tube_Depth, and Angle_of_Starting_Incline in Degrees. As you can see, the slopes are pretty small for reasonable trips on earth, assuming that we don't have a practical way to build long straight tunnels penetrating 10 miles deep into the earth (bottom entry). If we did, however, the payoff might be worth 45 minutes of boring scenery at speeds into the 1000 km/hour range...

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Created by Mathematica (February 13, 2007)