GraviTubes:  A term found in "Lost in a Good Book" by Jasper Fforde, refers to straight and nearly frictionless tunnels through the earth designed to enable rapid travel from point A to B at minimal energy cost (i.e. only the energy needed to overcome the friction of straight-line motion is required).  Gravitubes thus allow travelers to conveniently "fall" between any two points at the same distance from earth's center.  As we'll show here for planets with a density of around 5.5 g/cc (like that of earth), the time for any trip between points at the same distance from planetary center is about 40 minutes.  





Geosynchronous Orbits and Units

Acceleration g as a function of radius r from planetary center

outline and special cases

∮∮g◦A = 4π G M_in (* compare to ∮∮E◦A = 4π k Q_in *)

...where g is the local vector acceleration due to gravity, dA is the infinitesimal area vector pointing in the direction of the surface normal, the integration is over any close:d "Gaussian" surface, and M_in is the mass contained inside that surface.  This gravitational flux is equated, on the right hand side, to the mass inside the closed surface times a suitable constant.  When the mass and the surface share spherical symmetry, two special cases are...

In the classical example of a spherically-symmetric mass (or charge) distribution, a concentric spherical Gaussian surface of radius R will have a constant radial g field whose magnitude on integration obeys g(4πR^2)=4πGM_in, so that g = GM_in/R^2.  In the electric field case, this instead gives us the Coulomb field E = kQ_in/R^2.  Thus a form of Gauss's Law may be implicated in all forces that fall off as one over radial distance squared.

When r > R, g = GM/r^2.

When r<R, g = GMr/R^3.

yielding general rules for gravitational acceleration [plot],

gravitational potential [plot],

and escape velocity [plot].

Energy gained on a fall to the center: GmM/(2R).

Kinetic Energy GmM(R^2 - r^2)/(2R^3) and Speed (G M (R^2 - r^2)/R^3)^(1/2)as a function of radius r.

Time-elapsed, or why all trips are around (3π)/(16Gρ)^(1/2) = 42.2415 minutes...

On a trip to the center

If you know the radial velocity = dr/dt as a function of radius, you can integrate its reciprocal (dt/dr) over a range of r values to get elapsed time, as shown  below...

FullSimplify[∫_R^0 -1/v[r, R, G, M] r, Assumptions→R>0] (* Integrate (dt/dr) dr to get Δt *)

(π R^(3/2))/(2 (G M)^(1/2))

Here's a function to calculate time to center...


t0[R_, G_, M_] := π/2R^3/(G M)^(1/2) (* trip to the center *)

Here we note that R cubed over M is inversely proportional to density ρ, so that the time is also a function simply of ρ...

Solve[{ρ == M/(4/3π R^3), t == t0[R, G, M]}, t, M] (* this can be rewritten even more simply in terms of the planet ' s effective "uniform" density ρ *)

{{t→ -(3 π)^(1/2)/(4 G^(1/2) ρ^(1/2))}, {t→ (3 π)^(1/2)/(4 G^(1/2) ρ^(1/2))}}

Here we check on earth's effective density in this context...

M/(4/3Pi R^3)/. {R->6378140, M->5.9742 * 10^24} (* for example, the average density of the earth is 5.496 g/cc or 5496 kg/m^3 *)


This gives time elapsed on a fall to the center in terms of density alone...

t00[ρ_, G_] := Sqrt[(3 Pi)/(16 G ρ)] (* depends only on planet density, works for level - to - level trajectories IN or ON any uniform - density planetoid *)

Here we test this latter function for the earth's effective density...

Convert[t00[5.49679 * Gram/Centimeter^3, GravitationalConstant], Second] (* trip to center *)

1267.24 Second

The time elapsed falling through the earth's center to the other side (this time using the t0 instead of t00) would be...

Convert[2 * t0[EarthRadius, GravitationalConstant, EarthMass], Second]

2534.49 Second

In minutes, this is...

Convert[2 * t0[EarthRadius, GravitationalConstant, EarthMass], Minute]

42.2415 Minute

It turns out (see the section on chords NOT through the center below) that π/2R^3/(G M)^(1/2)= 42.2415 Minutes is also the elapsed time (in this uniform density approximation) for all gravitube "slides" between any two points on the earth at the same distance from earth's center.  To determine the time elapsed partway into a radial dive, one integrates only to rf...

FullSimplify[∫_R^rf -1/v[r, R, G, M] r, Assumptions→ {R>0, rf>0, rf<R, G>0, M>0}]

(R^(3/2) (π - 2 ArcTan[rf/(R^2 - rf^2)^(1/2)]))/(2 (G M)^(1/2))

Thus an analytic function for time-elapsed partway through a radial fall can also be defined...


tRadial[rf_, R_, G_, M_] := R^(3/2)/(2 (G M)^(1/2)) (π - 2 ArcTan[rf/(R^2 - rf^2)^(1/2)])

Distance, velocity, and acceleration vs time on the way

Along a chord not through the center (42.2415 minutes derived)

Radial distance, velocity and acceleration vs time along the chord

Numerical Solutions and Plots

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