Jul 9, 2014

The Gravity Train Revisited


The gravity train is a conceptual of design of transportation making use of the gravity of the planet, first proposed by 17th century scientist Robert Hooke in a letter to Issac Newton. It features a tunnel through the Earth.

Suppose a straight tunnel was drill through the Earth. A train operating solely on gravity starts from one end of the tunnel. Assuming there is no friction inside the tunnel, and the density of the Earth is uniform. The problem asks how long it takes for the train to arrive at the other end.

In the classic solution, the spin (rotation) of the Earth is not considered. (Strictly speaking, the rotation effect can be factorized in by using a locally measured gravitational acceleration $g$). The answer then is $t = \pi \sqrt{\frac{R_E}{g}} \simeq 42 \mathrm{min.}$, regardless of the exact location of the tunnel, where $R_E \simeq 6371 \mathrm{km}$ is the radius of the Earth, $g \simeq 9.8 \mathrm{m/s^2}$ is the gravitational acceleration. The rotation of the planet introduces a second time scale: $T_s \simeq 24 \mathrm{hr} \simeq 17 T_G$, where  $T_G = 2\pi \sqrt{\frac{R_E}{g}}$. The separation of the scales validates the neglecting of the Earth spin in the classical solution. Nevertheless, in this essay we want to look at the correction effect introduced by the spin of the Earth.

Fig. 1

General Settings

Let the center of the Earth be the origin of our coordinate system. The tunnel is characterized by the position of the center of the tunnel $\vec d$, and the orientation of the tunnel $\hat \tau$, $\vec d \cdot \hat \tau = 0$. Let $\vec{r} = \vec{d} + x \hat{\tau}$ be the position of the train, where $x$ is the distance of the train from the center of the tunnel, $\vec{v} = \dot x\hat\tau$ the velocity of the train. The spin of the Earth is described by an angular velocity vector $\vec\Omega$. Suppose $\Omega \cos \theta \equiv \vec \Omega \cdot \hat \tau$, $\Omega d \cos\phi = \vec\Omega \cdot \vec d$.

In the reference frame of the Earth (a non-inertial frame), there are four forces acting on the train: the gravity $\vec F_G = -\frac{GM_Em}{R^3_E} \vec r \equiv -mg \frac{\vec r}{R_E}$, the centrifugal force $\vec F_e = - m \vec\Omega \times (\vec\Omega \times \vec r)$, the Coriolis force $\vec F_c = 2 m \vec v \times \vec \Omega$, and the contact force. Since there is no friction, the contact force simply counter all forces in the direction that perpendicular to the tunnel. We only need to consider the force along the tunnel. $\hat \tau \cdot \vec F_G = -m g \frac{x}{R_E}$, $\hat \tau \cdot \vec F_e = m \Omega^2 x - m \Omega^2 (d \cos\phi + x \cos \theta)$, $\hat \tau \cdot \vec F_c = 0$. The total force along the tunnel is $F = -mg\frac{x}{R_E} + m \Omega^2 ( (1-\cos\theta)x - d\cos\phi)$.

The equation of motion thus reads, \[
\ddot x = \left( -\frac{g}{R_E} + \Omega^2 (1-\cos\theta) \right) x - \Omega^2 d \cos\phi
\] The solution is a harmonic oscillator with angular frequency $\omega = \frac{g}{R_E} - \Omega^2 (1-\cos\theta)$ or period $T = 2\pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}$. The center of the harmonic oscillator is not in the middle of the tunnel. In stead, it is shifted away from there by $\Delta = \frac{\Omega^2dR_E\cos\phi}{g - \Omega^2 R_E(1-\cos\theta)}$.

To calculate the travel time, we need to consider two cases. If the train starts from the high altitude side, it arrives at the low altitude terminal in time $t = \pi \sqrt{\frac{R_E}{g - \Omega^2R_E(1-\cos\theta)}}\left(1 - \frac{1}{\pi}\arccos\frac{\sqrt{R_E^2-d^2}-\Delta}{\sqrt{R_E^2-d^2}+\Delta} \right)$. If the train starts from low altitude terminal (farther from the axis of the Earth) with zero velocity, it would never reach the high altitude side. Some minimum velocity is needed in order to arrive the other side, $v_{\min} = 2 \Omega \sqrt{\left( g - \Omega^2R_E (1-\cos\theta) \right) d\cos\phi} \sqrt[4]{1-\frac{d^2}{R^2_E}}$. by doing so, it also takes time $t$ to arrive at the destination.


The spin of the Earth affects the problem in two ways. The correction in the period of the oscillator is about $\sim \frac{\Omega^2 R_E}{g}\sin^2\frac{\theta}{2} = 0.345\% \sin^2\frac{\theta}{2}$. The second correction is through the displacement of the center of the harmonic oscillator. We compare $\Delta$ and $l = \sqrt{R_E^2 - d^2}$ half the length of the tunnel. $\Delta \simeq \frac{\Omega^2 R_E}{g} \cos\phi d = 0.345371\% \cos\phi \sqrt{R_E^2 - l^2}$. Compare $\Delta_{\max}$ (with $\phi = 0$) and $l$ (see Fig. 2). The correction from $\Delta$ is only small for tunnels either of thousands of kilometers long or well oriented (i.e. choosing a good $\phi$). Thousands of kilometers long tunnel would be for transcontinental express.

Fig. 2, the numbers for the tunnel length $l$ and maximum shift of the harmonic oscillator center $\Delta_{\max}$ are in kilometers.

We have only considered the straight tunnels. It turns out, straight tunnels are not the fastest ones for gravity train. The fastest tunnels are the brachistochrone tunnels. The correction due to the Earth rotation can be considered accordingly.

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