Gravity train

In the case of a straight line through the center of the Earth, the acceleration of the body is equal to that of gravity: it is falling freely straight down. We start falling at the surface, so at time ${\displaystyle t}$ (treating acceleration and velocity as positive downwards):

${\displaystyle r_{t}=R-\int _{0}^{t}v_{t}\,dt=R-\int _{0}^{t}\int _{0}^{t}a_{t}\,dt\,dt}$

Differentiating twice:

${\displaystyle {\frac {d^{2}r}{dt^{2}}}=-a_{t}=-{\frac {r}{R}}\,g=-\omega ^{2}\,r}$

where ${\displaystyle \omega ={\sqrt {\frac {g}{R}}}}$. This class of problems, where there is a restoring force proportional to the displacement away from zero, has general solutions of the form ${\displaystyle r=k\cos(\omega t+\varphi )}$, and describes simple harmonic motion such as in a spring or pendulum.

In this case ${\displaystyle r_{t}=R\cos {\sqrt {\frac {g}{R}}}\,t}$ so that ${\displaystyle r_{0}=R}$, we begin at the surface at time zero, and oscillate back and forth forever.

The travel time to the antipodes is half of one cycle of this oscillator, that is the time for the argument to ${\displaystyle \cos {\sqrt {\frac {g}{R}}}\,t}$ to sweep out ${\displaystyle {\pi }}$ radians. Using simple approximations of ${\displaystyle g=10{\text{ m}}/{\text{s}}^{2},R=6500{\text{ km}}}$ that time is

${\displaystyle T={\frac {\pi }{\omega }}={\frac {\pi }{\sqrt {\frac {g}{R}}}}\approx {\frac {3.1415926}{\sqrt {\frac {10}{6500000}}}}\approx 2532{\text{ s}}}$

Path of gravity train

For the more general case of the straight line path between any two points on the surface of a sphere we calculate the acceleration of the body as it moves frictionlessly along its straight path.

The body travels along AOB, O being the midpoint of the path, and the closest point to the center of the Earth on this path. At distance ${\displaystyle r}$ along this path, the force of gravity depends on distance ${\displaystyle x}$ to the center of the Earth as above. Using the shorthand ${\displaystyle b=R\sin \theta }$ for length OC:

${\displaystyle g_{r}={\frac {x}{R}}\,g={\frac {\sqrt {r^{2}+b^{2}}}{R}}\,g}$

The resulting acceleration on the body, because is it on a frictionless inclined surface, is ${\displaystyle g_{r}\cos \varphi }$:

Diagram of forces on a gravity train on non-diametrical straight line path

${\displaystyle a_{r}=g_{r}\cos \varphi ={\frac {\sqrt {r^{2}+b^{2}}}{R}}\,g\cos \varphi }$

But ${\displaystyle \cos \varphi }$ is ${\displaystyle r/x={\frac {r}{\sqrt {r^{2}+b^{2}}}}}$, so substituting:

${\displaystyle a_{r}={\frac {\sqrt {r^{2}+b^{2}}}{R}}\,g\,{\frac {r}{\sqrt {r^{2}+b^{2}}}}={\frac {r}{R}}\,g}$

which is exactly the same for this new ${\displaystyle r}$, distance along AOB away from O, as for the ${\displaystyle r}$ in the diametric case along ACD. So the remaining analysis is the same, accommodating the initial condition that the maximal ${\displaystyle r}$ is ${\displaystyle R\cos \theta =AO}$ the complete equation of motion is

${\displaystyle r_{t}=R\cos \theta \cos {\sqrt {\frac {g}{R}}}\,t}$

The time constant ${\displaystyle \omega ={\sqrt {\frac {g}{R}}}}$ is the same as in the diametric case so the journey time is still 42 minutes; it's just that all the distances and speeds are scaled by the constant ${\displaystyle \cos \theta }$.

The time constant ${\displaystyle \omega }$ depends only on ${\displaystyle {\frac {g}{R}}}$ so if we expand that we get

${\displaystyle {\frac {g}{R}}={\frac {GM/R^{2}}{R}}={\frac {GM}{R^{3}}}={\frac {G\rho \,V}{R^{3}}}={\frac {G\rho \,{\frac {4}{3}}\pi \,R^{3}}{R^{3}}}=G\rho \,{\frac {4}{3}}\pi }$

which depends only on the gravitational constant and ${\displaystyle \rho }$ the density of the planet. The size of the planet is immaterial; the journey time is the same if the density is the same.