If you dig straight down, you reach the centre of the Earth 6400 km below the surface, but what happens if you dig at a shallow but constant angle? Intuitively the result would be a spiral, and it’s easy to show. We just need to assume some constant digging “speed” and a constant ratio of the radial and tangential velocity components given by the angle’s tangent.
In radial coordinates:
\[\begin{align*} \dot{r} & =-v\sin\beta\\ r\dot{\phi} & =v\cos\beta. \end{align*}\]The first equation is trivial to integrate
\[r(t)=r_{0}-v\sin\beta t.\]The second is also pretty easy
\[\begin{align*} \dot{\phi} & =\frac{v\cos\beta}{r_{0}-v\sin\beta t}\\ \phi(t) & =\phi_{0}-\cot\beta\log\left(1-\frac{v\sin\beta}{r_{0}}t\right). \end{align*}\]We can now also show that it’s a logarithmic spiral by isolating $t$ from $r(t)$ and substituting in $\phi(t)$. We get
\[\begin{align*} \phi(r) & =\phi_{0}-\cot\beta\log\left(\frac{r}{r_{0}}\right)\\ r(\phi) & =r_{0}\exp\left[(\phi_{0}-\phi)\tan\beta\right]. \end{align*}\]The length element is just $v\mathrm{d}t$ and from here it’s clear that the length of the tunnel to the center of the Earth is
\[L=\frac{r_{0}}{\sin\beta}\]PREVIOUSMars