Gravity Train

Since the gravitational field inside the planet is proportional to the radius $r$ , conservation of energy tells us that $\tfrac12mv^2 + \tfrac{1}{2R}mgr^2 \; = \; \tfrac12mgR \hspace{2cm} 0 < r < R$ so that $v^2 = \omega^2(R^2 - r^2)$ for $0 < r < R$ , where $\omega^2 = \tfrac{g}{R}$ . A gravity train path is a function $r\,:\, [0,\tfrac14\pi] \to \mathbb{R}$ such that $0 < r(\theta) \le R$ for all $0 \le \theta \le \tfrac14\pi$ , and such that $r(0) = r(\tfrac14\pi) = R$ . The time taken to traverse such a path is $T[r] \; = \; \int \frac{ds}{v} \; = \; \frac{1}{\omega}\int_0^{\frac14\pi} \sqrt{\frac{(r')^2 + r^2}{R^2 - r^2}}\,d\theta$ and we want to find the path that minimizes $T[r]$ . Standard Calculus of Variations considerations tell us that $\mathcal{L} - r'\tfrac{\partial \mathcal{L}}{\partial r'}$ is constant, where $\mathcal{L} = \sqrt{\frac{(r')^2 + r^2}{R^2-r^2}}$ . If we assume that the minimum value of $r$ throughout the motion is $\rho R$ , we deduce that $\frac{r^2}{\sqrt{((r')^2 + r^2)(R^2-r^2)}} \; = \; \frac{\rho}{\sqrt{1-\rho^2}}$ and hence $r^{-2}(r')^2 \; = \; \frac{r^2 - \rho^2R^2}{\rho^2(R^2 - r^2)}$ It is clear that $r$ will be a symmetric function in $\theta$ , so that $r' \; = \; \left\{ \begin{array}{lll} -\frac{r}{\rho}\sqrt{\frac{r^2 - \rho^2R^2}{R^2 - r^2}} & \hspace{1cm} 0 < \theta < \tfrac18 \pi \\ \frac{r}{\rho}\sqrt{\frac{r^2 - \rho^2R^2}{R^2 - r^2}} & \hspace{1cm} \tfrac18\pi < \theta < \tfrac14 \pi \end{array}\right.$ Thus we obtain the following integral expression for $\theta$ as a function of $r$ (in the range $0 < \theta < \tfrac18\pi$ to begin with): $\theta \; = \; \int_r^R \frac{\rho}{r}\sqrt{\frac{R^2-r^2}{r^2 - \rho^2R^2}}\,dr \hspace{2cm} \rho R \le r \le R$ The substitution $r^2 = \tfrac12R^2(1+\rho^2) + \tfrac12R^2(1-\rho^2)\cos\phi$ , for $0 \le \phi \le \pi$ , gives us $\theta \; = \; \tfrac12\int_0^\phi \frac{\rho(1-\rho^2)\,\tan\frac12\phi\,\sin\phi\,d\phi}{1+\rho^2 + (1 - \rho^2)\cos\phi}$ and then the substitution $u = \tan\tfrac12\phi$ gives $\theta \; = \; \rho(1-\rho^2)\int_0^u \frac{u^2\,du}{(1 + \rho^2u^2)(1 + u^2)} \; = \; \rho\int_0^u\left(\frac{1}{1 + \rho^2u^2} - \frac{1}{1+u^2}\right)\,du \; = \; \tan^{-1}(\rho u) - \rho \tan^{-1}u$ Thus the first half of the path is described by the parametric equations: $r^2 \; = \; \tfrac{1}{2}R^2\big[1 + \rho^2 + (1- \rho^2)\cos\phi\big] \hspace{1cm} \theta \; = \; \tan^{-1}\big(\rho\tan\tfrac12\phi\big) -\tfrac12\rho\phi$ for $0 \le \phi \le \pi$ . To ensure that $\theta$ varies between $0$ and $\tfrac18\pi$ throughout this motion, we deduce that $\rho = \tfrac34$ . It can be shown that the entire path, for $0 \le \theta \le \tfrac14\pi$ , is a segment of the hypocycloid obtained by a circle of radius $\tfrac18R$ rolling inside a circle of radius $R$ (the angle $\phi$ is then the rolling angle).

Running through the above equations, the time taken to traverse this optimal gravity train track is $\begin{aligned} T & = \; \tfrac{2}{\omega}\int_0^{\frac18\pi} \sqrt{\frac{(r')^2 + r^2}{R^2-r^2}}\,d\theta \; = \; \tfrac{2}{\omega}\int_0^{\frac18\pi} \frac{\sqrt{1-\rho^2}}{\rho} \frac{r^2}{R^2-r^2}\,d\theta \\ & = \; \tfrac{2\sqrt{1-\rho^2}}{\omega \rho}\int_{\rho R}^R \frac{r^2}{R^2-r^2} \times \frac{\rho}{r} \sqrt{\frac{R^2-r^2}{r^2-\rho^2R^2}}\,dr \; = \; \tfrac{2\sqrt{1-\rho^2}}{\omega}\int_{\rho R}^R \frac{r\,dr}{\sqrt{(R^2-r^2)(r^2-\rho^2R^2)}} \\ & = \; \tfrac{\pi}{\omega}\sqrt{1-\rho^2} \; = \; \boxed{\tfrac14\pi\sqrt{\tfrac{7R}{g}}} \end{aligned}$

The not all that mathematically able engineer works it out like this...

It is known that the time taken for a transit along a chord tunnel between any 2 places on the surface is $\frac{\pi}{4}\sqrt{\frac{16R}{g}}$ .

It is known that the least time path is a cycloid path, at least in a uniform parallel gravity field, so a partial cycloid curve will be plausible. A bit of sketching of a cycloid path between two point 45° apart looks like it gets to be more than twice the depth of the chord and but not as much as thrice. Thrice the depth gives at most $\sqrt{3}$ the speed. So $\frac{\pi}{4}\sqrt{\frac{16R}{3g}}$ would be a lower bound and that leaves the remaining choice of $\boxed{\frac{\pi}{4}\sqrt{\frac{7R}{g}}}$

Really at this level giving multi choice answers lets less able people such as myself an escape route!

Try similar problems Gravity Train 2 and 3 .