## Duration of a trajectory

Suppose we launch an object straight up. We wish to find how long it will take to return. Suppose we launch it up at a speed $v_0$. It is well known that the classical escape velocity is given by $v_e=\sqrt{\frac{2MG}{R}}$ By examining the energy equation, we find that the speed when the object is a distance r from the center of the planet is given by: $v(r)=v_e\sqrt{\frac{R}{r}-\gamma}$ Where $\gamma=1-\frac{v_0^2}{v_e^2}$ To find the travel time, we integrate: $T=2\int_{R}^{R/\gamma}\frac{dr}{v(r)}=\frac{2}{v_e}\int_{R}^{R/\gamma}\frac{dr}{\sqrt{\frac{R}{r}-\gamma}}=\frac{2R}{v_e}\int_{\gamma}^{1}\frac{du}{u^2\sqrt{u-\gamma}}$ $T=\frac{2R}{v_e}\frac{\tan^{-1}\left ( \sqrt{\frac{1}{\gamma}-1} \right )+\sqrt{\gamma-\gamma^2}}{\gamma^{3/2}}=\frac{2R}{v_e}\frac{\sin^{-1}(u)+u\sqrt{1-u^2}}{(1-u^2)^{3/2}}$ Where $u=v_0/v_e$.

## Optimal Path through a Planet

We want to find the best path through a planet of radius R, connecting two points $2\alpha$ radians apart (great circle angle). We assume the planet is of uniform density. As is well known, the acceleration due to gravity a radius r from the center of the planet is given by: $a=-gr/R$ Where $g$ is the surface gravitational acceleration. Thus, if it falls from the surface along a path through the planet, its speed at a distance r from the center will be given by $\tfrac{1}{2}mv^2=\tfrac{1}{2}m\frac{g}{R}\left ( R^2-r^2 \right )$ $v(r)=\sqrt{\frac{g}{R}} \sqrt{R^2-r^2}$ Let us suppose it falls along the path specified by the function $r(\theta)$, where r is even and $r(\pm\alpha)=R$. The total time is given by $T=2\int_{0}^{\alpha}\frac{d\ell}{v}=2\sqrt{\frac{R}{g}}\int_{0}^{\alpha}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}d\theta$ In order to obtain conditions for the optimal path, then, we use calculus of variations. The Lagrangian is $L(r,r',\theta)=\frac{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}$ Using the Beltrami Identity, we find: $\frac{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}-\frac{r'^2}{{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}}=\frac{r^2}{{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}}=C$ Let $1+ \tfrac{1}{C^2}=1/q^2$. Rearranging, we find: $r'=r\sqrt{\frac{\left (1+ \tfrac{1}{C^2} \right )r^2-R^2}{R^2-r^2}}=\frac{r}{q}\sqrt{\frac{r^2-R^2q^2}{R^2-r^2}}$ As $r'(0)=0$, this implies that $r(0)=Rq$ $r=\frac{r(0)}{R}$ Let us make the change of variables: $u=r^2/R^2$. This then gives: $u'=2u\sqrt{\frac{\tfrac{1}{q^2}u-1}{1-u}}$ $u(0)=q^2$ In order to determine this value, we can integrate the differential equation: $\frac{1}{2u} \sqrt{\frac{1-u}{\tfrac{1}{q^2}u-1}}du=d\theta$ $\int_{q^2}^{1}\frac{1}{2u} \sqrt{\frac{1-u}{\tfrac{1}{q^2}u-1}}du=\frac{\pi}{2}(1-q)=\int_{0}^{\alpha}d\theta=\alpha$ Thus $q=1-\frac{2\alpha}{\pi}$ We can then find the total travel time: $T=2\sqrt{\frac{R}{g}}\int_{0}^{\alpha}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}d\theta=2\sqrt{\frac{R}{g}}\int_{Rq}^{R}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}\frac{1}{r'}dr$ $T=2\sqrt{\frac{R}{g}}\int_{Rq}^{R} \frac{q}{r} \sqrt{r^2+\frac{r^2}{q^2}{\frac{r^2-R^2q^2}{R^2-r^2}}}\frac{dr}{\sqrt{r^2-R^2q^2}}$ $T=\sqrt{\frac{R}{g}}\sqrt{1-q^2}\int_{Rq}^{R} \frac{2rdr}{\sqrt{R^2-r^2} \sqrt{r^2-R^2q^2}}$ $T=\sqrt{\frac{R}{g}}\sqrt{1-q^2}\int_{q^2}^{1} \frac{dx}{\sqrt{1-x^2} \sqrt{x^2-q^2}}$ $T=\pi \sqrt{\frac{R}{g}}\sqrt{1-q^2}=2\sqrt{\frac{R}{g}}\sqrt{\pi\alpha-\alpha^2}$ Below we show several trajectories along the optimal path for several values of alpha:

In fact, these solutions are hypocycloids.