 Solution

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.