= Solution
For variations vanishing at the endpoints, the <Euler-Lagrange equations> are
$$
\boxed{\frac d{dt}F_{\dot r}-F_r=0,\qquad
\frac d{dt}F_{\dot\theta}-F_\theta=0.}
$$
Differentiate the stated energy expression using the <chain rule>. The acceleration terms cancel, leaving
$$
\begin{aligned}
\frac{dH}{dt}
&=\dot r\left(\frac d{dt}F_{\dot r}-F_r\right)
+\dot\theta\left(\frac d{dt}F_{\dot\theta}-F_\theta\right)-F_t.
\end{aligned}
$$
Along 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 $T=\tfrac12m(\dot r^2+r^2\dot\theta^2)$. An inverse-square central <gravitational force> has <potential energy> proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, giving
$$
F=T-V=\frac m2(\dot r^2+r^2\dot\theta^2)-\frac{km}{r}.
$$
For attraction to a star of mass $M_\star$, this convention has $k=-GM_\star<0$. 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 $F_{\dot\theta}=mr^2\dot\theta$, the equations of motion are
$$
\boxed{\ddot r-r\dot\theta^2=\frac{k}{r^2},\qquad
\frac d{dt}(mr^2\dot\theta)=0.}
$$
The second equation gives the conserved <angular momentum> $J=mr^2\dot\theta$; equivalently $r\ddot\theta+2\dot r\dot\theta=0$. The autonomous-functional identity gives a second conserved quantity,
$$
\boxed{H=\dot rF_{\dot r}+\dot\theta F_{\dot\theta}-F
=\frac m2(\dot r^2+r^2\dot\theta^2)+\frac{km}{r}=T+V.}
$$
This is the total <mechanical energy>; for attractive gravity the potential contribution is negative because $k<0$.
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