= Solution
{c}
Initially $\dot r=0$ also gives $\dot z=H'(b)\dot r=0$. Thus the initial <velocity> is azimuthal, with $|h|=bv$. By <conservation of energy>, the supplied <mechanical energy> divided by $m/2$ gives
$$
[1+(H'(r))^2]\dot r^2+\frac{b^2v^2}{r^2}+2gH(r)=v^2+2gH(b).
$$
The first term is nonnegative. Consequently
$$
\boxed{2gH(r)+\frac{b^2v^2}{r^2}\leq2gH(b)+v^2.}
$$
For a positive launch <speed> $v>0$, introduce twice the <radial effective potential on a surface of revolution> per unit <mass>,
$$
S(r)=2gH(r)+\frac{b^2v^2}{r^2},\qquad B=S(b)=2gH(b)+v^2.
$$
It satisfies
$$
S''(r)=2gH''(r)+\frac{6b^2v^2}{r^4}>0,
$$
so it is a <strictly convex function>. Also $S(r)\to\infty$ as $r\to\infty$. To justify the other endpoint carefully, <convexity> gives $H(r)\geq H(b)+H'(b)(r-b)$ for $0<r<b$. This supplies a finite lower bound near $0$, and the positive $r^{-2}$ term consequently forces $S(r)\to\infty$ there.
Thus $S$ has a unique <minimum> at some $r_m>0$. Its <derivative> is strictly increasing and zero there, so $S$ decreases to $r_m$ and increases afterwards. Its <sublevel set> $S(r)\leq B$ is a single interval $[b_1,b_2]$, with $0<b_1\leq b\leq b_2<\infty$. If $B>S(r_m)$, the two boundary <equations> $S(b_i)=B$ have distinct solutions; if $B=S(r_m)$, both boundaries are $r_m$. The <conservation of energy> identity therefore proves
$$
\boxed{b_1\leq r(t)\leq b_2.}
$$
The two <radii> coincide exactly when $b=r_m$, equivalently $S'(b)=0$. This gives
$$
\boxed{b_1=b_2=b\quad\Longleftrightarrow\quad v^2=gbH'(b).}
$$
The <equation> in <5/b> then has the constant solution $r=b$, with $\dot\theta=\pm v/b$: this is a <circular orbit>.
\b[A positive inner bound requires $v>0$.] If zero launch <speed> is allowed, $h=0$ and the displayed <inequality> only gives $r\leq b$ because $H$ is strictly increasing; one may take the limiting inner bound $b_1=0$. For example, $H(r)=r^2$ satisfies all the stated hypotheses but permits inward motion from rest all the way towards the excluded axis. The two-positive-turning-radii argument therefore uses the natural interpretation of “launched with <speed> $v$” as $v>0$.
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