= Solution
{c}
Take $a>0$ and the intended horizontal launch around a parallel of the <surface of revolution>: $\dot r(0)=\dot z(0)=0$, $a\dot\theta(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot z=f'(r)\dot r$, the conserved <energy> becomes
$$
\frac12[1+f'(r)^2]\dot r^2+U(r)=E,\qquad
U(r)=\frac{a^2v^2}{2r^2}+gf(r).
$$
Here $U$ is the <radial effective potential on a surface of revolution>. Rearrangement gives
$$
\boxed{D(r):=\frac{v^2}{2g}(r^2-a^2)+r^2[f(a)-f(r)]
=\frac{r^2}{2g}[1+f'(r)^2]\dot r^2\geq0.}
$$
For a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalently
$$
\boxed{\frac{v^2}{g}=\frac{2b^2[f(b)-f(a)]}{b^2-a^2}.}
$$
The interval between the radial bounds must be dynamically accessible: $D\geq0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\leq E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the <centrifugal barrier> already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.
The initial radial <acceleration>, obtained by projecting <Newton's second law> along $\mathbf e_r+f'(a)\mathbf e_z$, is
$$
\ddot r(0)=\frac{v^2/a-gf'(a)}{1+f'(a)^2}.
$$
Thus $v^2/g>af'(a)$ starts the particle outward, and $v^2/g<af'(a)$ starts it inward. Equality gives the exact <circular orbit> $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this <separatrix> is bounded, although the radial motion is not periodic. If $v=0$, the <centrifugal barrier> disappears and the particle may pass through the axis, still with bounded $r$.
The source leaves a direction ambiguity at points where $f'(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f'(a)\ne0$ that direction is forced by horizontality and tangency.
Back to article page
