= Solution
The equation has the one-dimensional <effective potential> per unit mass
$$
\boxed{\Phi(z)=-\frac{2mk}{\sqrt{z^2+a^2}}},
\qquad
\ddot z=-\Phi'(z).
$$
Thus the conserved specific energy is
$$
\mathcal E=\frac12\dot z^2+\Phi(z).
$$
The potential has its minimum $-2mk/a$ at $z=0$ and tends to $0$ from below as $|z|\to\infty$. Therefore $-2mk/a<\mathcal E<0$ gives bounded oscillation between two <classical turning point>[turning points]; $\mathcal E=0$ gives marginal escape with asymptotic speed zero; and $\mathcal E>0$ gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium $z=0$.
Back to article page
