= Solution
{c}
For motion in a <central potential>, <conservation of angular momentum> gives $L=mr^2\dot\theta$. In <polar coordinates>, the conserved <mechanical energy> is
$$
E=\frac m2\dot r^2+\frac m2r^2\dot\theta^2-G(r)
=\frac m2\dot r^2+\frac{L^2}{2mr^2}-G(r).
$$
Consequently, for $L\ne0$,
$$
\dot r=\pm\sqrt{\frac{2}{m}[E+G(r)]-\frac{L^2}{m^2r^2}},\qquad
\boxed{\frac{dr}{d\theta}=\frac{mr^2}{L}\dot r
=\pm\frac{r^2}{L}\sqrt{2m[E+G(r)]-\frac{L^2}{r^2}}.}
$$
The sign in $\dot r$ distinguishes inward and outward motion. Choosing the orientation so $L>0$, it is also the sign of $dr/d\theta$. The expression under the <square root> must be nonnegative. If $L=0$, the motion is radial and $\theta$ is constant, so $r$ cannot be parametrized by $\theta$ this way.
Back to article page
