= Solution
{c}
For <polar coordinates>, the rotating <unit vectors> satisfy $\dot{\mathbf e}_r=\dot\theta\mathbf e_\theta$ and $\dot{\mathbf e}_\theta=-\dot\theta\mathbf e_r$. Applying the <product rule> twice to $\mathbf r=r\mathbf e_r$ gives the <velocity> and <acceleration>
$$
\dot{\mathbf r}=\dot r\mathbf e_r+r\dot\theta\mathbf e_\theta,\qquad \ddot{\mathbf r}=(\ddot r-r\dot\theta^2)\mathbf e_r+(r\ddot\theta+2\dot r\dot\theta)\mathbf e_\theta.
$$
<Newton's second law> for a <central force> now gives
$$
m(\ddot r-r\dot\theta^2)=F(r),\qquad m(r\ddot\theta+2\dot r\dot\theta)=0.
$$
Multiplying the azimuthal <force> equation by $r/m$ yields
$$
\frac{d}{dt}(r^2\dot\theta)=0,\qquad \boxed{r^2\dot\theta=h.}
$$
This is conservation of <specific angular momentum>. Here $F(r)$ is the signed radial component of the <force>; a negative value points inward.
When $h\ne0$, $\theta$ is a valid local parameter along the <trajectory>. Put $u=1/r$ and let primes denote <derivatives> with respect to $\theta$. The <chain rule> yields
$$
\dot\theta=hu^2,\qquad \dot r=-\frac{u'}{u^2}\dot\theta=-hu',\qquad \ddot r=-h^2u^2u'',\qquad r\dot\theta^2=h^2u^3.
$$
Substituting into the radial <Newton's second law> equation and dividing by $-mh^2u^2$ proves the <Binet equation>:
$$
\boxed{u''+u=-\frac{F(1/u)}{mh^2u^2}.}
$$
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