= Solution
{c}
Divide <Newton's second law> by the <mass> and resolve along the radial and transverse unit <vectors> in <polar coordinates>. A <central force> has no transverse component, so
$$
\ddot r-r\dot\theta^2=-F(r),\qquad2\dot r\dot\theta+r\ddot\theta=0.
$$
Multiplication of the second equation by $r$ shows that $d(r^2\dot\theta)/dt=0$. Thus
$$
\boxed{\ddot r-r\dot\theta^2=-F(r),\qquad r^2\dot\theta=h,}
$$
where $h$ is the signed <specific angular momentum> perpendicular to the <orbital plane>.
If $h\ne0$ and $r>0$, the <polar angle> is locally a valid independent variable. For $u=1/r$, use primes for <derivatives> with respect to $\theta$. The <chain rule> gives
$$
\dot\theta=hu^2,\qquad\dot r=-hu',\qquad\ddot r=-h^2u^2u'',\qquad r\dot\theta^2=h^2u^3.
$$
Substitution into the radial equation gives the <Binet equation>
$$
\boxed{u''+u=\frac{F(1/u)}{h^2u^2}.}
$$
If $h=0$, the <orbit> is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot\theta=0$ away from the origin.
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