= Solution
Write $k=GM$ and $r=|\mathbf r|$. The gravitational <central force> has zero <torque>, so the specific <angular momentum>
$$
\mathbf h=\mathbf r\times\dot{\mathbf r},\qquad
\dot{\mathbf h}=\mathbf r\times\ddot{\mathbf r}=0
$$
is constant. For a nonradial orbit $h=|\mathbf h|>0$, motion lies in the plane normal to $\mathbf h$, and plane polar coordinates give $r^2\dot\theta=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,
$$
\dot r=-hu',\qquad \ddot r=-h^2u^2u'',\qquad r\dot\theta^2=h^2u^3.
$$
The radial equation $\ddot r-r\dot\theta^2=-k/r^2$ is therefore the <Binet equation>
$$
u''+u=\frac{k}{h^2}.
$$
Its general solution, after choosing the polar origin of angle, is
$$
u=\frac{k}{h^2}(1+e\cos\theta),\qquad
\boxed{r=\frac{p}{1+e\cos\theta},\quad p=\frac{h^2}{GM}.}
$$
The origin is a focus of this <conic section>. To identify which <conic section> occurs, use the conserved specific <energy>
$$
\varepsilon=\frac12(\dot r^2+r^2\dot\theta^2)-\frac{k}{r}
=\frac{h^2}{2}\left[(u')^2+u^2\right]-ku
=\frac{k^2}{2h^2}(e^2-1).
$$
A bound, nonradial orbit has $\varepsilon<0$, hence $0\leq e<1$ and is an <ellipse>. For example, putting $x=r\cos\theta$, $y=r\sin\theta$ in $r+ex=p$ gives
$$
(1-e^2)\left(x+\frac{ep}{1-e^2}\right)^2+y^2=\frac{p^2}{1-e^2},
$$
an explicit <ellipse> equation. This is the intended low, bound <Kepler orbit>. The inverse-square equation alone does not imply an <ellipse>: an initially tangential <speed> larger than $\sqrt{2GM/r}$ gives $\varepsilon>0$, $e>1$ and a <hyperbolic Kepler orbit>. Zero <angular momentum> instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.
In a circular orbit, centripetal <acceleration> balances the gravitational <acceleration>:
$$
r_0\Omega^2=\frac{GM}{r_0^2},\qquad
\boxed{T=\frac{2\pi}{\Omega}=2\pi\sqrt{\frac{r_0^3}{GM}}.}
$$
Here $\Omega$ is the positive <angular speed>; either sense of circulation has this <period>.
With the stated drag <force>, use the full <angular momentum> $\mathbf L=m\mathbf r\times\dot{\mathbf r}$. Gravitational <torque> is still zero, while the drag <torque> is
$$
\dot{\mathbf L}=\mathbf r\times(-A\dot{\mathbf r})
=-\frac{A}{m}\mathbf L.
$$
Therefore $\boxed{\mathbf L(t)=\mathbf L(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has <exponential decay> with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed <ellipse>.
Back to article page
