How to analytically solve: $\ddot{r} = \frac{k}{m}\frac{1}{r^3}$ I'm reading through Susskind-Hrabovsky's Theoretical Minimum.  I've come across this problem (Ex3): 
Here is what I've done so far:
$$F = - \nabla V$$
choosing polar coordinates:
$$ma = - \frac{\partial V}{\partial r}$$
$$m\ddot{r} = \frac{k}{r^3}$$
Now how can I solve this? Is there an analytical $r(t)$?
 A: $$\frac{d^2r}{dt^2} =\frac{A}{r^3}.$$ Multiply by $2 \frac{dr}{dt}$ on both sides to have $$ 2 \frac{dr}{dt}\frac{d^2r}{dt^2} =2\frac{A}{r^3}.\frac{dr}{dt}.$$ We can rewtite this as
$$ \frac{d}{dt} \left( \frac{dr}{dt}\right)^2=2\frac{A}{r^3}.\frac{dr}{dt}.$$ Now integrate w.r.t. $t$ both sides as
$$ \left( \frac{dr}{dt}\right)^2= \int 2\frac{A}{r^3} dr=-\frac{A}{r^2} +C$$ Here $C$ is the constant of integration. Next, $$\frac{dr}{dt}=\pm \sqrt{C-\frac{A}{r^2}} \implies  \int \frac{dr}{\sqrt{C-\frac{A}{r^2}}}= \pm t+ D.$$ the constanta $C$ and $D$ will be determined y thr initial value of $r$ and $\frac{dr}{dt}$ at the initial time $t=t_0$ This integral is easily doable by letting $r^2=u$.
We get $$\sqrt{Cu-A}=\pm Ct+CD \implies Cu-A=(\pm Ct+CD)^2 \implies r^2=\frac{C}{ (\pm Ct-CD)^2+A}.$$
A: The vector equation of motion
The equation of motion is
$\newcommand{\fx}{\mathfrak{x}}$
$$
m\ddot\fx=\frac{k\,\fx}{r^4}
$$
Circular solutions, all constants positive
A circular solution has the form $\fx(t)=(r\cos(u(t)), r\sin(u(t)))$ with some angle function $u(t)$. We get by the chain and product rule
$$
-m\dot u^2\,\fx+m\ddot u\,J\fx=\frac{k\,\fx}{r^4}
$$
where $J$ is a $90^\circ$ rotation matrix. As $\fx$ and $J\fx$ are linearly independent, this implies $\ddot u=0$ and $mr^4\dot u^2=-k$. This last identity can not be satisfied in real numbers, the left side is positive, the right side negative.

Radial force field, pointing outward
This is conform with the physical intuition that the potential is not a well but a peak at the origin, thus its force field is repelling and can thus not provide the centripetal force that continuously curves the trajectory back towards the origin, which is required for a bounded orbit.

Circular solutions, general constants
If the constants are assumed to allow negative values, then for $k<0$ set $r_c^4=\frac{-k}m$. Then circular orbits of every radius exist, with a constant angular speed of $\dot u=\pm\frac{r_c^2}{r^2}$.
General solutions
$\newcommand{\fs}{\mathfrak{s}}$
Let $\fs(\alpha)=(\cos(α),\sin(α))$. Then the polar decomposition of the trajectory is $\fx(t)=r(t)\fs(u(t))$. The derivatives are $\dot \fx(t)=\dot r(t)\fs(u(t))+r(t)\fs'(u(t))\dot u(t)$ and 
$$
\ddot \fx(t)=\ddot r(t)\fs(u(t))+2\dot r(t)\fs'(u(t))\dot u(t)-r(t)\fs(u(t))\dot u(t)^2+r(t)\fs'(u(t))\ddot u(t)=-\frac{r_c^4\fs(u(t))}{r(t)^3}
$$
Separating the coefficients of the orthogonal directions $\fs,\fs'$ gives
\begin{align}
0&=\ddot r(t)-r(t)\dot u(t)^2+\frac{r_c^4}{r(t)^3}\\
0&=r(t)\ddot u(t)+2\dot r(t)\dot u(t)&&\implies r(t)^2\dot u(t)=K
\end{align}
which means that the second Kepler law remains in action, the angular momentum is preserved. Now the first equation reduces to
$$
0=\ddot r(t)+\frac{r_c^4-K^2}{r(t)^3}
$$
which in the end leads back to the original differential equation in the question, which is extensively discussed in the other answer. The solution found there is
$$
\sqrt{Cr(t)^2+r_c^4-K^2}=\pm t+D,~~~ C=\dot r_0^2-\frac{r_c^4-K^2}{r_0^2}
$$
So the conclusion is that apart from the case $C=0$ there are no closed orbits, all other solutions spiral inwards or outwards.
