Newton-Coulomb differential equation solution

Question

I am trying to solve the following DE: $$ m \mathbf{\ddot{r}}=\frac{Q_1Q_2}{4\pi\epsilon_0\mathbf{r}\cdot\mathbf{r}}\mathbf{\hat{r}} $$ Which I know is a tough one, so I settled for the simpler version: $$ m \ddot{r}=\frac{Q_1Q_2}{4\pi\epsilon_0r^2} $$ But I still couldn't figure out an exact solution to this DE, and can't really recall any method I know that would help but analytical solution which is not the goal, so I inserted it to Mathematica and got this equation after setting $c_1=1$ and $c_2=0$ for sake of simplicity: $$ r \sqrt{1-\frac{\alpha }{r}}+\alpha \tanh ^{-1}\left(\sqrt{1-\frac{\alpha }{r}}\right)=t $$ where: $$ \alpha = \frac{Q_1Q_2}{2m\pi\epsilon_0} $$ but the program fails at solving this equation and I too can't figure out a way to extract an exact solution. So, if anyone can help, much appreciated!

Answer 1

That is an exact (albeit implicit) solution. Sorry, you're not going to get a closed form expression for $r(t)$ as a function of $t$.

EDIT: The way to get the implicit solution is as follows. This being an autonomous second-order differential equation, you can write a first-order equation for $v = dr/dt$ as a function of $r$:

$$ \dfrac{dv}{dr} = \dfrac{dv/dt}{dr/dt} = \dfrac{\alpha/(2r^2)}{v} = \dfrac{\alpha}{2 r^2 v}$$

This is a separable equation, so you get

$$\eqalign{2 v \; dv & = \alpha \;\dfrac{ dr}{r^2} \cr v^2 &= -\frac{\alpha}{r} + c_1\cr v &= \pm \sqrt{c_1 - \frac{\alpha}{r}}}$$ Writing $v = dr/dt$ again, that becomes another separable equation, and so $$ t = \pm\int \dfrac{dr}{\sqrt{c_1 - \alpha/r}} + c_2 $$

and that rather complicated integral (in the case $c_1 = 1$) gives you Mathematica's solution.

Most integrals, and most differential equations, don't have closed-form solutions. The question is not "why doesn't it?" but rather "why does it?" You're lucky to get even this much of a solution.