Solution of ODE for electrons repulsive force

Question

Force between two electrons is defined by Coulomb law: $$ F(t)=k_e\frac{q_e^2}{r^2(t)} $$ where $r(t)$ is distance between electrons and $k_e=\left(4\pi\epsilon_0\right)^{-1}=8.99\cdot 10^9$, $\epsilon_0= 8.8542 \cdot 10^{−12} $ and $q_e=−1.602\cdot 10^{-19}$ is physical constants. Inertial force is defined by Newton's law as $$ F(t)=m_e\cdot \frac{d^2}{dt^2}r(t) $$ where $a(t)=\frac{d^2}{dt^2}r(t)$ is acceleration and $m_e=9.11\cdot 10^{-31}$. So following differential equation can be written: $$ \frac{d^2}{dt^2}r(t) = \frac{K}{r^2(t)} $$ where $K={q_e^2}\cdot\left(4\pi\epsilon_0\cdot m_e\right)^{-1}$ and initial conditions are $r(0)=R$, $\frac{d}{dt}r(0)=0$. This differential equation can be represented as system of equations using speed $v(t)=\frac{d}{dt}r(t)$: $$ \begin{cases} \frac{d}{dt}v(t)=\frac{K}{r^2(t)}\\ \frac{d}{dt}r(t)= v(t)\end{cases} $$ $\mathbf {Question}$: For given $v(0)=0$ and $r(0)=R$ initial conditions whether there is a closed solution for the system of ODE especially for speed $v(t)$? If initial distance $R$ is equal to classical electron radius $R=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_e^2}{m_e\cdot c^2}$ then steady state of speed is equal to light velocity $\lim_{t\to\infty} v(t)=c$. Whether this limit is to be found for no known closed form of speed? Numerical experiment for speed is shown below.

Answer 1

That is a simple equation:

$$ r'' = \frac{K}{r^2} $$

There the $'$ denotes the time derivative.

To solve it, multiply both sides by $r'$ to get

$$r' r'' = \frac{Kr'}{r^2}$$

Notice that $$\frac{r'}{r^2} = -\left(\frac{1}{r}\right)'$$

Using this, integrate both sides to get

$$\frac{K}{r} = -\int r' r'' \ \text{d}t = -\int r'\ \text{d}{r'} $$

This gives you

$$\frac{K}{r} = -\frac{(r')^2}{2} + C$$

which implies

$$r' = \sqrt{C - \frac{2K}{r}}$$

This equation can be readily integrated to obtain

$$\int \frac{\text{d}r}{\sqrt{C - 2K/r}} = \int \text{d}t$$

The integral is not difficult to compute (use the substitution $z = \sqrt{C - 2K/r}$).

Finally

$$ \frac{K \ln \left(\sqrt{C} r \sqrt{C-\frac{2 K}{r}}+C r-K\right)}{C^{3/2}}+\frac{r \sqrt{C-\frac{2 K}{r}}}{C} + C_2 = t$$