Here's the problem I am trying to find a solution to:
Consider a heavy point particle of mass $m_2$ fixed at a point (preferably the origin) and another light point particle of mass $m_1$ at an initial separation of $r_i$ from the fixed particle. There are no other bodies that can interfere with their interaction. Given any time $>t$, what will be the separation $r$ of the two particles at that time?
I tried to solve this myself with the following procedure:
$$E = K_E + P_E$$ where $E$ is the constant total energy, $K_E$ and $P_E$ the kinetic and potential energies of the light particle respectively. Elaborately, $$\frac{-Gm_2m_1}{r_i}=\frac{1}{2}m_1v^2-\frac{Gm_2m_1}{r}$$ Solving for $v$ gives me $$v=\sqrt{2Gm_2(\frac{r-r_i}{rr_i})}$$ after that I get to a halt when this appears $$t=\frac{1}{\sqrt{2Gm_2}}\int{\sqrt{\frac{rr_i}{r_i-r}}dr}$$ which I cannot solve further, So I post this question in Mathematics Stack Exchange, and a satisfying answer to the integral they give is $$t=r_i\sqrt{r_i}\ln\left(\sqrt{r}+\sqrt{r-r_i}\right)+\sqrt{r(r+r_i)}-\frac{r_i\sqrt{r_i}\ln\left(\sqrt{r_i}\right)}{\sqrt{2Gm_2}}-(r-r_i)^2$$ which doesn't seem correct by the way. Differentiating this again must get me back to where i started but I don't get there. Is there any other way to solve such a question or is this the best that can be achieved?
