Consider the differential equation 

$$(1) \space\space\space\space\space\ddot{r} = - \gamma \frac{M}{r^2}$$

with $r, M$ and $\gamma > 0$. Also, consider the initial value problem $r(0) = R > 0; \dot{r}(0) = v\_0 > 0$.

I first want to show: an *unbounded* solution $r > 0$ for this initial value problem can only exist if

$$v\_0 ≥ v\_F := \sqrt{\frac{2\gamma M}{R}}$$

Next, I want to explicitly find a solution for the IVP, in case that $v\_0 = v\_F$.

I first didn't really know how to approach this problem. While searching for a way to solve (1), I found the "trick" to multiply both sides of the equation with $\dot{r}$, giving us

$$\ddot{r}\dot{r} = - \gamma \frac{M \dot{r}}{r^2}$$

which can then be integrated on both sides, giving us

$$\frac{1}{2}\dot{r}^2 = \frac{\gamma M}{r} + c, c \in \mathbb{R}$$

$$<=> \dot{r} = \sqrt{\frac{2 \gamma M}{r} + c}\space\space\space\space (2)$$

at which point I got stuck. I see that $(2)$ contains the $\sqrt{\frac{2 \gamma M}{r}}$, that I need for the first part of the task. But how exactly can I continue from there; how can I prove the statement about the unbounded solution using $(1)$ or $(2)$?

Also, I'm not sure if the $c$ in $(2)$ is really required, or if I can somehow get rid of it? If I want to solve the equation for $r$ for the second part of the task, it'll probably get confusing the carry on with the $c$.

(As a background: the differential equation is said to describe the distance $r$ inbetween two points of mass, which the larger of both having the mass $M$, and $\gamma$ being the constant of gravity.)

Edit: I think the intuition behind this problem is that these two points of mass can only "escape" each other if their initial movement is faster than a certain "speed", with this speed being $\sqrt{\frac{2\gamma M}{R}}$, whereas otherwise, their gravitational attraction is too strong and they will eventually collide.