The solution of a differential equation being unbounded under a certain condition 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.
 A: Let's start with the equation 
$$\frac{1}{2}\dot{r}^2 = \frac{\gamma M}{r} + c.$$
Since $r(0) = R$ and $\dot{r}(0) = v_0$, then 
$$\frac{1}{2}v_0^2 = \frac{\gamma M}{R} + c,\quad \text{or}\quad c = \frac{v_0^2}{2} - \frac{\gamma M}{R}.$$
So then
$$\dot{r}^2 = \frac{2\gamma M}{r} + \left(v_0^2 - \frac{2\gamma M}{R}\right).\tag{*}$$
If $v_0 \ge v_F$, then the expression in parentheses in (*) is nonnegative, making 
$$\dot{r}^2 \ge \frac{2\gamma M}{r}.$$
That is, 
$$\dot{r} \ge \sqrt{\frac{2\gamma M}{r}}.\tag{**}$$
Let $s$ be a fixed positive time. Multiplying both sides of (**) by $\sqrt{r}$ and integrating from $0$ to $s$, we get
$$\int_0^s \dot{r}\sqrt{r}\, dt \ge \int_0^s \sqrt{2\gamma M}\, dt,$$
which results in 
$$\frac{2}{3}r^{3/2}(s) \ge \frac{2}{3}v_0^{3/2} + s\sqrt{2\gamma M},$$
that is, 
$$r(s) \ge \left(v_0^{3/2} + \frac{3\sqrt{2\gamma M}}{2}s\right)^{2/3}.\tag{***}$$
Since the right-hand side of (***) increases without bound as $s\to \infty$, then the solution $r(s)$ is unbounded.
Edit: Here I will address the OP's second question

Do you maybe have an idea though how I could show that, in case $v_0 < \sqrt{\frac{2\gamma M}{R}}$, $r$ is bounded?

If $v_0 < \sqrt{\frac{2\gamma M}{R}}$, then we may set 
$$v_0^2 - \frac{2\gamma M}{R} = -\alpha$$
for some positive number $\alpha$. Then (*) becomes
$$\dot{r}^2 = \frac{2\gamma M}{r} - \alpha.$$
Thus
$$\frac{2\gamma M}{r} = \alpha + \dot{r}^2 \ge \alpha,$$
which implies 
$$0 < r(s) \le \frac{2\gamma M}{\alpha}$$
for all $s$. Hence $r$ is bounded.
A: Problem Statement: 
$${\ddot {\bf{r}}} =  - {{k} \over {{{\left\| {\bf{r}} \right\|}^3}}}{\bf r}\qquad k\gt 0$$ 
First of all notice that:
$${\bf{r}} \times {\ddot {\bf{r}}} = {\bf{r}} \times {{d{\dot{\bf{r}}}}\over{dt}}={d\over {dt}}\left({\bf{r}} \times {\dot{\bf{r}}} \right)=0$$
So we have:
$${\bf{r}} \times {\dot{\bf{r}}}=constant\,vector:=\bf c$$
Above equation tells that particle movement is planar!
Suppose
$${\bf A}:=\left(\|{\dot{\bf{r}}}\|^2-{k\over\|{\bf r}\|}\right){\bf r}-\left({\bf r}.{\dot{\bf{r}}}\right){\dot{\bf{r}}}$$
It is easy to prove that ${\dot{\bf{A}}}\equiv 0$ and so we have ${\bf{A}}=constant\,vector={\bf{A}}(0)$.
We have also:
$${\bf{A}}.{\bf r}=\|{\bf r}\|^2\|{\dot{\bf{r}}}\|^2-k\|{\bf r}\|-\left({\bf r}.{\dot{\bf{r}}}\right)^2=-k\|{\bf r}\|+\left({\bf{r}} \times {\dot{\bf{r}}} \right).\left({\bf{r}} \times {\dot{\bf{r}}} \right)=-k\|{\bf r}\|+\|{\bf c}\|^2$$
And we know:
$${\bf{A}}(0)=\left(\|{\dot{\bf{r}}}(0)\|^2-{k\over\|{\bf r}(0)\|}\right){\bf r}(0)-\left({\bf r}(0).{\dot{\bf{r}}}(0)\right){\dot{\bf{r}}}(0)$$
So:
$$\left(\left(\|{\dot{\bf{{\bf r}}}}(0)\|^2-{k\over\|{\bf r}(0)\|}\right){\bf r}(0)-\left({\bf r}(0).{\dot{\bf{r}}}(0)\right){\dot{\bf{r}}}(0)\right).{\bf r}+k\|{\bf r}\|=\|{\bf c}\|^2$$
You can now derive what you want from this equation.
For example if:
$$\|{\dot{\bf{{\bf r}}}}(0)\|^2-{k\over\|{\bf r}(0)\|}=0\,\,\,and\,\,\,{\bf r}(0).{\dot{\bf{r}}}(0)=0$$
We have:
$$\|{\bf r}\|=constant$$
That means path is a part of circle.
