Let us reformulate the question(v1) as a 1D kinematic mouse-and-cat optimal control problem. The masses are irrelevant for the kinematic problem and hence put to
$$m~=~1.$$
1) Consider first the cat. The goal for the cat is to obtain the position and velocity(!) of the mouse as quickly as possible. The cat can accelerate
$$|a|~\leq~ a_{0},$$
where $a_{0}>0$ is a maximal acceleration. (Update: This type of problem is in optimal control theory known as a double integrator. See also the textbook by H.P. Geering, Optimal Control with Engineering Applications, Springer, 2007, Section 2.1.4.) We want to show that, ideally speaking, there is an optimal strategy so that the acceleration of the cat is at all times either the maximally allowed or none,
$$ a(t)~\in~\left\{ -a_0,0 , a_0 \right\}, $$
i.e., the control parameter $a$ has the bang-bang property.
Let us define a signed kinetic energy
$$K~:=~\frac{mv|v|}{2}~=~\frac{v|v|}{2}~=~T ~{\rm sgn}(v), \qquad \qquad T~:=~\frac{mv^2}{2}~=~\frac{v^2}{2}~=~|K| . $$
It is convenient to consider a $(x,K)$ coordinate system. One may view it as a configuration space (or phase space) of the system, because the map $v \leftrightarrow K$ is a bijection: $\mathbb{R}\to\mathbb{R}$. In particular, one may plot the trajectories of the cat and the mouse in a $(x,K)$ diagram. From the work energy theorem, the slope of the trajectory is (up to a sign) the acceleration
$$ a~=~ma~=~\frac{dT}{dx}~=~ \frac{dK}{dx} {\rm sgn}(v) .$$
Thus the cat in initial state $(x_0,K_0)$ must proceed within a cone $C(x_0,K_0)$ as indicated in red in Figures 1, 2 and 3. The cat can exit the cone $C(x_0,K_0)$ through the $x$-axis $K=0$ only, and turn around, in order to reach a final state $(x,K)$ outside the cone.
$\uparrow$ Figure 1. Case $K_0>0$. The red region denotes the cone $C(x_0,K_0)$. The black oriented paths indicate optimal strategies for the cat to reach 3 different final states $(x,K)$.
$\uparrow$ Figure 2. The cone $C(x_0,K_0)$ marked with red in the case $K_0=0$.
$\uparrow$ Figure 3. The cone $C(x_0,K_0)$ marked with red in the case $K_0<0$.
In mathematical detail, the cone $C(x_0,K_0)$ is
$$ C(x_0,K_0)~:=~\left\{ \begin{array}{lcc} C_{+}(x_0,K_0)&{\rm for}& K_0>0, \cr\cr C_{+}(x_0,K_0)\cup C_{-}(x_0,K_0)&{\rm for}& K_0=0,\cr\cr C_{-}(x_0,K_0)&{\rm for}& K_0<0, \end{array} \right. $$
where we have defined the positive and negative cones as
$$ C_{\pm}(x_0,K_0)~:=~ \left\{(x,K)\in\mathbb{R}^2 \mid \pm a_0 (x-x_0) \geq |K-K_0| \wedge \pm K\geq 0\right\} .$$
For the cat to go from state $(x_0,K_0)$ to state $(x,K)$, there is an optimal strategy that leads to minimal time consumption $\tau(x,K;x_0,K_0)$, which we have tried to indicate in Figure 1. Roughly speaking, the cat should choose a route as far away from the $x$-axis $K=0$ as possible, as it is most costly in terms of time to have small speed. If the final state $(x,K) \in C(x_0,K_0)$ is in the cone, then two legs are needed (one maximal acceleration and one maximal de-acceleration). It is straightforward to calculate that the minimal time consumption $\tau(x,K;x_0,K_0)$ for $(x,K) \in C(x_0,K_0)$ is
$$ \tau(x,K;x_0,K_0) ~=~ \frac{2\sqrt{|K|+|K_0|+a_0|x-x_0|}
-\sqrt{2|K|}-\sqrt{2|K_0|}}{a_0} .$$
There are similar expressions for $\tau(x,K;x_0,K_0)$ in various cases where $(x,K) \notin C(x_0,K_0)$ but with more legs/terms, which we will leave as an exercise to determine.
2) Next consider the mouse. Let us assume that the full future trajectory of the mouse $t\mapsto x_1(t)$, $t\geq 0$, is known to an all-knowing cat. (There are other possible rules of the game, but this setup seems to be closest to what OP is after.) Let the velocity and the signed kinetic energy of the mouse be denoted
$$v_1(t)~=~\frac{dx_1}{dt} \qquad {\rm and}\qquad K_1(t)~=~\frac{v_1(t)|v_1(t)|}{2},$$
respectively. For each future time $t\geq 0$, define the difference
$$\Delta\tau(t)~:=~ \tau\left(x_1(t),K_1(t);x_0,K_0\right) - t$$
between the time the cat could be at the mouse state $(x_1(t),K_1(t))$ (if the cat made a run for it), and the time $t$ the mouse would be there.
If the two initial states of cat and mouse are different,
$$(x_0,K_0)~\neq~(x_1(t=0),K_1(t=0)),$$
then $\Delta\tau(t=0)>0$. The first instant $t_*$ that the cat can obtain the $(x,K)$ state of the mouse is the first time that $\Delta\tau(t)$ turns non-positive,
$$t_*~=~\inf \left\{ t\in \mathbb{R}_+ \mid \Delta\tau(t) \leq 0 \right\}. $$
This is the answer to how quickly the cat can obtain the position and velocity of the mouse.
Best Answer
I'm certain that it isn't possible to write out $r$ and $v$ as functions of $t$, at least explicitly, meaning you can't isolate $r$ or $v$. The problem lies in the form of the final solution that you get after integration.
So, you're stuck on how to integrate from here:
$\int \ddot{r}\dot{r} dt = -GM \int \frac{\dot{r}}{r^2} dt$
Use the following equalities:
$\frac{d}{dt} \bigg[\frac{\dot{r}^2}{2}\bigg] = \dot{r}\ddot{r}$
$\frac{d}{dt} \bigg[\frac{1}{r}\bigg] = -\frac{\dot{r}}{r^2}$
After, it shouldn't be difficult to obtain the following relation, which is shown in one of the posts you've mentioned.
$\dot{r} = \sqrt{v_0^2 + 2GM\big[\frac{1}{r} - \frac{1}{r_0}\big]}$
What you would want to do next is divide both sides by the square root and integrate over $t$, noting that the integration variable should then change as:
$\int \frac{\dot{r}}{\sqrt{v_0^2 + 2GM\big[\frac{1}{r} - \frac{1}{r_0}\big]}} dt = \int \frac{1}{\sqrt{v_0^2 + 2GM\big[\frac{1}{r} - \frac{1}{r_0}\big]}} dr$
This is the part where our jobs become difficult because although I know the solution, and I will share it with you, I do not know the steps to take to get it. I just thought I'd share it so you can get a feel for implicit solutions, the ones where you can't isolate what you want, namely $r(t)$.
$\frac{2GM}{(v_0^2-\frac{2GM}{r_0})^\frac{3}{2}} \tanh^{-1} \Bigg[\frac{\sqrt{v_0^2 + 2GM\big[\frac{1}{r} - \frac{1}{r_0}\big]}}{\sqrt{v_0^2 - \frac{2GM}{r_0}}}\Bigg] + r\frac{\sqrt{v_0^2 + 2GM\big[\frac{1}{r} - \frac{1}{r_0}\big]}}{v_0^2-\frac{2GM}{r_0}} = t + c_1$
I hope I helped a bit!