The equation $$\ddot{s}(t) = -\gamma M \cdot \frac{1}{s(t)^2}$$ models a free fall trajectory under the influence of gravity from a large distance ($\gamma$ and $M$ are constant). One of the solutions to this equation has the form $$\overline{s}(t) = at^{2/3}$$
I want to show that if $s(t) > 0$ for all $t\in \{t\in\mathbb{R}:t\ge 0\}$, $s(t_0) = \overline{s}(t_0)$, $0<\dot{s}(t_0)<\dot{\overline{s}}(t)$, then $s(t) < \overline{s}(t)$ and $\dot{s}(t)<\dot{\overline{s}}(t)$ for all $t>t_0$. Moreover, $\dot{s}(t)$ has one zero.
I'm including my questions within my solution attempt below, and I'm asking you to kindly answer these questions.
My attempt:
Let $d(t):= \overline{s}(t)-s(t)=at^{2/3}-s(t)$, then $d(t_0)=0$. $\dot{d}(t)=\dot{\overline{s}}(t)-\dot{s}(t)=\frac{2a}{3t^{1/3}}-\dot{s}(t)$ and $\dot{d}(t_0)>0$. Also, $\ddot{d}(t_0)=0$. Thus we have that there is a $t'>t_0$ in the $\varepsilon$-neightbourhood of $t_0$ such that $\dot{d}(t')>0$, so that ${d}(t')>0$ and $\ddot{d}(t')>0$.
I'm stuck here. Can we rigorously conclude from the above that $\dot{d}(t)>0$ for all $t$?
For the second part of the question, if we have that $\lim\limits_{t\to \infty} \dot{s}(t)=0$ and $\dot{s}(t)<\dot{\overline{s}}(t)$, then does this imply that $\lim\limits_{t\to\infty} \dot{s}(t) <0$? I think this only implies that $\lim\limits_{t\to\infty} \dot{s}(t) \le 0$. Also, taking this limit does not necessarily imply $\dot{s}(t)$ has a root somewhere.
Would appreciate some help with this.