Finding velocity $v$ and position $r$, given a time $t$ under the acceleration of a gravitational force I was messing with the maths, when I tried to find the velocity as a function of time, $v(t)$, and the position, also, as a function of time, $r(t)$ under the gravity force.
$$ m \ddot{r} = -G \frac{Mm}{r^2}$$
$$ \ddot{r} = - \frac{GM}{r^2} $$
In order to find out the velocity, you will have to integrate over both sides.
$$\dot{r} = \int{ -\frac{GM}{r^2} \; \mathrm d t}$$
$$\dot{r} = -GM \int{ \frac{1}{r^2} \; \mathrm d t}$$
The tricky thing you have to take in account is thar $r$, is a function of time, $r(t)$, so I'm not sure how to continue the integration from here. I've came across this post and also this one.
It talks about how to solve this exact problem by multiplying both sides of the equations by $\dot{r}$ and then, integrate.
$$\ddot{r} \dot{r} = - \frac{GM}{r^2} \dot{r}$$
$$\int{ \ddot{r} \dot{r} \; \mathrm d t }= - GM \int{\frac{\dot{r}}{r^2} 
 \; \mathrm d t}$$
The thing is, I don't understand exactly how the steps (in the other posts) are done, it seems to me there are big jumps in between the steps, so I can't follow very well the integration process.
It would be nice, if someone could explain me, in depth, how to get the position $r(t)$ and $v(t)$ by integrating over this formulas.
 A: 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!
