Suppose we have a force which is dependent on distance from some point. It is easy to find the acceleration based on the distance from that point, but in a previous question, I asked they told me that acceleration as a function of distance is meaningless because you can't use it to calculate anything with it.
If we have the Newtonian gravity
$$F \propto \frac{1}{r^2}$$
how can we find the acceleration with respect to time for such a force?
Best Answer
Let's say we are actually talking about gravity:
$$\vec{F}=\frac{GMm}{r^2}\hat{r}$$ $$\vec{F}=m\vec{a}$$ $$\rightarrow \vec{a}=\ddot{\vec{x}}=\frac{GM}{r^2}\hat{r}$$
Which is really easy to obtain; but it provides us with a system of 3 Differential Equations that describe acceleration at all times, and don't actually give acceleration as a function of time. So finding the acceleration as a function of time, much like finding the position as a function of time becomes a task of solving the Differential Equations.
Luckily this can be simplified with geometry, and once we understand the shape of orbits, we can find the velocity of objects going through that orbit, and ofcourse the acceleration. Bodies that experience an Inverse Square Central Force Field($F \propto \frac{1}{r^2}$) move in paths that correspond to conic sections i.e. ellipses, parabolas and hyperbolas. You can find a very direct and clear proof of that statement here.
If an orbit is closed its path is elliptical (a circle is a special case of an ellipse). This is Keppler's 2nd Law.