homework and exercises - Integrating radial free fall in Newtonian gravity

I thought this would be a simple question, but I'm having trouble figuring it out. Not a homework assignment btw. I am a physics student and am just genuinely interested in physics problems involving math, which would be all of them.

So lets say we drop an object from a height, $R+r$, it falls toward earth. This height, $R+r$, is far enough away that the $g$ it experiences is a fraction of $g$ at sea-level. Let's just say that air resistance is negligible, and it wouldn't be that complicated to just integrate from $0$ velocity to the terminal velocity piece wise and deal with the rest of it later.

So the key here is that acceleration is changing with time. I thought I could simplify this by saying it changes with distance, and it has nothing to do with time, but this didn't really help, my guess is that maybe time is important (doh).

I tried integrating acceleration with time, and ended up nowhere. I tried integrating $a=GM/R^2$ with respect to $R$ from $R+r$ to $R$ and ended up with a negative function.

I saw somewhere someone tried to expand with taylor series, they even have something similar on hyperphysics, but I can't figure out how to obtain the polynomials that precede the variables.

http://hyperphysics.phy-astr.gsu.edu/hbase/images/avari.gif

This is the hyperphysics site where they use polynomials to find the distance. http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html#c1

Maybe I can't solve this because I haven't taken a course in differential equations yet. What I want to know is how to calculate the distance at any time.

Answer

So you were on the right track with integrating over r and over t. Here's how you could do it:
The acceleration at any radius, r (if we assume Earth is a point mass) is: $$a=-{GM\over r^2}$$ The minus sign is because the acceleration is anti-radial. Then you can do the following: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ $$\Delta r~=~\lim_{\Delta t\rightarrow0}-{GM\over r^2}\Delta t^2$$ $$then$$ $$r^2\Delta r~=~-GM\Delta t^2$$ I dropped the limit in the last part because it is implied. If we now use $\lim_{\Delta t\rightarrow0}\Delta t=dt$ and integrate: $$\int_{R_o}^{r_f}r^2dr~=~\iint_0^t-GMd^2t$$ $R_o$ is any initial radius and $r_f$ is any final radius (although because this derivation assumed a zero initial velocity, if $r_f>R_o$ it all breaks down). And after some razzmatazz algebra: $$r_f(t)~=~\sqrt[3]{R_o^3-{3\over2}GMt^2}$$ And if you want to check it, type this into Wolfram, differentiate it twice, plug in the radius of earth, its mass, and $t=0$ and you'll find it says the acceleration is $-9.8{m\over s^2}$