Change of velocity of a falling object over height (acceleration changing with height) I've been trying to solve this by myself without success. I am trying to find the velocity of a falling object as its height varies (change in velocity as it falls). Although it is often assumed that the acceleration is constant, I know that it isn't due to Newton's Law of Gravitation, where a higher height will cause acceleration to be lower. So I started off with this.
$$a=G\frac{M}{\left(R+h\right)^2}$$
Then, I got a relation with velocity:
$$a=\frac{\text{d}v}{\text{d}t}=\frac{\text{d}v}{\text{d}h}\times\frac{\text{d}h}{\text{d}t}=v\times\frac{\text{d}v}{\text{d}h}$$
So to find velocity in terms of h, I integrated both sides w.r.t. h and came to
$$\int_{}^{}a(h)dh=\frac{1}{2}v(h)^2+C$$
But I didn't know what to do with that constant and after doing some research I realised that I had to make it in a definite integral form, with set values. So I had final position $h_f=0$ since that would be when the object comes to a stop (hits the ground) with $h$ as a changing variable and initial velocity $v_0=0$ since at release, the velocity is 0. So I tried:
$$\int_{h}^{0}a(h)dh=\frac 12\left(v_0^2-v(h)^2\right)=-\frac{1}{2}v(h)^2$$
I integrated $a$ from above:
$$\int_{h}^{0}\frac{GM}{(R+h)^2}dh=\left[-\frac{GM}{R+h}\right]_h^0=-\frac{GM}{R}+\frac{GM}{R+h}$$
Putting this with the equation of velocity above gave:
$$v(h)=±\sqrt{2\left(\frac{GM}{R}-\frac{GM}{R+h}\right)}$$
I assumed that the velocity would be negative since the object is falling. But when I tried to graph this, I realised that as $h$ gets smaller, or as the object falls, the velocity also decreases. This doesn't really make sense to me because wouldn't the velocity increase as an object keeps falling, meaning that as $h$ decreases, the velocity should increase? Does this velocity represent something else? Or did I simply do something wrong?
Also, how would I get to obtaining a height or velocity function against time?

EDIT: Okay, so I've tried solving this with an initial height set in mind and adding a negative in front of the acceleration. So, $h_0=50\times10^3$m and I put the final height as an unknown (negative values would be ignored). Doing this allowed me to 
$$\int_{50\times10^3}^{h}-\frac{GM}{(R+h)^2}dh=\left[\frac{GM}{R+h}\right]_{50\times10^3}^h=-\frac{GM}{R+h}+\frac{GM}{R+50\times10^3}
=\left[\frac{1}{2}v(h)^2\right]_{50\times10^3}^h=\frac{1}{2}v(h)^2$$
This gave me a new equation of,
$$v(h)=\sqrt{2\left(\frac{GM}{R+h}-\frac{GM}{R+50\times10^3}\right)}$$
This new equation starts off at 0 velocity when $h=50\times10^3$ and increases as $h$ decreases. This seems more of a reasonable result to me. Does this mean that my equation needs to have another variable, $h_0$, that also affects the velocity?
 A: From energy conservation
$$\dfrac{mv^2}{2}=\frac{GMm}{R+h}-\frac{GMm}{R+h_o}$$
We immediately obtain your solution
$$v=\sqrt{2GM\left(\frac{1}{R+h}-\frac{1}{R+h_o}\right)}\tag{A}$$
Or
$$-\frac{dh}{dt}=\sqrt{\frac{2GM}{R+h_o}}\cdot\sqrt{\dfrac{h_0-h}{R+h}}$$
By integrating
$$-\int_{h_o}^h\sqrt{\frac{R+h}{h_0-h}}\cdot dh=\sqrt{\frac{2GM}{R+h_o}}\space \int_{t_o}^t dt$$
We obtain the exact (but not explicit) solution for $h(t)$
$$(R+h_0)\arctan\left(\sqrt{\dfrac{h_0-h}{R+h}}\right)+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\space(t-t_o)\tag{B}$$
For a verification, simplify for $\space h\ll R$
Remember $\space\arctan(x)\approx x\space$ for $\space x\ll1$
$$\left(R+h_0\right)\sqrt{\dfrac{h_0-h}{R+h}}+\sqrt{(h_0-h)(R+h)}=\sqrt{\frac{2GM}{R+h_o}}\left(t-t_o\right)$$
And also $\space R+h\approx R\space$  for $\space h\ll R\space$ and $\space R+h_o\approx R\space$ for $\space h_o\ll R$
$$R\space\sqrt{\dfrac{h_0-h}{R}}+\sqrt{(h_0-h)R}=\sqrt{\frac{2GM}{R}}\left(t-t_o\right)$$
We obtain
$$\sqrt{(h_0-h)}=\sqrt{\frac{GM}{2R^2}}(t-t_o)$$
Or
$$h_0-h=\frac{g\space(t-t_o)^2}{2}$$
This is a well known formula where $g$ is the gravitational acceleration
$$\space g=\dfrac{GM}{R^2}$$
The solution for $v(t)$ consists of $(A)$ and $(B)$. It is not explicit, but is exact and can be solved numerically to any degree of precision.

EDIT: Integration steps
First, lets derive a specific reduction rule for integration. Note that
$$\int\dfrac{dx}{x^2+1}=\arctan{(x)}+C_1=-arctan{\frac{1}{x}}+C_2\tag{1}$$
Take a derivative
$$\dfrac{d}{dx}\left(\frac{x}{x^2+1}\right)=\frac{1}{x^2+1}-\frac{2x^2}{(x^2+1)^2}$$
And integrate it
$$\dfrac{x}{x^2+1}=\int\frac{dx}{x^2+1}-\int\frac{2x^2dx}{(x^2+1)^2}+C^{'}$$
By rearranging and using $(1)$, we obtain the reduction rule
$$-\int\dfrac{2x^2dx}{(x^2+1)^2}=\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\tag{2}$$
Now define
$$x=\sqrt{\dfrac{R+h}{h_o-h}}\tag{3}$$
Solve for h
$$h_o-h=\dfrac{R+h_o}{x^2+1}\tag{4}$$
Differentiate (3)
$$\dfrac{dx}{dh}=\frac{1}{2x}\left(\frac{R+h}{(h_o-h)^2}+\frac{1}{h_o-h}\right)$$
Combine with (3) and (4)
$$\dfrac{dx}{dh}=\frac{1}{2x}\cdot\frac{x^2+1}{R+h_o}(x^2+1)$$
Rearrange
$$dh=(R+h_o)\dfrac{2xdx}{(x^2+1)^2}\tag{5}$$
From (3) and (5)
$$\int\sqrt{\dfrac{R+h}{h_o-h}}dh=\int xdh=(R+h_o)\int\dfrac{2x^2dx}{(x^2+1)^2}$$
By using (2) we obtain
$$-\int\sqrt{\dfrac{R+h}{h_o-h}}dh=(R+h_o)\left(\arctan\dfrac{1}{x}+\frac{x}{x^2+1}+C\right)$$
Finally by using (3) we obtain the solution (B). The lower integration limit $h=h_o$ defines $C=0$.
A: Let us start with
$$\ddot r = -\frac{GM}{r^2}$$
Now multiply with $\dot r$:
$$\dot r \ddot r = -\frac{GM \dot r}{r^2}$$
The antiderivatives are
$$\frac12 \dot r^2 = \frac{GM}{r} + C$$
where $C$ is some constant.
Thus,
$$\dot r = \pm \sqrt{\frac{2GM}{r} + 2C}$$
where sign $+$ is valid for a rising body and sign $-$ is valid for a falling body.
Setting $r = R+h$ then gives
$$\dot h = \pm \sqrt{\frac{2GM}{R+h} + 2C}$$
If $\dot h_0=0$ we have the condition
$$0 = \sqrt{\frac{2GM}{R+h_0} + 2C}$$
i.e.
$$2C = -\frac{2GM}{R+h_0}$$
so we end up with
$$
\dot h = \pm \sqrt{\frac{2GM}{R+h} - \frac{2GM}{R+h_0}}
= \pm \sqrt{2GM} \sqrt{\frac{1}{R+h} - \frac{1}{R+h_0}} \\
= \pm \sqrt{2GM} \sqrt{\frac{h_0-h}{(R+h)(R+h_0)}}
\approx \pm \sqrt{\frac{2GM}{R^2}} \sqrt{h_0-h} \\
= \pm \sqrt{2g(h_0-h)}
$$
