Is gravitational potential energy proportional or inversely proportional to distance? We know that if an object has been lifted a distance $h$ from the ground then it has a potential energy change:
$$\Delta U = mgh $$
so $h$ is proportional to $\Delta U$.
However, we have also the gravitational potential energy law:
$$ U= -\frac{G M m}{r} $$
where the distance is inversely proportional to the potential energy.
What did I miss? Is the distance of the object proportional or inversely proportional to the potential energy? 
 A: The formula:
$$ \Delta U = mgh $$
is an approximation that applies when the distance $h$ is small enough that changes in $g$ can be ignored. As you say, the expression for $U$ is:
$$ U= -\frac{G M m}{r} $$
So the change when moving a distance $h$ upwards is:
$$ \Delta U = \frac{GMm}{r} - \frac{GMm}{r + h} $$
We rearrange this to get:
$$\begin{align}
\Delta U &= GMm \left( \frac{1}{r} - \frac{1}{r + h} \right) \\
         &= GMm \frac{h}{r^2 + rh} \\
         &= \frac{GM}{r^2} m \frac{h}{1 + h/r} \\
         &\approx \frac{GM}{r^2} m h
\end{align}$$
where the last approximation is because $h \ll r$ so $1 + h/r \approx 1$. And since $GM/r^2$ is just the gravitational acceleration $g$ at a distance $r$, we get:
$$ \Delta U = g m h $$
A: Your first potential energy arises from the approximation that the graviational field is approximately constant for "small heights" , i.e.
$$\frac{GMm}{r^2} \approx mg$$
The full law leads to your second formula, the approximation to the first. For heights above the earth, it is justified, as we can see by taylor expanding $\frac{1}{r^2}$ around the earth's radius $R$:
$$ \frac{1}{r^2} = \frac{1}{R^2} - \frac{2(r - R)}{R^3} + \mathcal{O}((r-R)^2)$$
Here, $h = r - R$, so
$$ \frac{1}{r^2} = \frac{1}{R^2}(1 - \frac{2h}{R}) + \mathcal{O}(h^2)$$
The term with $h$ is certainly neglegible for $h \ll R$.
