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?
Answer
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:
$$ \Delta U = g m h $$
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