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?


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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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 $h \ll R$.

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