$mgh$ assumes the field $g$ stays the same. For rockets and satellites it doesn't: $g$ weakens with distance from Earth's center. Adding up the work against the full law of gravitation gives a formula that works at any distance, and it comes with a minus sign.
Key idea
For two masses a distance $r$ apart (center to center), $U = -\dfrac{GMm}{r}$, choosing $U = 0$ when they are infinitely far apart. $U$ rises toward zero as they separate.
Why it works
Pulling the masses apart takes positive work, because gravity pulls them together, and that work raises $U$. Going outward, the force weakens as $1/r^2$, and the area under the force-distance graph from $r$ out to infinity is finite: $\dfrac{GMm}{r}$. Since $U$ must rise by that much to reach $0$ at infinity, it starts at $-\dfrac{GMm}{r}$. Near the surface, the change between $R$ and $R + h$ is very close to $mgh$.
Worked examples
Worked example: A satellite's potential energy
Find $U$ for a $1{,}000\ \text{kg}$ satellite $400\ \text{km}$ up ($M = 5.972 \times 10^{24}\ \text{kg}$, $R = 6.371 \times 10^{6}\ \text{m}$).
- $r = 6.771 \times 10^{6}\ \text{m}$.
- $$U = -\dfrac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{6.771 \times 10^{6}} \approx -5.89 \times 10^{10}\ \text{J}.$$
Answer: about $-5.89 \times 10^{10}\ \text{J}$
Worked example: Raising a payload high
How much does $U$ increase when that $1{,}000\ \text{kg}$ payload rises from the surface to $6{,}371\ \text{km}$ up (one Earth radius)?
- $\Delta U = GMm\left(\dfrac{1}{R} - \dfrac{1}{2R}\right) = \dfrac{GMm}{2R}$.
- $\dfrac{GMm}{R} \approx 6.26 \times 10^{10}\ \text{J}$, so $\Delta U \approx 3.13 \times 10^{10}\ \text{J}$.
- $mgh$ would give $6.24 \times 10^{10}\ \text{J}$, about twice too much, because $g$ is only a quarter as strong at that height.
Answer: about $3.13 \times 10^{10}\ \text{J}$
Worked example: Leaving Earth
How much energy must be added to a $1\ \text{kg}$ object at rest on the surface to take it infinitely far away?
- $U$ goes from $-\dfrac{GM}{R}$ to $0$, so the energy needed is $$\dfrac{GM}{R} = \dfrac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.371 \times 10^{6}}.$$
- That is about $6.26 \times 10^{7}\ \text{J}$, or $62.6\ \text{MJ}$ per kilogram.
Answer: about $6.26 \times 10^{7}\ \text{J}$
Common mistake
Being puzzled by the negative sign, or using $mgh$ when $h$ is comparable to Earth's radius. The sign only reflects the choice of zero at infinity: every finite separation has less energy than that. And $mgh$ assumes constant $g$, which fails far from the surface. Check: $U$ should always increase (become less negative) as $r$ grows.
Try it in the Science Lab's Rocket Builder, where the climb out of Earth's gravity sets how much fuel a rocket needs.