Gravitational Potential Energy
Gravitational potential energy is the energy stored by an object because of its height, equal to mass times g times height (PE = mgh).
Formula
PE = mgh
Definition
Gravitational potential energy is the energy an object stores by being up high: the higher it is and the heavier it is, the more energy it has waiting to be released when it falls. It is calculated as $PE = mgh$, where $m$ is mass in kilograms, $g = 9.8\ \text{m/s}^2$, and $h$ is height in meters above a reference level, measured in joules; the reference level where $h = 0$ can be chosen freely, because only changes in height matter. The formula $U = mgh$ is valid when $g$ is effectively constant near Earth's surface. The general expression for two masses is $U = -Gm_1m_2/r$, which is zero at infinite separation and negative for bound systems, and the near-surface form is the first-order expansion of it. Because gravity is conservative, $U$ depends only on position, and $\vec{F} = -\nabla U$ recovers the force.
Example
A skier at the top of a mountain has lots of gravitational potential energy, and as she skis down, that stored energy turns into speed. A $3\ \text{kg}$ rock on a $5\ \text{m}$ ledge has $PE = 3 \times 9.8 \times 5 = 147\ \text{J}$ relative to the ground, and ignoring air resistance it would hit the ground with $147\ \text{J}$ of kinetic energy. A $0.5\ \text{kg}$ pendulum bob pulled aside until it is $0.20\ \text{m}$ above its lowest point and released arrives at the bottom with $mgh = 0.5 \times 9.8 \times 0.20 = 0.98\ \text{J}$ of kinetic energy, so $v = \sqrt{2gh} = \sqrt{3.92} \approx 1.98\ \text{m/s}$. The mass cancels; the speed depends on height alone.
Key Insight
Lifting something is a way of "charging" it with energy; everything you carry upstairs is storing the work you did to get it there. The work needed to lift an object equals the gravitational potential energy it gains no matter the path, so carrying a box up a ramp or straight up a ladder stores the same energy. Escape velocity comes straight from this energy. Setting $\tfrac{1}{2}mv^2 = GMm/R$ gives $v_{\text{esc}} = \sqrt{2GM/R} \approx 11.2\ \text{km/s}$ for Earth, the speed at which total energy reaches zero and the object is no longer bound.