Editing Potential energy (section)

== Potential energy for near-Earth gravity ==
[[File:Trebuchet.jpg|thumb|A [[trebuchet]] uses the gravitational potential energy of the [[counterweight]] to throw projectiles over two hundred meters]]

For small height changes, gravitational potential energy can be computed using
<math display="block">U_\text{g} = mgh ,</math>
where ''m'' is the mass in kilograms, ''g'' is the local gravitational field (9.8 metres per second squared on Earth), ''h'' is the height above a reference level in metres, and ''U'' is the energy in joules.

In classical physics, gravity exerts a constant downward force {{math|1='''''F''''' = (0, 0, ''F''<sub>z</sub>)}} on the center of mass of a body moving near the surface of the Earth.  The work of gravity on a body moving along a trajectory {{math|1='''''r'''''(''t'') = (''x''(''t''), ''y''(''t''), ''z''(''t''))}}, such as the track of a roller coaster is calculated using its velocity, {{math|1='''''v''''' = (''v''<sub>x</sub>, ''v''<sub>y</sub>, ''v''<sub>z</sub>)}}, to obtain
<math display="block">W = \int_{t_1}^{t_2} \boldsymbol{F} \cdot \boldsymbol{v} \, dt = \int_{t_1}^{t_2} F_\text{z} v_\text{z} \, dt = F_\text{z}\Delta z. </math>
where the integral of the vertical component of velocity is the vertical distance. The work of gravity depends only on the vertical movement of the curve {{math|'''''r'''''(''t'')}}.