Near-Earth Gravitational Potential Energy

Near the surface the gravitational force on a mass m is F = −mg ĵ at every point, with ĵ pointing up. Its work along a path is W_g = ∫F⋅dr, and with dr = dx î + dy ĵ the dot product is just −mg dy: every horizontal step contributes nothing. So W_g = −mg∫dy = −mg(y_f − yᵢ), whatever the route. Carry a 2.50 kg bag up a 4.20 m flight of stairs, up a long wheelchair ramp to the same landing, or in a lift, and gravity does −103 J of work each time. Work fixed by the end points alone can be booked as stored energy, so define the change in gravitational potential energy as ΔU_g = −W_g = mgΔy. Negative work by gravity on the way up means energy is being stored; positive work on the way down means it is being released.

Potential energy belongs to an interaction, so it lives in a system that contains both partners. That leaves two honest bookkeepings for a ball moving near the ground. Take the ball alone and gravity is an external force: it does work W_g = −mgΔy, there is no U_g, and ΔK = W_g + Wₒₜₕₑᵣ. Take ball plus Earth and gravity is internal: its effect appears as ΔU_g = mgΔy, and ΔK + ΔU_g = Wₒₜₕₑᵣ. The two give the same ΔK. Why, then, is it safe to speak of the ball's potential energy? Starting from rest, the pair acquires equal and opposite momenta, and K = p²/(2m) makes the Earth's share of the kinetic energy smaller than the ball's by the factor m/M_E. For a 1.0 kg ball that is 1.0/(5.97 × 10²⁴) = 1.7 × 10⁻²⁵: essentially all the released energy goes to the ball. The shorthand is harmless as long as you still know which system you chose.

Because only ΔU_g is physical, U_g = mgy + C is defined only up to a constant, and choosing C means choosing a reference level y₀ where U_g = 0. A 0.500 kg book slides off a shelf 1.80 m above the floor and lands on a desk 0.750 m high. With the floor as zero, U_g goes from 8.83 J to 3.68 J; with the desk as zero, from 5.15 J to 0; with the shelf as zero, from 0 to −5.15 J. Three sets of values, one change: ΔU_g = −5.15 J every time. A negative U_g says only that the object is below the zero. Two rules keep the freedom safe. Choose y₀ once and use it for every state and every object in the problem. And fix the axis direction: with y measured downward the same physics reads U_g = −mgy, and mixing the two conventions in one equation flips a sign without warning.

Most slips in this topic are not in mgΔy but in Δy. Measure it along the vertical, from the object's starting height to its final height, and ignore every other feature of the route. On an incline at angle θ, a displacement s along the slope changes the height by s sinθ, so 6.00 m up a 25.0° ramp is a rise of 2.54 m, not 6.00 m. A pendulum of length L pulled out to angle θ sits L(1 − cosθ) above its lowest point, which for 1.20 m at 30.0° is 0.161 m. A roller-coaster car that climbs a hill and plunges through a dip before stopping level with its start has ΔU_g = 0 for the trip, however much U_g rose and fell on the way. Path length still matters for other forces, since the work done by friction grows with distance travelled, but gravity's bookkeeping sees only the two heights.

A ladder, a rope or a tank of water has mass at many heights. Slice the body into elements dm at height y, each storing g y dm, and add them: U_g = g∫y dm. By the definition of the centre of mass, ∫y dm = M y_cm, so U_g = Mg y_cm and ΔU_g = MgΔy_cm. A uniform 3.00 m ladder of mass 15.0 kg lying on the ground, its thickness ignored, has its centre of mass at ground level. Standing it vertical raises that point 1.50 m and stores 15.0 × 9.81 × 1.50 = 221 J; leaning it at 60.0° raises it only 1.50 sin60.0° = 1.30 m and stores 191 J. Lifting the whole 15.0 kg through 3.00 m, a tempting shortcut, would claim 441 J. When the mass is not evenly spread, write dm = λ(y) dy and do the integral: the centre of mass sits where the calculus puts it, not at the midpoint.

The value 9.81 m s⁻² is GM_E/R² at the surface, and it falls with height as g(h) = g(R/(R + h))². Integrating the inverse-square force from r = R to r = R + h gives ΔU_g = GM_E m(1/R − 1/(R + h)) = mghR/(R + h), so mgh divided by the exact change is 1 + h/R. The constant-g model overstates the energy by the fraction h/R, with R = 6.37 × 10⁶ m: 0.016% for a 1.0 km hill, 0.16% for an airliner at 10 km, 6.3% at the 400 km of a space-station orbit. A second limit hides in the word near. Surface g itself runs from about 9.78 m s⁻² at the equator to 9.83 m s⁻² at the poles, a 0.5% spread that outweighs the altitude correction for any rise below about 30 km. Use the local g when three figures matter, and switch to U_g = −GM_E m/r once h/R exceeds the precision you need.