What gravitational potential energy is
Gravitational potential energy is the work you had to do against gravity to put a mass where it is, and it is the work gravity will do for you if you let it come back down. Lift a 20 kg sack onto a shelf 2 m up and you have transferred 20 × 9.80665 × 2 = 392 J into the system. Drop it and you get every one of those joules back as kinetic energy.
The word potential is doing real work in that sentence. The energy is not stored in the object — it is stored in the configuration of the object and the Earth, in their separation. That is why potential energy always belongs to a system of two or more bodies and why you can never say what its absolute value is, only how much it changed.
That last point is the one that confuses people, and it has a simple resolution: you pick a reference level, and every height is measured from there. Put the datum at the floor and a book on a 1 m table has 1 m worth of potential energy; put the datum at the table and the same book has zero. Both are correct, because every physical prediction depends on the difference in U between two positions, and that difference is the same whichever datum you chose.
Potential energy is the reason a hydro dam, a pendulum clock, a counterweighted lift and a grandfather clock's weights all work. All of them are the same trick: put mass up high, take the energy back on the way down.
Where mgh comes from, and when it stops being right
Potential energy is defined as the work done against a conservative force. Lifting a mass at constant speed means applying an upward force equal to its weight, mg, through a vertical distance h. Work is force times distance, so the energy transferred is mgh. That is the entire derivation.
It relies on the weight mg being the same at every point in the lift. Gravity actually falls off with the square of the distance from the Earth's centre, so g decreases with altitude — but only by about 0.003% per hundred metres near the surface, which is negligible against every other uncertainty in a real problem. The free fall calculator relies on the same approximation for the same reason.
The general expression is U = −GMm/r, which sets the datum at infinite separation and is therefore negative everywhere else. Take the difference of that expression between two nearby radii and it collapses to mgh with g = GM/r², which is exactly the near-surface formula. Use the general form for orbits, escape velocity and anything that leaves the atmosphere; use mgh for everything else.
Notice the symmetry with kinetic energy. Setting mgh = ½mv² and cancelling the mass gives v = √(2gh) — the speed the object reaches if it falls the whole way with nothing resisting it. That single line is conservation of energy for a falling body, and it is why this calculator reports the fall speed alongside the energy.
Worked example: a pumped-storage reservoir
An upper reservoir holds 500,000 m³ of water with an average head of 300 m above the turbine hall. How much energy is stored, and what does that mean in household terms?
- Convert volume to mass. Fresh water has a density near 1,000 kg/m³, so 500,000 m³ is m = 5.0 × 10⁸ kg.
- Apply the formula. U = mgh = 5.0 × 10⁸ × 9.80665 × 300 = 1.471 × 10¹² J.
- Convert to kilowatt-hours. One kWh is 3.6 × 10⁶ J, so U = 1.471 × 10¹² ÷ 3.6 × 10⁶ = 408,600 kWh, or about 409 MWh.
- Sanity-check per cubic metre. One cubic metre at 300 m head stores 1000 × 9.80665 × 300 = 2.942 MJ = 0.817 kWh. Multiply by 500,000 and you get the same 408,600 kWh, which confirms the arithmetic.
- Discount for reality. That is the ideal figure. A real plant loses energy in the penstock, the turbine, the generator and the transformer on the way out, and again in the pumps and motors on the way in, so the electricity you get back is a fraction of it. Use your plant's own measured round-trip efficiency rather than any generic number.
Turn the calculation round to see the scale of the engineering. To store 1 kWh at a 300 m head you need 3.6 × 10⁶ ÷ (9.80665 × 300) = 1,224 kg of water — well over a tonne per kilowatt-hour. Gravity storage is cheap per unit of energy stored and enormous per unit of mass moved, which is exactly why it is built with mountains and lakes rather than with tanks.
How to read the number you get
Compare the joules against something familiar before you trust them. A 1 kg book on a 1 m table stores 9.8 J — roughly the energy in a single second of a small LED lamp. A 100 kg person at the top of a 10 m diving board stores 9,807 J, which at 4,184 J per food calorie is 2.34 kcal — two thirds of a teaspoon of sugar. A tonne raised 100 m stores 0.27 kWh, less than a fan heater consumes in twenty minutes. Gravitational storage is diffuse, and seeing that in numbers is the most useful thing this calculator does.
The fall speed is the other reality check. Because v = √(2gh), a height that stores twice the energy produces only 1.41 times the speed. If you are thinking about impact rather than storage, energy is the quantity that scales with height and speed is the one that does not.
When you solve for height or mass, watch the two cases with no answer. A target energy in a zero mass has no height that satisfies it, and a target energy at zero height has no mass that satisfies it. The calculator reports those as undefined rather than as infinity, because infinity is not the answer — the answer is that the configuration you asked for cannot exist.
Finally, treat every figure as the ideal. Real systems that store or recover gravitational energy lose some of it in friction, turbulence, electrical resistance and heat. The ideal number is the ceiling, and it is a useful one precisely because nothing can beat it.
Energy stored per cubic metre of water at different heads
| Head (m) | Energy per m³ (MJ) | Energy per m³ (kWh) | Mass per stored kWh (kg) |
|---|---|---|---|
| 10 | 0.098 | 0.0272 | 36,710 |
| 25 | 0.245 | 0.0681 | 14,684 |
| 50 | 0.490 | 0.1362 | 7,342 |
| 100 | 0.981 | 0.2724 | 3,671 |
| 200 | 1.961 | 0.5448 | 1,835 |
| 300 | 2.942 | 0.8172 | 1,224 |
| 500 | 4.903 | 1.3620 | 734 |
| 1,000 | 9.807 | 2.7241 | 367 |
These are ideal values with no losses. Doubling the head halves the mass needed for a given stored energy, which is why high-head sites are so much more valuable than large-volume low-head ones.
Choosing the reference level
You may put the zero of potential energy anywhere, and the choice never changes a physical prediction — only the bookkeeping. The convenient choice is usually the lowest point the object reaches in the problem, because then every potential energy is positive and every energy balance reads naturally. If your object goes below the datum, its potential energy is negative with respect to that datum, which is perfectly valid: the calculator will report it and note why. What you must not do is change the datum halfway through a problem.
Assumptions and common mistakes
- Only the vertical rise counts. Carrying a load 50 m horizontally does no work against gravity at all. The h in mgh is the vertical component of the displacement and nothing else.
- The path taken is irrelevant. Gravity is a conservative force, so a mass hauled up a winding ramp and one lifted straight up to the same height store identical potential energy. Extra work done on the ramp goes into friction, not into U.
- Mass, not weight. Enter kilograms. If you have a figure in kilograms-force or pounds-force you have a force, and you should divide by g first or use the weight directly as U = Wh.
- mgh is a near-surface approximation. Above roughly 20 km it starts to matter that g falls with altitude, and for orbits you need U = −GMm/r instead.
- The stored energy is not the recoverable energy. Every real conversion loses some. Treat the calculator's output as the thermodynamic ceiling.
- Water volume is not water mass. One cubic metre of fresh water is about 1,000 kg; seawater is denser at roughly 1,025 kg/m³. Use the mass, or the m³-of-water unit option which applies the 1,000 kg/m³ conversion for you.
How this connects to the rest of mechanics
Potential energy is one half of the mechanical energy of a system; the other half is kinetic energy. In the absence of friction their sum is constant, which is the single most useful shortcut in mechanics: you can find the speed at the bottom of any smooth track from the height alone, with no need to know its shape, its length or how long the descent takes. The free fall calculator is that shortcut applied to the simplest possible track.
Getting energy into the system means doing work against gravity, and the work calculator handles the case where the lifting force is not vertical or where friction takes a cut. Getting it back out at a useful rate is a question of power: energy divided by time. A 1,000 kg lift car raised 30 m gains 294,200 J, and doing it in 20 s needs at least 14.7 kW at the drum before any losses.
On planetary scales the near-surface formula gives way to U = −GMm/r, and the interesting quantities become escape velocity and orbital energy rather than mgh. The bridge between the two is worth seeing at least once: differentiate the general form with respect to r and you recover the inverse-square force law, and evaluate the difference over a small height near the surface and you recover mgh exactly.