The formula
How to calculate potential energy
Gravitational potential energy is the energy stored in an object because of its height. Let it fall and that energy converts to kinetic energy, which is why the impact speed depends only on the height, not the mass.
Height is measured from whatever reference you choose — only differences in potential energy have physical meaning. The impact speed assumes a vacuum; air resistance reduces it substantially for light objects and long falls.
Here is what each field means:
- Mass (kg)
- Height (m)
- Gravitational acceleration (m/s²)
The result updates on every keystroke. The URL updates too, which makes the filled-in version easy to bookmark or send to someone else.
Units matter more here than the arithmetic itself: the formula assumes a specific set of units for each input, stated next to the field, and converting into those units first is usually the difference between a correct result and one that is wrong by a clean power of ten.
Why potential energy matters
A potential energy calculation gets used both to check work already done by hand and to explore how a formula behaves without redoing the algebra every time an input changes — this page exists for both, since the underlying arithmetic is the same either way.
Beyond a single check, the same calculation is worth rerunning whenever a measured input changes — a new reading, a corrected value, an updated assumption — since the result here always reflects exactly what is currently in the fields above rather than a value calculated once and then left stale.
A formula like this one is rarely the last step in a piece of work — the figure it produces usually feeds into a further calculation, a comparison against a published value, or a write-up that needs to state both the result and how confident it is. Getting this step right the first time, rather than propagating a small arithmetic slip through several more steps, is the main practical reason to check a manual calculation against a tool like this one before building on top of it.
It is worth keeping a note of which inputs were used to produce a given result, particularly where the figure is going into a report or a further calculation — reproducing a result later, or explaining how it was reached, is far easier with the original inputs to hand than by trying to reverse-engineer them from the output alone.
Worked example
A concrete run-through, using the values already in the fields:
- Mass: 75 kg
- Height: 10 m
- Gravitational acceleration: 9.8067 m/s²
That gives:
- Potential energy: 7,354.9875 J
- Speed on impact if dropped: 14.0047 m/s
- In kilowatt-hours: 0.002 kWh
The figures above are the calculator's own default values, shown purely so the working is visible rather than hidden — the same steps apply exactly to your own numbers, entered in the fields at the top of this page.
Reading the result
The kilowatt-hour figure puts the scale in perspective. Lifting a person ten metres stores about 0.002 kWh — a fifth of a penny's worth of electricity, which is why gravity storage needs enormous masses and heights to be useful.
Where this goes wrong. Assuming a heavier object hits harder because it falls faster. It does not — it arrives at the same speed and carries more energy because of its mass, not its velocity.
A result that is wrong by an exact factor of ten, a hundred or a similar round number is almost always a units error rather than a mistake in the formula itself — checking each input against the unit stated beside it is the fastest way to track it down.
Not in a vacuum. Gravitational acceleration is the same for all masses, which is why a feather and a hammer land together on the Moon. Air resistance is what separates them on Earth.
About 1.62 m/s² on the Moon, 3.72 on Mars, 24.79 on Jupiter. Changing the gravity input recalculates both the stored energy and the impact speed.
The headline figure is potential energy. With 75 kg mass, 10 m height and 9.8067 m/s² gravitational acceleration, that comes to 7,354.9875 J. Change any field and the figure moves with it.
Generally, no more than the least precise input justifies — a result reported to six decimal places from inputs measured to two significant figures is implying a precision the calculation does not actually have. The calculator shows full precision so you can round appropriately for your own use.
Yes — the equation shown in the formula section above is the standard form used in textbooks and reference material for this calculation, not a simplified or approximate version.
Yes, in the sense that it applies the correct standard formula and returns an accurate result for the inputs given — but check your own course or publication's requirements for how results should be rounded, presented and referenced, since those conventions vary and are not something a calculator can know on your behalf.