Potential Energy Calculator

How to use this calculator

1. Choose the unknown firstThe calculator rearranges the formula for you. Rearranging is where marks are most often lost in exams, and seeing the rearranged form written out is half the point of using it.
2. Work in SI units throughoutKilograms, metres and newtons per kilogram. Grams and centimetres are the single most common source of a wrong answer, and the working shows the substitution so a unit slip is visible.
3. Check which value of g your course uses9.81 is the standard, but many exam boards use 9.8 and some use 10 for mental arithmetic. Use whichever your mark scheme expects.
4. Look at the extra rowsWhen solving for energy, the breakdown also gives free-fall speed and fall time — because energy questions and kinematics questions are usually two halves of the same problem.

How the calculation works

Gravitational potential energy is the work done against gravity in lifting something. Work is force multiplied by distance; the force needed is the object's weight, mg; the distance is the height h. Multiply and you have mgh.

That derivation is worth carrying because it shows what the formula assumes. It treats g as constant over the height involved — an excellent approximation near the Earth's surface and a poor one over any distance where the field strength changes appreciably.

The reference level is the conceptual point students most often miss. There is no absolute potential energy, only differences. Setting the zero at the floor, at the bench, or at sea level changes every value you compute and changes no physics at all, because every question depends on the change in height rather than the height itself.

Rearranging is straightforward algebra. Solving for height is the same operation as asking how far something falls to release a given energy, which is why the calculator shows the free-fall figures alongside — v = √(2gh) and t = √(2h/g) both follow directly from equating mgh with ½mv².

Worked example: a book on a shelf

A 0.5 kg book lifted onto a shelf 1.8 m above the floor, taking the floor as the zero level.

U = mgh = 0.5 × 9.81 × 1.8 = 8.829 J. That is the work done against gravity in lifting it, and the energy that would be released if it fell.

If it does fall, all of that converts to kinetic energy just before impact. Setting ½mv² equal to 8.829 and solving gives v = √(2 × 9.81 × 1.8) = 5.94 m/s. Notice that the mass cancels — a 0.5 kg book and a 5 kg one strike the ground at the same speed from the same height, ignoring air resistance.

The fall takes √(2 × 1.8 ÷ 9.81) = 0.61 seconds.

Now change the reference level to the shelf itself. The book on the shelf has U = 0, and on the floor it has U = −8.829 J. Every number changed sign and magnitude; the 8.829 J released in falling did not. That is the reference level being arbitrary in action.

Gravitational field strength elsewhere

Earth's own value varies with latitude and altitude — roughly 9.78 at the equator and 9.83 at the poles, because of the planet's rotation and its non-spherical shape. Exam questions almost always use 9.81, and the variation is well below the precision most questions require.

Where mgh stops being the right formula

The simple form holds wherever g can be treated as constant. Near the Earth's surface that is true to well within any measurement you are likely to make — even a kilometre of altitude changes g by about 0.03 per cent.

Over larger distances it fails. Gravitational field strength falls with the square of distance from the centre of mass, so anything at orbital altitude needs the full expression U = −GMm/r. The negative sign there is not an error: potential energy is conventionally defined as zero at infinite separation, which makes every gravitationally bound state negative.

The formula also covers gravitational potential energy only. Elastic potential energy stored in a spring is ½kx², and chemical, electrical and nuclear potential energies follow entirely different relationships. A question asking simply for potential energy almost always means the gravitational kind, but it is worth reading carefully.

One subtlety that catches people in energy questions: mgh gives the work done against gravity, which equals the energy stored only if the object arrives at rest. Lift something and let it arrive moving, and some of the work you did went into kinetic energy instead.

Assumptions and limitations

• Assumes g is constant over the height involved — valid near a planetary surface, not over orbital distances
• Covers gravitational potential energy only, not elastic, chemical or electrical
• Height is measured from a reference level you choose; only differences carry physical meaning
• Free-fall speed and time assume no air resistance
• Requires SI units throughout — kilograms, metres and newtons per kilogram
• Gives work done against gravity, which equals stored energy only if the object arrives at rest

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