Potential Energy Calculator

How to use this calculator

1. Choose the unknownThe calculator rearranges the formula for you. Solving for height or mass is just as common in problem sets as solving for energy, and the rearrangement is where marks are lost.
2. Use SI units throughoutKilograms, metres and metres per second squared. Grams and centimetres are the most common source of a wrong answer, and the working shows the units so a mistake is visible.
3. Set g for the body you are on9.81 on Earth. The Moon is 1.62 and Mars 3.72, and problem sets use them precisely to check whether you understood that g is not a universal constant.
4. Decide your reference levelHeight is measured from a zero point you choose. Only changes in height are physically meaningful, so pick a convenient reference and stay consistent.

How the calculation works

Gravitational potential energy is the work done against gravity to lift an object. Work is force times distance; the force needed is the object's weight, mg; the distance is the height h. Multiply them and you have mgh.

That derivation matters because it shows what the formula assumes. It treats g as constant over the height involved, which is an excellent approximation near a planetary surface and a poor one over orbital distances. For a satellite you need the full inverse-square expression, U = −GMm/r, and the two only agree for small height changes.

The reference level is arbitrary, and this is the conceptual point students most often miss. There is no absolute potential energy — only differences. Putting the zero at the floor, at the bench, or at sea level changes every value you compute and changes none of the physics, because every problem depends on ΔU rather than U.

Rearranging is straightforward algebra, but it is worth noting that solving for height is the same operation as asking how far something falls to release a given energy, which is why the calculator also shows the free-fall speed and time.

Worked example: a book on a shelf

A 2 kg book is lifted onto a shelf 5 metres above the floor, on Earth.

U = mgh = 2 × 9.81 × 5 = 98.1 joules. That is the work done against gravity in lifting it, and the energy that would be released if it fell back.

If it does fall, all of that converts to kinetic energy at the moment before impact. Setting ½mv² equal to 98.1 and solving gives v = √(2gh) = √(2 × 9.81 × 5) = 9.9 m/s. The mass cancels entirely — which is why a 2 kg book and a 20 kg one hit the ground at the same speed from the same height, ignoring air resistance.

The fall takes √(2h/g) = √(10/9.81) = 1.01 seconds. Both figures appear in the breakdown, because energy questions and kinematics questions are usually two halves of the same problem.

Gravitational field strength elsewhere

Earth's own value varies slightly with latitude and altitude — roughly 9.78 at the equator and 9.83 at the poles. Problem sets almost always use 9.81, and some textbooks use 9.8 or even 10 for mental arithmetic. Use whichever your course specifies, because the difference will show in a marked answer.

Where the simple formula stops working

U = mgh holds where g can be treated as constant. Near Earth's surface that is true to well within measurement error for any height you are likely to encounter in a problem set — even a kilometre changes g by about 0.03%.

Over larger distances it fails. Gravitational field strength falls with the square of the distance from the centre of mass, so lifting something to orbital altitude requires 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 bound state negative.

The formula also covers only gravitational potential energy. Elastic potential energy in a spring is ½kx², and chemical, electrical and nuclear potential energy are governed by entirely different relationships. A question asking for potential energy without qualification almost always means the gravitational kind, but it is worth checking.

One final subtlety: mgh gives the work done against gravity, which equals the energy stored only if the object is raised without gaining speed. Lift it and let it arrive moving and some of the work 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 are physically meaningful
• Free-fall speed and time assume no air resistance
• Requires SI units throughout — kilograms, metres and metres per second squared
• The formula gives work done against gravity, which equals stored energy only if the object arrives at rest

Questions people ask about this in Philippines