Gravitational Potential Energy Calculator
Calculate gravitational potential energy GPE = mgh stored in an object at a given height above a reference level. Useful for energy-conservation problems, hydroelectric and roller-coaster design, and any analysis where elevation translates into stored energy.
Last updated: May 2026
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About this calculator
Gravitational potential energy (GPE) is the energy an object stores by virtue of its position in a gravitational field. The formula is GPE = m·g·h, where m is mass in kilograms, g is local gravitational acceleration in m/s² (9.81 near Earth’s surface), and h is height above a chosen reference level in metres. The result is in joules (J). Physically, GPE represents the work done against gravity to raise the object to that height, and equivalently the energy that converts to kinetic energy as it falls back. The reference level (h = 0) is arbitrary: ground, tabletop, sea level, or the centre of the Earth all work — only the difference in GPE between two states is physically meaningful, so absolute GPE values can be positive, zero, or negative depending on where you put h = 0. The formula assumes constant g, which holds as long as the height change is small relative to Earth’s radius (~6,370 km); for satellites or interplanetary problems use the general expression GPE = −G·M·m/r. Edge cases: g varies slightly with latitude (9.78 at the equator, 9.83 at the poles) and altitude (−0.3% per 10 km up), so 9.81 is an Earth-surface average. On other bodies use the local value — Moon ≈ 1.62, Mars ≈ 3.71, Jupiter ≈ 24.79. Negative heights are valid and represent positions below the reference; GPE simply becomes negative. The formula ignores rotational PE, elastic PE, and relativistic corrections, all of which require separate treatments.
How to use
Example 1 — Boulder on a cliff. A 50 kg boulder sits on a cliff 30 m above the valley floor with g = 9.81 m/s²: GPE = 50 × 9.81 × 30 = 14,715 J ≈ 14.7 kJ. ✓ If the boulder falls and air resistance is negligible, all 14.7 kJ converts to KE at impact: ½ × 50 × v² = 14,715 → v² = 588.6 → v ≈ 24.3 m/s (about 87 km/h). Example 2 — Water in an elevated reservoir. A hydroelectric dam holds 1,000,000 kg of water (a million litres) at an average height of 60 m above the turbines. With g = 9.81: GPE = 1,000,000 × 9.81 × 60 = 588,600,000 J ≈ 589 MJ. ✓ At 90% turbine efficiency, releasing all of this water delivers about 530 MJ ≈ 147 kWh of electricity — enough to power roughly five average European households for a day. This is exactly the principle behind every hydroelectric plant from a local dam to the Three Gorges.
Frequently asked questions
What does it mean physically that gravitational potential energy is ‘relative’?
GPE is defined only with respect to a chosen reference level where h = 0. Pick the floor and a book on the table has positive GPE; pick the table as h = 0 and the same book has zero GPE; pick the ceiling and it has negative GPE. The absolute number changes, but physically nothing about the book has changed — what matters in any real calculation is the change in GPE between two states (initial and final). This is what governs how much kinetic energy will appear if the book falls, or how much work was required to lift it. The reference choice is purely a matter of bookkeeping convenience; choose the level that makes the math simplest for the problem at hand. This is also why GPE values can be negative — they simply indicate a position below the reference.
Why does the mgh formula break down at high altitudes?
The formula assumes g is constant, but it isn’t — Newton’s law of universal gravitation gives g = G·M/r², so g weakens as r (distance from Earth’s centre) grows. At Earth’s surface g ≈ 9.81 m/s². At the ISS orbit (~400 km up) g has dropped to about 8.7 m/s², roughly 11% lower. At geostationary orbit (~36,000 km) g is only ~0.22 m/s². For height changes within a few kilometres of the surface, the variation introduces less than 0.1% error — fine for engineering. But for ballistic missiles, satellites, or planetary escape problems you must integrate the variable gravity field: GPE = −G·M·m/r. The negative sign indicates that GPE is taken as zero at infinity and becomes more negative as you approach the planet.
How does GPE relate to energy conservation in falling objects?
In a closed system without friction or air resistance, mechanical energy (KE + GPE) is conserved — every joule of GPE lost converts to a joule of KE gained, and vice versa. This is why a pendulum swings forever in an ideal model: at the top of the swing it has maximum GPE and zero KE; at the bottom it has zero GPE and maximum KE; the total stays constant. Setting GPE₀ + KE₀ = GPE_f + KE_f lets you solve for unknown speeds, heights, or both in one equation without ever computing forces or accelerations. In any real system with drag, however, mechanical energy leaks away as heat — a real pendulum slows down because air resistance dissipates energy per swing. For frictionless problems, energy conservation is the fastest route to an answer; for messy real-world problems, you must model the dissipation explicitly.
What are the most common mistakes people make with GPE problems?
The first is unit confusion: entering mass in grams or height in centimetres while expecting joules — always work in SI (kg, m/s², m). The second is forgetting that GPE is relative and trying to use absolute GPE values from different reference levels in the same equation. The third is applying mgh at altitudes where g has changed meaningfully (above ~100 km) — use the general −GMm/r form. The fourth is treating GPE as an intrinsic property of the object rather than of the object-planet system; this matters for advanced problems transferring an object between gravitational fields. The fifth is using Earth’s g (9.81) for problems on other planets — Moon 1.62, Mars 3.71, Jupiter 24.79. And in energy-conservation problems, people often forget that air resistance, friction, and inelastic collisions all silently drain mechanical energy into heat — the simple KE + GPE = constant equation only holds in the conservative-force idealisation.
When should I not use this calculator?
Skip it for orbits and space missions — at altitudes where g is not constant (above about 100 km, or more than a few percent of the planet’s radius), use the general gravitational potential −G·M·m/r instead. Avoid it for objects within strong gravitational gradients (near black holes, neutron stars, or in any general-relativistic regime) where Newtonian gravity itself breaks down. Skip it for elastic potential energy stored in springs or stretched materials — that needs the ½kx² formula with the spring constant k, not mgh. Do not apply it to bound systems like atoms or molecules where the potential is electromagnetic, not gravitational. For everyday tasks like lifting a box, throwing a ball, draining a reservoir, or designing a roller coaster, mgh fits perfectly — but the moment your problem involves space, large vertical scales, or non-gravitational forces, reach for the appropriate generalisation instead.