Gravitational Potential Energy Calculator Eₚ = mgh

The gravitational potential energy calculator solves E_p = mgh: given a mass m, the gravitational acceleration g and a height h above some reference level, it finds the object's gravitational potential energy; it can also solve for height or mass from the energy—how much energy a 70 kg person gains climbing to a third-floor landing 3 m up, how high 1000 J can lift that person, and how much energy each tonne of water stores in a reservoir 100 m high.

Gravitational Potential Energy Calculator Eₚ = mgh

About this calculator

The gravitational potential energy calculator solves $E_p = mgh$ : given a mass $m$ , the gravitational acceleration $g$ and a height $h$ above some reference level, it finds the object's gravitational potential energy; it can also solve for height or mass from the energy—how much energy a 70 kg person gains climbing to a third-floor landing 3 m up, how high 1000 J can lift that person, and how much energy each tonne of water stores in a reservoir 100 m high.

$E_p = mgh$ is shorthand for “the negative work gravity does while an object is lifted steadily from the reference level to height $h$ ”. It is the most common entry point to energy conservation: as an object falls, potential energy becomes kinetic energy ( $mgh = \tfrac12 mv^2$ ), with the kinetic-energy formula on the Kinetic Energy Calculator; where $g$ comes from, its units and its values are covered by Newton's Second Law. The reference level may be chosen freely; changing it shifts the numerical value of $E_p$ but never the energy difference between two points—a point this page stresses throughout.

What it does not do: it does not handle heights approaching the Earth's radius (that needs $-GMm/r$ ; this page gives an error estimate); it does not solve for $g$ ; it does not compute kinetic energy or speed; and it does not convert units (see the examples for kcal and kWh conversions).

Formula and variables

E_p = m\,g\,h

The inverse forms (two implemented by the engine):

h = \frac{E_p}{m\,g},\qquad m = \frac{E_p}{g\,h}

Symbol	Meaning	Calculator field	Unit	Range and notes
$m$	mass	Mass m (kg)	kg	$m > 0$ ; default 1
$g$	gravitational acceleration	Gravitational acceleration g (m/s²)	m/s²	default 9.80665 (the standard value $g_0$ ); the Moon 1.62, Mars 3.72. When cleared, the engine calculates with 9.8
$h$	height above the reference level	Height h (m)	m	any real number; negative below the reference level. No default
$E_p$	gravitational potential energy	Potential energy Eₚ (J)	J	output; an input when solving in reverse. $1\ \text{J} = 1\ \text{kg·m}^2/\text{s}^2$

Dimensional check: $\text{kg} \times \text{m/s}^2 \times \text{m} = \text{N·m} = \text{J}$ . Useful conversions: $1\ \text{kcal} = 4184\ \text{J}$ , $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$ .

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Gravitational potential energy” (/#/mathtools/formula?calculator=calculator.phys-potential). On open, “Mass m” = 1 and “Gravitational acceleration g” = 9.80665 are already filled in and “Height h” is empty; both “Height” and “Potential energy” then carry a “calculate” marker and show 0—the engine treats the empty $h$ as 0. This is not a result.
Find the potential energyenter the height in metres and the potential energy updates immediately. Enter a negative number below the reference level for a negative potential energy.
Change gset “Gravitational acceleration g” to 9.8, 9.81 or another body's value. If you clear g, the engine does not fall back to 9.80665 but to 9.8 (the field then shows empty); to use the standard value, enter 9.80665 explicitly.
Solve for heightclear “Height h” and enter “Potential energy Eₚ”; the height field switches to “calculate” and shows $E_p/(mg)$ .
Solve for massclear “Mass m” and enter $g$ , $h$ and $E_p$ ; the mass field shows $E_p/(gh)$ .
Read the resultvalues keep at most 8 decimal places with trailing zeros removed; convert J to kcal by dividing by 4184, and to kWh by dividing by 3 600 000.
Copy the formulathe copy button at the right of the panel title puts the LaTeX E_p = mgh on the clipboard.

Worked examples

All four examples were recomputed by the engine's phys-potential compute routine; the display convention is at most 8 decimal places with trailing zeros removed, in units kg, m/s², m and J.

Example 1: lifting 1 kg by 1 m

Inputs: $m = 1$ , $g = 9.80665$ , $h = 1$ . $E_p = 1 \times 9.80665 \times 1 = 9.80665$ . Interface shows: potential energy Eₚ 9.80665—about 10 J. This is another way to read $g$ : 9.8 joules per kilogram per metre. A 500 mL bottle of water carried from the floor to a table (0.75 m) gains about 3.7 J; an AA battery stores roughly 10 000 J, enough to lift that bottle by 2 km.

Example 2: stairs, a bookshelf and a basement

Climbing stairs: a 70 kg person goes up to a landing 3 m high. $m = 70$ , $g = 9.80665$ , $h = 3$ : $E_p = 70 \times 9.80665 \times 3 = 2059.3965$ . Interface shows: potential energy Eₚ 2059.3965—about 0.49 kcal. Change $g$ to 9.8 and the interface shows 2058; the two values differ by 0.07%. Climbing 10 floors is about 5 kcal, less than a sugar cube (about 16 kcal): the “cost” of climbing is dominated by muscle efficiency (about 20%–25%) and maintaining body temperature, not by the potential energy itself.

A book on a shelf: a 0.5 kg book on a shelf 1.2 m up. $m = 0.5$ , $h = 1.2$ : $E_p = 0.5 \times 9.80665 \times 1.2 = 5.88399$ . Interface shows: 5.88399.

A basement: an object of 2 kg 1.5 m below ground, with the ground as the reference level. $m = 2$ , $h = -1.5$ : $E_p = -29.41995$ . With $h = -3$ the interface shows −58.8399. The minus sign means “below the reference level”, not “not enough energy”: choose the basement floor as the reference level and this value becomes 0, while the difference from the ground is always $2 \times 9.80665 \times 1.5 = 29.4\ \text{J}$ .

Figure 1: three heights under one reference level (Example 2). Walking up stairs, taking a lift or climbing a rope to the same floor gives the same increase in potential energy

Example 3: solving for height and mass from the energy

Solving for height: how high can 1000 J lift a 70 kg person? Clear $h$ and enter $m = 70$ , $g = 9.80665$ , $E_p = 1000$ : $h = 1000 / 686.4655 = 1.45673745$ . Interface shows: height h 1.45673745—less than one storey. 1 kcal (4184 J) corresponds to 6.09 m, about two storeys.

Solving for mass: how heavy is an object that stores 500 J at a height of 10 m? Clear $m$ and enter $g = 9.80665$ , $h = 10$ , $E_p = 500$ : $m = 500 / 98.0665 = 5.09858106$ . Interface shows: mass m 5.09858106.

Example 4: a reservoir, the Moon and Everest

Pumped storage: 1000 kg (1 m³) of water raised 100 m. $m = 1000$ , $g = 9.80665$ , $h = 100$ : $E_p = 980665$ . Interface shows: potential energy Eₚ 980665—0.272 kWh. Conversely, 1 kWh can lift 1 m³ of water by 367 m. A pumped-storage plant with a 100 m head and a reservoir of $10^7\ \text{m}^3$ stores about 2.7 GWh—this is how the grid stores electricity by “pumping water uphill”.

The Moon: the same person climbs the same 3 m with $g = 1.62$ : $E_p = 70 \times 1.62 \times 3 = 340.2$ . Interface shows: 340.2, one sixth of the value on Earth.

Everest: a 70 kg person from sea level to 8849 m, $E_p = 70 \times 9.80665 \times 8849 \approx 6.07 \times 10^6\ \text{J} \approx 1452\ \text{kcal}$ —only about one day's food. The real cost of mountaineering is far higher, for the same reason as with stairs: potential energy is only the net effect. At that altitude $g$ is already about 0.28% smaller than at sea level, and the relative error of $mgh$ against the exact $GMm(1/r_1 - 1/r_2)$ is about $h/R_\oplus = 0.14\%$ , still acceptable.

The inputs of Example 2 can be entered directly into the panel to reproduce it; change the height to 6, 9 and 30 and watch the potential energy grow linearly; change g to 1.62 (the Moon), 3.72 (Mars) and 24.79 (Jupiter) to compare the same 3 m.

Principle and derivation

From the work done by gravity to potential energy

Lifting a mass $m$ steadily by a height $h$ requires an upward force $F = mg$ (Newton's second law; at zero acceleration the applied force equals gravity), doing work $W = Fh = mgh$ . That work does not disappear but is “stored” in the system of object plus Earth and can be recovered as kinetic energy on the way down—so gravitational potential energy is defined as $E_p = mgh$ . More rigorously, $E_p$ is the negative of the work done by gravity: as the object rises, gravity does work $-mgh$ and the potential energy increases by $mgh$ .

Independent of the path

Gravity is a conservative force: the work it does from A to B depends only on the height difference between A and B, not on the route. Stairs, a lift, a rope, a winding mountain road—so long as the start and end heights are the same, the increase in potential energy is the same. That is why $h$ only needs to be the vertical height, and why the length of a slope or a rope never appears in the formula. The three scenarios in Figure 1 can be compared on a single height scale.

Reference levels and “negative potential energy”

The $h$ in $E_p = mgh$ is measured relative to some agreed $h = 0$ surface. Changing the reference level shifts the $E_p$ of every point by the same constant, leaving differences unchanged. Physical laws involve only differences (work, energy conservation), so the reference level is free. The negative values for the basement in Example 2 come from exactly this. In celestial mechanics the convention is to set the potential energy at infinity to zero, which makes every bound orbit negative—that is a different convention and does not conflict with $mgh$ .

Conservation of mechanical energy and $\sqrt{2gh}$

Ignoring drag, an object falling from rest through a height $h$ loses potential energy $mgh$ and gains kinetic energy $\tfrac12 mv^2$ ; equating them gives $v = \sqrt{2gh}$ , independent of mass. The person in Example 2 jumping down from 3 m lands at $\sqrt{2 \times 9.80665 \times 3} = 7.67\ \text{m/s}$ (about 28 km/h). This is also the energy version of $v^2 = 2as$ from the uniform acceleration displacement calculator.

Which value of $g$

$g_0 = 9.806\,65\ \text{m/s}^2$ is the standard value defined by the 3rd General Conference on Weights and Measures in 1901, used to define the kilogram-force among other units, and it is the engine's default. The actual gravitational acceleration varies with latitude (9.780 at the equator, 9.832 at the poles) and altitude (about 0.03% less per kilometre of ascent). Textbooks commonly use 9.8 or 9.81; the three values differ by less than 0.1%, far less than typical measurement error. When several calculators are cross-checked, use one value consistently.

When $mgh$ is not enough

$mgh$ assumes $g$ is constant over the whole height range, i.e. $h \ll R_\oplus = 6371\ \text{km}$ . The exact potential-energy difference is $GMm\left(\dfrac1{r_1} - \dfrac1{r_2}\right)$ , with relative error about $h/R_\oplus$ : 0.14% for Everest, 6% for the International Space Station (400 km), and completely unusable for geostationary orbit (35 786 km). Rockets and satellites need the law of universal gravitation.

Assumptions

$g$ is constant: the height is far smaller than the Earth's radius.
A reference level has been chosen: $h$ is the vertical height relative to it and may be positive or negative.
Mass is constant: relativistic effects are ignored, as are mass changes such as fuel consumption.
Only gravitational potential energy is computed: elastic and electric potential energy are not included.
SI units: kg, m/s², m, J; no conversion is performed.
Interface conventions: an unfilled $h$ participates as 0, so the potential energy shows 0 when the panel opens; when g is cleared the engine falls back to 9.8, not to the default 9.80665.

Scope and limitations

Can calculate: any one of $m$ , $g$ , $h$ , $E_p$ from the other three, or $h$ or $m$ from $E_p$ ; negative heights and negative potential energies are supported.
Approximate only: above a few tens of kilometres the error grows like $h/R_\oplus$ ; the value of $g$ itself carries a version difference of order 0.1%.
Cannot calculate: solve for $g$ (hand-calculate $g = E_p/(mh)$ ); falling speed (use $\sqrt{2gh}$ or the kinetic energy calculator); orbital energy (use $-GMm/r$ ).
Does not: convert units (kcal, kWh, kgf·m).
Numerics: inverse results keep 8 decimal places, so substituting them back leaves a rounding tail in the last digit.

Common pitfalls

Reading negative potential energy as “no energy”: the minus sign only says “below the reference level” (the basement in Example 2).
Entering the length of a slope as $h$ : $h$ is the vertical height; walking 10 m along a 30° slope only rises 5 m.
Confusing weight with mass: enter $m$ in kg; if the weight in N is known, divide by $g$ to get the mass.
Clearing g expecting the default: after clearing, the engine uses 9.8, not 9.80665 (step 3 of “How to use”).
Using mgh to estimate the “cost” of climbing: potential energy is only the net effect; the human body is about 20%–25% efficient, so the real cost is four to five times larger.
Still using mgh at huge heights: satellites and rockets need gravitational potential energy from the universal law.
Comparing potential energies against different reference levels: use one reference level when comparing two objects.

Typical use cases

Teaching

The definition of potential energy, the choice of reference level and negative potential energy; conservation of mechanical energy $mgh = \tfrac12 mv^2$ ; the work–energy relationship; the value of $g$ .

Energy

Storage and output estimates for pumped storage and hydropower (together with flow rate and efficiency); lower bounds on the energy used by cranes and lifts.

Sport and health

Potential-energy gains and calorie conversion for stairs and mountaineering (mind the efficiency); upper bounds on entry speed for diving and bungee jumping.

Engineering and safety

Impact-energy estimates for falling objects ( $mgh$ is the kinetic energy at landing); the potential-energy term in stacking and rack-load assessments.

FAQ

Why is the potential energy 0 when the panel opens?

The height has not been filled in, and the engine treats $h$ as 0. Enter a height and it is normal.

Can potential energy be negative?

Yes. When $h$ is below the reference level $E_p$ is negative, meaning only that it is lower than the reference level; the energy difference between two points is unaffected.

Should g be 9.8, 9.81 or 9.80665?

Use the value given by the problem or the specification; the engine defaults to 9.80665 (the standard value $g_0$ ), and the three differ by less than 0.1%. Note that after g is cleared the engine uses 9.8.

Is the potential-energy gain the same going up stairs and taking a lift to the same floor?

Yes. Gravity is conservative and potential energy depends only on the height difference, not on the path.

How do I calculate the speed at which something lands?

Ignoring drag, $v = \sqrt{2gh}$ , independent of mass; or enter $E_p$ into the kinetic energy calculator and solve for $v$ .

From what height does mgh stop being accurate?

The relative error is about $h/R_\oplus$ : about 0.16% at 10 km and 6% at 400 km. Satellite orbits need $-GMm/r$ .

References and further reading

The phys-potential definition in the CalcX engine source src/data/formulasScience.ts (compute(known, solveFor) for E / h / m, with g defaulting to 9.80665 and falling back to 9.8 when cleared); every figure on this page was recomputed by that engine.
Wikipedia, Gravitational energy （访问日期：2026-09-09）—the relationship between $mgh$ and $-GMm/r$ and their ranges of validity.
Wikipedia, Potential energy — Gravitational potential energy （访问日期：2026-09-09）—conservative forces, path independence and reference levels.
NIST, CODATA 2018: standard acceleration of gravity （访问日期：2026-09-09）— $g_0 = 9.806\,65\ \text{m/s}^2$ .
BIPM, The International System of Units (SI), 9th edition （访问日期：2026-09-09）—the definition of the joule; the basis of the calorie and kilowatt-hour conversions.
Wikipedia, Pumped-storage hydroelectricity （访问日期：2026-09-09）—engineering applications and efficiency of potential-energy storage.

Privacy

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Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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