Gravitational Potential Energy

Equivalent forms

U_{\text{grav}} = m g h

\Delta U = m g \Delta h

Energy hidden in altitude — the simplest 'battery' in the universe.

Symbol	Name	SI unit	Dimension	Range
$U$	Potential Energyoutput Gravitational potential energy relative to a reference height	J	ML^2T^-2	0 – 10000
$m$	Mass Mass of the object	kg	M	0.1 – 100
$g$	Gravitational Acceleration Local gravitational acceleration (Earth surface = 9.8 m/s²)	m/s²	LT^-2	0 – 25
$h$	Height Height above the chosen reference level	m	L	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$U$	J	erg	ft·lbf
$m$	kg	g	slug
$g$	m/s²	cm/s²	ft/s²
$h$	m	cm	ft

Where it holds

Valid near Earth's surface where g is approximately constant. For h comparable to Earth's radius,

use U = -GMm/r

instead.

Dimensional analysis

U \to [ML^{2}T^{-2}]

m\cdot g\cdot h \to [M]\cdot [LT^{-2}]\cdot [L] = [ML^{2}T^{-2}]

Both sides are Joules.

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Discovery

Gottfried Leibniz / James Joule · 1845

Leibniz proposed 'vis viva' (living force) in 1686; Joule's mechanical-equivalent-of-heat experiments tied falling weights to thermal energy, formalizing potential energy.

Try this

How much energy is stored in a 1 kg book sitting on a 2 m shelf?

Calculate the gravitational potential energy of a 1 kg textbook lifted to 2 m above the floor near Earth's surface. Use U = mgh with g = 9.8.

Research status: stable

Real-world applications

Hydroelectric power generation
Pumped-storage batteries
Pile drivers and demolition wrecking balls
Pendulum clocks and grandfather clock weights

Common misconceptions

Potential energy belongs to the system (object + Earth), not the object alone
The reference 'zero' is arbitrary — only differences matter
g varies slightly with latitude and altitude (9.78 to $9.83\,\mathrm{m/s}^{2})$

Experimental verification

Hydroelectric dams convert U directly to electrical energy. Pendulum amplitude-height correlations and ballistic pendulum data verify mgh to high precision.

Derivation

Work done against gravity lifting mass m by height h:

W = F\cdot h = (mg)\cdot h

.

Defining U as the work-storage capacity gives

U = mgh

.

The reference height is arbitrary; only

\Delta U

has physical meaning.

Limiting cases

h \to 0

⟶

U \to 0

At the reference level, potential energy is zero by convention.

g \to 0

⟶

U \to 0

In free fall or deep space, height carries no potential energy.

h >> R_Earth⟶

U \to -GMm/r

Far from the surface, replace mgh with the universal form

-GMm/r

.

What if…

What if h doubles?

Potential energy doubles linearly. Dropping from 4 m hits with $\sqrt{2} \approx 1.41\times the$ speed (since $v = \sqrt{2gh})$ .

What if you're on the Moon?

g_Moon $\approx 1.62\,\mathrm{m/s}^{2}$ , so U is about $6\times$ smaller for the same height — that's why Apollo astronauts could leap so high.

What if h is several Earth radii?

$U = mgh$ breaks down. $Use U = -GMm/r$ where $G = 6.674e-11$ to track the inverse-square decline.

1

Book on a shelf

Given ·

m:: 1
g:: 9.8
h:: 2

Find · U

Steps

Apply $U = mgh$ with $m = 1\,\mathrm{kg}$ , $g = 9.8\,\mathrm{m/s}^{2}$ , $h = 2\,\mathrm{m}$
$U = 1 \times 9.8 \times 2$
$U = 19.6\,\mathrm{J}$ — enough to power a $1-W LED for \sim 20$ seconds

Result ·

U = 1 \times 9.8 \times 2 = 19.6\,\mathrm{J}

2

Hydroelectric reservoir

Given ·

m:: 1000000
g:: 9.8
h:: 100

Find · U

Steps

Mass of $1000\,\mathrm{m}^{3} of$ water: $m = 1$ ,000,000 kg
$U = 10^{6} \times 9.8 \times 100 = 9.8 \times 10^{8} J$
Convert: $9.8 \times 10^{8} J \div 3.6 \times 10^{6} \approx 272\,\mathrm{kWh}$

Result ·

U = 9.8 \times 10^{8} J \approx 272\,\mathrm{kWh}

Kinetic Energy→Work-Energy Theorem→newtons law of universal gravitation