2.6.3.1. Gravitational Potential Energy

Table of Contents

Meaning and Definition

Gravitational potential energy is the energy associated with the position of an object in a gravitational field. If you lift an object, you do work against gravity. That work is stored as gravitational potential energy.

Near Earth's surface, this idea is familiar. A book held on a shelf has more gravitational potential energy than the same book on the floor. If it falls, that stored energy can turn into kinetic energy.

For gravity, gravitational potential energy is especially important because gravity is a conservative force. That means the change in potential energy depends only on the starting and ending positions, not on the path taken between them.

The change in gravitational potential energy is related to the work done by gravity by
$$\Delta U = -W_{\text{gravity}}$$
If gravity does positive work, gravitational potential energy decreases.

Near Earth's Surface

Close to Earth's surface, the gravitational force on an object of mass $m$ is approximately constant:

$$F_g = mg$$

If the object moves vertically by a height change $\Delta y$, the change in gravitational potential energy is

$$\Delta U = mg \Delta y$$

If we choose the zero level of potential energy at some reference height, then the gravitational potential energy can be written as

$$U = mgy$$

where $y$ is the vertical position measured from that chosen reference level.

This formula works well when the height change is small compared with Earth's radius.

Near Earth's surface,
$$U = mgy$$
and
$$\Delta U = mg \Delta y$$
This expression depends on a chosen zero level. Only changes in potential energy have direct physical meaning.

Choosing the Zero Level

Potential energy is not absolute. You are free to choose where $U = 0$. For example, you might choose the floor, the ground, or the top of a table as the zero level.

Different choices give different numerical values of $U$, but the change in potential energy between two points stays the same.

This is why physics problems often say things like "take the ground as zero potential energy."

Sign of the Change

Suppose an object rises by a height $\Delta y > 0$. Then

$$\Delta U = mg \Delta y > 0$$

Its gravitational potential energy increases.

If the object falls, then $\Delta y < 0$, so

$$\Delta U < 0$$

Its gravitational potential energy decreases.

This sign tells you whether energy is being stored in the gravitational system or released from it.

Relation to Work

If you lift an object slowly upward, your applied force balances the weight. The work you do on the object increases its gravitational potential energy.

For an upward displacement $h$,

$$W_{\text{you}} = +mgh$$

and gravity does

$$W_{\text{gravity}} = -mgh$$

So the change in gravitational potential energy is

$$\Delta U = +mgh$$

If the object moves downward, gravity does positive work and the potential energy decreases.

Gravitational Potential Energy Far from Earth

When distances become very large, the simple formula $U = mgy$ is no longer accurate. Then we must use Newton's law of gravitation.

For two masses, $M$ and $m$, separated by a distance $r$, the gravitational potential energy is

$$U(r) = -\frac{GMm}{r}$$

where $G$ is the gravitational constant.

This formula is used for planets, satellites, moons, and stars.

The zero of potential energy is usually chosen at infinite separation, where

$$U(\infty) = 0$$

Because of that choice, gravitational potential energy for bound systems is negative.

For two masses separated by distance $r$,
$$U(r) = -\frac{GMm}{r}$$
with the standard choice
$$U(\infty) = 0$$
Gravitational potential energy is negative because gravity is attractive.

Why the Energy Is Negative

The negative sign means that work must be done to separate the two masses to an infinite distance. In other words, a bound gravitational system has less energy than two masses infinitely far apart.

If an object moves farther away from Earth, $r$ increases and $U$ becomes less negative, which means $U$ increases. If it moves closer, $r$ decreases and $U$ becomes more negative, which means $U$ decreases.

Comparing the Two Formulas

The two common expressions for gravitational potential energy apply in different situations.

Situation	Formula	Use
Near Earth's surface	$U = mgy$	Small height changes
General two-body gravity	$U = -\dfrac{GMm}{r}$	Large distances, planets, satellites

The near Earth formula is an approximation of the more general one when the height above Earth's surface is small compared with Earth's radius.

Example Near Earth

Take a mass $m = 2.0\ \text{kg}$ lifted by $h = 3.0\ \text{m}$ near Earth's surface. Using $g = 9.8\ \text{m/s}^2$,

$$\Delta U = mgh = (2.0)(9.8)(3.0) = 58.8\ \text{J}$$

So the object's gravitational potential energy increases by $58.8\ \text{J}$.

Example for Large Distance

A satellite of mass $m$ at distance $r$ from Earth's center has gravitational potential energy

$$U = -\frac{GM_E m}{r}$$

If the satellite moves to a larger orbit, $r$ increases, so the value of $U$ becomes less negative. That means the gravitational potential energy increases.

For example, changing from

$$U = -5000\ \text{J}$$

$$U = -3000\ \text{J}$$

is an increase in potential energy, even though both values are negative.

Visual Picture

You can think of gravitational potential energy as describing an energy landscape. Near Earth, increasing height means increasing potential energy. In the general case, moving farther from a massive body raises the potential energy toward zero.

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Gravitational potential energy near Earth's surface

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Gravitational potential energy versus distance

Key Ideas to Remember

Gravitational potential energy measures how position in a gravitational field stores energy. Near Earth, use $U = mgy$ or $\Delta U = mg \Delta y$. For large distances from a massive body, use $U = -GMm/r$. The change in gravitational potential energy is always the negative of the work done by gravity.

Important summary:
$$\Delta U = -W_{\text{gravity}}$$
Near Earth:
$$\Delta U = mg\Delta y,\qquad U = mgy$$
General gravitational case:
$$U(r) = -\frac{GMm}{r}$$
Only differences in potential energy are physically important.

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