Gravitational Potential Energy

Lift a book onto a high shelf and you have given it the ability to do something later: nudge it off and it will fall, speeding up all the way down. The energy you stored in it by lifting it is its gravitational potential energy — the energy an object has because of its position in a gravitational field, that is, because it has been raised up. This store is at the heart of how rollercoasters, pendulums, hydroelectric dams and falling objects work, because as something falls its gravitational potential energy is steadily transferred into kinetic energy. This lesson, part of Topic P7 (Energy) of OCR Gateway Science A, defines gravitational potential energy, introduces and rearranges $E_p = mgh$ , and shows how to solve problems where gravitational potential energy converts into kinetic energy.

By the end of this lesson you should be able to define gravitational potential energy, use and rearrange $E_p = mgh$ with $g = 9.8\text{ N/kg}$ , carry out calculations, and solve problems in which gravitational potential energy is transferred to kinetic energy using $mgh = \dfrac{1}{2}mv^2$ .

What Is Gravitational Potential Energy?

Gravitational potential energy (often written GPE) is the energy stored in an object because of its height above a chosen level (usually the ground) in a gravitational field. Lifting an object higher does work against gravity and transfers energy to its gravitational potential store; letting it fall transfers that energy back out again. The higher you lift something, and the heavier it is, the more gravitational potential energy it gains.

Gravitational potential energy depends on three things:

the mass of the object ( $m$ , in kg) — heavier objects store more;
the gravitational field strength ( $g$ ) — on Earth, $g = 9.8\text{ N/kg}$ ;
the height raised ( $h$ , in m) — the higher it goes, the more it stores.

Exam Tip: Gravitational potential energy is the energy of an object due to its height in a gravitational field. On Earth, always use $g = 9.8\text{ N/kg}$ unless the question tells you otherwise. The height $h$ is the vertical height, measured from your chosen reference level.

The Gravitational Potential Energy Equation

The gravitational potential energy of an object is given by:

$E_p = mgh$

where $E_p$ is the gravitational potential energy (in joules, $\text{J}$ ), $m$ is the mass (in kilograms, $\text{kg}$ ), $g$ is the gravitational field strength (in newtons per kilogram, $\text{N/kg}$ , equal to $9.8\text{ N/kg}$ on Earth) and $h$ is the height (in metres, $\text{m}$ ).

The equation rearranges to make any quantity the subject:

$E_p = mgh \qquad m = \dfrac{E_p}{gh} \qquad h = \dfrac{E_p}{mg}$

Worked Example 1 — Lifting a box

A box of mass $20 \text{ kg}$ is lifted onto a shelf $2.5 \text{ m}$ high. Calculate the gain in gravitational potential energy. ( $g = 9.8\text{ N/kg}$ .)

Step 1 — write the equation: $E_p = mgh$ .

Step 2 — substitute: $E_p = 20 \times 9.8 \times 2.5$ .

Step 3 — calculate: $E_p = 490 \text{ J}$ .

Answer: the box gains $490 \text{ J}$ of gravitational potential energy.

Worked Example 2 — A climber

A climber of mass $65 \text{ kg}$ ascends a cliff of height $40 \text{ m}$ . How much gravitational potential energy do they gain? ( $g = 9.8\text{ N/kg}$ .)

Step 1 — write the equation: $E_p = mgh$ .

Step 2 — substitute: $E_p = 65 \times 9.8 \times 40$ .

Step 3 — calculate: $E_p = 25\,480 \text{ J}$ .

Answer: the climber gains $25\,480 \text{ J}$ (about $25.5 \text{ kJ}$ ) of gravitational potential energy.

Worked Example 3 — Rearranging for height

A $2 \text{ kg}$ object gains $98 \text{ J}$ of gravitational potential energy when lifted. How high was it raised? ( $g = 9.8\text{ N/kg}$ .)

Step 1 — rearrange for height: $h = \dfrac{E_p}{mg}$ .

Step 2 — substitute: $h = \dfrac{98}{2 \times 9.8} = \dfrac{98}{19.6}$ .

Step 3 — calculate: $h = 5 \text{ m}$ .

Answer: the object was raised $5 \text{ m}$ .

Exam Tip: Keep the mass in kilograms and the height in metres, and use $g = 9.8\text{ N/kg}$ . The height is the vertical rise — for an object pulled up a slope, it is the height gained, not the distance along the slope.

Transfer Between Gravitational and Kinetic Stores

The real power of gravitational potential energy is what happens when an object falls. As it drops, its height falls, so its gravitational potential store empties; at the same time it speeds up, so its kinetic store fills. Energy is transferred mechanically (gravity is the force doing the work) from the gravitational potential store to the kinetic store. This is exactly what happens for:

a falling object — gravitational potential energy becomes kinetic energy on the way down;
a pendulum — at the top of each swing all the energy is gravitational potential; at the bottom it is all kinetic; the energy swaps back and forth;
a rollercoaster — the car is hauled to the top of the first hill (a big gravitational potential store), then this is transferred to kinetic energy as it races down, and back to gravitational potential as it climbs the next hill.

graph LR
  A[Top of fall: full gravitational potential store, zero kinetic] -->|object falls, gravity does work| B[Bottom: zero gravitational potential store, full kinetic store]

If we ignore air resistance (and friction), no energy is dissipated, so the gravitational potential energy lost is exactly equal to the kinetic energy gained:

$E_p \text{ lost} = E_k \text{ gained}$ $mgh = \dfrac{1}{2}mv^2$

This single equation lets you work out how fast a falling object is going, or how high it will rise, without needing to know the time taken.

Exam Tip: When an object falls (ignoring air resistance), gravitational potential energy lost = kinetic energy gained, so $mgh = \dfrac{1}{2}mv^2$ . Notice the mass cancels from both sides — the final speed of a freely falling object does not depend on its mass.

Solving `mgh = ½mv²` Problems

When all the gravitational potential energy converts to kinetic energy, set $mgh = \dfrac{1}{2}mv^2$ . Because $m$ appears on both sides, it cancels, leaving:

$gh = \dfrac{1}{2}v^2 \qquad \Rightarrow \qquad v = \sqrt{2gh}$

So the speed of a falling object depends only on $g$ and the height fallen — not on its mass.

Worked Example 4 — Speed of a falling object

A stone is dropped from a height of $5 \text{ m}$ . Calculate its speed just before it hits the ground, ignoring air resistance. ( $g = 9.8\text{ N/kg}$ .)

Step 1 — equate the stores: $mgh = \dfrac{1}{2}mv^2$ ; the mass cancels, giving $v = \sqrt{2gh}$ .

Step 2 — substitute: $v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98}$ .

Step 3 — calculate: $v = 9.9 \text{ m/s}$ (to 2 significant figures).

Answer: the stone is travelling at about $9.9 \text{ m/s}$ just before impact.

Worked Example 5 — A pendulum

A pendulum bob is pulled to one side so that it is raised $0.2 \text{ m}$ above its lowest point. Calculate its speed at the lowest point of the swing, ignoring air resistance. ( $g = 9.8\text{ N/kg}$ .)

Step 1 — at the lowest point all the gravitational potential energy has become kinetic energy: $v = \sqrt{2gh}$ .

Gravitational Potential Energy

Gravitational Potential Energy

What Is Gravitational Potential Energy?

The Gravitational Potential Energy Equation

Worked Example 1 — Lifting a box

Worked Example 2 — A climber

Worked Example 3 — Rearranging for height

Transfer Between Gravitational and Kinetic Stores

Solving mgh = ½mv² Problems

Worked Example 4 — Speed of a falling object

Worked Example 5 — A pendulum

More in Physics

Solving `mgh = ½mv²` Problems