Gravitational Potential Energy

This lesson covers the gravitational potential energy (GPE) store and how to calculate it using the equation $GPE = mgh$GPE=mgh, as required by the Edexcel GCSE Combined Science specification (1SC0). You will learn how to apply the equation and link it to energy transfers.

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What Is Gravitational Potential Energy?

Gravitational potential energy (GPE) is the energy stored in an object because of its position in a gravitational field — typically its height above the ground.

• Lifting an object increases its GPE.
• Lowering an object decreases its GPE.
• The higher the object, the greater its GPE.

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The GPE Equation

$GPE = mgh$GPE=mgh

where:

• GPE = gravitational potential energy in joules (J)
• m = mass in kilograms (kg)
• g = gravitational field strength in newtons per kilogram (N/kg)
• h = height in metres (m)

On Earth, g = 9.8 N/kg (often approximated as 10 N/kg in calculations).

Quantity	Unit	Symbol
GPE	joule	J
Mass	kilogram	kg
Gravitational field strength	N/kg	g
Height	metre	m

Exam Tip: The question will tell you the value of g to use. If it says g = 10 N/kg, use that value. If it says g = 9.8 N/kg, use 9.8. Never assume — always read the question carefully.

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Worked Examples

Worked Example 1

A 2 kg book is lifted 1.5 m onto a shelf. Calculate the increase in GPE. (g = 9.8 N/kg)

1. Write the equation: $GPE = mgh$GPE=mgh
2. Substitute: $GPE = 2 \times 9.8 \times 1.5$GPE=2×9.8×1.5
3. Calculate: $GPE = 29.4 \text{ J}$GPE=29.4 J
4. Answer: GPE = 29.4 J

Worked Example 2

A 60 kg climber ascends 250 m. Calculate the gain in GPE. (g = 9.8 N/kg)

1. $GPE = mgh = 60 \times 9.8 \times 250$GPE=mgh=60×9.8×250
2. $GPE = 147\,000 \text{ J} = 147 \text{ kJ}$GPE=147000 J=147 kJ
3. Answer: GPE = 147 000 J (147 kJ)

Worked Example 3

A crane lifts a 500 kg steel beam. The gain in GPE is 24 500 J. What height was it raised to? (g = 9.8 N/kg)

1. Rearrange: $h = \frac{GPE}{mg}$h=mgGPE​
2. Substitute: $h = \frac{24\,500}{500 \times 9.8} = \frac{24\,500}{4900}$h=500×9.824500​=490024500​
3. $h = 5 \text{ m}$h=5 m
4. Answer: h = 5 m

Exam Tip: Always check your answer makes physical sense. A crane lifting a beam 5 m is realistic. If you calculated 5000 m, you have probably made a unit error.

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Changes in Height

GPE depends on the change in height (Δh), not the absolute position. It does not matter what path the object takes — only the vertical height change matters.

flowchart TB
    A["Object at height h₂"] -->|"Drops"| B["Object at height h₁"]
    style A fill:#4dabf7,color:#000
    style B fill:#69db7c,color:#000

$\Delta GPE = mg\Delta h = mg(h_2 - h_1)$ΔGPE=mgΔh=mg(h2​−h1​)

Worked Example 4

A 0.3 kg ball rolls off a table that is 0.8 m high and falls to the floor. Calculate the decrease in GPE. (g = 10 N/kg)

1. $\Delta GPE = mg\Delta h = 0.3 \times 10 \times 0.8$ΔGPE=mgΔh=0.3×10×0.8
2. $\Delta GPE = 2.4 \text{ J}$ΔGPE=2.4 J
3. Answer: The GPE decreases by 2.4 J

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Linking GPE and KE

When an object falls freely (ignoring air resistance), its GPE is converted into KE:

$\text{Decrease in GPE} = \text{Increase in KE}$Decrease in GPE=Increase in KE

$mgh = \frac{1}{2}mv^2$mgh=21​mv2

Worked Example 5

A 0.5 kg ball is dropped from a height of 20 m. What is its speed just before hitting the ground? (g = 10 N/kg, ignore air resistance)

1. GPE lost = KE gained: $mgh = \frac{1}{2}mv^2$mgh=21​mv2
2. Cancel m from both sides: $gh = \frac{1}{2}v^2$gh=21​v2
3. Rearrange: $v^2 = 2gh = 2 \times 10 \times 20 = 400$v2=2gh=2×10×20=400
4. $v = \sqrt{400} = 20 \text{ m/s}$v=400​=20 m/s
5. Answer: v = 20 m/s

Exam Tip: When GPE is fully converted to KE (no air resistance), the mass cancels. This means the speed of a falling object does not depend on its mass — a key insight that often appears in exam questions.

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Rearranging the Equation

To find	Rearranged equation
GPE	$GPE = mgh$GPE=mgh
m	$m = \frac{GPE}{gh}$m=ghGPE​
h	$h = \frac{GPE}{mg}$h=mgGPE​
g	$g = \frac{GPE}{mh}$g=mhGPE​

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GPE on Other Planets

The value of g varies between planets and moons:

Location	g (N/kg)
Earth	9.8
Moon	1.6
Mars	3.7
Jupiter	24.8

A 50 kg person at 2 m height would have:

• On Earth: GPE = 50 × 9.8 × 2 = 980 J
• On the Moon: GPE = 50 × 1.6 × 2 = 160 J

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Summary

• Gravitational potential energy is the energy stored due to an object's height in a gravitational field.
• The equation is $GPE = mgh$GPE=mgh.
• GPE depends on mass, gravitational field strength and height.
• Only the change in height matters, not the path taken.
• When an object falls freely, GPE is transferred to the kinetic store.
• On Earth, g = 9.8 N/kg (use the value given in the question).

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Extended Worked Examples

Worked Example 6 — A Diver on a 10 m Board

A 65 kg diver climbs to a 10 m board. Calculate the gravitational potential energy relative to the pool (g = 9.8 N/kg) and then the diver's speed just before entering the water.

1. $GPE = mgh = 65 \times 9.8 \times 10 = 6370 \text{ J}$GPE=mgh=65×9.8×10=6370 J
2. Ignoring air resistance, all GPE becomes kinetic: $\tfrac{1}{2}mv^2 = 6370$21​mv2=6370
3. $v^2 = \frac{2 \times 6370}{65} = 196$v2=652×6370​=196
4. $v = \sqrt{196} = 14 \text{ m/s}$v=196​=14 m/s

Worked Example 7 — A Skier

A 75 kg skier descends a vertical drop of 350 m. (g = 10 N/kg.)