Gravitational potential energy — Physics

Before this

This leads to

Conservation of energy — the big idea

Where did Ravi's energy go?

Two lectures ago, Ravi lifted a 5.0 kg box at steady speed onto a shelf 1.6 m high. He did 78.4 J of work on it. Gravity did −78.4 J. The box's kinetic energy did not change, because it started and ended at rest. The last lecture asked the obvious question: where did Ravi's 78.4 J go?

Here is a way to find out. Suppose the box slides off the edge of the shelf. As it falls 1.6 m to the floor, gravity does positive work on it, $49 \times 1.6 = 78.4$ J. By the work-energy theorem, the box gains 78.4 J of kinetic energy. Its speed just before it lands is about 5.6 m/s.

Look at what happened. Ravi put 78.4 J in by lifting. The fall gave exactly 78.4 J back as motion. The energy was not lost on the shelf. It was waiting there, stored in the box's height. This lecture names that store, finds its formula, and shows why the formula is allowed to exist at all.

A store of energy because of height

The energy an object has because of its height in a gravitational field is called gravitational potential energy. Many books shorten it to GPE. We write it $E_{p}$ , like $E_{k}$ for kinetic energy.

To find a formula, follow Ravi's reasoning with letters. An object of mass $m$ has weight $m g$ . Lift it straight up at steady speed through a height $Δ h$ . Because the speed is steady, the resultant force is zero, so your upward force equals the weight, $m g$ . The work you do is force × displacement:

W = m g \times Δ h

The kinetic energy has not changed. So all of this work has gone into the store. This gives the change in gravitational potential energy:

Gravitational potential energy

$Δ E_{p} = m g Δ h$

$m$ is the mass in kg, $g$ is the gravitational field strength (9.8 N/kg near the Earth's surface), and $Δ h$ is the change in vertical height in m. The energy is in joules.

A note on $g$ . It is called the gravitational field strength, measured in N/kg: the force of gravity on each kilogram. The same number is the free-fall acceleration in m/s². The two units are the same thing, because 1 N = 1 kg m/s². So 9.8 N/kg and 9.8 m/s² mean the same $g$ . Some questions round it to 10; use whatever your question gives.

Why this store can exist at all

Here is a question most books skip. Why can we say that a store of energy depends only on height? You could lift the box straight up, carry it up a ramp, or take it up a winding staircase. Those paths are very different. Why should they all store the same energy?

The lecture on work answered the key part. It found that gravity's work on an object depends only on the change in height, not on the length or shape of the path. A steep 1.5 m climb and a gentle 5.0 m ramp to the same height gave the same −588 J. That lecture promised that this fact is the reason gravitational potential energy can be defined at all. Here is why.

Suppose gravity's work did depend on the path. Then you could lift a box up a path where gravity takes little energy, and let it fall back down a path where gravity gives a lot. You would end where you started with more energy than you began with, and you could repeat it forever. No such machine exists. Because gravity's work depends only on where you start and where you finish, every height has one fixed amount of stored energy. A force with this property is called conservative, and only a conservative force can have a potential energy.

That is it. That is the whole idea. Gravity's work depends only on the change in height, so the energy stored depends only on the height. Friction is different. Its work depends on the path, so there is no "friction potential energy".

Test the idea on two slides. Commit to an answer first.

Predict — a steep slide and a gentle slide

Two children of the same mass start from rest at the top of two slides, both 2.5 m high. One slide is steep and short. The other is gentle and long. Ignore friction and air resistance. Which child is moving faster at the bottom?

Both children lose the same height, 2.5 m, so gravity does the same work on each, and each gains the same kinetic energy. Both reach the bottom at the same speed, $2 \times 9.8 \times 2.5 = 7.0$ m/s. The steep slide gets there sooner, but not faster.

The speed at the bottom depends only on the drop in height. The shape of the slide decides how long the trip takes, not how fast it ends. This is the property of gravity at work.

Only changes matter: choosing a reference level

The formula gives a change, $Δ E_{p}$ . To give a single object a value of $E_{p}$ , you must choose where $E_{p} = 0$ . This is the reference level, and you are free to choose it: the floor, the table top, the ground outside. Then $E_{p} = m g h$ , where $h$ is the height above that level.

Different choices give different values of $E_{p}$ . But they always give the same change, and the change is the only thing that ever enters a calculation.

A 0.50 kg book is lifted from a table to a shelf. Measured from the floor, its height goes from 0.80 m to 2.0 m. Measured from the table, it goes from 0 to 1.2 m. The rise, 1.2 m, is the same either way, and so is the change in

E_{p}

0.50 \times 9.8 \times 1.2 = 5.88

From the floor, the book's $E_{p}$ goes from $0.50 \times 9.8 \times 0.80 = 3.92$ J to $0.50 \times 9.8 \times 2.0 = 9.8$ J. The change is $9.8 - 3.92 = 5.88$ J. From the table, it goes from 0 to $0.50 \times 9.8 \times 1.2 = 5.88$ J. The change is again 5.88 J. Choose the reference level that makes the numbers easiest, state it, and keep it for the whole problem. Below the reference level, $h$ is negative, so $E_{p}$ is negative. That is not a problem. It only means "less than at the reference level".

Falling: height becomes speed

Now let an object fall freely from rest through a height $h$ , with no air resistance. It loses $m g h$ of potential energy, and gravity's work gives it the same amount of kinetic energy:

m g h = \frac{1}{2} m v^{2}

The mass appears on both sides, so it cancels:

g h = \frac{1}{2} v^{2} \Rightarrow v = 2 g h

The landing speed does not depend on the mass. A heavier object stores more energy at the same height, but it also needs more energy for each m/s of speed, in exactly the same proportion. The two cancel.

You have seen this result before, from a different direction. For free fall, the SUVAT equation $v^{2} = u^{2} + 2 a s$ with $u = 0$ , $a = g$ and $s = h$ gives $v^{2} = 2 g h$ . Energy and SUVAT agree, because the work-energy theorem was built from SUVAT. But the energy route does more. It works on a curved slide, where SUVAT along a straight line cannot be used.

On Earth, air resistance makes a feather fall slowly, so the result is hard to see. On the Moon there is no air. On 2 August 1971, the Apollo 15 commander David Scott dropped a 1.32 kg geological hammer and a feather together on the Moon's surface, in front of a television camera. They landed together (NASA).

Stop and think

A ball dropped from 5 m lands at about 9.9 m/s. From what height must it be dropped to land at twice that speed?

Show the answer

Not 10 m. Speed depends on the square root of the height: $v = 2 g h$ . To double $v$ , you must multiply $h$ by 4, so the height is 20 m. Check: $2 \times 9.8 \times 20 \approx 19.8$ m/s, twice 9.9 ✓. This is the same "square" you met in kinetic energy, seen from the other side. Twice the speed needs four times the energy, so four times the height.

Watching the energy move

Throw a ball straight up and the energy moves both ways: from kinetic to potential on the way up, and back on the way down. Follow it in the explorer below.

Explore — a ball thrown straight up

A 0.30 kg ball is thrown straight up from the ground (the reference level). Choose a launch speed. Then move the ball's height with the slider, from the ground to the top of its flight. Watch the two energy bars, and the total.

Launch speed:

Height h = 0

A 0.30 kg ball thrown straight up at 8 m/s from the ground, with $g = 9.8$ m/s² and no air resistance:

Height (m)	$E_{p}$ (J)	$E_{k}$ (J)	Total (J)	Speed (m/s)
0	0	9.6	9.6	8.0
1.0	2.94	6.66	9.6	≈ 6.7
2.0	5.88	3.72	9.6	≈ 5.0
3.0	8.82	0.78	9.6	≈ 2.3
≈ 3.27 (top)	9.6	0	9.6	0

At every height, the potential energy gained equals the kinetic energy lost, so the total stays at 9.6 J. On the way down, the same table is read from the bottom up.

What did you see? The two bars always trade. Whatever the potential energy bar gains, the kinetic energy bar loses, and the total does not move. A faster throw starts with more energy, so the ball climbs higher before the trade is complete. Notice too that at each height the speed is the same on the way up and on the way down. Energy does not care about direction, because $v^{2}$ has no sign. The next lecture turns this into the principle of conservation of energy, and adds what happens when friction or air resistance takes some energy away.

Worked examples

Take $g = 9.8$ N/kg and ignore air resistance throughout.

Worked Example 1 — A dropped ball, two ways

A 2.0 kg ball is held 5.0 m above the ground. Find its $E_{p}$ relative to the ground, and its speed just before it lands if it is dropped.

Energy. $E_{p} = m g h = 2.0 \times 9.8 \times 5.0 = 98$ J.

Speed, the long way. All 98 J becomes kinetic energy: $\frac{1}{2} \times 2.0 \times v^{2} = 98$ , so $v^{2} = 98$ and $v \approx 9.9$ m/s.

Speed, the short way. $v = 2 g h = 2 \times 9.8 \times 5.0 = 98 \approx 9.9$ m/s. The mass was never needed.

Check with SUVAT. $v^{2} = 0 + 2 \times 9.8 \times 5.0 = 98$ ✓.

Worked Example 2 — A ball thrown up, tracked at three points

A 0.30 kg ball is thrown straight up at 8.0 m/s from the ground. Take the ground as the reference level. Find the energies at launch (A), at the top (B), and back at the ground on the way down (C). How high does it go?

A, launch. $E_{k} = \frac{1}{2} \times 0.30 \times 8. 0^{2} = 9.6$ J. $E_{p} = 0$ . Total 9.6 J.

B, the top. The ball stops for an instant, so $E_{k} = 0$ , and all 9.6 J is potential energy. Height: $h = \frac{E _{p}}{m g} = \frac{9.6}{0.30 \times 9.8} \approx 3.27$ m.

C, back at the ground. $E_{p} = 0$ again, so $E_{k} = 9.6$ J and the speed is 8.0 m/s again, now downwards.

Check. With SUVAT, $0 = 8. 0^{2} - 2 \times 9.8 \times h$ gives $h = 64 \div 19.6 \approx 3.27$ m ✓. The mass cancelled again: any ball thrown up at 8.0 m/s reaches the same height.

Worked Example 3 — Climbing the stairs

Aiko, mass 60 kg, climbs stairs that rise 4.0 m. How much gravitational potential energy does she gain? Compare it with the energy in food.

$Δ E_{p} = 60 \times 9.8 \times 4.0 = 2352$ J, about 2.4 kJ.

Compare. Raw banana provides 371 kJ per 100 g (US Department of Agriculture data). A medium banana of about 118 g therefore holds about 438 kJ. The climb stores only about 0.5% of that. Your body actually uses several times more than 2.4 kJ for the climb, because muscles also turn much of their energy into heat. Following where that energy goes is the job of the next lectures.

Worked Example 4 — Two reference levels, one answer

The 0.50 kg book in the figure falls from the 2.0 m shelf back onto the 0.80 m table. Find its speed as it reaches the table, using the floor as the reference level. Then repeat using the table top.

Floor as reference. $E_{p}$ falls from 9.8 J to 3.92 J, a loss of 5.88 J. So $\frac{1}{2} \times 0.50 \times v^{2} = 5.88$ , $v^{2} = 23.52$ , $v \approx 4.85$ m/s.

Table as reference. $E_{p}$ falls from 5.88 J to 0, a loss of 5.88 J. The same equation, the same $v \approx 4.85$ m/s.

Check. $v = 2 \times 9.8 \times 1.2 \approx 4.85$ m/s ✓. Only the drop of 1.2 m matters.

Three classic mistakes

Trap 1 — Using the distance along a slope

For a ramp 5.0 m long that rises 1.5 m, the $Δ h$ in $m g Δ h$ is 1.5 m, not 5.0 m. Gravity only "sees" the vertical change. Draw the vertical height on your diagram before you calculate.

Trap 2 — Multiplying by g twice

If a question gives the weight, say 49 N, then $m g$ is already 49 N, and $Δ E_{p} = 49 \times Δ h$ . Writing $49 \times 9.8 \times Δ h$ multiplies by $g$ twice. Ask yourself: is this number a mass in kg or a weight in N? Also convert grams to kilograms first.

Trap 3 — Changing the reference level halfway

Any reference level works, but only one per problem. Measuring the start from the floor and the end from the table gives a wrong change. State your reference level at the start, as you state the positive direction.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Take $g = 9.8$ N/kg and ignore air resistance and friction.

Kwame, mass 45 kg, climbs a 12 m high lookout tower. How much gravitational potential energy does he gain?

Hint 1
Use $Δ E_{p} = m g Δ h$ with the vertical height of the tower.

Hint 2
$45 \times 9.8 \times 12$ .

Full answer
$Δ E_{p} = 45 \times 9.8 \times 12 = 5292$ J, about 5.3 kJ. Check: the weight is $45 \times 9.8 = 441$ N, and 441 N through 12 m is $441 \times 12 = 5292$ J ✓. The route up the tower, stairs or ladder, does not change the answer.
A stone is dropped from a bridge 20 m above a river. How fast is it moving when it reaches the water? Do you need its mass?

Hint 1
Set the potential energy lost equal to the kinetic energy gained, and look for what cancels.

Hint 2
$v = 2 g h$ .

Full answer
$m g h = \frac{1}{2} m v^{2}$ , so the mass cancels and $v = 2 \times 9.8 \times 20 = 392 \approx 19.8$ m/s. The mass is not needed. Check with SUVAT: $v^{2} = 0 + 2 \times 9.8 \times 20 = 392$ ✓.
A child starts from rest at the top of a curved water slide 3.2 m high. How fast is she moving at the bottom? How fast is she moving at a point halfway down in height, 1.6 m above the bottom? Why can you not use SUVAT here?

Hint 1
Only the vertical drop from the start matters, not the shape of the slide.

Hint 2
At the halfway point she has dropped 1.6 m. Use $v = 2 g \times drop$ .

Full answer
Bottom: $v = 2 \times 9.8 \times 3.2 \approx 7.9$ m/s. Halfway: $v = 2 \times 9.8 \times 1.6 = 5.6$ m/s. Half the drop gives $1 \div 2$ , about 71%, of the final speed, not half of it: $7.9 \times 0.707 \approx 5.6$ ✓. SUVAT needs a constant acceleration along a straight line. On a curved slide, the acceleration keeps changing direction and size, but the energy method does not care about the path.
A ball is thrown straight down at 5.0 m/s from a window 10 m above the ground. How fast is it moving when it lands? Then: would the answer change if it were thrown straight up at 5.0 m/s from the same window?

Hint 1
This time the ball starts with kinetic energy as well as potential energy. Take the ground as the reference level and add the two at the start.

Hint 2
$\frac{1}{2} m u^{2} + m g h = \frac{1}{2} m v^{2}$ . Divide by $m$ and multiply by 2: $v^{2} = u^{2} + 2 g h$ .

Full answer
Start: kinetic energy $\frac{1}{2} m \times 5. 0^{2}$ and potential energy $m \times 9.8 \times 10$ . At the ground all of it is kinetic. So $v^{2} = 5. 0^{2} + 2 \times 9.8 \times 10 = 25 + 196 = 221$ , and $v \approx 14.9$ m/s. Thrown up at 5.0 m/s, the answer is the same. The ball starts with the same kinetic energy, because energy does not depend on direction. It rises, stops, and passes the window again at 5.0 m/s downwards, so from there the fall is identical. Check with SUVAT for the downward throw: $v^{2} = 5. 0^{2} + 2 \times 9.8 \times 10 = 221$ ✓.

The bigger idea

Whose energy is gravitational potential energy? Not the ball's alone. A ball far out in space, with no planet nearby, has none. The energy belongs to the ball and the Earth together, a system pulled towards each other. Potential energy is always energy of an arrangement: how far apart things are that attract or repel. The same idea will return as the energy stored in a stretched spring, in the electric attraction inside atoms, and in the bonds of chemistry.

The formula $m g Δ h$ also has a limit. It assumes that $g$ is the same at every height. Near the ground that is very nearly true. But $g$ weakens with distance from the centre of the Earth. Space stations orbit about 400 km above the surface. There, $g$ is only about 8.7 N/kg, because 400 km is a noticeable fraction of the Earth's radius of about 6400 km. For rockets and satellites, physicists use a more general formula. For anything you can climb, $m g Δ h$ is excellent.

People use this store on a very large scale. Hydroelectric power stations let water fall from a reservoir high up and turn its lost potential energy into electricity. Some even pump water uphill when electricity is cheap and let it fall back when it is needed, using a mountain lake like a giant battery.

Key concepts in this lesson

Gravitational potential energy is energy stored because of height in a gravitational field, and its change is $Δ E_{p} = m g Δ h$ .
The formula comes from the work done lifting an object at steady speed: force $m g$ through a height $Δ h$ .
The store can be defined only because gravity's work depends on the change in height, not on the path taken.
Only changes in $E_{p}$ matter, so any reference level can be chosen, as long as one level is used for the whole problem.
An object falling freely from rest through a height $h$ reaches $v = 2 g h$ , whatever its mass, because the mass cancels.