Gravitational Potential Energy: Definition, Formula, Units (W/ Examples)

Most people know about the conservation of energy. In a nutshell, it says that energy is conserved; it isn't created and it isn't destroyed, and it simply changes from one form to another.

So if you hold a ball completely still, two meters above the ground, and then release it, where does the energy it gains come from? How can something completely still gain so much kinetic energy before it hits the ground?

The answer is that the still ball posses a form of stored energy called ​gravitational potential energy​, or GPE for short. This is one of the most important forms of stored energy a high school student will encounter in physics.

GPE is a form of mechanical energy caused by the height of the object above the surface of the Earth (or indeed, any other source of a gravitational field). Any object that isn't at the lowest-energy point in such a system has some gravitational potential energy, and if released (i.e., allowed to fall freely), it will accelerate towards the center of the gravitational field until something stops it.

Although the process of finding the gravitational potential energy of an object is quite straightforward mathematically, the concept is extraordinarily useful when it comes to calculating other quantities. For example, learning about the concept of GPE makes it really easy to calculate the kinetic energy and the final speed of a falling object.

GPE depends on two key factors: the object's position relative to a gravitational field and the mass of the object. The center of mass of the body creating the gravitational field (on Earth, the center of the planet) is the lowest-energy point in the field (although in practice the actual body will stop the falling before this point, as the Earth's surface does), and the farther from this point an object is, the more stored energy it has due to its position. The amount of stored energy also increases if the object is more massive.

You can understand the basic definition of gravitational potential energy if you think about a book resting on top of a bookshelf. The book has the potential to fall to the floor because of its elevated position relative to the ground, but one that starts out on the floor can't fall, because it's already on the surface: The book on the shelf has GPE, but the one on the ground doesn't.

Intuition will also tell you that a book that's twice as thick will make twice as big a thud when it hits the ground; this is because the mass of the object is directly proportional to the amount of gravitational potential energy an object has.

The formula for gravitational potential energy (GPE) is really simple, and it relates mass ​m​, the acceleration due to gravity on the Earth ​g​) and height above the Earth's surface ​h​ to the stored energy due to gravity:

\(GPE=mgh\)

As is common in physics, there are many potential different symbols for gravitational potential energy, including ​U​_g, ​PE​_grav and others. GPE is a measure of energy, so the result of this calculation will be a value in joules (J).

The acceleration due to Earth's gravity has a (roughly) constant value anywhere on the surface and points directly to the center of mass of the planet: g = 9.81 m/s². Given this constant value, the only things you need to calculate GPE are the mass of the object and the height of the object above the surface.

So what do you do if you need to calculate how much gravitational potential energy an object has? In essence, you can simply define the height of the object based on a simple reference point (the ground usually works just fine) and multiply this by its mass ​m​ and the terrestrial gravitational constant ​g​ to find the GPE.

For example, imagine a 10-kg mass suspended a height of 5 meters above the ground by a pulley system. How much gravitational potential energy does it have?

Using the equation and substituting the known values gives:

\(\begin{aligned}
GPE&=mgh \
&= 10 \;\text{kg} × 9.81 \;\text{m/s}^2 × 5 \;\text{m}\
&= 490.5 \;\text{J}
\end{aligned}\)

However, if you've been thinking about the concept while reading this article, you might have considered an interesting question: If the gravitational potential energy of an object on Earth is only truly zero if it's at the center of the mass (i.e., inside the Earth's core), why do you calculate it as if the surface of the Earth is ​h​ = 0?

The truth is that the choice of the "zero" point for height is arbitrary, and it's usually done to simplify the problem at hand. Whenever you calculate GPE, you're really more concerned about gravitational potential energy ​changes​ rather than any sort of absolute measure of the stored energy.

In essence, it doesn't matter if you decide to call a tabletop ​h​ = 0 rather than the Earth's surface because you're always ​actually​ talking about changes in potential energy related to changes in height.

Consider, then, somebody lifting a 1.5-kg physics textbook from the surface of a desk, raising it 50 cm (i.e., 0.5 m) above the surface. What is the gravitational potential energy change (denoted ∆​GPE​) for the book as it is lifted?

The trick, of course, is to call the table the reference point, with a height of ​h​ = 0, or equivalently, to consider the change in height (∆​h​) from the initial position. In either case, you get:

\(\begin{aligned}
∆GPE &= mg∆h \
&= 1.5 \;\text{kg} × 9.81 \;\text{m/s}^2 × 0.5 \;\text{m}\
&= 7.36 \;\text{J}
\end{aligned}\)

The precise value for gravitational acceleration ​g​ in the GPE equation has a big impact on the gravitational potential energy of an object raised a certain distance above a source of a gravitational field. On the surface of Mars, for instance, the value of ​g​ is about three times smaller than on the surface of the Earth, so if you lift the same object the same distance from the surface of Mars, it would have about three times less stored energy than it would on Earth.

Similarly, although you can approximate the value of ​g​ as 9.81 m/s² across the Earth's surface at sea level, it's actually smaller if you move a substantial distance away from the surface. For example, if you were on a Mt. Everest, which rises up 8,848 m (8.848 km) above the Earth's surface, being so far away from the center of mass of the planet would reduce the value of ​g​ slightly, so you would have ​g​ = 9.79 m/s² at the peak.

If you had successfully climbed the mountain and lifted a 2-kg mass 2 m from the peak of the mountain into the air, what would be the change in GPE?

Like calculating GPE on another planet with a different value of ​g​, you simply input the value for ​g​ that suits the situation and go through the same process as above:

\(\begin{aligned}
∆GPE &= mg∆h \
&= 2 \;\text{kg} × 9.79\;\text{m/s}^2 × 2 \;\text{m}\
&= 39.16 \;\text{J}
\end{aligned}\)

At sea level on Earth, with ​g​ = 9.81 m/s², lifting the same mass would change the GPE by:

\(\begin{aligned}
∆GPE &= mg∆h \
&= 2 \;\text{kg} × 9.81\;\text{m/s}^2 × 2 \;\text{m}\
&= 39.24 \;\text{J}
\end{aligned}\)

This isn't a huge difference, but it clearly shows that altitude does affect the change in GPE when you perform the same lifting motion. And on the surface of Mars, where ​g​ = 3.75 m/s² it would be:

\(\begin{aligned}
∆GPE &= mg∆h \
&= 2 \;\text{kg} × 3.75\;\text{m/s}^2 × 2 \;\text{m}\
&= 15 \;\text{J}
\end{aligned}\)

As you can see, the value of ​g​ is very important to the result you get. Performing the same lifting motion in deep space, far away from any influence from the force of gravity, there would be essentially no change in gravitational potential energy.

The conservation of energy can be used alongside the concept of GPE to simplify ​many​ calculations in physics. In short, under the influence of a "conservative" force, total energy (including kinetic energy, gravitational potential energy and all other forms of energy) is conserved.

A conservative force is one where the amount of work done against the force to move an object between two points doesn't depend on the path taken. So gravity is conservative because lifting an object from a reference point to a height ​h​ changes the gravitational potential energy by ​mgh​, but it doesn't make a difference whether you move it in an S-shaped path or a straight line – it always just changes by ​mgh​.

Now imagine a situation where you're dropping a 500-g (0.5-kg) ball from a height of 15 meters. Ignoring the effect of air resistance and assuming it doesn't rotate during its fall, how much kinetic energy will the ball have at the instant before it contacts with the ground?

The key to this problem is the fact that total energy is conserved, so all of the kinetic energy comes from the GPE, and so the kinetic energy ​E​_k at its maximum value must equal the GPE at its maximum value, or ​GPE​ = ​E​_k. So you can solve the problem easily:

\(\begin{aligned}
E_k &= GPE \
&= mgh\
&= 0.5 \;\text{kg} × 9.81\;\text{m/s}^2 × 15 \;\text{m}\
&= 73.58 \;\text{J}
\end{aligned}\)

The conservation of energy simplifies many other calculations involving gravitational potential energy, too. Think about the ball from the previous example: now that you know the total kinetic energy based on its gravitational potential energy at its highest point, what is the final speed of the ball at the instant before it hits the Earth's surface? You can work this out based on the standard equation for kinetic energy:

\(E_k=\frac{1}{2}mv^2\)

With the value of ​E​_k known, you can re-arrange the equation and solve for the speed ​v​:

\(\begin{aligned}
v&=\sqrt{\frac{2E_k}{m}} \
&=\sqrt{\frac{2 × 73.575 \;\text{J}}{0.5\;\text{kg}}} \
&=17.16 \;\text{m/s}
\end{aligned}\)

However, you can use the conservation of energy to derive an equation that applies to ​any​ falling object, by first noting that in situations like this, -∆​GPE​ = ∆​E​_k, and so:

\(mgh = \frac{1}{2}mv^2\)

Cancelling ​m​ from both sides and re-arranging gives:

\(gh = \frac{1}{2}v^2 \
\text{Therefore} \;v= \sqrt{2gh}\)

Note that this equation shows that, ignoring air resistance, mass doesn't affect the final speed ​v​, so if you drop any two objects from the same height, they will hit the ground at exactly the same time and fall at the same speed. You can also check the result obtained using the simpler, two-step method and show that this new equation does indeed produce the same result with the correct units.

Finally, the previous equation also gives you a way to calculate ​g​ on other planets. Imagine that you dropped the 0.5-kg ball from 10 m above the surface of Mars, and recorded a final speed (just before it hit the surface) of 8.66 m/s. What is the value of ​g​ on Mars?

Starting from an earlier stage in the re-arrangement:

\(gh = \frac{1}{2}v^2\)

You see that:

\(\begin{aligned}
g &= \frac{v^2}{2h} \
&= \frac{(8.66 \;\text{m/s})^2}{2 × 10 \;\text{m}} \
&= 3.75 \;\text{m/s}^2
\end{aligned}\)

The conservation of energy, in combination with the equations for gravitational potential energy and kinetic energy, has ​many​ uses, and when you get used to exploiting the relationships, you'll be able to solve a huge range of classical physics problems with ease.