# Gravitational potential energy

#### Prerequisites

- Weight: Gravitational forces
- Flat-earth gravity
- Energy of place -- potential energy
- Reading the content in the Work-Energy theorem

## The work done by gravity

An object's kinetic energy, ½mv^{2}, is a scalar measure of an object's motion — one that only depends on the mass and the speed, not the direction of the velocity. Our work-energy theorem tells us that what matters in changing an object's kinetic energy is the work — the component of force along the direction of motion.

The situation is shown by the vectors in the figure at the right. The force of gravity (labeled F and in red) points down. An arbitrary displacement is shown in blue. The work done by the gravitational force on the object as it moves through the displacement is

$$W = F_{\parallel} \Delta r = F\Delta r \cos \theta = mg \Delta r \cos \theta$$

To interpret this, it makes more sense to associate the cosine factor with the displacement rather than the force. From the second figure at the right, it's clear that the displacement times the cosine factor is the change in the height. This gives us:

the work done by the force of gravity is just mg times the change in height.

We have to be a little careful to get the sign right since the geometry only tells us magnitudes. If the object is going down, we know that the force of gravity is acting to speed it up. This means that this work increases the KE. If we are going down, therefore, the work should be positive. But if we keep our use of Delta — that it means the change, so the final minus the initial — if we define positive height as up, the final height is less than the initial. This would make the change in height negative when the object falls. So we want to write the work as

$$W = mg (h_{initial} - h_{final}) = -mg \Delta h $$

This makes sense. If the object goes up (the height increases so $Δh$ is positive) then the work done by gravity is negative and tries to make the object's KE decrease. If the object goes down (the height decreases so $Δh$ is negative), the work done by gravity is positive so it tries to make the object's KE increase.

Note that we have been careful with our wording here in case there are other forces acting on the object as well. We are only looking here at what the effect of the force of gravity is. The object could be doing the opposite of what gravity is trying to do if there are other forces acting on the object in addition to gravity. However the object is moving, if it is changing its height, the gravitational force is doing work on it equal to $-mg \Delta h$.

## Gravitational PE

Notice that the work done only depends on the initial and final position of the object — the difference in the heights. This means that we can get away with defining the work as a change in gravitational potential energy — and this leads us to define a potential energy function for gravity:

$$\Delta U_{grav} = -W = mg \Delta h = \Delta (mgh)$$

so for flat-earth gravity, we can choose

$$U_{grav} = mgh$$

## Making sense — free fall

To make sense of how gravitational PE works, let's consider the special case of free-fall in flat earth gravity; in the case where there are no other forces acting on the object. In this case the only force acting on the object is gravity so the only work done is by gravity. If there are no other forces, the WE Theorem becomes

$$\Delta \big( {1 \over 2} mv^2 + mgh \big) = 0$$

$$ {1 \over 2} mv_i^2 + mgh_i = {1 \over 2} mv_f^2 + mgh_f $$

We've moved the PE onto the left and put it together with the KE. Our WE Theorem becomes a conservation law — there is no change in the KE + PE of the object as it moves. This means that the initial (i) and final (f) value of the terms together are equal.

Consider a ball thrown upward. It starts with an upward velocity at a height which we will take to be 0. When it gets to the top, its velocity is 0. This gives the relation:

$$ {1 \over 2} mv_0^2 + 0 = 0 + mgh $$

or just

$$ {1 \over 2} mv_0^2 = mgh $$

where we have now simplified our notation to have $v_0$ be the initial velocity and $h$ to be the height to which it rises. This makes it very easy to see how high an object goes if you throw it up with a certain velocity. Or, if you know how high it went, you can figure out how fast it was going when it started upward.

The picture expressed by our equation helps us make sense of what's happening: the ball starts upward with a certain KE. As it rises, the PE grows so the KE must drop. When the KE has fallen to 0, the speed goes to 0 and the object is at its turn around point. As is begins to descend, the PE falls so the KE can grow.

This approach can be very useful in describing any free-fall situation — even in 2D or 3D.

## Where do you start?

One of the issues that is often confusing for students is the question of where do you start? Our PE came from work, which was referring to a change in position. We discovered that it was the height part of the position that mattered. We then defined a full-fledged potential energy, not just a change — $mgh$. But where do you start the $h$? Where is $h$ equal to 0?

The answer is that it doesn't matter. Since it is always only changes we will be calculating with, you can start anywhere. If we are bouncing an object on the floor, it might be convenient to choose the floor as 0. But if we have a table, we might want to choose the table top as 0. If we then drop an object, the $h$ goes negative! THIS IS NOT A PROBLEM. The *h* is a coordinate of a position. It can be positive or negative. It is only the **changes** in it that matter.

Joe Redish 11/2/11

Last Modified: February 20, 2019