# Proving that gravitational potential is work done by the object against gravity while KE increases and PE reduces

There's a person in my class who thinks that the formula for gravitational potential (-GM/r) represents the work done by gravity to move an object from infinity to any point in the gravitational field.

I know it's actually the work done by the object as it goes towards the earth. That's consistent with the formula W=Fd (gravity would be doing positive work, hence negative work by the object itself) and the fact that the object would be gaining kinetic energy as it moves towards the earth, so it's doing negative work (since work represents the transfer of energy).

None of these explanations seems to fly for the dude, and one problem he has is this:

If the object is gaining kinetic energy as it moves towards the earth, it is also losing gravitational potential energy (since potential reduces as the object goes towards the earth). So, the net change in energy wold be 0. Hence, the formula -GM/r can only represent the work done by gravity, not the object itself. Otherwise, gravity isn't doing any work in the first place. Ergo, gravity is doing negative work as it is attracting an object from infinity to any point r.

Could you please help me clear up this confusion he has?

## 3 Answers

So you are both wrong, but the other person is off by a sign while you seem to have some larger misconceptions.

I know it's actually the work done by the object as it goes towards the earth. That's consistent with the formula W=Fd (gravity would be doing positive work, hence negative work by the object itself)

This doesn't make sense. As the object moves in the gravitational field, it isn't doing any work at all. You can only talk about the work done by forces, and the only force present here is gravity.

The confusion is understandable. We usually talk fast and loose like this. For example, if I push on a box, you would probably here discussion about "the work I do on the box". But really the more precise language is to talk about the work the force I apply does.

Another misconception you seem to have is that there needs to be no net work done. This is only the case if the ball is moving at a constant speed$$^*$$. But since it is "falling" in this field, this is not the case.

Really, the potential energy is the negative work done by the conservative force. Technically the more general statement is $$W=-\Delta U$$ but since you are asking about when we start at infinity where $$U=0$$ in this case, we can say for this specific process $$W=-U$$

But this isn't needed at all to think about the sign of the work done by gravity. The force is always acting in the direction of the displacement of the ball. So the work done by gravity is always positive.

So in this case the net work isn't zero? Isn't work the transfer of energy? If that's the case, shouldn't net work be zero all the time (since energy has to come from somewhere)? Isn't the net energy change for the object falling into the gravitational field positive?

Saying work is the transfer of energy is kind of misleading. Really what you should be thinking is that the work done on an object changes its kinetic energy. You might be familiar with this: $$W_{net}=\Delta K$$ So when gravity does work on our object as it falls, it gains kinetic energy. We can stop right here and never even think about potential energy. If we do this, we are treating gravity as an external force. All we see is our object and a force acting on it. This force changes its kinetic energy.

However, there are special forces called conservative forces which we can associate a scalar potential energy with. This means (as I somewhat explained above) instead of directly determining how much work the conservative force does, we can just look at the change in this potential energy. If we go this route, we don't worry about the work done by gravity anymore. We instead look at the total mechanical energy ($$E=K+U$$) and we see that it does not change during this process. $$\Delta E=0$$ does not mean there is no net work being done. $$\Delta K=0$$ means there is no net work being done. What $$\Delta E=0$$ means is that (assuming we have taken into account all conservative forces) there are no other external, forces acting on our ball.

Long story short: Gravity (the only force acting on our object) does positive work on our ball, which increases its kinetic energy ($$W=\Delta K>0$$). If we decide to work with potential energy as well, we can say that $$\Delta E = \Delta K+\Delta U=0$$, or $$\Delta K = -\Delta U = W_{grav}$$ This is the usual statement of "conservation of energy" (without external/non-conservative forces). Notice how work can still be done by conservative forces and $$\Delta E$$ is still going to be $$0$$.

$$^*$$This might also be where your confusion lies. Typically you hear people say the potential energy is "the work done to move a mass from infinity to that point." But this statement doesn't say it's assumptions. What this is considering is if I were to also be applying a force to the ball equal to the gravitational force but in the opposite direction as it moves to the point of interest from infinity. Therefore the net work done on the ball is in fact $$0$$. Therefore I can say the work my force does is negative the work done by gravity. i.e. $$W_{me}=-W_{grav}=-(-\Delta U)=\Delta U$$ Add this is the work that particular statement is referring to. The work my force does in this specific scenario.

• So in this case the net work isn't zero? Isn't work the transfer of energy? If that's the case, shouldn't net work be zero all the time (since energy has to come from somewhere)? Isn't the net energy change for the object falling into the gravitational field positive? – Main Man Andy Feb 8 at 13:33
• @MainManAndy The only force acting on the ball is gravity, and it is doing positive work. When I have time I can add more detail to the answer. But essentially you are getting confused trying to consider the work done by gravity and potential energy at the same time, when really they are two sides of the same thing depending on what you consider to be part of your system. – Aaron Stevens Feb 8 at 13:50
• 🤔I see. Eager to see your extra details if you add 'em. – Main Man Andy Feb 8 at 14:01
• @MainManAndy I have added more to address your concerns about what net work really means in terms of the energies we are talking about (which was mentioned in your main question as well). Please let me know if something still doesn't make sense. This is something I had to struggle through as well when learning all of this. These are great questions, and wrestling through them is the right path to take to understanding these things at a deeper level. – Aaron Stevens Feb 8 at 15:56

My answer is in two parts.
The first part tries to explain in terms of energy and work done what is going on when a mass falls towards the Earth and the second part is a commentary on the statements made by the OP in the question.

There are two things that you should be clear about.

In such a discussion you must define the system under consideration.
In this case is it the object or the object and the Earth?
This is important because you need to be able to identify internal (to the system and coming in Newton third law pairs) forces and external forces.

There is a world of difference between the gravitational potential at a point and the potential energy of a system of objects.
The gravitational potential at a point is the work done by an external force in taking unit mass from a position of zero potential to the point.
The gravitational potential energy of a system of objects is the work done by external forces in taking the objects from old positions where the potential energy is zero to their new positions.

Consider an object of mass $$m$$ as the system and the object is situated in the Earth's gravitation field which for the present assume to be uniform and have a strength $$g$$.

There is only one external force on the mass which is the gravitational attraction of the Earth of magnitude $$mg$$ and directed downwards.

If the mass starts from rest and falls a distance $$h$$ to reach the surface of the Earth then the work done by the gravitational field (external force) on the mass is $$+mgh$$.
It is a positive quantity because the external force and the direction of travel of the mass are both in the same direction.
The work energy theorem tells you that this external work done on the mass results in a change (increase) in the kinetic energy of the mass.
Note that potential energy and potential have not been mentioned.

If the gravitational field is not constant then the initial magnitude of the force on the mass $$m$$ is $$\dfrac{GMm}{(R+h)^2}$$ where $$M$$ is the mass of the Earth and $$R$$ its radius.
The final magnitude of the force is $$\dfrac{GMm}{R^2}$$ so to evaluated the work done one must do an integration.

The work done by external force on the mass is $$GMm\left[ \dfrac 1 R -\dfrac {1}{R+h} \right] = \dfrac {GMm}{R}\left[ 1 - \left( 1+\frac hR \right )^{-1} \right] \approx m\, g\, h$$ if $$R\gg h$$ and the gravitational field strength $$g =\dfrac{GM}{R^2}$$.

Now this could have been done by using the idea that the mass $$m$$ finds itself in a gravitational field due to the Earth and at a distance $$r$$ from the centre of the Earth the potential is $$- \dfrac{GMm}{r}$$ having taken the zero of potential to be when $$r$$ is infinity.

The potential energy of the mass $$m$$ changes from $$- \dfrac{GMm}{R+h}$$ to $$- \dfrac{GMm}{r}$$and so the change in potential energy of the mass is $$GMm\left[ \dfrac 1 R -\dfrac {1}{R+h} \right]$$ the same value as the work done by the "external" force acting on the mass $$m$$.

However we now have a system of two masses, the Earth and the mass, and the gravitational forces of attraction (there are two - force on mass due to Earth and force on Earth due to mass) are internal forces but because $$M\gg m$$ the Earth does not move only the work done by the force on the mass due to the Earth is considered.
It is the mass and the Earth system that have gravitational potential energy.

The statement "that work done by the internal gravitational force on the mass due to the Earth" can be put in another way - "the mass (and Earth) system lose gravitational potential energy"

Note that I have added words and symbols to some of the statements as indicated by [square brackets].

If the object is gaining kinetic energy as it moves towards the earth, it is also losing gravitational potential [energy] (since potential [due to the Earth] reduces as the object goes towards the earth).

This statement is correct.

So, the net change in energy would be 0.

This statement is correct if by energy it is meant the sum of the kinetic energy and the gravitational potential energy.

Hence, the formula -GM[m]/r can only represent the work done by gravity, not the object itself.

This statement is "correct" if by gravity it is meant "the force on the mass due to the gravitational attraction of the Earth" as the gravitational field of the mass cannot attract the mass itself but the negative sign should not be there as the gravitational attractive force on the mass due to the Earth is in the same direction as the movement of the mass.

Otherwise, gravity isn't doing any work in the first place. Ergo, gravity is doing negative work as it is attracting an object from infinity to any point r.

This statement is not correct as the gravitational force on the mass due to the Earth is downwards and the mass is moving downwards so the work done by this gravitational force must be positive.

One problem with what is written is the interpretation of the word gravity.
Is it a force or a field?

There's a person in my class who thinks that the formula for gravitational potential (-GM/r) represents the work done by gravity to move an object from infinity to any point in the gravitational field.

-GM/r is the gravitational potential at a distance $$r$$ from the centre of the Earth and is the work done by an external force in moving unit mass from infinity (zero of potential) to a distance $$r$$ from the centre of the Earth.
The work done by the gravitational attraction on the unit mass due to the Earth (gravity?) is positive as the force and the movement of the mass are in the same direction.

I know it's actually the work done by the object as it goes towards the earth. That's consistent with the formula W=Fd (gravity would be doing positive work, hence negative work by the object itself) and the fact that the object would be gaining kinetic energy as it moves towards the earth, so it's doing negative work (since work represents the transfer of energy).

I found this statement very difficult to unravel.

I know it's actually the work done by the object as it goes towards the earth.

It would have been better to add a word to make it . . . . actually the [negative] work done by the object . . . .

That's consistent with the formula W=Fd (gravity would be doing positive work, hence negative work by the object itself)

This concept of negative work done by an object is not really needed in this case.

and the fact that the object would be gaining kinetic energy as it moves towards the earth, so it's doing negative work (since work represents the transfer of energy).

Here this idea of negative work done by an object is continued to explain the increase in the kinetic energy of the object.

You can see the gravitational potential as the work done by the field to move a unit point mass from infinity to the point r. The energy inside the gravitational field (because a conservative field) is conserved so that E=U+K is a constant:

$$E=-GMm/r+ \frac{1}{2}mv^2$$

where U(r) is potential gravitational energy: $$U(r) =\frac{-GMm}{r}$$

for $$r \to \infty$$, $$U(r) \to 0$$ so that as $$m$$ approaches $$M$$, its kinetic energy increases while its potential energy decreases