# Question about the calculation of potential energy

While taking my physics classes, I used to think that the formula mgh calculates the gravitational potential energy of a single object h distance off the ground.

Recently, I learned however that it is unreasonable to talk about the gravitational potential energy of a single object. So mgh actually calculates the gravitational potential energy of the system made up of the earth and ball.

This however gave me some confusion about the derivation of mgh.

So:

the change in the potential energy = - (dot product of force and displacement)

I can consider lifting a ball off the ground to a height h.

While I lift the ball, the earth exerts a gravitational force of mg downwards and the ball has a displacement of h upwards.

The dot product of force and displacement is -mgh.

So consequently the change in the potential energy of the earth and ball is -(-mgh) or mgh.

In the previous example, we considered the case in which the earth exerts a force on the ball.

We should be able to get the same answer if we consider the case in which the ball exerts a force on the earth.

So:

the change in the potential energy = - (dot product of force and displacement)

I can consider lifting a ball off the ground to a height h.

While I lift the ball, the ball exerts a gravitational force of mg upwards and the earth has a displacement of 0.

The dot product of force and displacement is 0.

So consequently the change in the potential energy of the earth and ball is 0.

I'm not sure what I did wrong here.

When we consider displacement, are we supposed to consider the relative displacement of one object to another?

Any help would be appreciated.

• Could you please tell me then what is the displacement of the earth? Also is the displacement we are considering a relative displacement between the earth and soccer ball? – Pranav Jain Aug 10 at 21:35

Recently, I learned however that it is unreasonable to talk about the gravitational potential energy of a single object. So mgh actually calculates the gravitational potential energy of the system made up of the earth and ball.

That is absolutely correct. Potential energy (of any kind) is a property of a system and not a single object, because that energy depends on the position of an object relative to something else.

So consequently the change in the potential energy of the earth and ball is 0.

The conclusion is incorrect. Without trying to address each step you took to arrive at this conclusion, let me just summarize what is happening. Perhaps that will help you see why you reached the wrong conclusion.

When you lift an object from the ground initially at rest to a height $$h$$ at rest above the ground, you do positive work on the object equal to $$mgh$$. At the same time, however, gravity does an equal amount of negative work equal to $$-mgh$$ because its force is opposite the direction of the displacement, so that the net work done on the object is zero. All this means, according to the work energy principle, is the change in kinetic energy between the ground at rest to the height $$h$$ at rest is zero, which it obviously correct if it started and ended at rest. So where did the energy you put into the object raising it go?

The answer is gravity, by doing an equal amount of negative work, took the energy you transferred to the ball and stored it as gravitational potential energy of the ball-earth system, as you initially learned.

You do make a very intuitive argument about why the formula mgh is correct. However, could you please point out what I did incorrectly in my derivation.

OK. So let's look at your analysis step by step:

So:

the change in the potential energy = - (dot product of force and displacement)

Yes. Because the change in potential energy is due to gravity. The dot product of the force of gravity and the displacement is negative. The negative of that negative that is positive.

While I lift the ball, the ball exerts a gravitational force of mg upwards and the earth has a displacement of 0. The dot product of force and displacement is 0.

Technically, the displacement of the earth is not zero, but it is close enough to zero to say this is true with respect to the dot product of the displacement of the earth and the force of gravity being zero. In any case, that is only the change in potential energy of the earth's portion of the ball-earth system.

So consequently the change in the potential energy of the earth and ball is 0.

This is where you are going wrong. Just because the change potential energy of the earth is zero, doesn't mean the change in potential energy of the ball-earth system is zero. That change equals $$mgh$$ where $$m$$ is the mass of the ball.

Hope this helps.

• Thanks for your reply. You do make a very intuitive argument about why the formula mgh is correct. However, could you please point out what I did incorrectly in my derivation. Thanks again. – Pranav Jain Aug 10 at 21:36
• @PranavJain Since my answer didn't lead to you knowing why your derivation was incorrect, I will need more time to go through your derivation. Please be patient. – Bob D Aug 10 at 21:39
• No problem, thank you for your time. – Pranav Jain Aug 10 at 21:40
• @PranavJain I have updated my answer. Hope it helps. – Bob D Aug 10 at 22:41
• Thank you very much. This is a very clear explanation. So you are essentially stating: change in PE of system = change in PE of earth (with respect to ball) + change in PE of ball (with respect to earth). Because the change in PE of earth (with respect to ball) is essentially 0, we only have to consider change in PE of ball (with respect to earth). If this is correct, please let me know and thank you again! – Pranav Jain Aug 10 at 22:46