Gravitational potential energy

[Chenkel]

This is exactly the same logic as used by many geocentric natural philosophers and clergy. Of course its intuitive power would seem less to anyone not on earth. The more descriptive and less neutral term is prejudiced, not intuitive.

mgh is more local to earth, than say mars, so I can see your point about geocentric natural philosophy. I would say intuition of approach depends on the problem at hand, for objects near earth, mgh, at least to me, seems to be far more intuitive. I don't think I'm biased in which formula I use, but I think elegance comes with simplicity and common sense, and knowing which formula to use for the problem at hand. I think it's hard to say a method is prejudice without knowing what the specific physics problem is at hand. I'm not sure what you mean by less neutral, perhaps you mean less universal? And finally I wonder what you mean by more descriptive, what quality makes something more descriptive?

 

[kuruman]

$\dots$ it's the amount of work you would have to do to move the mass from infinity far away in the sky to the surface of earth.

The definition of potential energy is much more general than that and in any case it is not related to the work "you have to do." A correct definition of the potential energy of a system consisting of the Earth and a mass is the negative of the work done by the force of gravity on the mass as the mass moves from a reference point to another point. Note that
1. It is the conservative force of gravity that does the work, not "you" and infinity as a reference point is too restrictive.
2. It is not the mass that "has" potential energy. It is the configuration, i.e. the relative position, of the Earth-mass system that is associated with potential energy. A change in the relative position results in a change of potential energy. If you exclude the Earth from the system, the mass can only have kinetic energy. If gravity does work on it, the result can only be a change in this system's kinetic energy, nothing else.
3. The choice of the zero of potential energy is arbitrary. One can choose it anywhere one pleases, usually where it's most convenient. For masses near the surface of the Earth, it is convenient to use the surface of the Earth as zero. In fact all zeroes of energy are arbitrary and that includes kinetic energy. We say that an object at rest on the surface of the Earth has zero kinetic energy because we chose tit to be so. If we choose the center of the Earth as the zero, the same object will have quite a bit of kinetic energy that will vary with latitude.

 

[hutchphd]

Descriptive referred to the word "prejudiced". It better describes the situation than "intuitive" I think. Intuitive implies a universality of thought whereas prejudice implies a more parochial mindset. Not really important.

 

[Delta2]

2. It is not the mass that "has" potential energy. It is the configuration,

According to another interpretation it is the gravitational field that has the energy. See that thread of mine with the self potential energy of a sphere, posts by Orodruin.

 

[kuruman]

According to another interpretation it is the gravitational field that has the energy. See that thread of mine with the self potential energy of a sphere, posts by Orodruin.

Yes, indeed. However one must attempt to match the level of the answer to the level of the person who asked the question.

 

[Chenkel]

The definition of potential energy is much more general than that and in any case it is not related to the work "you have to do." A correct definition of the potential energy of a system consisting of the Earth and a mass is the negative of the work done by the force of gravity on the mass as the mass moves from a reference point to another point. Note that
1. It is the conservative force of gravity that does the work, not "you" and infinity as a reference point is too restrictive.
2. It is not the mass that "has" potential energy. It is the configuration, i.e. the relative position, of the Earth-mass system that is associated with potential energy. A change in the relative position results in a change of potential energy. If you exclude the Earth from the system, the mass can only have kinetic energy. If gravity does work on it, the result can only be a change in this system's kinetic energy, nothing else.
3. The choice of the zero of potential energy is arbitrary. One can choose it anywhere one pleases, usually where it's most convenient. For masses near the surface of the Earth, it is convenient to use the surface of the Earth as zero. In fact all zeroes of energy are arbitrary and that includes kinetic energy. We say that an object at rest on the surface of the Earth has zero kinetic energy because we chose tit to be so. If we choose the center of the Earth as the zero, the same object will have quite a bit of kinetic energy that will vary with latitude.

So just trying to understand what you're saying with a practical example, if I have a mass m positioned h units above the surface of the earth, and it is falling to the ground, without any upward velocity, then the work done by gravity is mgh because the force is in the direction of displacement, furthermore the potential energy of the system (not the mass) is negative mgh (based on your logic), so the mass doesn't have potential energy, but the system does, and it's negative? I fail to see how a mass does not possesses potential energy, is the word 'configuration' interchangeable with the word 'system' in your example?

 

[jbriggs444]

furthermore the potential energy of the system (not the mass) is negative mgh

For an object of mass $m$ near the Earth's surface where gravitational acceleration has magnitude $g$ falling through a height difference of $h$, the potential energy of the Earth-mass system has decreased by $mgh$ indeed.

That is not the same as saying that the potential energy of the system was $-mgh$ either before or after the fall.

 

[hutchphd]

I fail to see how a mass does not possesses potential energy, is the word 'configuration' interchangeable with the word 'system' in your example?

Because the mass absent the rest of the system (ie the earth} has no gravitational potential energy. It is a system property, and depends upon system variables

 

[Delta2]

What happens in high school books, is that we put a reference point or reference height for which we say it is the zero potential energy height and we then say "The potential energy of the mass is mgh" where h=H-H' where H the absolute height of the mass (relative to Earth surface) and H' the reference height (relative to Earth surface again).

Where we should actually say "The potential energy difference of the mass-earth system relative to the reference height is mgh"

 

[kuruman]

Practical example. Please study it.
Mass $m$ is released from rest and falls distance $h$
Case I Choose the two component Earth plus mass as the isolated system.
(a) Choose the zero of potential energy at the initial position of the mass
Initially, the kinetic energy is $K_i=0$ and the potential energy is $U_i=0$.
The initial mechanical energy is the sum of the two $ME_i=K_i+U_i=0$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$ and the potential energy is $U_f=-mgh$.
The final mechanical energy is the sum of the two $ME_f=\frac{1}{2}mv^2-mgh$.
According to energy conservation the final mechanical energy is equal to the initial
$\frac{1}{2}mv^2-mgh=0.$

(b) Choose the zero of potential energy at the final position of the mass.
Initially, the kinetic energy is $K_i=0$ and the potential energy is $U_i=mgh$.
The initial mechanical energy is the sum of the two $ME_i=K_i+U_i=mgh$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$ and the potential energy is $U_f=0$.
The final mechanical energy is the sum of the two $ME_f=\frac{1}{2}mv^2$.
According to energy conservation the final mechanical energy is equal to the initial
$\frac{1}{2}mv^2=mgh.$

You get the same result regardless of where you choose the zero of the potential energy.

Case II Choose just mass $m$ as the isolated system.
(a) There is no potential energy of the mass
Initially, the kinetic energy is $K_i=0$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$
The force of gravity does work $W_g=mgh$ on the mass.
The work energy theorem says that the work done on the system by all external forces is equal to the change in kinetic energy
$mgh=\frac{1}{2}mv^2-0.$

Still the same equation.

 

[Chenkel]

Practical example. Please study it.
Mass $m$ is released from rest and falls distance $h$
Case I Choose the two component Earth plus mass as the isolated system.
(a) Choose the zero of potential energy at the initial position of the mass
Initially, the kinetic energy is $K_i=0$ and the potential energy is $U_i=0$.
The initial mechanical energy is the sum of the two $ME_i=K_i+U_i=0$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$ and the potential energy is $U_f=-mgh$.
The final mechanical energy is the sum of the two $ME_f=\frac{1}{2}mv^2-mgh$.
According to energy conservation the final mechanical energy is equal to the initial
$\frac{1}{2}mv^2-mgh=0.$

(b) Choose the zero of potential energy at the final position of the mass.
Initially, the kinetic energy is $K_i=0$ and the potential energy is $U_i=mgh$.
The initial mechanical energy is the sum of the two $ME_i=K_i+U_i=mgh$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$ and the potential energy is $U_f=0$.
The final mechanical energy is the sum of the two $ME_f=\frac{1}{2}mv^2$.
According to energy conservation the final mechanical energy is equal to the initial
$\frac{1}{2}mv^2=mgh.$

You get the same result regardless of where you choose the zero of the potential energy.

Case II Choose just mass $m$ as the isolated system.
(a) There is no potential energy of the mass
Initially, the kinetic energy is $K_i=0$.
Finally the kinetic energy is $K_f=\frac{1}{2}mv^2$
The force of gravity does work $W_g=mgh$ on the mass.
The work energy theorem says that the work done on the system by all external forces is equal to the change in kinetic energy
$mgh=\frac{1}{2}mv^2-0.$

Still the same equation.

Makes sense to me, thank you.

 

[Chenkel]

What happens in high school books, is that we put a reference point or reference height for which we say it is the zero potential energy height and we then say "The potential energy of the mass is mgh" where h=H-H' where H the absolute height of the mass (relative to Earth surface) and H' the reference height (relative to Earth surface again).

Where we should actually say "The potential energy difference of the mass-earth system relative to the reference height is mgh"

Is there something inherently wrong with saying the mass has potential energy?

Because the mass absent the rest of the system (ie the earth} has no gravitational potential energy. It is a system property, and depends upon system variables

So the mass absent the Earth has no potential energy, does that mean in the presence of Earth it does have potential energy?

Also I can imagine as long as the system is closed and we know all forces are conservative we might be able to write an equation stating the mechanical energy in terms of kinetic energy and potential energy, without saying what mass 'has' potential energy, and just refer to the potential energy of the system and think of potential energy as a 'system property.' So I suppose I can make a little sense out of this approach.

 

[Delta2]

Is there something inherently wrong with saying the mass has potential energy?

No, but I think it is more accurate to say that the system mass-earth has potential energy.

 

[Delta2]

There is also the potential energy that a mass has due to its own gravitational field(given that it is not a point particle), you might find the topic confusing but let me locate a relative thread
https://www.physicsforums.com/threads/potential-energy-of-a-sphere-in-the-field-of-itself.1016348/

 

[kuruman]

So the mass absent the Earth has no potential energy, does that mean in the presence of Earth it does have potential energy?

It's more subtle than that. It's not the presence or absence of the Earth that affect the potential energy of the Earth. Look at cases I and II in post #40.

In case I the system which is Earth + mass has potential energy and can be considered isolated, i.e. no external forces act on it. One system component (the mass) gains kinetic energy at the expense of the potential energy of the Earth-mass system.

In case II the Earth is still there but the system now is only the mass. We have a one-component system for which potential energy is meaningless because it takes two to tango. However, this one component system is not isolated because the Earth, by our choice of system, is now exerting an external force on the system that does work on the system and increases its kinetic energy.

Let's say that with equations
Case I
Mechanical energy conservation says that the change in potential energy plus the change in kinetic energy is zero of the two component system is zero: $$\begin{align}\Delta U+\Delta K=0.\end{align}$$ Case II
The work-energy theorem says that the change in kinetic energy is the work done by gravity:$$\begin{align}\Delta K=W_g~.\end{align}$$The definition of gravitational potential energy says that the change in potential energy is equal to the negative of the work done by gravity:$$\begin{align}\Delta U_g=-W_g~.\end{align}$$If you add equations (2) and (3), you get equation (1). Two different systems, two different approaches, same result for the mass falling by height $h$.

 

[Chenkel]

It's more subtle than that. It's not the presence or absence of the Earth that affect the potential energy of the Earth. Look at cases I and II in post #40.

In case I the system which is Earth + mass has potential energy and can be considered isolated, i.e. no external forces act on it. One system component (the mass) gains kinetic energy at the expense of the potential energy of the Earth-mass system.

In case II the Earth is still there but the system now is only the mass. We have a one-component system for which potential energy is meaningless because it takes two to tango. However, this one component system is not isolated because the Earth, by our choice of system, is now exerting an external force on the system that does work on the system and increases its kinetic energy.

Let's say that with equations
Case I
Mechanical energy conservation says that the change in potential energy plus the change in kinetic energy is zero of the two component system is zero: $$\begin{align}\Delta U+\Delta K=0.\end{align}$$ Case II
The work-energy theorem says that the change in kinetic energy is the work done by gravity:$$\begin{align}\Delta K=W_g~.\end{align}$$The definition of gravitational potential energy says that the change in potential energy is equal to the negative of the work done by gravity:$$\begin{align}\Delta U_g=-W_g~.\end{align}$$If you add equations (2) and (3), you get equation (1). Two different systems, two different approaches, same result for the mass falling by height $h$.

So I think what you are saying is that the choice of system can effect if the object has potential energy or not, even in the presence of another object outside the system.

I think based on what I read there is a time and place to say that a mass has potential energy. I believe that there might even be certain situations where it is necessary to refer to objects as having potential energy, for example a system of three objects, earth, and two more objects A and B in a gravitational field, the potential energy of object A is separate from the potential energy of object B, the system here has potential energy but it's the sum of the potential energy of A and the potential energy of B.

 

[hutchphd]

I can't even remember which Newton equation is which (to my students horror). I think this requirement for labels is not that useful. But even Feynman


admitted the one time they are of use is when communicating to another human.