# Talk:Potential energy

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This article can use some cleanups (which I might just do sometime soon). Potential energy is not 'stored into the body'. That is just the wrong idea! —Preceding unsigned comment added by Angadude (talkcontribs) 09:26, 10 May 2010 (UTC)

I assume electrical potential energy refers mainly to the energy gained by placing charges closer or farther apart than they would naturally be inclined to go? (You could probably state this more generally in terms of fields.) Does it also refer to electromagnetic radiation, e.g. light, or would that be "actual energy" instead of "potential energy"? --Ryguasu 00:19 Nov 21, 2002 (UTC)

## What the?

"When the book is raised from the floor to the table, the gravitational force does negative work." Bullshit. The gravitational force does not lift the book. —Preceding unsigned comment added by 124.183.115.231 (talk) 10:10, 6 October 2007 (UTC)

You forget that the work done on the book is simply the applied force multiplied by the distance the object has travelled. In this way of defining work, the gravitational force does indeed do work on the book, it just does negative work, i.e. opposes the motion, which is why you have to do extra work to make up the deficit, i.e. lift harder against gravity. —Preceding unsigned comment added by 134.226.1.229 (talk) 17:20, 22 February 2008 (UTC)

## Gravitational nitpicking

I just added a paragraph pointing out that gravitational potential energy is also imparted to the Earth.

I also commented out this:

[We may also want to link to an explanation of that second term (the gravitational forces created by hollow spherical shells)]

PhilHibbs | talk 18:23, 6 September 2005 (UTC)

## Electrical potential energy = electrostatic potential energy = coulomb energy?

Are these terms equivalent? If so, they should be redirected and mentioned in the article. -- Kjkolb 09:24, 25 November 2005 (UTC)

## Magnetic potential energy

Should magnetic potential energy be included here? -- Kjkolb 09:57, 27 November 2005 (UTC)

I believe that is covered by electrical potential energy, since electricity and magnetism are two aspects of the same fundamental force. Perhaps the section on electrical potential energy should be modified to say electromagnetic potential energy. Wbrameld 23:22, 25 January 2006 (UTC)

## Does Mass-Energy Equivalence Apply?

Does potential energy have mass? For example, take any two stars in our galaxy, let's say Polaris and Rigel, to pick two at random. The gravitational attraction and physical distance between Polaris and Rigel embody an enormous amount of potential energy. If you solve E=mc^2 for mass:
m = E/c^2
and if you plug in the gravitational potential energy between Polaris and Rigel as E in that equation, you get a significant mass value. But does this mass make any real-world sense? Does the mass of that potential energy have its own gravitational field, and if so, what shape does it take? (And to get really way out there, has this been considered by people working on the problem of 'dark' matter?) Wbrameld 22:55, 25 January 2006 (UTC)

It does apply, but doesn't solve the dark matter problem (in fact, it was one of the original reasons for conjecturing that there was a large amount of unseen matter). Among other things, this is why the nuclei of atoms weigh less than their constituent protons and neutrons would in isolation (they're stuck way down at the bottom of a Strong-force potential well, and the mass difference is equal to the energy needed to pull them up out of that well). Cosmologically, the flatness problem relates to the difference between the gravitational potential energy of the universe and the mass of all objects within it. See that article for more information on the subject. --Christopher Thomas 21:26, 26 April 2006 (UTC)

On the same subject, does a stretched spring have more mass than an unstretched spring? (by virtue of its greater potential energy?). If so, it might be worth mentioning in the article, given that a) it is unintuitive and b)if this wasn't the case, then it would seem to break the law of the conservation of energy (at least to a layman). Tomandlu (talk) 12:59, 12 December 2007 (UTC)

Energy is related to mass, exhibits inertia, and exhibits gravitational force, so how could you tell? If the spring is released, its potential energy goes into kinetic energy, and then to thermal energy. Put it in a black box with perfectly insulated walls and the inertia of the box and its apparant mass do not change through the processes. If you let heat through the walls and allow the spring to cool to its starting temperature, energy is lost and the mass of the box decreases. Stretch the spring again using energy derived from an external source and the spring gets just a little more massive.Pvkeller (talk) 17:33, 11 November 2008 (UTC)
The confusion arises from thinking loosly about mass and trying to separate it from energy. Mass can be defined as total energy. See Invariant mass. When you talk about potential energy of one kind or another, you are choosing a special way to account for some of the mass/energy in the system, a choice that makes sense because that mass/energy is available in a particular way.Pvkeller (talk) 17:33, 11 November 2008 (UTC)
Take the case of a star. You can calulate its total thermal energy and the gravitational force exerted by that energy, but why have you separated that gravitational force from thermal energy from the rest of the gravitational force? The star has a mass that includes the contributions from all its forms of energy.Pvkeller (talk) 17:33, 11 November 2008 (UTC)

## Cleanup

This article has been edited in bits by many people, and has ended up as a random collection of facts and approaches. The overall organization needs to be redone, I think. -- SCZenz 20:38, 26 April 2006 (UTC)

## Gravitational potential integral

The removal of this math was reviewed at Wikipedia talk:WikiProject Physics the first time around. User:Enormousdude thought it was in error, and removed it. User:SCZenz thought it was in error, and agreed with the removal. I doublechecked it, and concluded that it was, indeed, in error.

It turns out we were wrong.

After spending an hour or more trying to figure out what the heck they did, it turns out it's valid, but (almost, but not quite) the most confusing possible way of working that math. As a result, I'm leaving it out of the article. By all means replace it with something else that makes the same point less confusingly. The reasoning, as near as I can tell, is below. I'm including an alternate way of working the problem that's much easier to follow. --Christopher Thomas 06:54, 27 April 2006 (UTC)

Yes, I see now. I'm impressed you went through the effort of clarifying the situation. I don't think any information of significance is gained by the more complex approach, but certainly the easier way should be added to the article. -- SCZenz 07:08, 27 April 2006 (UTC)

### Method originally presented

Original equation:

${\displaystyle U_{g}=\int _{h_{0}}^{h}{GmM \over r^{2}}dr+\int _{0}^{h_{0}}{GmM \over h_{0}^{2}}{r \over h_{0}}dr}$

What's happening is that they're trying to integrate force (across distance) on a test particle (with mass ${\displaystyle m}$) moved from the centre of a uniform sphere (with mass ${\displaystyle M}$) outwards. They move the test particle from radius 0 to radius ${\displaystyle h_{0}}$ (the sphere's radius), integrating force applied by the mass within radius ${\displaystyle r}$, then as a separate integral integrated from radius ${\displaystyle h_{0}}$ out to the final radius ${\displaystyle h}$.

The first term of the equation above represents the integration outside the sphere. I don't think anyone disputed that.

The second term represents integration inside the sphere. This starts off expressed as:

${\displaystyle +\int _{0}^{h_{0}}{\left({\frac {Gm}{r^{2}}}\right)\left({\frac {r}{h_{0}}}\right)^{3}M}dr}$

Cancelling gives:

${\displaystyle +\int _{0}^{h_{0}}{{\frac {GmMr}{h_{0}^{3}}}dr}}$

...And then they grouped the terms within this integral strangely.

Integrating gives:

${\displaystyle U_{g}=GmM\left[-\left({\frac {1}{r}}\right)\vert _{h_{0}}^{h}\right]+\left({\frac {GmM}{h_{0}^{3}}}\right)\left[\left({\frac {1}{2}}\right)r^{2}\vert _{0}^{h_{0}}\right]}$

${\displaystyle U_{g}=GmM\left[-{\frac {1}{h}}+{\frac {1}{h_{0}}}\right]+\left({\frac {GmM}{h_{0}^{3}}}\right)\left[\left({\frac {1}{2}}\right)h_{0}^{2}-0\right]}$

${\displaystyle U_{g}=GmM\left[{\frac {1}{h_{0}}}-{\frac {1}{h}}\right]+GmM\left[\left({\frac {1}{2}}\right){\frac {1}{h_{0}}}\right]}$

${\displaystyle U_{g}=GmM\left[\left({\frac {3}{2}}\right){\frac {1}{h_{0}}}-{\frac {1}{h}}\right]}$

...So the equations as-given were correct, just incredibly non-intuitive.

Setting the final distance ${\displaystyle h}$ as infinity makes ${\displaystyle {\frac {1}{h}}}$ vanish, giving:

${\displaystyle U_{g}(\infty )=GmM\left({\frac {3}{2}}\right){\frac {1}{h_{0}}}}$

Declaring ${\displaystyle U_{g}(\infty )}$ to be zero gives:

${\displaystyle U_{g}(h)=GmM\left[\left({\frac {3}{2}}\right){\frac {1}{h_{0}}}-{\frac {1}{h}}\right]-GmM\left({\frac {3}{2}}\right){\frac {1}{h_{0}}}}$

${\displaystyle U_{g}(h)=-GmM\left({\frac {1}{h}}\right)}$

As mentioned above, this is pretty close to being the most complicated way of deriving the gravitational potential energy. --Christopher Thomas 06:54, 27 April 2006 (UTC)

### A simpler way

If you want to use a "work to take a particle from point A to point B" approach to get gravitational potential energy, and you know the particle will always be outside the sphere, just integrate outside the sphere, and consider it a point mass with mass ${\displaystyle M}$. Here, we're taking a test mass with mass ${\displaystyle m}$ from radius ${\displaystyle r_{1}}$ to radius ${\displaystyle r_{2}}$, as measured from the centre of the sphere:

${\displaystyle U_{g}(r_{2})-U_{g}(r_{1})=\int _{r_{1}}^{r_{2}}{\frac {GmM}{r^{2}}}dr}$

${\displaystyle U_{g}(r_{2})-U_{g}(r_{1})=GmM\left[-\left({\frac {1}{r}}\right)\vert _{r_{1}}^{r_{2}}\right]}$

${\displaystyle U_{g}(r_{2})-U_{g}(r_{1})=GmM\left({\frac {1}{r_{1}}}-{\frac {1}{r_{2}}}\right)}$

We want a common reference point for potential energy. We could set it to the radius of the sphere (making potential energy relative to the surface), but that isn't very good as a common reference point, because our spheres may vary in size and density. Setting it at infinity works, though, as making radius tend towards positive infinity doesn't cause our equations to blow up. We can therefore define gravitational potential energy as GPE relative to ${\displaystyle U_{g}(+\infty )}$:

${\displaystyle U_{g}(r)\equiv U_{g}(r)-U_{g}(+\infty )}$

${\displaystyle U_{g}(r)\equiv GmM\left({\frac {1}{+\infty }}-{\frac {1}{r}}\right)}$

${\displaystyle U_{g}(r)\equiv -{\frac {GmM}{r}}}$

...This gives the usual convention for defining gravitational potential energy using Newtonian gravitation. --Christopher Thomas 06:54, 27 April 2006 (UTC)

Couldn't you also make it simpler by just saying that the radius from the center of a sphere of mass ${\displaystyle M}$ of a particle with some mass ${\displaystyle m}$ is changing by the small amount ${\displaystyle dr}$? Then
${\displaystyle U_{g}=\int {\frac {GMm}{r^{2}}}dr}$
${\displaystyle U_{g}=GMm\int {\frac {1}{r^{2}}}dr}$
${\displaystyle U_{g}=-{\frac {GMm}{r}}}$
It might take away from the conceptual nature a little, but it makes the math more straightforward. Ginogrz (talk) 00:18, 7 July 2009 (UTC)

## Revision of gravitational potential energy section

This section was a horrible mess, so although I'm not an expert I've attempted a tidy-up on the basis that I could hardly make it any worse. In an ideal world someone might review it and make any necessary corrections. In particular, the original text contained the statement "Note that the potential energy of both objects is the same, so the potential energy of the whole system is 2×${\displaystyle U_{g}}$ (as defined above)." It seems to me that this statement is untrue, so I have rewritten that paragraph based on the asumption that ${\displaystyle U_{g}}$, as defined earlier, measures the total potential energy of the system.

Matt 00:19, 28 July 2006 (UTC).

## Rest Mass Energy

By the definition given

• Potential energy is the of energy that is by virtue of the relative positions (configurations) of the objects within a physical system

How is rest mass an example potential energy? Note that${\displaystyle \nabla E_{0}=0}$, It doesn't depend whatsoever on the position relative to other object. 131.215.7.198 06:09, 2 December 2006 (UTC)

Ahh.. Never mind I figure that out that consider A -> B + C and if we label state by number of A, B and C, then chemical potential(relativistically correct) will be non-relativistic+mc^2131.215.7.198 11:37, 2 December 2006 (UTC)

## The Kinetic Energy page needs help

Hello all, this page is great. I left a message on the kinetic talk page about how the kinetic format is not nearly what the format here is:

"The information here is commendable, as is the discussion on this page. However, it seems reasonable that since there are relationships between kinetic and potential energies, there should be a relationship between their two respective articles here at Wikipedia. The potential energy article lists, explains, and hypertext references all the types of potential energy; this article does not, which reduces its audience to only people who already fully understand the principles." —Preceding unsigned comment added by John van v (talkcontribs) 19:17, 10 September 2007 (UTC)

## energy good or bad life

its something about energy that make pepole well understand things about kinetic energy or potential energy just the world so great make so little like good but more of bad who would think one day there will be no clean water or trees or food but some do for me as an exaple you life wonts last forever here are stuff on earth is disapering just right under are noses thats how some pepole konw by factores pollting the air by useing good enegry and turning it into bad so what do think about enegry good or bad its life like this that can make a diffense to us all so just as a waring think about what you are youing every day

You don't know what you're talking about. —Preceding unsigned comment added by 66.65.67.197 (talk) 03:45, 19 June 2010 (UTC)

## Potential

There should be a more conclusive entry on potential (as opposed to potential energy). This is an important subject and the Wikipedia entry on "potential" is very limited. Anyone want to do it? 128.113.195.102 (talk) 00:27, 14 February 2008 (UTC)

## Formatting

The text on this page, beginning in the Overview section, is centered, rather than left-justified. I can't figure out why or how to fix it, but the appearance is kind of disconcerting. 66.142.96.200 (talk) —Preceding comment was added at 05:42, 22 February 2008 (UTC)

## Shared property

There should be some words in the article emphasizing that for instance gravitational potential energy is a shared property of both a body and e.g. the earth. It is not correct to say that PE=mgh is gravitational potential energy of a particular body. This since GPE's source is Newton's law of gravity which involves two masses. See Young & Freedman, University Physics (in the 9th ed, p. 195). Mårten Berglund (talk) 18:42, 22 August 2008 (UTC)

See remarks under [Does Mass-Energy Equivalence Apply?]. Potential energy is just a way of accounting for the possibility of doing work using the force that draws massive objects together. Total energy of two distant objects, as determined from their masses, is no greater than the sum of the energies of the two objects considered individually.Pvkeller (talk) 18:02, 11 November 2008 (UTC)

## Image needs replacement

Hello all...

You have your mission, take your camera and start clicking.--Jordan 1972 (talk) 17:07, 3 October 2008 (UTC)

## Woolly opening sentence

Potential energy can be thought of as energy stored within a physical system.

Rather than saying how it can be "thought of", can we say what it actually is? —Preceding unsigned comment added by 86.137.136.32 (talk) 02:49, 20 November 2008 (UTC)

"It is called potential energy because it has the potential to be converted into other forms of energy, such as kinetic energy, and to do work in the process"... I don't understand it... Kinetic energy also has the potential to be converted into other forms of energy such as heat energy... so what so speacial about potential energy?

thanks--89.139.9.247 (talk) 09:56, 3 March 2009 (UTC)

Kinetic energy is typically being spent. Potential energy isn't. You spend potential energy by converting it into other forms. You just spend the other forms directly. Chris Cunningham (not at work) - talk 10:35, 3 March 2009 (UTC)
More exact, kinetic energy is in motion. The name itself is more of a misnomer, since motion without resistance consumes no energy, kinetic energy is actually stored energy within an object until it comes into contact with another object. Potential energy is the result of kinetic energy meeting resistance, (eg: the result of work performed). In the example from the section below this one, the hand bending the spring must exert force on the spring until the time of release. Work is being performed, and energy must constantly be consumed to keep the spring bent against the constant "restoring force." Kinetic energy is simply the transmission of energy through space, via motion. Zaereth (talk) 00:03, 16 June 2009 (UTC)

Energy is energy. Energy cannot be "stored" or "spent". Energy exists in one form or another. It existed in the past and will continue to exist in the future. It is never created or destroyed. —Preceding unsigned comment added by 66.65.67.197 (talk) 03:51, 19 June 2010 (UTC)

Hold on a minute! Most of the different categories of energy can be stored, hence are called "potential" energy. Jet fuel being burned in a jet engine is "chemical" energy. Jet fuel stored in a tank at the airport could be called "potential" energy. A spinning flywheel slowing as it powers a piston through an intake stroke and a compression stroke is rotational "kinetic" energy. A flywheel freely spinning awaiting the engagement of a clutch in a (for example) punch-brake metal sheer is "potential" energy being stored until the steel plate is in position for the next cut. As soon as the clutch engages, the "kinetic" energy of the flywheel supplies "mechanical" energy to slam the cutting surfaces past each other, cutting the steel plate like a piece of paper.
One intriging aspect of compressed air is (if my information is correct) the fact that energy is lost from the compression of the air to its utilization, even though there are no leaks. Is this lost energy called "entropy"?
The three preceeding paragraphs were submitted by Jeff Jones ('jefojones') in the hope that someone can add needed information about characteristics of compressed air. — Preceding unsigned comment added by Jefojones (talkcontribs) 05:22, 18 October 2014 (UTC)
A spinning flywheel is kinetic energy, not potential. For the example of the punch, perhaps it would be easiest to use a river and water wheel as an example. The flowing of the river is synonymous with the spinning of the flywheel. When you lower your water wheel in the river it begins spinning. This can be compared to the engaging of a clutch. As the water wheel spins, it extracts gravitational kinetic-energy from the river to perform the work of milling your grain. Both the flywheel and the river"s energy are kinetic.
Compressed air is a form of elastic potential-energy. It is the elastic properties of the gas that allow the storage of energy, and those same elastic properties provide the restoring force. When people talk about "losing air," even though there were no leaks, this is due to friction. If you have 100 psi of pressure at the compressor, after plumbing it through 100 feet of 1/4" hose, you will probably have around 90 psi at your impact wrench. This is the result of something called "static friction." The air doesn't need to be moving to experience friction. The static friction limits its ability to generate the full pressure at the end of the line. (The same thing happens with electricity, or any other power source.) The energy that is lost is absorbed along the way by this friction, generating heat.
The losses due to friction are part of entropy, but, mostly, entropy can be thought of as the energy needed just to overcome the constant restoring-force. When you boil water, it turns to steam. A huge amount of energy is absorbed by the water just turning it into steam, just to overcome the cohesive forces of the water. Energy must continually be added just to keep the water in the form of steam, thus, the entropy constantly increases. Only after it turns to steam can the water be used to provide energy for a steam engine. The excess energy can be extracted from the steam, but all that energy needed to create the steam in the first place is lost when the steam turns back to water. In example, if I burn wet, green wood in my wood stove, most of the energy is absorbed by the water, turning it into steam. This energy is lost just trying to get the steam out of the chimney, and so the wet wood produces far more entropy losses than would dry wood.
If you need some formulas or info relating to compressed air in particular, I would suggest the Compressed Air and Gas Handbook by the Compressed Air and Gas Institute. It will contain nearly all of the information you could ever need. Zaereth (talk) 09:16, 18 October 2014 (UTC)
Colleagues, my understanding is that the term "potential" is used to define a specific kind of work done by forces in a system. It is work that is conservative, and is done by forces that can be derived from a scalar potential function. While the term potential can be applied to many things that have the potential to do work, it is my understanding that this article is intended to focus on those cases that are associated with a scalar potential function also called a potential energy. Prof McCarthy (talk) 23:27, 18 October 2014 (UTC)
The purpose of the article is to explain potential energy in a way that will be understandable to a general audience. No offense, but what you just said is extremely vague without some serious background knowledge of the particular jargon you are using. For me, and many others, I need to understand things in terms of practical use, so the purely mathematical explanations make little sense to me. Math is an extremely rudimentary language compared to English, which is far advanced, so it is often very difficult to translate exactly what the math means into English. Therefore, subjects like this can be very difficult for the non-mathematicians to understand. Zaereth (talk) 23:49, 18 October 2014 (UTC)
True enough, so the challenge is to write the article in form that says what Prof McCarthy said (which might be jargon, but is nevertheless correct) in an understandable form, at least in the lead. Certain fields (e.g. gravitation, electric field) give rise to potential energy, whereas stress, compression, chemical, kinetic etc. are generally not regarded as potential energy. With potential energy, there is a direct relationship between position and energy. —Quondum 23:57, 18 October 2014 (UTC)
Yes, exactly. As I understand it, the compression of air itself does not give rise to a potential, but it is the position of the air between two states. Same with a capacitor, or the bow and arrow. Zaereth (talk) 00:08, 19 October 2014 (UTC)
The definition can be a bit fuzzy, but the example of compressed air does not fit into the picture of potential energy at all well. You cannot say that for each unit of air, the amount of energy depends only on its position; it is not conservative, and it depends on all sorts of factors such as temperature. It is far better treated as the kinetic energy of its constituent particles. The energy stored in a pulled bow is also not a great example: the energy does not depend specifically on the position of something, but or the relative position of different parts of the bow. A capacitor also has similar problems. So none of these should be regarded as potential energy. The closest example is a spring attached to a fixed point, where the position of the other end correspond to a specific energy. —Quondum 00:44, 19 October 2014 (UTC)
And yet, compressed air is a spring attached to a fixed point. I can store that air for very long periods of time, and then release it to do very real work. The potential of which is measured in pressure. So perhaps I am confused as to how the energy can be kinetic. Zaereth (talk) 01:07, 19 October 2014 (UTC)
In the very artificial and constrained context where the compression is adiabatic, a pneumatic cylinder could be thought of as a spring in which there is elastic potential energy. Without essentially perfectly insulation (no heat exchange between the gas and the cylinder walls, for example), the restoring force is not conservative, though: compress the gas, which becomes hotter in adiabatic compression and then loses some of this heat to the cylinder walls, and the restoring force is then lower. So, storing the energy for a very long time is not conservative here. The kinetic energy is in the thermal motion of the atoms and molecules of the gas. —Quondum 06:11, 19 October 2014 (UTC)
While that makes perfect sense on an atomic scale, it provides no real value for a person designing an air system. Yes, the air loses heat, but eventually reaches an equilibrium with the surrounding environment. Yet the pressure remains unaltered, except to the most accurate of gauges. The losses from heat then become insignificant. Zaereth (talk) 09:01, 19 October 2014 (UTC)
In the study of gasses it is usually assumed that work is not conservative and therefore the forces are not derivable from a potential function. This is because heat is generated and absorbed in the expansion and compression of a gas separately from the work that is done. For this reason I believe it is a mistake trying to describe a compressed gas as a potential energy. Prof McCarthy (talk) 20:08, 19 October 2014 (UTC)
I understand what you are saying. But what confuses me is that there is no heat "generated" during compression. It is merely adiabatic heating that occurs. If the air is immediately released, expanding it to atmospheric pressure, the temperature returns to normal. The compression merely concentrates the heat already in the air. The heat is not generated by the wattage of the motor, but the energy from the motor is stored as pressure. When you use that energy (unlike a steam engine) temperature plays very little role in the amount of electricity you are able to generate with an air motor. Even if the air is allowed time to cool and the excess heat is lost (redistributed to the environment), when the air expands the adiabatic cooling simply means it will be colder than room temperature.
Perhaps my confusion comes from the fact that I can use a chemical charge to fire a bullet. I can use compressed air to fire a bullet. I can use a spring or even a capacitor to fire a bullet. These are all different methods of storing energy to perform the same work. Zaereth (talk) 01:44, 21 October 2014 (UTC)
Maybe this has to do with the term "potential energy"? The temptation is to interpret this as energy that has the potential to become other forms of energy, or, in a sense, stored energy. However, as with many technical terms, this is not the meaning of the term in this article; it has a more specific meaning. And with regard to adiabatic compression, if there is no heat exchange, it could be regarded as a spring, but when the heat is lost to the environment, it very definitely does not adhere to this more specific definition, since not only is the air cooler, the mechanical energy returned is less than that used to compress the gas in the first place, and so the force is not conservative. —Quondum 03:05, 21 October 2014 (UTC)
Potential energy is a particular type of work associated with specific type of force field. It is not a generic term for stored energy that can be released later.Prof McCarthy (talk) 03:22, 21 October 2014 (UTC)
Well, that goes against everything I've read on the subject. To demonstrate, here is a quote from the book Fundamentals of Materials for Energy and Environmental Sustainability: "Work must be performed on gas to compress it. As the air is compressed, the input work is converted into elastic (pressure) potential energy and heat." The fact that it takes more energy to produce it than you get back seems no different than any other form work. If I shoot an arrow straight into the sky, it's potential energy rises as its speed falls, until it reaches its apex, but that energy still will never equal the amount of energy used up by my arm. perhaps if you explained exacly what the special forces and types of work are, it would help to set everything straight. Zaereth (talk) 03:55, 21 October 2014 (UTC)
This quote seems to use the term potential energy in a more general way than what is presented in this article and what I think is the common definition. In all physics courses I have taken and taught potential energy has been connected to a conservative force as in this article. As written above compressed gas could be considered to have potential energy if there is not heat transfer, but this is usually not a good approximation, so mentioning this would probably only be confusing. In your example with the arrow there are energy losses such as air resistance, but that doesn't stop the gravitational force from having a potential connected to it. The gravitational force is independent of the air resistance, but for compressed gas the force depends on the temperature which is likely to change over time. Ulflund (talk) 04:54, 21 October 2014 (UTC)
To be honest, I never advocated using compressed air as an example. I was merely trying to explain to the original poster that it was (or is considered in many books) to be a form of elastic potential energy. As this conversation bloomed it simply became more and more interesting to me, so I guess I was trying to see if I could learn something. (I know, this is becoming more of a forum and should probably end, but perhaps it could also be seen as an exercise in phrasing, if it helps make the article more accessible to the general reader.) Just to note, however, according to everything I've read, as long as the compression and expansion are purely adiabatic or isothermal, the energy is conserved. Like the arrow, however, in the real world this is never the case, and potential energy is also used in real-world calculations to produce actual results. (I should know. Pneumatics is my specialty.) Zaereth (talk) 05:14, 21 October 2014 (UTC)

## Gravitational Potential Energy

If an object is in space, can it be said to have gravitational potential energy with respect to all the planets at the same time? If it moves closer to Mars and away from the Earth, does it lose potential energy with respect to Mars but gain potential energy with respect to the Earth?

Why do we speak of energy being stored in the object in this case? Doesn't it make more sense to say it has the abstract potential to be given energy by any gravitational field it comes in contact with? And isn't this abstract potential something entirely different to actual energy like heat and momentum? Don't you think it's strange that we say that a spacecraft with the potential to fly toward Mars, or to fly toward Earth has energy by virtue of this abstract potential?

It seems like everyone who studies physics just takes the Laws of Physics at face value and never critically analyzes them to form their own opinion about whether they are valid. Are you sure energy is never created or destroyed? Have you considered whether potential energy is merely an arbitrary addition to the law of conservation of energy to make it seem true? Just because an object might some day gain energy from a gravitational or magnetic field it moves into, doesn't mean it has a reservoir of stored energy. —Preceding unsigned comment added by 203.211.92.253 (talk) 18:36, 7 April 2009 (UTC)

I was just about to comment on this, when I find someone else beat me to it. What is probably most confusing to this user, and others, is exemplified by the last line of paragraph #1 in the 'Overview' section: "...and gravity in the case of the weight." I think to clear up this confusion it would be important to note that gravity itself is not a source of energy, (of any kind), nor is it a type of energy itself. The potential energy of the weight is stored in the form of altitude, and not gravity, and this is done during the lifting process. For example, according to the U.S. Naval air combat manual, http://www.tpub.com/content/aviation2/P-821/P-8210201.htm , a fighter plane flying with a kinetic energy (speed) of .9 Mach and a potential energy (altitude) of 5000 feet has the same specific energy as the same plane traveling at only .1 Mach at 20,000 feet. But the energy is not supplied by gravity, but by the fuel that allowed the aircraft to reach that altitude.
Similarly, as we revolve around the sun, we do not feel its effects because our energy is kinetic. As our spaceship stores massive amounts of energy into altitude while breaking the Earth's gravity, we can change that potential energy to kinetic energy by diving toward the Sun, or we can store even further potential energy by moving away from it. As we approach Mars, in this hypothetical, all of this potential energy we have stored up fighting the Sun is tranfered to kinetic energy as we approach Mars' orbital speed, (and once again we become weightless compared to the sun), and and the potential energy we stored up climbing away from Earth literally reaches "ground state" when we touch down on Mars.
I hope that helps answer your question in sufficient detail. In short, gravity is neither energy, nor a place to store it in itself, and I think some simple language changes could help emphasize this. Perhaps a change to the first paragraph of the 'Overview' section, something like this:
Potential energy is energy that is stored within a system. It exists when there is a force that tends to pull an object back towards some original position when the object is displaced. This force is often called a restoring force. For example, when a spring is stretched to the left, it exerts a force to the right so as to return to its original, unstretched position. Similarly, when a weight is lifted up, the force of gravity will try to bring it back down to its original position. The energy from lifting up the weight is stored in its altitude, while similarly, the energy it took to bend the spring is stored in the distance of the bend. The initial steps of stretching the spring or lifting the weight both require energy to perform. According to the principle of conservation of energy, energy cannot be created or destroyed; hence this energy cannot disappear. Instead, it is stored as potential energy. If the spring is released or the weight is dropped, this stored energy will be converted into kinetic energy by the restoring forceelasticity in the case of the spring, and gravity in the case of the weight.
The suggested changes are marked in bold. Zaereth (talk) 00:06, 13 June 2009 (UTC)
Actually, I have made some changes to my above suggestion, since the last line is referring to the "restoring force", (resistance), and not energy, then it would be correct as it is. The suggested addition is still marked in bold, however, since my expertize is in mechanical engineering and not physics, I would like to get some input from a physicist before I make any such alteration. Zaereth (talk) 18:36, 15 June 2009 (UTC)

## Potential mass

It would be interesting if physicists could theorize the existence of potential mass, when you take Einstein's equation of mass-energy equivalence into account. ADM (talk) 04:09, 3 July 2009 (UTC)

I think it would go something like this: An object with a certain weight, (say 100 pounds) suspended at an arbitrary altitude will have the potential to strike the ground with a force greater than its weight, (say an impact force of 1000 pounds). Zaereth (talk) 03:45, 4 July 2009 (UTC)

"Potential mass"? Do you have any knowledge of physics whatsoever? That's a completely nonsensical statement. —Preceding unsigned comment added by 66.65.67.197 (talk) 03:52, 19 June 2010 (UTC)

## Remove sections Thermal energy and Rest mass energy?

As I understand it, neither thermal energy nor rest-mass energy are normally considered to be forms of potential energy. There are articles on each of these topics that describe them better than they're described here - would anyone object to deleting these two sections? Djr32 (talk) 11:14, 10 October 2009 (UTC)

Done. Djr32 (talk) 11:53, 17 October 2009 (UTC)

## Small change requested

The second to lat sentence in the article is not grammatically correct. However, I don't understand the physics, so I don't feel qualified to fix it. Someone please give this little sentence a bit of attention. —Preceding unsigned comment added by 152.23.202.135 (talk) 03:27, 1 March 2010 (UTC)

Fixed; thanks for the note. --Christopher Thomas (talk) 06:19, 1 March 2010 (UTC)

## This is the worst Wikipedia entry I have ever read

The word "potential" meant something different in the 19th Century. Potentia is Latin for power. When "potential" was used in physics, it meant energy (the ability to do work). When it was used outside of physics it meant the ability to become something different in the future. Do not confuse the nonscientific notion of potential with the scientific notion of potential. The two meanings diverged in the early 20th Century (or late 19th Century).

Potential energy is the ability to do work (to exert a force over a distance). Kinetic energy is also the ability to do work (to exert a force over a distance). The former is due to position. The latter is due to motion. All forms of energy are identical based on the definition of energy as the ability to do work. All forms of energy are inter-convertible. There is no such thing as "stored" energy. There is energy due to position (or arrangement) and energy due to motion. Energy can be "stored" in the position of objects or the motion of objects.

Actually, the whole notion of "stored" energy is silly. A system "posses" energy due to its motion or configuration. Neither form is more energy than the other. When energy is transformed or transferred, work has been done. Since energy can never be destroyed, the notion of energy being "stored" is somewhat silly. Energy just is.

How much does this Wikipedia entry bother me? To no end, as they say. Someone should fix this. I'd do it, but I normally get paid to do this kind of stuff. If you amateurs don't understand the topic, you should avoid writing about it. If you're as pissed as I am and as knowledgeable as I am and as free as a bird (free in the monetary sense) then you should invest the time and energy and do the right thing. Write the real Wikipedia entry for Potential Energy.

Someone who knows the topic and doesn't mind throwing away a day's work should rewrite the entry -- top to bottom. It is really that bad. It is also really important to fix it. Battalions of students will plagiarize this webpage and throw it at their teachers unedited. A smaller army will try to learn the material as presented and will be mislead. Neither of these options is acceptable. —Preceding unsigned comment added by 66.65.67.197 (talk) 03:42, 19 June 2010 (UTC)

Thanks for your comments about the Wikipedia entry for potential energy. Wikipedia is not intended to be written by experts, but by people who have some understanding of a subject and are experts at combing the best available sources of information about that subject and writing an encyclopedia article to reflect the current knowledge about the subject. A commercial encyclopedia will employ professionals and experts like yourself to write its articles. Wikipedia is not such an encyclopedia - it makes use of the efforts of volunteers, and the result is not only freely available but it can be edited by anyone.
The article on Wikipedia is conspicuously lacking in in-line citations to point to the sources that were used to obtain the information found therein. I have added a banner at the top of the article to alert editors that more references and sources are required. Wikipedia is based on the expectation that as volunteers work to locate more and better references and sources, the article will grow and mature, and the result will be a higher quality article. I see your time is limited to the extent that you are unable to contribute to potential energy but you could assist in as little as a few seconds by using this thread, or this Discussion page, to alert myself and other interested editors to one or more good sources of information about the current knowledge regarding potential energy. Sources that are readily accessible via the worldwide web are particularly useful and an expert like yourself will be able to quickly provide links to those sites.
Wikipedia is a hobby for many thousands of volunteers. It isn't a commercial project and so it operates very differently to commercial encyclopedias. It works best when volunteers are encouraged to participate strongly, using positive motivation. Best regards. Dolphin (t) 05:46, 19 June 2010 (UTC)
While I agree with the assessment that this article could use more references, I disagree with the IP in that energy cannot be stored. A simple example of stored energy is a capacitor. When the capacitor is charged, a potential is created between two planes and stored until the system is discharged. Energy can be transferred. A simple example is found in game of pool. When the cue-ball strikes the eight-ball squarely, it stops. Its energy is transferred to the eight-ball, which continues the motion at the same velocity. The eight-ball has now consumed the kinetic energy possesed by the cue ball, not destroyed it. Of course, while I am no expert in this field, I do take a great interest in it. I can look to find some sources for this article, but it probably won't be for a few months. (And they'll most likely be books rather than online sources.) Zaereth (talk) 21:52, 21 June 2010 (UTC)
Of course energy can be stored. According to the anon potential/kinetic energy depends on position/motion. So by changing the position/motion of an object, we can increase or decrease its PE/KE. We call the former "storing energy" and the latter "releasing energy". DVdm (talk) 05:33, 22 June 2010 (UTC)

I have reverted Dolphin's take on the opening sentence of the lead. Rationale in my edit summary.

In his Lectures on Physics Feynman writes (Volume 1, page 4-4):

• The general name of energy which has to do with location relative to something else is called potential energy".

I think that is what an opening lead sentence should say, in this case perhaps rephrased as:

• Potential energy is the general name of energy which has to do with location relative to something else".

FIne details can be covered within the article. DVdm (talk) 13:05, 25 June 2010 (UTC)

That sounds fine and perfectly correct to me. From reading the section above, it might be worth mentioning here, on talk, that there is often confusion over the meanings of such words as "potential," "energy," "power," and "work." Kinesis simply means "motion," whereas potential means, "possessing the capability of, but not yet in motion" Energy is the ability to do work. Power is the ability to do work within a certain amount of time. Work is simply the transfer of energy, and requires resistance to perform. I don't know if that helps or not, but thought I'd point that out. Zaereth (talk) 21:56, 25 June 2010 (UTC)
Ok, in that case I think we can safely keep the current phrasing,
• 'In Physics, potential energy is energy stored within a physical system as a result of the position or configuration of the different parts of that system.,
as it is a close paraphrasing of Feynman's above quote. If someone prefers Feynman's quote, we can put it in place - with a reference to the lectures. DVdm (talk) 22:15, 25 June 2010 (UTC)
DVdm appears to have missed one of the fundamental principles of Wikipedia. The article Potential energy is conspicuously short of in-line citations and references to point to sources. I re-worked the lead sentence in accordance with a reputable introductory physics textbook and provided an in-line citation to allow other Users to independently verify what I had written. DVdm reverted my work, including my citation, and restored unsourced material. I can't think of a quicker way to start edit wars than reverting cited material and replacing it with uncited material. DVdm should immediately familiarise himself or herself with WP:Verifiability. Dolphin (t) 06:48, 26 June 2010 (UTC)
How about trying to address the concerns with your version of the lede, rather than assuming bad faith or ignorance of Wikipedia policies? User:DVdm is a long-time member of good standing in WikiProject Physics, having contributed usefully to many articles. --Christopher Thomas (talk) 08:37, 26 June 2010 (UTC)
A few remarks:
• I reverted your edit in perfect accordance with WP:BRD: you were bold and I reverted with a clear edit summary and started a discussion on the talk page.
• Lead sentences don't require references to sources. According to WP:LEADCITE, "the necessity for citations in a lead should be determined on a case-by-case basis by editorial consensus." That is the purpose of this section. I provided a clear way to keep the lead intro simple and properly sourced if someone would insist to have it cited anyway.
• According to WP:Lead_section#First_sentence, "The article should begin with a declarative sentence, answering two questions for the nonspecialist reader: "What (or who) is the subject?" and "Why is this subject notable?"" I think that the sentence satisfies the purpose and it looks like a few other editors agree.
• On the kinetic energy talk page you wrote: "The opening sentence needs to be the simplest thing that can be accurately stated about kinetic energy. The opening sentence may be the only thing read by young readers and newcomers to science. Whether they read further will depend on how meaningful the opening sentence is." As I replied there, I think that what you did here will immediately scare every young reader and newcomer to science away.
• On my talk page you ask: "Do you agree that potential energy is only defined for conservative forces?". I will answer by quoting my original edit summary again: "Please find some place *within* the article for this." Does that answer the question? Surely we can find a suitable place in the article to elaborate about conservative forces. What you wrote is good, but I don't think that it belongs early in the lead.
DVdm (talk) 10:16, 26 June 2010 (UTC)

I agree that lead sentences don't require citation because lead sentences summarise material that appears later in the article, and that material which appears later is properly cited. The article Potential energy now only contains two in-line citations and neither of them supports the sentence in contention. The sentence which DVdm restored presently has no more status than original research. My re-work was supported by an in-line citation but that has now been deleted. Does anyone really think this is the way to make progress? Dolphin (t) 12:17, 26 June 2010 (UTC)

If you really insist on having a sourced in-line citation, I propose we take Feynman's - see above. I also looked for your cited sentence ("The gravitational field does not vary with time so weight is an example of a conservative force, and hence the gravitational potential energy of a body or a system") in my copy (1966) of Resnick and Halliday. I can't find this in section 8-3, nor can I find the word "weight" in there. Are you sure about the section? Which edition do you have? DVdm (talk) 13:03, 26 June 2010 (UTC)
Wikipedia does indeed insist on having in-line citations - lots of them. (Today's Featured Article has 145 in-line citations.) I agree that Feynman's work would be ideal as a reference and in-line citations for Potential energy and similar articles. No-one could doubt Feynman's reliability!
My copy of R&H is also dated 1966. In section 8-3, third-from-last para, it states Here the gravitational force (=mg) is essentially constant ... I agree it doesn't use the word weight but elsewhere R&H show that mg is equal to weight. The word weight is used in section 8-2.
I was in error in citing 8-3 for the notion that The gravitational field does not vary with time ... In Section 8-2 it claims numerous times that the gravitational force is conservative but without explaining fully why that claim is made. R&H explain that path independence is essential for a force to be conservative but I now can't find any place in R&H where time invariance is mentioned. Dolphin (t) 08:25, 27 June 2010 (UTC)
Indeed, no-one could doubt Feynman's reliability, although I would be a bit reluctant quoting the Feynman Lectures on relativistic mass, in which they present a rather old-fashioned view.
Anyway, while still thinking that we could keep the current phrasing by virtue of its accuracy, simplicity and closeness to the Feynman quote, I did some searching and decided to replace the current opening sentence with one that resembles it extremely closely, and that also mentions the concept force field:
• "In physics, Potential energy is the energy stored in a body or in a system due to its position in a force field or due to its configuration."
Of course we also should work on adding inline citations in the body of the article itself. Google Books is Your Friend :-) DVdm (talk) 10:50, 27 June 2010 (UTC)
Very good. I like the simplicity of Jain's prose. Dolphin (t) 12:53, 27 June 2010 (UTC)

## Catapult *and* Trebuchet?

I think the inclusion of a picture of a catapult *and* a trebuchet is excessive. Potential energy is not intrinsically related to medieval siege warfare, and either a picture of a catapult will help explain the concept to you or it won't-- the trebuchet adds no value. — Preceding unsigned comment added by Ohnoezitasploded (talkcontribs) 02:29, 23 December 2010 (UTC)

Done. Agree about the medieval warfare bias. O.t.o.h. the trebuchet uses gravitational energy whereas the catapult uses elastic, so I would not remove one, but rather replace the catapult with, for instance, a spring.
Springs are used for storing elastic potential energy
DVdm (talk) 11:11, 23 December 2010 (UTC)

## Comparison table and some example numbers

How much potential energy is stored in a cubic meter of water 10 meters high, and how does it compare to a liter of gasoline used for combustion, both in finite values and in practical use. What are the typical costs for setting up and for using this energy? At small scale and large scale? Thanks פשוט pashute ♫ (talk) 14:15, 28 February 2011 (UTC)

Depends on how you're comparing it. (Apples to apples.) The gravitational potential energy of water at that height, 98,000 Joules. The gravitational potential energy of a cubic meter of gasoline at the same height, 70,560 Joules. (Apples to oranges.) The chemical potential energy for a cubic meter of gasoline at any height is 32,976,000,000 Joules. Zaereth (talk) 00:15, 1 March 2011 (UTC)

## An anomaly

If a spacecraft is at an altitude just outside the effective influence of Earth's gravity, it's gravitational energy is equivalent to the energy of the fuel used to elevate it to that altitude, and it could be converted back to kinetic energy by slightly moving the craft into the area of effective influence of the Earth's gravity.

(As an aside, I don't say gravitational "potential" energy because, as another user pointed out on this page, the idea of potential is already implied by the definition of the word "energy", that is the ability to do work, and thus it is redundant to call some types of energy "potential" energy. The item doesn't "potentially" have the ability to do work, it just "has" it, regardless of the type of energy. )

Anyway, let's say that instead of moving towards Earth, the craft rockets toward Mars. It maintains its gravitational energy with respect to Earth (which it will convert to kinetic energy later when it comes back to Earth) and it also gains kinetic energy now as a result of the rocket blast.

But let's suppose the craft engages it's retro-rockets so that it comes to a rest just before it enters Mars's area of effective gravitational influence. So far all the energy transfers have balanced out, and no energy has been created or destroyed.

If it makes one more little push (which could be an insignificant amount of energy depending on just how close the craft is to Mars's area of effective gravitational influence). It will now gain a massively disproportionate amount of kinetic energy as Mars's gravity accelerates the spacecraft towards Mars.

Does anyone want to argue that this energy is not being created? That the spacecraft always had the ability to do this work since it was first created on Earth? You would also be arguing that the spacecraft always had and still has gravitational energy with respect to every star and planet in the universe. This would be a mind-boggling amount of energy, and unmeasurable since we haven't even discovered most of the objects in the universe. This would make all other kinds of energy the spacecraft has seem totally insignificant in comparison. Someone making that argument would be going to extreme lengths to defend the idea of conservation of energy. When the spacecraft rockets up away from Earth, it has energy in the form of altitude imparted to it by its jet fuel. But where does its energy with respect to other stars and planets come from?

Isn't it far simpler to say that the energy was imparted to the spacecraft by Mars's gravitational influence?

The Mars system has not lost any energy in doing this, so doesn't this show that total energy is not conserved in this case? Gcsnelgar (talk) 21:25, 22 December 2011 (UTC)

The simple answer is that all of this energy came from the big bang; that force which spiraled the universe out from some central point. What you're missing in your equation, locally, is the influence of the Sun. Outside the influence of Earth's gravity, we are still well within the influence of the Sun's. As long as this spaceship remains in Earth's orbit --at Earth's orbital speed-- our energy with respect to the Sun remains kinetic instead of potential. In other words, our energy with respect to the Sun's gravitational field is in the form of speed. (Kinetic means: Of motion.)
Now if you want to come to a full-stop with respect to Earth's orbit, so that your position within the solar system remains stationary, you will fall into the Sun. Just like if you were hovering above Earth, you will need to point your thrusters toward the Sun and hover above it to maintain your potential energy; your "stopped energy." (Potential simply means: possessing the capability of motion, but not in motion.) In this case, you will feel weight in the presence of the Sun's gravity. Climbing away from the Sun in this case is just like climbing away from Earth. You have this energy, right now, with respect to the Sun but, in orbit, you are weightless because the energy is in the form of speed.
If you get far enough away from the Sun that its gravity no longer has a noticeable effect, then you will still be within the gravitational field of the galaxy. Right now, we are in orbit around it at some incredible velocity (kinetic enery). I don't know if anyone has a clue what would happen if we leave the galaxy, or what it might be in orbit around, or how fast it's hurtling through the universe, etc... Zaereth (talk) 23:55, 22 December 2011 (UTC)
Gcsnelgar you said, "The Mars system has not lost any energy in doing this," but actually, it has! The amount of energy the spacecraft "steals" from the Mars system is enough to boost the spacecraft's kinetic energy significantly and yet its effect on the Mars system is insignificant (since the spacecraft has such a small mass compared to Mars). This is the principle behind any and all "gravity assists". Check out: http://en.wikipedia.org/wiki/Gravitational_assist simkn
That's true. What many people don't know is that, once you leave the atmosphere, it makes no sense to climb straight away from a gravity well; it uses too much fuel, causes too much stress on the craft, and, if something goes wrong, there is no place to go but straight down. An easier way is to achieve orbital speed, and then keep gaining speed, throwing the craft out into higher orbits, basically spiraling away. When you reach the height of another body, (like the the moon or, in this example, Mars), you can pull in on its six, dive into it's gravity well, and it will pull the craft to its orbital speed. Relative to the Sun, the craft has gained speed and Mars has lost speed, reaching an equillibrium, but the loss to Mars is insignificant. At a higher speed, the craft can begin climbing to the next higher orbit. However, relative to Mars, the craft simply dove into its well, gaining speed, and then climbed back out, returning to its previous speed. (You can start to see why we needed a theory of relativity.)Zaereth (talk) 23:42, 11 July 2012 (UTC)

## Potential energy v. entropy

Entropy can be understood as a form of potential energy which falls outside the regular definition of the latter.[1] I think the article should explore the cohesion between the two concepts. Does anyone have any better sources than wisegeek.com? Narssarssuaq (talk) 09:18, 11 July 2012 (UTC)

Well, entropy is simply wasted energy; both potential and kinetic. I think a brief description could benefit this article, but I would avoid going into too much detail, because, although related, they are different concepts. For a better ref, I believe the book Entropy And Energy: A Universal Competition, by Ingo Müller and Wolf Weiss, is available on google books. Zaereth (talk) 16:27, 11 July 2012 (UTC)
A good way to think of entropy is probably in an incandescent light-bulb. In these bulbs, roughly 10% of the energy (the ability to do work) is making visible light (useful work). 70% is typically producing IR and near-IR (radiation heat), about 10% is lost in thermal conduction (heat) and the last 10% just gone (used to heat the filiment or to push the electricity across the resistance, etc...). This unusble work is called entropy, and 90% of the bulb's energy simply goes to waste this way.
In some cases, entropy can be considered "anti-work." In hot climates, for instance, the wasted heat from the incandescent bulb causes air-conditioning to work harder. However, in colder climates, the radiated heat is actually recoverable energy, which helps to heat the house. No system, though, is 100% efficient, for some energy is always lost to entropy. Zaereth (talk) 20:05, 13 July 2012 (UTC)
I don't see how entropy can bee seen as potential energy. They are totaly different concepts and have different physical dimensions. If there is such analog the best place for it is probably in the Introduction_to_entropy article since noone trying to get a grip on potential energy will have a good understanding of entropy. Ulflund (talk) 14:13, 19 July 2012 (UTC)
Entropy as potential energy pretty much has to do with recoverable energy, or "conservation of entropy." You tend to find this discussed mostly in books like Thermal Energy Storage: Systems and Applications. This brings about fancy terms like negative entropy or negentropy, (which is more or less "some other work"). For instance, I can power my cabin with a generator, which is its designed purpose, but I can conserve some of the entropy by using the coolant to heat a large tank of water. I can then use that water to heat the cabin, even at night after the generator has been shut down. Zaereth (talk) 23:36, 19 July 2012 (UTC)

## Correcting the Positive vs. Negative Gravitational Energy section

The argument appearing in this section involving comparing U(q) at r=0 and r=infinity is incorrect. It is not reasonable to allow the gravitational potential U(q) to approach 0 as r approaches 0, because regardless of the sign of U(q), it diverges as we approach r=0.

Perhaps a better way to think of the sign of U(q) is to calculate it using the force F describing Newton's law of gravity. The force vector on a test particle is directed toward r=0, because that's where the massive body is located. It makes sense to adopt a convention in which this force is positive: there is a positive force of gravity pulling the test particle toward the massive body. But by definition, the force is equal to the *negative* gradient of the potential energy. A positive force thus corresponds to a negative potential, on this convention.

How does this sound? Any other suggestions from the talky-folk about how to improve this subsection? Bryan (talk) 15:31, 13 September 2012 (UTC)

Well, you can choose any convention you like. You can make infinity zero, or you can make the center of Earth zero, or the surface of Earth, or even the surface of your desk. In the former, the energy is negative; in the latter, it's positive. However, cosmologists almost always use infinity as zero, which makes gravitational potential energy negative. This section is not meant to be a place to debate which is better, just to explain why most people use that particular convention. This can information can be found in many reliable sources. For instance, here is a quote from the book An Introduction to Galaxies and Cosmology:
The gravitational potential energy of each star is a negative quantity because, by convention, the gravitational potential energy of a star is taken to be zero when it is so far from the galaxy that it is eff3ectively free of the gravitational pull of the galaxy. Since any star would have to be given a positive amount of energy to enable it to attain this zero energy state, it must be the case that each star in the assembled galaxy has a negative gravitational potential energy.
The total gravitational potential energy (Eg) of a galaxy can be defined in a similar way. By convention, the total gravitational energy is taken to be zero when all the stars are so widely dispersed that each is effectively free of the gravitational influence of all the others. If you can imagine a process i which stars are removed one by one from the galaxy, then it's easy to see that some positive energy will be required to achieve the total disassembly that corresponds to the state of zero gravitational potential energy. The total gravitation potential energy of the assembled galaxy will be the negative of this disassembly energy.
Another example comes from the book Quantum World:
At the heart of inflation theory is: (1) the idea of negative energy, and (2) the equivalence of mass and energy. If you shoot an arrow into the air, it requires energy. When the arrow reaches its maximum height, it possesses a certain amount of gravitational potential energy. Gravitational potential energy is considered negative energy. When the arrow starts to fall back to earth, some of that potential energy is converted into the kinetic energy of the arrow. Kinetic energy is considered positive energy. As the kinetic energy become more and more positive, the gravitational potential energy becomes more and more negative so that the total energy never changes, In other words, total energy is conserved in accordance with the first law of thermodynamics.
The latter explanation is the very heart of theories such as Energy-Maneuverability theory, or the Big Bang theory. Personally, I think we could use a simpler explanation like these to go along with the mathematical explanations, so, in that way, it would be more easily understood. Zaereth (talk) 17:18, 13 September 2012 (UTC)

## Common symbols

This edit removed V from the list of common symbols with the rationale "V" denotes velocity or volume, never potential energy. I don't agree with that statement. V is commonly used for potential energy in contexts such as Lagrangians, Hamiltonians or the Virial theorem. I think we should leave it in the infobox. — HHHIPPO 12:05, 17 February 2013 (UTC)

I agree, it shouild remain in the list. V is a common symbol for potential energy as well as velocity and volume. It is also a common symbol for voltage (and so is "E" for "electromotive force"). It is often necessary in math equations to specify what the letter stands for. Zaereth (talk) 16:15, 17 February 2013 (UTC)
I have no objection to V being restored, provided it is genuinely in common usage in reliable published sources. It has not been in long-term usage in our article Potential energy, having only been added about ten days ago - see this diff. Dolphin (t) 00:01, 18 February 2013 (UTC)
Like "V" for voltage or "A" for amperage, the "V" for potential energy is somewhat archaic, but can be found still in common usage. Two examples which come to mind are Textbook of applied physics (here is the google books link) or Principles of engineering mechanics (google link). My personal opinion is that listing both symbols is useful to help people who may encounter books which use "V," because many of them do still exist. Zaereth (talk) 01:56, 18 February 2013 (UTC)
Just using the same query without specifying a book gives tons of hits. If the symbol wasn't mentioned here until recently that's overdue, since also a number of Wikipedia articles use it that way. — HHHIPPO 20:13, 18 February 2013 (UTC)
It looks like we've reached a concensus. I don't have time right now to make the change, or I would, so if someone would like to do the honors, it would be appreciated. I gotta go, I'm late. (No time to say hello, goodbye...) Zaereth (talk) 02:09, 21 February 2013 (UTC)
Done — HHHIPPO 07:34, 21 February 2013 (UTC)

## Zero point for gravitational energy

The presented reasoning for setting U=0 at infinity is somewhat peculiar/wrong. U would never become infinite, wherever the zero point was set, because delta U(distance) is always finite.

If two bodys do not interact with each other, they logically cannot have a potential energy with respect to each other. Thus U must vanish in infinity. How could anything gain energy by something in infinite distance becomings heavier? That´d be really counterintuitive.

--129.13.72.198 (talk) 13:14, 11 October 2013 (UTC)

See the discussion two sections above this one. At infinite distance, gravitational attraction is infinitely small, but it is there. If you make U=0 at Earth, the gravitational energy becomes positive. This means that all other forms of energy become negative. This means that, as my airplane climbs in altitude, the fuel and the thrust would actually be removing energy. As my plane gains speed, it would also be losing energy. In this scenario, my plane would be at its highest energy state when it is fully stopped on the ground, which becomes even more counterintuitive. Zaereth (talk) 13:52, 11 October 2013 (UTC)
Yes, saw that other discussion now. I don´t understand what you are trying to say, though.
I object to the reasoning, not to the conclusion that U must rise with r and be zero at r=inf (and thus always be negative).
The presented reasoning is, i think, too complicated and either misunderstandable or just wrong.
If something has left the gravitational field of another body, it makes no sense to define a potential energy any more. The only way to have that is let U be zero. I don´t think that it needs more explanation. --129.13.72.198 (talk) 15:18, 11 October 2013 (UTC)
That's assuming that you can ever actually leave a gravitational field. On Earth, we don't directly feel the effects of the Sun's gravity, but we haven't left its attraction. To make this work in a cosmological sense, we would need to find some gravitational canter of the universe, but no such center can be found. That leads to the idea of "inflation" (the universe didn't explode from a central point, but rather inflated like a balloon), and so you get the current theory of physics which most people now accept. Wikipedia is not really a place to debate the theory, but just to report it. Zaereth (talk) 15:31, 11 October 2013 (UTC)

## trebuchet

A picture of the trebuchet has a caption:

"A trebuchet uses the gravitational potential energy of the counterweight to throw projectiles over long distances."

By comparison with the distance that modern projectiles travel I think it misleading to say that a trebuchet throws projectiles over "long" distances. The maximum range of a trebuchet is probably about 250m which is not impressive or "long" by my 21st century understanding of a good range. The trebuchet compares poorly with a modern projectile such as an artillery gun shell which will have a range of tens of kilometers.

I would suggest changing "long" to read "short". 194.176.105.141 (talk) 13:23, 6 January 2014 (UTC)

Added "relatively" to the caption ... as for their historical use the range was long even though by modern terms not. P'raps could be tweaked further re the time period of invention/usage. Vsmith (talk) 15:09, 6 January 2014 (UTC)
Surely when we write an article it uses modern words and modern standards? If I was writing about stone aged people I would not describe bronze tools as advanced without providing a context. Furthermore I would say that by the standards of the time the range was short compared to a well thrown spear, an arrow, a crossbow bolt or a sling shot.194.176.105.141 (talk) 12:56, 7 January 2014 (UTC)
When speaking about the trebuchet, we are not comparing it to all of the other weapons previously developed. This article is about potential energy, not weapons. The movement that the projectile travels is a relatively long distance when compared to the movement of the counterweight. Comparing it to the range of an arrow, or a bullet, or even an intercontinental ballistic missile makes no sense here. Zaereth (talk) 17:26, 7 January 2014 (UTC)
I might add, however, that one of the benefits of a war engine ("engine" being a machine that changes potential energy into kinetic energy) is that it allowed devastating attacks from a relatively safe distance. At your estimate of nearly 2 1/2 football fields, that puts it well out of range for most conventional weapons at the time. Zaereth (talk) 19:30, 7 January 2014 (UTC)

## Potential energy in a person's arm?

The Intro to Potential energy includes: If the work of a force field acting on a body that moves from a start to an end position is determined only by these two positions, and does not depend on the trajectory of the body, then there is a function known as potential energy that can be evaluated at the two positions to determine this work. The image of an archer with a bow and arrow has a caption beginning: In the case of a bow and arrow, the energy is converted from the potential energy in the archer's arm to the potential energy in the bent limbs of the bow when the string is drawn back. In what sense is the force exerted by a person’s arm, and the work done by that force, independent of position?

I think the caption is incorrect because it refers to the potential energy in the archer’s arm. The work done by a person’s arm or any other muscle is accompanied by an identical reduction in the chemical energy of the person’s body, so energy is conserved. That work is strongly dependent on trajectory so it cannot be described as potential energy. Dolphin (t) 01:10, 12 August 2014 (UTC)

I agree, the wording seems a bit confused. The work done by the arm is kinetic energy. The bow acts as the restoring force (elasticity), but the potential energy is that energy contained in the position of the arrow within the bow/arrow system. Work is performed by the arm during the draw and while holding the arrow drawn, but like all work, this is done by kinetic energy. (Just because the arm is not moving doesn't mean that the energy is potential. Chemical kinetic-energy is being used within the muscles to draw the arrow and keep it drawn, using up fuel, and the arm will soon tire.) When released, the restoring force changes the potential energy of the arrow into kinetic energy.
An analogy is a simple helicopter. The kinetic energy of the rotor works to lift the chopper into the air (due to the chemical kinetic-energy from the engine), and continues to work to hover at a certain altitude. The potential energy of the helicopter comes from it's vertical position in the gravitational field. If the rotors suddenly stop spinning, perhaps because it ran out of fuel, the restoring force (gravity) changes the chopper's potential energy into kinetic, causing it to fall and crash into the ground. Zaereth (talk) 06:06, 12 August 2014 (UTC)
Thanks Zaereth. There has been no challenge or objection after a week so I have amended the caption to eliminate the problem I described above. See my diff. Dolphin (t) 06:58, 19 August 2014 (UTC)
I like the wording better. However, the sentences do run-on a bit (no offense intended). I think a couple of minor grammatical changes will make it a little more readable. How does this sound? "In the case of a bow and arrow, when the archer does work on the bow, drawing the string back, some of the chemical energy of the archer's body is transformed into elastic potential-energy in the bent limbs of the bow. When the string is released, the force between the string and the arrow does work on the arrow. Thus, the potential energy in the bow limbs is transformed into the kinetic energy of the arrow as it takes flight." Zaereth (talk) 20:04, 21 August 2014 (UTC)
I like it and I have used it to replace the former caption. See my diff. Dolphin (t) 00:40, 22 August 2014 (UTC)

The lead sentence now reads "In physics, potential energy is energy stored in a system of forcefully interacting physical entities.". Although this is correct I don't think anyone who doesn't already know what potential energy is will understand what a "system of forcefully interacting physical entities" is. I much prefer the sentence agreed upon on this talk page Talk:Potential_energy#Lead_sentence:"In physics, Potential energy is the energy stored in a body or in a system due to its position in a force field or due to its configuration.", backed up by this book reference.

Furthermore I find it strange that there is no mention of conservative forces in the lead. Also I think more readers would understand the concept if there was a few examples of potential energy early on, specifically mentioning gravitational potential since this is something everyone is familiar with. I would also move the etymology out of the lead section since the lead should be a summary of the rest of the article and as it is now there is no etymology later on. Any thoughts on this? Ulflund (talk) 21:22, 24 August 2014 (UTC)

I agree that the lead sentence is not consistent with WP:Make technical articles understandable. Your proposals for improving the situation are good ones. Please go ahead. Dolphin (t) 04:43, 25 August 2014 (UTC)

## Unnecessary overcomplication

I made some changes yesterday in the article, primarily replacing the equation

${\displaystyle W=\int _{\mathbf {x} (t_{1})}^{\mathbf {x} (t_{2})}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot {\mathbf {v}}\,\mathrm {d} t=U(\mathbf {x} (t_{1}))-U(\mathbf {x} (t_{2}))}$

with the more direct statement

${\displaystyle W=\int _{\mathbf {x} _{1}}^{\mathbf {x} _{2}}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=U(\mathbf {x} _{1})-U(\mathbf {x} _{2})}$.

This was reverted, with the comment "It makes more sense to parameterize the trajectory, whether it is with time or some other parameters."

I think this argument is wrong, I think it is better to use the position as the independent variable, rather than to express the position as a function of time and integrate over ${\displaystyle {\mathbf {v}}\,\mathrm {d} t}$. This can be kept at the level of secondary-school physics: "work done is force times distance moved". Bringing in more complications doesn't make the physics any more accurate, it just means we have to grind through some more complicated maths in order to get rid of them again. I think it's also making the point harder for the reader to grasp: that potential energy only depends on position, not on time or on the route taken between ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$. Having t's in the final equation implies that they're doing something there.

The problem is even more acute later on in the section on the potential function for gravitational forces between two bodies, where we find

${\displaystyle W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot ({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t})\mathrm {d} t=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}\mathrm {d} t={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.}$

Again, this is far more complicated than it needs to be. The whole calculation can be made much simpler by recognising that the problem is spherically symmetric, so the potential energy only depends on r, so you only have to do a 1D integration of ${\displaystyle \int _{r_{1}}^{r_{2}}{\frac {1}{r^{2}}}\mathrm {d} r}$.

Opinions from others would be welcome. Djr32 (talk) 20:09, 26 August 2014 (UTC)

Thank you for your attention to this article. I understand your intuition that integrating over dx is easier than integrating over another parameter such as time. This is the way Work is often defined in elementary presentations. However, as you know Work is the result of a path integral and so it is important not to just define the start point and end point of the path but the trajectory between the two points. As far as I know the easiest way to specify the path between two points with with a path parameter, which then appears naturally in the Work integral as the scalar product of force with path tangent integrated over the parameter. If time is used this has the advantage of linking power to work and providing an intuitive way of understanding the importance of the component of force tangent to the trajectory. Finally, of critical importance for a potential field is the fact that this integration shows plainly that the result is independent of the path. In my opinion, this is the correct way to make this calculation and obtain a very important result. The recommended derivations abandon the idea that there is start point and an end point in the three dimensional space of the force field and that the Work is computed along a path between them. Instead a calculation is presented that conveniently yields a known result. This is not a simplification of an overly complicated derivation. It is a different derivation that eliminates presentation of the fundamental property of potential fields. Prof McCarthy (talk) 20:48, 26 August 2014 (UTC)
I prefer the simpler version of the integral as suggested by Djr32 since this is both more general and (in my opinion) easier to grasp. Parametrisation of the curve is a general way of solving the integral, but the nice thing with conservative forces is that we can instead calculate the potential and use this. This is not an derivation showing that the integral is independent of path. That is the assumption and the integral just states that mathematically. Both version are clear on the fact that there is a start and an end position. Both versions state the same thing, but the old version does it in a more complicated way. Ulflund (talk) 22:20, 26 August 2014 (UTC)
I would suggest replacing the "vector" integration endpoints (which is an abuse of notation) with a symbol representing a curve between the endpoints. The final potential energy difference can then be expressed as a function of vectors, which do not need to be parameterized:
${\displaystyle W=\int _{C}{\mathbf {F}}\cdot \mathrm {d} {\mathbf {x}}=\int _{t_{1}}^{t_{2}}{\mathbf {F}}\cdot {\mathbf {v}}\,\mathrm {d} t=U(\mathbf {x} _{1})-U(\mathbf {x} _{2}).}$
I think this also works better in that each integral seems to have a natural interpretation. —Quondum 22:42, 26 August 2014 (UTC)
I try to avoid discussions like this because I am by no means a mathematician, so I won't even pretend to know which formula is best. I couldn't even begin to translate what these say into English. In general, I'd say that in either language, the simplest, most elegant, and most concise statement is usually the best, provided that it precisely defines the subject.
In specific, I find it difficult to imagine how a definition of energy should involve time. Because energy is a measure of quantity, I've always taken it as being independent of time. When time is factored in, then I usually see that as referring to power rather than energy, which are two related but different things. Power is a measure of intensity and is very linked to time.
In example, if I charge a flashtube RLC-circuit to 100 joules, the amount of energy is the same regardless of charging time. If that energy is released in 1 second, it is 100 watts of power, and the energy level is still 100 joules. If I release the energy in 1 microsecond, the power is 100,000,000 watts, but the energy level is still only 100 joules. Either way, only 100 joules of work can be done. The energy level remains the same regardless of time. But once again, I really cannot interpret what these equations are saying, so I'll leave it up to you all to work this out. Zaereth (talk) 22:59, 29 August 2014 (UTC)
I agree with Quondum that it is better to write ${\displaystyle \int _{C}}$, and with Zaereth that energy is independent of time. Therefore including an integral over time in the equation that defines work seems odd. Some people will probably find it helpfull, but I think more people will find it confusing. Ulflund (talk) 23:18, 29 August 2014 (UTC)
I appreciate very much the careful comments on this issue. I think there is some misunderstanding of what is communicated by the calculation of this integral. I believe it is understood that Work is calculated as a force moves along a trajectory, so it makes sense to calculate Work for the movement of the point of application of the force from point A at time t1 along the curve C to point B at time t2. This is simply the definition of Work and is true for any force in any situation. What makes this particular calculation interesting and important is that for specific forces, such as gravity, this integral yields a surprising result, which is that the Work calculated by the integral is the same no matter what path is selected from point A to point B, and no matter how the parameter t is defined. This is not a prior assumption, it is a remarkable fact that arises directly from this calculation. This calculation is the source of the fundamental concept that a force field derivable from a potential function defines a potential energy. This calculation is the source of the very concept of potential energy. Again, there is no prior assumption. To say that this fact is already known and this calculation is overly complicated is to say that this rather simple proof of the defining feature of potential energy is unnecessary. Prof McCarthy (talk) 02:01, 30 August 2014 (UTC)
The equation we are discussing is not showing that the integral is path independent. The sentence preceding the equation starts "If the work for an applied force is independent of the path ...", so I would call it a prior assumption. Parametrization of the curve is a possible first step in showing that the is a potential in a specific case, but does not show anything in itself. Ulflund (talk) 02:39, 30 August 2014 (UTC)
While I agree that a conservative vector field results in the integral being path-independent, what is expressed makes a difference in what is understood. What is currently there is a result, not a derivation. It does not matter that the result is known to be path-independent, as it is expressed it comes across as a circular argument. If one wants to express the result of path independence, simply say ${\displaystyle W=U(\mathbf {x} _{1})-U(\mathbf {x} _{2})}$, and leave it at that. And my point that integration limits being written as vectors is nonstandard (and technically meaningless) has not been addressed, though Ulflund agreed with my suggested change. There is actually a further implicit assumption that is not stated: that the vector field is static. This is a highly unrealistic, or at least unstated, assumption, given that the system being studied is implied to have time as a dimension. —Quondum 05:17, 30 August 2014 (UTC)
Thanks, all, for the good comments, and sorry for not having been able to reply for 2 weeks.
I recognise Quondum's objection regarding notation, I have no objection to expressing the integral as a line integral.
I agree with Prof McCarthy that it is a remarkable fact that the answer is the same for any path, C, going from A to B, but, as noted by Ulflund and Quondum, that fact doesn't come out of the given equation, it's stated as an assumption. Djr32 (talk) 22:47, 6 September 2014 (UTC)
It is generally understood that the statement "if A, then B" identifies an equivalence not an assumption. The expression is correct as written, because there is no assumption, only the conditional that if the normal work integral results in a function that is path independent, then the resulting function is a potential energy. In this case, the statement that the work integral is independent of the path is equivalent to the mathematical expression which clearly shows that a line integral from point x1 to point x2 which is the definition for work, can result in a function that does not depend on the curve or the parameter t. This is shown by the fact that the value of integral is obtained by evaluating the function U only at two points in space the beginning and the end of the path. This expression including the integral is exactly what one must write to express that the work integral along a path from x1 to x2 results in a function that depends only on the end points x1 and x2 and not on the path between them. Alternatives that eliminate the integral do not define work properly, and those that ignore the path parameter do not represent integration along a path properly. The property of path independence is not stated as an assumption, it is stated as an equivalence, which is that when the work integral for a particular type of force can be calculated using a function that is evaluated only at the end points of the path, then the work of that force is independent of the path and defines a potential function. I cannot agree with the recommendations provided, because they seem to simply state path independence without connection to the definition of work. If the work integral that usually depends on the path is not shown, then how does one demonstrate the importance of the case when it does not depend on the path and defines a potential energy. This seems important. Prof McCarthy (talk) 04:37, 7 September 2014 (UTC)
You seem to misunderstand me, or else I entirely misunderstand you. In this context, a statement like "it is generally understood" does not apply; we need to be more explicit. I was not suggesting dropping the intermediate expressions, only trying (unsuccessfully) to point out that the presentation as it stands comes across as nonsensical and/or circular. Your arguments about being explicit about the path integral for the work seem to my suggestion of being explicit about the path in the integral. Please give a some weight to consensus. —Quondum 05:03, 7 September 2014 (UTC)
Prof McCarthy, your comments are contradictory. Previously you said "this integration shows plainly that the result is independent of the path" and "this integral yields a surprising result". Now you say "the statement that the work integral is independent of the path is equivalent to the mathematical expression which clearly shows that a line integral from point x1 to point x2 which is the definition for work, can result in a function that does not depend on the curve or the parameter t". In other words, we aren't going to show that the integral yields any results at all, we're just going to state what it is. That's exactly what you criticised the simpler expression for: "a calculation is presented that conveniently yields a known result". Both the old and new mathematical statements are just "given A, then B", neither proves what you claim the original does.
Nobody is recommending changes that "eliminate the integral" or that "do not represent integration along a path properly", the question is about whether it makes the article clearer or less clear to parameterise the expression in a time-dependent way. Simply put, it's a lot easier for the reader to grasp that the final expression is a function that does not depend on t if you don't put unnecessary t's in it.
To my mind the physically-motivated chain of reasoning goes "because the work done against certain real-world forces is independent of path, then we can define a potential function, and it's an interesting and useful thing". I would like to see the article make this argument more clearly, but I don't think it has anything to do with the change I originally proposed. Djr32 (talk) 15:58, 7 September 2014 (UTC)
I agree with Djr32. It is confusing to parametrise the expression with respect to t when it is independent of t. I don't see a reason to parametrise it at all. Ulflund (talk) 15:40, 8 September 2014 (UTC)

## On a quantum level

This article doesn't really explain what potential energy fundamentally *is* on a quantum level, which is what I wanted to find out. Also, the first sentence of the overview says "Potential energy is the stored or pent-up energy of an object". While this is often true, potential energy is better thought of as the energy stored in the *fields* (e.g. gravitational or electromagentic) of the *system*. For instance, in multi-particle examples, like a collection of point charges, you don't talk about "the potential" of each one, but rather you speak of the potential of the *whole system*. --AndreRD (talk) 09:19, 28 September 2015 (UTC)

To quote Richard Feynman, "It is important to realize that in physics today, we have no knowledge what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way." That's what we all want to find out. It is impossible for a field alone to have energy. When talking potential energy, there is always an object for that field to exert force upon, so it's natural to conceptualize it as belonging to the object. The concept of a system is a difficult one to explain to a general audience in a single sentence, so it's best to start simple and work up to the advanced stuff. From a fighter pilot's perspective, energy is simply the state of an object's mass (note: mass is not the same thing as matter, so the common visualization of something actually changing in size as it gains energy does not apply), so the real question becomes, "What is mass?" Zaereth (talk) 17:47, 28 September 2015 (UTC)

## Near Earth gravity section is too complicated

The whole point of near earth gravity potential energy is to simplify things. As it stands, that section is not accessible to most people. In other words, its not simple enough and doesn't explain itself well. Why is the final equation U(r) = mgh(r)? What is h(r)? Something definitely doesn't make sense there. Fresheneesz (talk) 02:14, 16 December 2015 (UTC)

Although I'm not 100% sure of the purpose of "(r)," U = mgh means that potential energy is the mass of an object times gravitational acceleration at the Earth's surface (9.8 meters per second per second squared) times the height above the ground. This is good for near-Earth calculations, where the gravitational acceleration is nearly constant; that is, stable to within a 1% margin of error to about 64 kilometers up. Beyond that, gravitational acceleration is not constant but changes (gets less and less the farther away from Earth you get), so the math gets a little more complicated to account for this. Zaereth (talk) 10:18, 16 December 2015 (UTC)