# What is potential energy?

1. Aug 6, 2015

### rogerk8

Hi!

I just wish you to help me derive two types of potential energy.

A) Gravitational Potential Energy
B) Electrical Potential Energy

A useful equation is

$$W=\int Fdr$$

Which I consider is the work that has to be done to the particle to move it a certain distance from the attractive centre of force (which then becomes it's PE, is my thought).

For A we have

$$F=G\frac{mM}{r^2}$$

Now

$$W=\int_a^hG\frac{mM}{r^2}=GmM\int_a^h \frac{1}{r^2}$$

Which means

$$W=GmM((-1/h)-(-1/a))=GmM(1/a-1/h)$$

Where a cannot be zero but maybe it could be R?

Then we have

$$W=\frac{GmM}{R}(1-\frac{R}{h})$$

And if we set

$$\frac{GM}{R}=g$$

We have

$$W=mg(1-\frac{R}{h})$$

But this till does not look like Ep=mgh :D

I get the same problem in the B-case.

Considering Hydrogen for simplicity

$$F=\frac{e^2}{kr^2}$$

then

$$W=\int_a^h \frac{e^2}{kr^2}=\frac{e^2}{k}\int_a^h \frac{1}{r^2}$$

which gives

$$W=\frac{e^2}{k}((-1/h-(-1/a))=\frac{e^2}{k}(1/a-1/h)$$

where once again a cannot be zero.

Can it be the shell radius, then?

There's no real use starting at a point inside the shell radius so let's play with r, this time.

This gives

$$W=\frac{e^2}{kr}(1-r/h)$$

Which is about the same result as for the Gravitational Potential Energy above.

Is this totally wrong?

Best regards, Roger
PS
It is interesting to note that PE increses with distance in both cases. If I'm right, that is :)

2. Aug 6, 2015

### DEvens

What is $h$? Is it supposed to be the height above the original location? The height above the surface of the Earth? The distance from the location a?

If I guess that $h$ is supposed to be the distance from $a$ to $b$, then $r$ at $a$ is $a$, right? And $r$ at $b$ is $b=a+h$, not $h$. Right?

So I think you got your integral a little wrong. I think you did something similar in the electric force case.

After that, the formula $E_p = mgh$ assumes a constant gravity force. To get that, you would need to use the fact that $h << R$ where $R$ is the radius of the surface of the Earth, and you are not moving very far from the surface.

If $h$ is large enough that you cannot consider the force to be constant over a change of location the size of $h$, then you should not get a potential proportional to $h$.

3. Aug 6, 2015

### Staff: Mentor

Usually we set $a=\infty$

Last edited: Aug 6, 2015
4. Aug 6, 2015

### rogerk8

I think it is quite obvious that h is something larger than a, otherwise the integral sign would be reversed.

Here you are being a bit technocratic while of course being right, upper integration limit should equal a+h. I however just see it like h is "much" larger than a (in the EPE case) so you might as well use h only. This is however not true for GPE, I now understand.

Sloppy of me :)

A kind of obvious statement but thank you for telling me.

I will now revise my integrals :)

Best regards, Roger

5. Aug 6, 2015

### Staff: Mentor

You are probably actually interested in:

$$W=\int_R^{R+h}G\frac{mM}{r^2} dr$$

which gives (you can confirm this by substituting in these limits in the expression you derived above)
$$\frac{GMmh}{R^2+hR}$$

Do a Taylor series expansion around h=0 to get
$$\frac{GMmh}{R^2} + O(h^2) \approx mgh$$

for $g=GM/R^2$

Last edited: Aug 6, 2015
6. Aug 6, 2015

### DEvens

Dale, I was trying not to do the entire question for him.

7. Aug 6, 2015

### Staff: Mentor

Oh, I thought you were unsure too. Oops.

8. Aug 6, 2015

### rogerk8

Can this be in the vicinity of being right?

Best regards, Roger

9. Aug 6, 2015

### rogerk8

Let's try:

$$W=GmM(\frac{1}{R}-\frac{1}{R+h})$$

$$W=GmM\frac{(R+h)-R}{R(R+h)}$$

$$W=GmM\frac{h}{R(R+h)}$$

I got it :D

I think I get your Taylor expansion too, DaleSpam!

While h is so small copared to R, h^2 terms does not matter.

But it is really a more approximate formula than I thought!

Best regards, Roger

10. Aug 6, 2015

### Staff: Mentor

Yes. The $mgh$ formula is an approximation. Of course, the $GMm/r^2$ formula is also an approximation. The corresponding general relativity equations are also approximations.

11. Aug 7, 2015

### rogerk8

I've decided to collect my formulas so they look nice and comprehensive:

This is an attempt in deriving two expressions for potential energy.

A) Gravitational Potential Energy (GPE)
B) Electrical Potential Energy (EPE)

A useful equation is

$$W=\int Fdr$$

Which I consider is the work that has to be done to the particle to move it a certain distance from the attractive centre of force (which then becomes it's PE, is my thought).

For A we have

$$F=G\frac{mM}{r^2}$$

Now

$$W=\int_a^{a+h}G\frac{mM}{r^2}dr=GmM\int_a^{a+h} \frac{1}{r^2}dr$$

Which means

$$W=GmM((-1/(a+h))-(-1/a))=GmM(1/a-1/(a+h))$$

Where a cannot be zero but maybe it could be R?

Then we have

$$W=GmM(\frac{1}{R}-\frac{1}{R+h})$$

Rearrangeing this equation we get

$$W=GmM\frac{(R+h)-R}{R(R+h)}$$

and finally

$$W=GmM\frac{h}{R(R+h)}$$

which for h<<R gives

$$W=\frac{GmMh}{R^2}$$

and putting

$$g=\frac{GM}{R^2}$$

gives

$$W=mgh$$

In the B-case we have:

Considering Hydrogen for simplicity

$$F=\frac{e^2}{kr^2}$$

then

$$W=\int_a^{a+h} \frac{e^2}{kr^2}dr=\frac{e^2}{k}\int_a^{a+h} \frac{1}{r^2}dr$$

which gives

$$W=\frac{e^2}{k}((-1/(a+h)-(-1/a))=\frac{e^2}{k}(1/a-1/(a+h))$$

where once again a cannot be zero.

Can it be the shell radius, then?

There's no real use starting at a point inside the shell radius so let's play with r, this time.

This gives In the same manner as for GPE above

$$W=\frac{e^2}{k}\frac{h}{r(r+h)}$$

or when h>>r

$$\frac{e^2}{kr}$$

Let's repeat two equations for convenience:

$$W=GmM\frac{h}{R(R+h)}$$

$$W=\frac{e^2}{k}\frac{h}{r(r+h)}$$

The first is GPE non-approximated, the second is EPE non-approximated.

It is interesting to note that they only differ in two aspects:

1) Different constants.
2) In the GPE case the normal use is for h<<R while in the EPE case the normal use is for h>>r

It is even more interesting to note that PE increases with h.

Best regards, Roger

12. Aug 8, 2015

### rogerk8

It is a bit fascinating what happens in extreme cases of GPE and EPE.

If we begin with GPE.

When h is much larger than R we have

$$Ep=\frac{mMG}{R}$$

Which means that the potential energy of a mass unit outside the gravity field cannot be higher than that at the surface of (earth).

And in the EPE case where h (or r) is much larger than the nuclei radius we have

$$Ep=\frac{e^2}{kr}$$

So the potential energy is related to the electron orbit radius when Ep is considered inside the atom, but the same if the electron is far from the electron radius.

This has to do with the same principle as above that is if the electron is far from its orbit, its energy is what I say but we could turn that around and say that if the electron really is in its orbit, the distance to the nuclei radius is "equally far" so the same principle may apply.

Right?

Best regards, Roger

Last edited: Aug 8, 2015
13. Aug 8, 2015

### Staff: Mentor

No, in that formula the PE = 0 at h = 0. Indeed, that is the meaning of setting the lower limit of the original integral to R. You are choosing to define the PE such that it is 0 at R.

14. Aug 9, 2015

### rogerk8

I don't agree.

In my humble opinion GPE in the farfield is exactly that as at the surface, R

Repeating for convenience

Nearfield GPE (NGPE):

$$Ep=GmM\frac{h}{R(R+h)}$$

Farfield GPE (FGPE):

$$Ep=\frac{mMG}{R}$$

If you look at NGPE ant use h>>R you get FGPE.

In FGPE the only variable (except m) is R and R is the radius of the Earth.

So I would call FGPE as I said, that is that a mass unit far from the Earth cannot have higher PE than what it has on the surface, thus R.

Sorry, I see now that I am wrong because at small h I need to use NGPE which is zero at h=0 (as you say).

It was just the fact that only R is present in the FGPE expression that fooled me.

Also, I have made terrible mistakes regarding EPE too.

I do not even want to go into them.

Well, I do :)

Repeating for convenience

Nearfield EPE (NEPE):

$$W=\frac{e^2}{k}\frac{h}{r(r+h)}$$

Farfield EPE (FEPE):

$$Ep=\frac{e^2}{kr}$$

Here we have chosen the boundaries to be from the electron orbital radius (r) to a hight (r+h).

In the NEPE-case, h is small (compared to r).

In the FEPE-case, h is large.

Both equations considers movement of electron outside of its shell.

But what happens if we try to use the equation inside the shell?

Well, in that case we have

$$r=r_n$$

and

$$h=r_e$$

Where n stands for nuclei and e for electron.

If we just imagine this we may reuse the formula for FEPE (while h>>r)

So that the PE of the orbiting electron becomes

$$Ep=\frac{e^2}{kr_n}$$

Which probably is not true :)

Best regards, Roger
PS
Wait a minute. What is the potential energy of an electron doing what it's born to do? Maybe we could set EPE to zero for h=0 in the same manner as we do for GPE?

15. Aug 9, 2015

Staff Emeritus

16. Aug 9, 2015

### rogerk8

Have you read my whole post?

Where I confess that I understand that I'm wrong?

Or is this the way you have made over 18000 posts?

17. Aug 9, 2015

### Staff: Mentor

No, this is wrong.

Both of these are OK. Now, at the surface of the earth, h = 0, so: $$E_p = GmM \frac{0}{R(R+0)} = GmM \frac{0}{R^2} = 0$$ which is not the same as the far-field GPE ($h \rightarrow \infty$).

18. Aug 9, 2015

### rogerk8

Thanks for correcting me!

Best regards, Roger

19. Aug 9, 2015

Staff Emeritus
No, I didn't. Once people say something wrong, there is no point in reading further. Because most people don't hit the "post reply" but after they have written what they themselves recognize is uttrer and complete nonsense, because most people are respectful enough of others to not deliberately waste their time. I was wrong, obviously.

20. Aug 9, 2015

### rogerk8

Thank you for getting back to me!

But I hope it is ok to elaborate own thoughts which may not be considered physics truths but just to lay them on the table.

I actually think this is a liberating way to attack a problem.

You state something, you reflect upon what you have said, you either stick to it or abandon it.

I think this way of reasoning is rather healthy.

But of course, I can always listen to you skilled guys telling me how things are.

But try and put yourself into my ignorant position, how much will I then really understand?

Best regards, Roger