# Gravitational Potential Energy Sign

1. Dec 31, 2014

### Jzhang27143

I understand that GPE is negative, but it does not come out this way when i try to derive it. I took the change in potential energy in bringing a particle from an infinite distance to a distance of b from another particle.

$\Delta U = - \int \vec F \cdot d \vec r$. Since the gravitational force is in the same direction as the path of the particle, $\Delta U = U(b) = - \int \vec F \cdot d \vec r = - \int F dr = - \int_\infty^b \frac{GMm}{r^2} dr = -GMm(-\frac{1}{b}) = \frac{GMm}{b}$. Where is my mistake?

2. Dec 31, 2014

### PWiz

To simplify it, this is how I did it:
$G_b = \frac{PE_b}{m}$ where $G_b$ is the gravitational potential at point b. Then $G_b = \frac {1}{m}\int_b^∞ F dr = \int_b^∞ \frac{GM}{r^2} dr = \frac{GM}{b}$. Now notice that since gravitational force is always attractive, there is a "loss" in potential energy rather than a "gain" when I bring an object from infinity to point b, so I put a -ve sign to account for the loss, then the equation becomes $\frac{-GM}{b}$ .

So you just missed the last step.

3. Dec 31, 2014

### Jzhang27143

Hm, Ok. I have another question. Since the force of gravitation is in the same direction as the particle's path as it is moved from infinity to a distance b away from another particle, the work done by gravity should be positive. However, the math shows that the work done by gravity is negative (which leads to gravitational potential energy being positive.) How would I obtain the correct result purely from math?

4. Jan 1, 2015

### PWiz

The definition of gravitational potential at a point is "the work done on a unit mass in bringing it from infinity to that point", not " the work done by a unit mass in displacing itself from infinity to that point". A positive value suggests that work must be done on the unit mass in bringing it to that point, so the unit mass gains energy. A negative value suggests the unit mass does work coming to that point, so it loses energy(quite like the sign conventions in internal energy changes). Hence, since the gravitational potential energy at any point in the system is less than 0,(work was done in bringing it to that point, not the other way around), you can modify the equation to: $G_b = \frac{-1}{m} \int_b^∞ F dr$, so you have a negative sign with the integral in the end.
Conventionally, a repulsive force is designated a +ve value (since work is done against it in bringing something form infinity to point b) and an attractive force (as is the case with gravity) is designated a -ve value (as work is done in taking something from point b to infinity). If a graph of force (y axis) and radial distance from the mass $M$ (x axis) is plotted, quite clearly, the graph would lie below the x axis(A form similar to $\frac{-k}{r^2}$). The area between the the x axis and the curve with the limits $b$ and ∞ represents the potential energy at $b$, therefore the gravitational potential becomes negative. (You can arrive at this result intuitively as well, but since you wanted a mathematical reason, this is the one. In a regular homework question, a simple explanation between the steps should suffice.)

5. Jan 1, 2015

### sophiecentaur

@PWiz
I can never understand why people seem to have a problem with what you just wrote. It must be by ignoring the word "on" and interpreting it as "by" and by not realising that the formal definition involves the sophisticated idea of 'infinity' - and not 'the Earth's surface'.

6. Jan 1, 2015

### PWiz

@sophiecentaur The concept of infinity in fields is actually very easy to understand if you consider that the potential at that point is either 0 or undefined, depending on whether there is attraction or repulsion. And yes, I also think textbooks should place special emphasis on which object exactly is work being done on. (In electric fields generated by negative charges, the same problem will be encountered if the conceptual understanding is not very clear.) Personally, I always find it easier to understand by drawing graphs.

7. Jan 1, 2015

### Philip Wood

I prefer to derive straight from the definition "Energy of a system is the work it can do". So I consider the work done gravitationally by point mass M on m as m goes (whether dragged or because it's been given KE) from a point R away from M to infinity. We then have…
$$mV_R\ =\ \int_{R}^{\infty}-\frac{GMm}{r^2} \ dr = -\frac{GMm}{R}.$$

Last edited: Jan 1, 2015
8. Jan 1, 2015

### sophiecentaur

I did not say it's impossible; I just suggested that is a possible reason for the difficulty that people often have. If you do not have difficulty then you are lucky. I could suggest that you may find it more difficult, the more you think about it; Philosophers and Mathematicians spend lifetimes on the topic.
Yes. it is acceptable, rather than understandable (understanding is a very high level of operating) once you have done it a few times. The calculations are all there - or, in your case, the graphs. But your graph cannot go out to infinity without tinkering with the scale - which is also 'sophisticated'.

9. Jan 1, 2015

### sophiecentaur

Imo, that wording is open to confusion and I would rather stick to the standard way that Potential is defined. What is the point of introducing KE into the definition of PE?
This is also difficult because it implies work done ON the system rather than, as you start off with, energy 'available FROM'.
Trying to re-write a perfectly well known and universally agreed definition is 'asking for trouble' at some stage. When you try that sort of thing you can't be sure when to stop.

10. Jan 1, 2015

### Philip Wood

I don't bring KE into the definition. The words you've made green can be left out. They were put in only as an aid to visualisation. They are inessential.

I would argue that starting from "the energy of a system is the work it can do" [specifically - and I should have said this - the potential energy of a system is the work it can do because of the relative positions of its parts] is the most fundamental approach.

I am calculating the work done by the system: the dot product of force exerted on m (by M) and displacement of that force on m. The work the system can do is (by definition) its energy. What's wrong with that?

The nice things about this approach are (1) it goes back to a simple, basic definition of energy for simple systems, (2) the calculations don't involve forces from outside the system (e.g. external forces needed to move the body). [This is nice because we're trying to calculate a property of the system itself], (3) It's very easy to use.

Last edited: Jan 2, 2015
11. Jan 1, 2015

### PWiz

I'll be frank here - I derived this entirely on my own while studying fields. My Phy book had no mention of the derivation of the formula, much less the fact that $\int_b^∞ F dr$ represents the PE at b, when it stated the formula for Gravitational potential(it just had the formula for gravitational force of attraction and the formal definition of gravitational potential). I daresay it was the first time I actually had to use definite integration(which we had learned only last year) to find the area under a graph to prove an equation,(by understanding the meaning of having infinity as one of the limits) so I had to come up with the graphical visualization method all on my own(and solve the -ve problem alongside). We haven't even started fields in school yet - I did all the self study to challenge myself to some decently difficult physics hurdles.

Last edited: Jan 1, 2015
12. Jan 2, 2015

### sophiecentaur

That is your definition. However, the Potential is defined in terms of bringing the unit charge / mass from infinity. The limits of your integral imply taking the charge from the point to infinity - the other way round. Furthermore, for an attractive situation, the force will be negative (if your radius is positive). So your sign will be non-standard.
All credit to you for developing this yourself but it should be easy enough for you to retrace your process back to bring yourself in step with the conventions of the rest of the community. It would make it much easier to interface with anything you read elsewhere. There is no ned to be reinventing every wheel you find a need for ('Shoulders of giants' and all that) and, even if you reforming ahead faster than your course is progressing, there are plenty sources for this kind of information. Hyperphysics pages are pretty reliable and easy to navigate around.

13. Jan 2, 2015

### sophiecentaur

That's not 'wrong' but it is just not the definition of Potential, which is Work done to bring a standard mass (/charge) from infinity. If you use the reverse definition then you are asking for confusion at some point. I keep reading comments by people who seem to insist on doing this thing 'their own way'. Why resist changing to the standard approach and definition, even if you already feel comfortable with your personal view? It would hardly be a big step for you to take and then everyone would be singing from the same hymn sheet.

Oh yes they do (by implication). You can't do any work with your system unless you have a reaction force to work against. Your wording just 'hides' it a bit better than does the conventional wording.

14. Jan 2, 2015

### sophiecentaur

Problem with orbits:
The way orbits are analysed is to say they have a constant amount of Energy, shared between PE and KE. How do you square this if PE is defined the other way round?

15. Jan 2, 2015

### PWiz

@sophiecentaur I understand and appreciate your response, but I have a feeling that you and Philip Wood are slightly deviating from the focus of this topic. Please understand that this debate, although healthy in its own respect, may end up confusing the author of the thread.

16. Jan 2, 2015

Staff Emeritus
PWiz, with all due respect, I think the place where the confusion entered into it was when you posted your own derivation using non-standard signs.

17. Jan 2, 2015

### Philip Wood

SC I think we must agree to differ on this one. To me it seems a straightforward application of the definition of potential energy to calculate the work done BY the field as the body goes from P to infinity [See post 7]. What seem to me to be using a reverse definition is to calculate the work needed by an external force to BRING an object to P.

The latter approach is not wrong – I've never claimed that, but I find it less direct.
I think this is a dangerous argument. There was a lot of muttering among my colleagues when some writers of SR textbooks stopped using relativistic mass (which had been a standard textbook concept for decades). Now few writers use the term. I know the analogy with the present case is not perfect, and, emphatically, I am not trying to start an argument about SR. But I do think that, occasionally, textbook approaches should be looked at critically.

Last edited: Jan 2, 2015
18. Jan 2, 2015

### PWiz

Even though the derivation has non-standard signs, it has been accounted for by adding the relevant sign(s) at appropriate stages using sufficient explanation that follows the "unconvential" logic. Of course, that doesn't mean that the standard derivation shouldn't be preferred, but similar mathematical and theoretical treatment has resulted in what I believe to be a scientifically sound argument. Additionally, I have checked back on this with my physics professor and he doesn't see why the derivation should be considered incorrect. The author of the thread is free to ask any doubts that they might have in my reasoning, and as of yet I'm entirely unaware of any confusion that might have arisen in his/her mind regarding my logic. Furthermore, I added a negative sign to the integral to account for the attractive force, which is entirely consistent with the case of electric fields generated by negative charges(where positive unit charges experience an attraction), so I don't see a problem with that.

19. Jan 2, 2015