# Gravitational Potential Energy Sign

• Jzhang27143

#### 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?

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.

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?

donaldparida
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?
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.)

donaldparida, Jzhang27143 and 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'.

@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.

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…
$$mV_R\ =\ \int_{R}^{\infty}-\frac{GMm}{r^2} \ dr = -\frac{GMm}{R}.$$

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@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.
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'.

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…
$$mV_r\ =\ \int_{R}^{\infty}-\frac{GMm}{r^2} \ dr = -\frac{GMm}{R}.$$

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.

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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.

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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'.
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.

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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.
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.

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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.

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.

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?

@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.

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.

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.
Why resist changing to the standard approach and definition, even if you already feel comfortable with your personal view?
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.

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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.
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.

PWiz, I repeat. The place where the confusion entered was when you posted your own derivation using non-standard sign conventions, and not when (as you said) others pointed out that these were non-standard. I would also maintain that if someone is having trouble understanding a sign convention, presenting him with a different conventions cannot possibly do anything more than confuse him.

@Vanadium 50 If that is the case, then I shall make an attempt to actively refrain from posting my own derivations in responses to such questions and only present the most commonly accepted one from a credible source. In any case though, I'd love to hear from the thread author himself about any confusion he/she might have and clear it, since it would weigh on my conscience if someone goes away more confused than they originally were after my input.

if someone goes away more confused than they originally were
I certainly did, if not exactly because of your input.

I came to this thread with the intention of helping the OP, only to realize half-way through writing a response that I don't understand the issue myself, and the subsequent posts failed to clear it up for me (and, I dare to presume, perhaps also for the OP).

For one, I don't see the issue @sophiecentaur has with the direction in the derivation - you get the same result for ##U(b)## regardless of the direction you move the test particle.

For m moving from infinity to b:
$$\Delta U= - \int_\infty^b \vec F \cdot d \vec r = - \int_\infty^b F dr = - \int_\infty^b \frac{GMm}{r^2} dr = -GMm(-\frac{1}{b}) = \frac{GMm}{b}$$
$$\Delta U = U(b) - U(\infty) = U(b)$$
$$U(b) = \frac{GMm}{b}$$
For m moving from b to infinity:
$$\Delta U = - \int_b^\infty \vec F \cdot d \vec r = - \int_b^\infty - F dr = - \int_b^\infty \frac{GMm}{r^2} dr = -GMm(\frac{1}{b}) = - \frac{GMm}{b}$$
$$\Delta U = U(\infty) - U(b) = - U(b)$$
$$-U(b) = - \frac{GMm}{b}$$
$$U(b) = \frac{GMm}{b}$$

And of course the sign for U at point b in both cases is positive, and I don't see the reason why it should come out negative. If it's a matter of convention, then it feels artificial to insert it at this stage. The only two sign conventions I'm aware of in this case is the negative sign for the force if it's not in the same direction as the displacement when switching from the vector form, and the ##\Delta U=-W##. Shouldn't these two alone net the correct result?

I saw this derivation in the wikibooks section:
http://en.wikibooks.org/wiki/Physic.../Gravitational_Potential_Energy#Another_proof
that nets a negative result, but I don't understand why they switch the integration limits for no apparent (to me at least) reason in the third line. They certainly do not mention any extra conventions being applied at any stage.

I checked in Kleppner-Kolenkow(p.171), and they just leave the derivation at
$$U(r)=\frac{A}{r}$$
again, with the positive sign.

So, to repeat the question the OP asked: where's the mistake?

O.k., never mind. I found the cock-up:
For m moving from b to infinity:
$$\Delta U = - \int_b^\infty \vec F \cdot d \vec r = - \int_b^\infty - F dr = - \int_b^\infty \frac{GMm}{r^2} dr = - \frac{GMm}{b}$$
I missed a minus sign. It should be:
$$\Delta U = - \int_b^\infty \vec F \cdot d \vec r = - \int_b^\infty - F dr = - \int_b^\infty - \frac{GMm}{r^2} dr = \frac{GMm}{b}$$
So while the ##\Delta U## is equal in both cases, the energy at ##U(b)## has opposite sign.

So the convention of bringing the mass from infinity to b rather than the other way around changes the sign of ##U(b)##, and the OP's mistake was to use the wrong convention with regards to the direction of motion.

Is that correct?

I still don't get what happened in the wikibooks derivation.

edit: Hmmm, and why does K-K use the other convention? (with initial position at infinity)

Welp. I'm still quite confused, I suppose.

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@Bandersnatch To resolve your confusion, let's disassemble the problem from the solution. Gravitational potential energy is given by ##\frac{-GM}{r}## , and since G.P. = G.P.E /mass , then it means that GPE is always negative(so let's ignore G.P now). Gravitational potential energy at a point is defined as "the work done on a mass in bringing it from infinity to that point." This definition is fixed, so note two important points that this implies:
1) It is a measure of the work done "on" the mass.
2) The mass is always brought from infinity.
Firstly, from point 2, we know that the limits of the definite integral will always be ##\int_b^∞##.
Second, gravity is always attractive, so no external work needs to be done on the mass to move it from infinity to a particular point - it naturally moves towards that point due to the gravitational field. This means that the mass does work in moving from infinity to a particular point, and since the work is not externally applied on the system, but rather being done "by" the system, G.P.E is negative. So how do we prove this mathematically?
Force is a vector quantity. A force which pushes the object away from the center of the field is considered positive, and a force which pulls the object towards the center of the field is considered negative. A force which causes work to be done against it to move something from infinity to that point is considered positive. A force which does the work in moving something from infinity to that point is considered negative. Hence, by convention, an attractive force in a field is always negative.

So G.P.E = ##\int_b^∞ -F dr = \int_b^∞ \frac{-GMm}{r^2} dr = \frac{-GMm}{b}## . Graphically, -F (y axis) and radial distance from ##M## (x axis) plotted on a graph will give you a curve similar to ##\frac{-k}{r^2}##, and you are simply required to find the area with the limits ∞ and ##b## between the graph and the x axis, which will always be negative. (I've mentioned a lot of this in my previous posts too.)

@PWiz that is not the point of my or the OP's confusion. It's not a question about physical meaning, but about maths. The derivations are all laid down already, and just rewriting them is not helping, I'm afraid.
What I don't understand is the reason for switching integration limits in the third line of the wikibooks link I gave, and for adopting the convention of bringing the mass from infinity to b in the Kleppner-Kolenkow derivation (which contradicts you supposedly hardline point #2).

@PWiz that is not the point of my or the OP's confusion. It's not a question about physical meaning, but about maths. The derivations are all laid down already, and just rewriting them is not helping, I'm afraid.
What I don't understand is the reason for switching integration limits in the third line of the wikibooks link I gave, and for adopting the convention of bringing the mass from infinity to b in the Kleppner-Kolenkow derivation (which contradicts you supposedly hardline point #2).
The answer is simple - reversing the limits does the same job as removing the -ve sign, as has been done in the set of equations you have provided in the link. Theoretically, there is no need to reverse the limits, as the negative sign can be taken out of the expression to make it ##-\int_b^∞ F dr##. Reversing the limits and removing the negative sign results in the same final equation, but it is unnecessary and confuses one. Therefore, there is no contradiction of point 2).

The answer is simple - reversing the limits does the same job as removing the -ve sign, as has been done in the set of equations you have provided in the link. .
You'll see that they don't do that. The minus sign from the misalignment of vectors is inserted in line four.

But no matter, I get it now. They used the work-energy theorem with the convention of ##\Delta U = -W##, and reversed the limits while getting rid of the minus by the ##W##

Which brings me to your derivation in post #23. It's incorrect.

You should start with ##\Delta U = -W## (by convention), and regardless of the direction you get a positive value because areas are never negative. Instead you wrote ##\Delta U = W##. You've missed a minus sign there.
So, actually, you can only get ##U(b)=-\frac{A}{b}## after splitting ##\Delta U = U_{final}-U_{initial}##, which for motion from b to infinity nets you ##\Delta U = -U(b)## and you get the much sought after minus sign.

But never mind that, I seem to be getting there. The only question left is why did Kleppner-Kolenkow use the convention of bringing the point from infinity to b, only to arrive at the positive value for ##U(b)## if doing the opposite is supposedly so hardline?

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After all this learned discussion, I rest my case (post 7)...

regardless of the direction you get a positive value because areas are never negative. Instead you wrote ΔU=W\Delta U = W. You've missed a minus sign there.
Areas are always positive, but integrals are not. I've already accounted for the - sign by specifying the nature of the force (attractive/repulsive). I'm not going by convention, but by definition. If work is done on the mass, then the value is is positive, if not, then it's negative. I feel a little tired having to repeat myself so many times, so if you're still disagreeing with what I say, then I believe it is best for me to leave the discussion here, since I have no explanations more exhaustive in nature than the ones I have provided.
Have a good day.