I usually think of gravitational potential energy as representing just what it sounds like: the energy that we could potentially gain, using gravity. However, the equation for it (derived by integrating Newton's law of gravitational force)...

$$PE_1 = -\frac{GMm}{r}$$

..has me thrown for a loop, especially after this answer.

  • If potential energy really meant what I thought it did, then it would always have to be non-negative... but this equation is always negative. So what does "negative potential energy" mean!?
  • If $KE + PE$ is always a constant, but PE is not only negative but becomes more negative as the particles attract, doesn't that mean the kinetic energy will become arbitrarily large? Shouldn't this mean all particles increase to infinite KE before a collision?
  • If we are near the surface of the earth, we can estimate PE as $$PE_2 = mgh$$ by treating Earth as a flat gravitational plane. However, $h$ in this equation plays exactly the same role as $r$ in the first equation, doesn't it?
    • So why is $PE_1$ negative while $PE_2$ is positive? Why does one increase with $h$ while the other increases inversely with $r$?
    • Do they both represent the same "form" of energy? Since $PE_2$ is just an approximation of $PE_1$, we should get nearly the same answer using either equation, if we were near Earth's surface and knew our distance to its center-of-mass. However, the two equations give completely different answers! What gives!?

Can anyone help clear up my confusion?

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    $\begingroup$ Energy is spent doing work. $\endgroup$ Commented Aug 13, 2018 at 4:24

7 Answers 7


About negative energies: they set no problem:

On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.

However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:

let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:

$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$

as expected: we lose $PE$ and win $KE$.

Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.

Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $ r = r_{earth}$.

The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).

By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:

$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.

As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$

Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:

$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.

Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.

So, from what I understood, your logic is totally correct, apart from two key points:

  • energy is defined apart of a constant value.

  • in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.

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    $\begingroup$ Ah, I see, the trick is that it's a relative value - I keep thinking of energy as something absolute (though I guess even kinetic energy changes, depending on your frame of reference). I suppose we'd like to set PE=0 when r=0, but unfortunately, according to the equation it would take infinite energy to pull the particles apart! So I guess PE=0 when r=∞ is the only other reasonable choice. It all makes sense now - thanks! $\endgroup$ Commented Nov 17, 2011 at 22:18
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    $\begingroup$ Also, the formula changes inside a non-point mass, so the $r\to 0$ limit is finite. $\endgroup$
    – J.G.
    Commented May 10, 2018 at 19:57

I will first (1) summarize the differences between the definitions of PE1 and PE2 and then I will (2) equate the two.

(1) First, as this answer to "Why is gravitational energy negative?" says, PE1 defines the potential energy of a body of mass m in the gravitational field of a mass M as the energy (work) required to take it from its current position $r$ to infinity. PE1 assumes $r=\infty$ is $PE=0$ $$PE1=\frac{−GMm}{r}$$

PE2, on the other hand, is defined as the negative of the work done by gravity to lift a body of mass m from the surface of a planet to a height h above the planet.

$$PE2=-W=-Fdcos\theta =mgh$$

PE2 has a different frame of reference than PE1, as it assumes $PE=0$ at $r=R$, or on the surface of the planet. Also, and very importantly, PE2 is only used when an object is close to the surface of a planet, when $h<<<R$ (R is radius of planet), and g can be assumed to be constant:

$$g=\frac{GM}{(R+h)^2} \approx \frac {GM}{R^2}$$

(2) OK, now on to equating the two. Though the frames of references for PE1 and PE2 are different, $|\Delta PE|$ between two points should surely be the same. For the sake of example, let's say the two points are the surface of the planet and height h above the planet.

PE1 says $|\Delta PE|=mgh-mg(0)=mgh$

PE2 says $|\Delta PE|=\frac {-GMm}{R+h}- \frac {-GMm}{R}=GMm\left(\frac{1}{R}-\frac{1}{R+h}\right)=GMm\left(\frac{h+R-R}{(R)(h+R)}\right)=\frac{GMmh}{(R)(R+h)}$

and because $h<<<R$, $\frac{GMmh}{(R)(R+h)} \approx \frac{GMmh}{R^2} =mgh$

And thus, PE1 and PE2 both represent the same form of energy, but we must keep in mind the frames of reference and the conditions of use when we use them.

Hope this helps!! Peace.


I think it's just a preference.

We could see gravitational potential energy as being positive, representing the energy "invested" in our position relative to a massive object. We can "regain" that energy (increase kinetic energy) by moving closer to the object, at which point we have lowered the amount of energy we could gain by moving further. So potential energy decreases as we move closer (approaching zero energy at zero distance), increases as we move further away, and the sum of PE and KE is constant.

But what value is the constant? When we are very very far away from the massive object we should have very very large potential energy. But even when we're quite close to the massive object, we're very very far away from every other massive object in the universe, and thus should have very very large gravitational potential energies relative to all those objects. We can approximately calculate a value for KE + PE by considering only the most relevant objects (the closest and/or largest ones), but our approximate value just grows and grows and grows as we try to get more accurate approximations by including smaller and more-distant objects in our category of "relevant" objects. So our KE + PE constant is some impossibly large value we can never really calculate or estimate as some specific value. In some ways it doesn't matter that we can't ever claim a value, since differences of energies are all we really need to work with, and we can still calculate those (by assuming that our PE relative to everything else in the universe has only changed negligably when we move around near the massive object we're considering). But it seems unsatisfying.

On the other hand, instead of considering PE as a positive quantity of energy "invested" in our position (energy we've already "spent" if we were moving away from the massive object, which we could gain by moving closer), we can instead consider it a negative quantity of energy we "owe" because of our position (energy we've gained "for free" if we moved closer to the object from infinity, which we would have to "spend" to escape to infinity again).

All the calculations of energy differences work out the same anyway. But now our PE relative to an object goes to zero as we get very very far away from the object. This mean that as we can calculate an approximation of our KE + PE constant by only considering the most relevant objects, and as we try to get better approximations by including smaller and more-distant objects in our calculation the effects of those additional objects gets closer and closer to zero. So we come up with an actual number we can justifiably say is the value for our KE + PE constant.


It is because gravitational force is attractive and work is done by gravitational force itself. When system does work itself energy is taken as negative and when work is done by external agency on system energy is take as positive.


Gravity is an acceleration. No negative involved.

However, when you use acceleration to find a velocity, since velocity is a vector quantity, you must describe a direction. It is convention that anything that accelerates up, is described as a positive(+) like "The ball accelerates at 20m/s^2", whereas gravity describing a downward acceleration is described as (-) "-9.8m/s^2".

This applies to anything accelerating on the X axis as well. "The car accelerates at 10m/s^s when you apply the gas" or "The car accelerates at -4m/s^2 when you apply the brakes."

I believe this is done to make things easier when making graphs.

However, if you were to just say "I have a ball. It will be displaced, how far will it be displaced? (Notice how its not 'displaced north, or to the left ')" In a situation like that, you would use the acceleration of gravity without the negative. "It will be displaced by 9.8m every second^2".

I hope this helps. Then again, I might have completely misread your question. Either way, have a good day!

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    $\begingroup$ This question is about potential energy, not acceleration vectors... $\endgroup$ Commented Aug 23, 2014 at 21:30

The fact that the gravitational potential energy as with all potential energies of attarctive forces are negative is based on the fact that we want to assume that when the particles are at infinity with respect to each other and at rest the system have zero total energy. Imagine if this was not the case and a system of two particles at infinite separation at rest would be taken to have a net energy then there would arise some confusion as to energy associated with the rest mass. The total energy of the system would then not be $E=Mc.c$ where $M$ is the sum of two masses. From where then would this extra energy come?


It is wrong to consider gravitational potential energy to be negative--tho common.

The big mistake is in assigning the P.E. at infinity = 0. This is clearly wrong--P.E. is clearly 0 at 0 separation, and large at great separations. The P.E. of objects far away from each other would have to be the summation of the P.E. for the first say 100' of separation plus the P.E. for the second 100' of separation plus---the P.E. for every 100' until the entire separation was accounted for. (I will express this as an integral after I brush up my calculus.) Viz, P.E. INCEASES as separation increases--starting at 0 at no separation.

Many people are making a big mistake in considering gravitational potential energy to be negative!

  • $\begingroup$ With the field from a point source obeying the inverse-square law, the force is proportional to $r^{-2}$ and the potential (and the potential energy) is therefore proportional to $r^{-1}$. The linear $P=mgh$ is just an approximation for small changes in distance. $\endgroup$
    – HDE 226868
    Commented May 10, 2018 at 20:00
  • $\begingroup$ @HDE226868 Did you mean to comment on a different answer? $\endgroup$
    – diracula
    Commented May 10, 2018 at 20:48
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    $\begingroup$ @diracula No - I should have made myself clearer. I was showing mathematically why the potential energy vanishes at infinity rather than growing to infinity; as $r\to\infty$, $r^{-1}$ goes to $0$. $\endgroup$
    – HDE 226868
    Commented May 10, 2018 at 20:49

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