# Why is gravitational potential energy negative, and what does that mean?

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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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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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! –  BlueRaja - Danny Pflughoeft Nov 17 '11 at 22:18

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.

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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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This question is about potential energy, not acceleration vectors... –  BlueRaja - Danny Pflughoeft Aug 23 '14 at 21:30