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NOTE: when I say potential energy I mean gravitational PE

The formula for potential energy is P.E = mgh.

What is h referring to? Height, obviously.

Consider the example: What is the potential energy of a 1kg mass lifted 2 metres off the ground?

m=1, g=9.8, h=2 => P.E=17.6J

My problem is this: why is h the height off the ground, this seems rather arbitrary. Would it not make more sense to have h as the height from the centre of gravity?

Suppose we repeat the experiment on top of a mountain, does the mass still have the same potential energy? I am pretty sure it doesn't.

I am foreseeing one or both of the following answers, so which one will it be?

  1. It doesnt make sense to talk about potential energy in absolute terms, only in terms of gain in potential energy

  2. Potential energy is defined in a way where h is the height off the ground (I dont buy this)

I am leaning towards the first one, but I am still generally uncomfortable with the idea that you can't have an absolute P.E

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    $\begingroup$ What's wrong with PE being not absolute? $\endgroup$
    – Kyle Kanos
    Commented Dec 30, 2014 at 3:19
  • $\begingroup$ @KyleKanos Nothing I suppose.. But if you take h to be the distance from the centre of mass of the earth to the object can't you have an absolute PE? $\endgroup$ Commented Dec 30, 2014 at 3:21
  • $\begingroup$ GPE represents the work required to move an object from one height to another, so it's really a "difference" that matters. $\endgroup$
    – Kyle Kanos
    Commented Dec 30, 2014 at 3:26

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It's the first one. This is a really excellent observation! It's a fascinating fact of physics.

Absolute potential energy is a silly idea. If you take a bunch of different objects, list their potential energies, and then add $100$ to each one, nothing will change about how the system behaves. We only talk about relative potential energy.

The kinetic energy an object gains in falling a certain height is equal to the potential energy it has lost. If we let an object fall from $h_1$ to $h_2$, we find that its change in kinetic energy is $\Delta KE = m g h_1 - m g h_2.$ If we add any arbitrary number $C$ to each of these potential energies, the difference is the same: $\Delta KE = (m g h_1 + C) - (m g h_2 + C) = m g h_1 - m g h_2.$

We often use the height off of the ground because it means that at ground level, $PE = mgh = mg \cdot 0 = 0$ for all objects, and since nothing can be lower than ground level in a simple system it makes sense to say that $0$ is the lowest possible potential energy anything can attain.

EDIT:

To add to this, let's look at a little extra mathematical formalism. It turns out that in classical mechanics, the nonexistence of "absolute potential energy" is a special case of something called gauge invariance.

For simplicity, let's talk about a one dimensional system - we have a ball that can only move back and forth along a single line. Let $x$ be the position of the ball.

Let $U(x)$ be the potential energy of the system as a function of the position of the ball. This could, for example, be a simple gravitational problem -- $x$ is the height of the ball above the ground, and $U(x) = m g x.$ But for generality, we won't specify what the form of $U$ is.

What we will say is that potential energy is the result of some force acting on an object. We know that the potential energy of an object at a certain position resulting from a given force is the work required to bring that object to that position. So if an object is acting under a force $F(x)$, the potential energy at position $x$ is $U(x) = W = - \int F(x)\, dx$ (the negative sign comes from the fact that we must work in the opposite direction of the force).

So from the fundamental theorem of calculus, if $U(x) = - \int F(x)\, dx$, then

$F(x) = - \frac{d U(x)}{d x}.$

Okay, this is interesting. In classical mechanics, we can completely describe the motion of a system if we know the forces acting on it (since we can then use Newton's law $F = ma$). But since we know $F(x) = - \frac{d U(x)}{d x}$, we can describe the motion of the system completely by knowing the potential energy.

Here's the payoff:

If the forces are the same for two different potential energy functions, then those potential energy functions result in the same physical behavior.

Mathematically:

If $U_1(x)$ and $U_2(x)$ are two potential energy functions such that $- \frac{d U_1(x)}{dx} = - \frac{d U_2(x)}{dx}$, then the potential energy functions result in the same physical behavior.

What does it mean if two functions have the same derivative? Well, it means that they differ by a constant.

Oh! That's where we wanted to get to, isn't it? If two potential energy functions differ by a constant, then they result in the same physical behavior. So it doesn't make sense to talk about "absolute potential energy", because no matter what we can add any constant we want and we'll obtain the same forces and thus the same physical behavior.

Hence it only makes sense to talk about changes in potential energy, not absolute potential energy.

(I said earlier that this is an example of a gauge invariance -- choosing a different constant to add to your potential energy function may be referred to as choosing a different "gauge" [which is a physical term]. The principle of gauge invariance states that the physical behavior of the system is the same regardless of which gauge you choose. In physics we often choose the gauge that makes our calculations the simplest -- which is why we choose the potential energy function $m g h$, where the potential energy is zero at ground level. This is an example of picking a useful gauge)

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  • $\begingroup$ If I were to bury something under the ground, wouldnt it still have potential energy? $\endgroup$ Commented Dec 30, 2014 at 3:14
  • $\begingroup$ It would! But its potential energy would be negative if you choose to let the ground level be 0 potential energy. I just added a bunch of mathematical formalism to the end of my answer that you may find useful, but it may be too obtuse. Let me know if you need further clarification. $\endgroup$
    – user_35
    Commented Dec 30, 2014 at 3:26
  • $\begingroup$ Your maths explanation really cleared it up, the bit with the derivatives and all enlightened me. Thanks heaps! $\endgroup$ Commented Dec 30, 2014 at 3:29
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"Why is h the height off the ground, this seems rather arbitrary?"

It is not arbitrary, it is useful and convenient because we live on the ground. But your intuition is still right.

P.E. = mgh, but this formula is only an approximation. It assumes that g is constant, which it is not, it depends on altitude. If you are at the height of the moon, g is very different. But if we are sticking with heights that are relatively close to the ground, this formula works well. If you want to be exact, you have to do the integral where you are using the correct value of g for each height.

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In brief: Actually$ h$ is the height of body obtained by subtracting it absolute distance from centre of earth by Radius of earth. You should look for the derivation of this relation which first uses the original gravitational potential formula and then by neglecting and binomial expansion etc derives this approximate relation.

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