Is it correct to say, any object has a huge amount of potential energy?

Question

Let's say if it is a $ 1 kg $ metal ball.

1. Now if we consider it together with a planet some 500 light years away (or if we consider Neptune), then there is potential energy between this metal ball and that planet

2. With the absence of anything else, that is, if we only consider two physical bodies, of this metal ball and the planet, the metal ball will accelerate towards that planet.

3. But since it is so small, let's consider that we the metal is only 1000 meter away from that planet

4. Now we can say, there is a certain amount of potential energy here, as the metal can accelerate towards the surface of the planet and even do some damage -- it is not a negligible amount

5. Let's just say it is 1 Joule of potential energy, to make it simple. Some people claim that let's say at 1000000 meter, the potential energy approaches zero, but there shouldn't be the case, because if it is 1 Joule at 1000 m, then it has to be more potential energy at 2000 m, and so forth. So at 1000000 m or 1 trillion meter, the potential energy is 1 Joule or greater.

6. Now if we consider this metal ball with the 100 trillion of planets in the universe. Between the metal ball and each of the planet, there is a non-negligible amount of potential energy -- at least 1 Joule.

7. So therefore, this metal ball has 1 Joule x 100 trillion = 100 trillion Joule of potential energy

8. And this applies to any object on Earth or in the universe at all

9. And in fact, since if we consider any two objects (or two planets) at one time, then we can use the $N$ choose 2 calculation, and see that the amount of potential energy has an $ n^2 $ amount, if we consider $ n $ objects, such as $ n $ being 100 trillion

Is the above true? If it is true, that means in the universe, there is a huge amount of potential energy, even if we are just to look at a $ 1 kg $ metal ball?

Answer 1

There are a couple of kinks in your logic that we need to straighten out.

The first is that when people say the potential energy at infinity is zero, they don't mean that it is less than the potential energy at 1000 metres from Neptune. It is just a convention to set the PE to zero at infinity, so the PE decreases (ie becomes more negative) as the ball approaches Neptune.

The second is that adding trillions of other planets/stars into the picture doesn't increase the PE trillions of times. If one planet is trying to pull the ball in one direction, and another is trying to pull the ball equally in the opposite direction, the effects cancel, they don't double the PE.

Answer 2

In response to your edits and comments ...

In the sense in which I think you mean the question -- you are right that the potential wells of all the stars and planets in the galaxy add to change the potential energy of the 1kg ball.

I think that the people here, including myself, are reading between the lines. They are responding to something that you did not say explicitly.

They think that you think that this potential energy is some kind of massive energy density stored in the 1kg ball: that the ball you see in front of you has this massive potential energy stored in it that might be realized into work on the Earth in some sense.

However, the other planets and stars also change the potential energy of the Earth - so this process does not change the potential energy of the ball with respect to the Earth.

It only changes it with respect to a datum that is not attached to the Earth. The datum at infinity in a flat Newtonian universe is one such datum. In a less trivial topology, even that option is not available. It becomes very difficult to find a universal standard datum for potential energy.

---

Consider two planets in an otherwise empty Newtonian universe with a single test particle. Infinity can be given as an objective reference for zero potential energy. Then the potential energy for the point anywhere is the energy it would take to move this test particle out to infinity (approaching zero velocity in the limit).

The energy required to move the particle to infinity if it was a long way from Neptune is relatively small and the potential energy near the Earth is fairly much the potential energy due to the Earth alone.

If however, you measure the potential energy rather subjectively from the bottom of the gravitational well of Neptune, then the particle does have a lot of potential energy. But this is not the sum of the potential energy from the nearby Earth plus that of the Neptune - in fact being near the Earth would reduce the potential energy with respect to the datum at Neptune.

For example, if you had two Neptunes far apart, then the potential energy on the surface of one Neptune is just once that potential energy and not twice it.

However, in the practical sense of the local behaviour of the particle, this is meaningless, as it is pushed not by the potential energy but the change in potential energy with position. The change in the potential energy due to Neptune, a long way from Neptune, is very small. So such a datum make little practical sense.

If you wanted to measure the potential energy from deep in the gravitational well of a neutron star, then the particle would have even more energy. But, again, it would not have any practical meaning if you were a long way away from the star.

For denser and denser planets in the Newtonian universe, the possible datums for the potential energy go to infinitely negative. And, in fact, you could just arbitrarily add any number to the potential energy and it would have no practical effect - you do not need another planet to excuse the addition.

In this way it is like measuring temperature in Celsius or Kelvin. Although in that case there is a kind of universal zero that is useful. In the same sense - the potential at infinity is often a useful datum for potential energy.

Answer 3

(For the E field) For generality, I'm going to explain potential.

Potential as a definition is

$-\int_{ref}^{r} \vec{E} \cdot \vec{dl}$

Notice that I have included the variable ref in the definition. This is a variable, that is chosen to set the 0 of potential at some chosen location. When the reference is chosen to be infinity, then the mathematics sets the 0 of potenti at infinity. There is an inherit freedom in the definition of potential, and you are free to set the 0 of potential at any location.

For all "ref"

$E=-\nabla V$

They are all equally valid.

The only thing that matters is the CHANGE IN POTENTIAL. The actual value doesn't matter, and is not unique, and depends on the reference chosen. So to answer your question, yes and no, the value cam be as small and as big as you want, but the change in potential is the only thing that matters.

Answer 4

Your conception of potential energy is wrong.One can only speak of differences of energy, so the amount of potential (or kinetic) energy depends, where you decide to put E=0. Since you metal ball will never leave earth it is meaningful to put his potential energy 0 at some point of i.e. surrounding.