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adjacent

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- #1

adjacent

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- #2

russ_watters

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- #3

adjacent

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Nugatory

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There's nothing wrong with negative potential energy, because the only thing that matters is the difference between the potential energy at different levels.

If I choose the zero point to be the bottom of the hole then the top of the hole will have positive potential energy ##mgh## and the energy released by dropping the weight into the hole will be ##E = E_{top}-E_{bot} = mgh - 0 = mgh##. If I choose the zero point to be the top of the hole then the bottom of the hole will have potential energy ##-mgh## and the energy released by dropping the weight into the hole will be ##E = E_{top}-E_{bot} = 0 - (-mgh)= mgh##. Either way, the energy released comes out to be ##mgh##.

In some problems it's actually convenient to set things up so that all the potentials are negative. If you're looking at objects moving in the Sun's gravitational field, we want to use Newton's law for the gravitational force ##F=\frac{Gm_1m_2}{r^2}## and that goes to infinity at ##r=0##. Thus, it's easiest to define the zero of potential energy to be an infinite distance away from the sun, and anything at any finite distance from the sun has a negative potential that gets ever more negative as you move closer to the sun.

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adjacent

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So,the h should be the difference between the two level.It makes sense.Thanks

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The point at which potential energy is zero is completely arbitrary and hence has no special meaning. All that matters how potential energy changes. Adding an arbitrary constant to some potential function has no impact on those changes in potential. You could make the potential at infinity be the zero point, and then anything at a finite distance will have a negative potential. In fact, making the zero be at an infinite distance is a very widely used choice for gravitational potential energy.

- #7

Nugatory

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I think the zero point should be the center of the earth.

That won't work. There are two problems:

1) It's not even slightly convenient to use that as the zero point for many problems, such as for example, the movement of our sun around the galactic center.

2) The gravitational force changes with distance from the center of the earth, so ##mgh## only works when you're reasonably close (within a few hundred kilometers) to the earth's surface so we don't need to worry about the changes in field strength. If you're going to extend the height change all the way to the center of the earth, you have to use the more general formula for the potential energy ##\frac{GmM_E}{r}## (##M_E## is the mass of the earth, ##r## is the distance from the center of the earth, and ##G## is Newton's gravitational constant - if you plug in the numbers you'll see that this gives you the ##mgh## that you expect at the surface of the earth). And that ##1/r## term means that you cannot use the center of the earth as the zero point - the potential isn't even defined there.

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Sure it is. You haven't taken Newton's shell theorem into account.And that ##1/r## term means that you cannot use the center of the earth as the zero point - the potential isn't even defined there.

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Nugatory

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Sure it is. You haven't taken Newton's shell theorem into account.

No, just assuming a point source for simplicity. There's no singularity at r=0 if you don't have a point source so can apply the shell theorem, but that doesn't make ##r=0## a natural place to set the zero point of potential energy in a central force problem... And I think that's the issue when OP (who seems to have figured all this out in the meantime) reaches for it as a natural zero point.

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