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It's said that potential energy is "energy of position." If an object is sitting on a shelf five feet above the floor, its potential energy can be thought of as equal to the amount of energy that would be involved in it falling off the shelf and onto the floor.

Conservation says that energy is neither created nor destroyed, only transformed. Therefore, the potential energy of an object has to be absolute, a constant in some sense. But that makes me wonder. If a meteor is drifting through space, and it ends up caught in Earth's gravitational field (and is coming in at the right angle, etc,) it will fall to the surface of the Earth, converting potential energy into possibly a few megatons of kinetic energy.

On the other hand, if that exact same meteor were to meet the exact same fate, except the planet in question was Jupiter, which has much stronger gravity, the impact would therefore be much more intense. So how does that work? If the potential energy of the meteor in space is constant, how can it be determined?

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    $\begingroup$ Potential energy is a scalar term so adding a constant term to its value will not affect anything. It always differences in potential that drive dynamics. The Earth's surface might set a local zero point for thinking about "dropping out of space", but it would not be the same "zero point" as needed for thinking about "zero" when dropping into Jupiter or the Sun. $\endgroup$ – DWin Dec 28 '13 at 17:57
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You say:

Therefore, the potential energy of an object has to be absolute, a constant in some sense.

The (gravitational) potential energy is a function of position so it varies as objects move around in space. The PE at some fixed point is sort of constant in the sense that all observers will agree on it's value, but note that PE has a global gauge symmetry - you can add a constant value to it without changing the physics because only changes of PE appear in the equations of motion.

We normally take the gravitational PE to be zero in flat spacetime far from any other masses. Therefore the PE near Earth's surface is negative, and the PE near whatever passes for Jupiter's surface is considerably more negative. The (positive) kinetic energy is therefore a lot greater near Jupiter's surface to keep the total energy constant.

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The potential energy you are asking about is called gravitational potential energy. The precise 'amount' of gravitational potential energy depends not only on the mass and position of the object in question (e.g., the meteor), but also the mass of the planet you're considering.

The most common form of gravitational potential energy you'll see is $$U_\text{grav}=-\frac{GMm}{r}.$$ In that expression, $M$ is the mass of the planet, $m$ is the mass of your object/meteor, and $r$ is the distance between the centers of mass. A few things that one can take away from this expression are:

  • The potential energy is indeed position dependent; there's no velocity or other kinematic term
  • The more massive the object $m$, the greater the change in $U_\text{grav}$.
  • The more massive the planet $M$, the greater the change in $U_\text{grav}$.

So, yes, gravitational potential energy is position dependent, but that is not the only variable that influences its value.

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protected by Qmechanic Mar 21 '17 at 10:36

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