0
$\begingroup$

"Potential energy: The energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors."

in other words, I think that potential energy is caused by gravity because of its position and distance from the surface of the earth, so will you have no potential energy if you were at the core? Or will you have the same potential energy as the earth due to the gravitational pull of the sun?

$\endgroup$
2
$\begingroup$

In many cases, rather than asking if energy "exists", it is better to ask what the difference in energy is between two configurations.

An object at the center of the earth would have less potential energy than it would some distance away (like on the surface). You could even calculate the difference for a given mass. Further, you could say that the energy of the earth-object system is minimized at this location, or that you can't extract any energy from the potential of the object at this position.

But I wouldn't feel comfortable saying that the position didn't have potential energy. Rather that it was a certain amount less there than it is in some other location.

$\endgroup$
1
$\begingroup$

In Newtonian physics, only differences of potential energy are meaningful. But the world is not Newtonian.

In relativistic theories, potential energy is well-defined and is zero at infinite separation. How do we know? Because otherwise the nonzero potential energy between every pair of masses in the universe would create huge unobserved spacetime curvature.

If you complain that there is no gravitational potential energy in General Relativity, then work in the post-Newtonian approximation where there is (and it’s negative, and it gravitates negatively). The dynamics of the solar system don’t work out unless the gravitational potential energy goes to zero at infinite separation.

So my answer is that the potential energy of an object at the center of the Earth is negative, absolutely.

The actual value is $$U=-\frac{3GMm}{2R^3},$$ where $M$ is the mass of the Earth, $m$ is the mass of the object at the center of the Earth, and $R$ is the radius of the Earth. This comes from the more general expression

$$U=-\frac{GMm}{2R}\left(3-\frac{r^2}{R^2}\right)$$

for when $m$ is inside the Earth at $r<R$. (This assumes uniform density.)

You can check that this is the correct potential energy inside the Earth because at the surface it becomes $-GMm/R$ (matching the outside potential energy $-GMm/r$ for $r>R$) and its negative gradient gives the correct linear-with-$r$ force inside, which is determined by the mass within radius $r$. (The mass further out does not contribute to the gravitational force, by the Shell Theorem.)

$\endgroup$
0
$\begingroup$

In addition to the answers already given, it's worth noting that we are only discussing gravitational potential energy here. As you mentioned, potential energy can also be due to internal stresses, chemical or electrical potential, just to name a few.

$\endgroup$
0
$\begingroup$

Energy is always dealt in changes. In your extraction "by virtue of it's position relative to the other" it also means that energy is changed(lost or gained) when an object shifts it's orientation relative to the others. In the center of the earth the GRAVITATIONAL potential energy is at the lowest but whether it's zero or a million is up to you. What ever value you assign wouldn't change your calculations because you are calculating a change. In my studies it's common practice to assign a 'zero potential level' beforehand. That is if you are not using calculus. Consider this. You throw a ball up from a negligible height from the surface of the earth. And if you assign the value of zero to the energy at the initial point then when it's at a height of h from the starting point then it has gained a potential of mgh (again roughly)(let m be the mass of the ball.) But when you assign a value to the initial position then you will get a different value for the potential at an h height but the difference will still be the same. That is what actually counts.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.