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Let's suppose the earth is a perfect sphere and let's ignore its rotation and movement.

What would happen if I would be in the center of the earth? Would the gravity be zero in any direction so I wouldn't feel any gravity force? Or would there be 'pulling' forces (of half of the gravity force on the surface) that would attract me in every direction.

Or in other words - If I would be in the center of the sun, would I be in a weightlessness state or would I be torn to pieces (except of course, having a serios sunburn)?

I know there are similar questions here:
Effect of gravity at center of Earth
Would you be weightless at the center of the Earth?

But they are all related to whether there would be no force or there would be forces to all direction. Some answers imply that the forces would be the same but, in every direction, so they would 'cancel' each other. But I'm not sure what that means? If I'm dragging a paper to different directions the forces are not zeroed, the paper is torn apart.

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    $\begingroup$ A square is a 2-dimensional shape? The Earth's rotation and (uniform) motion through space are irrelevant to your question. $\endgroup$
    – ProfRob
    Commented Aug 26, 2023 at 10:17
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    $\begingroup$ I think the OP meant "perfect sphere" $\endgroup$ Commented Aug 26, 2023 at 10:26
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    $\begingroup$ "If I'm dragging a paper to different directions the forces are not zeroed, the paper is torn apart" - the paper is only torn apart if your pulling forces exceed the forces that hold the paper together. Otherwise, the paper is not moving (or accelerating), and we say that the net force (the sum of all forces) is zero. Another difference compared to being inside a shell that's acting on you gravitationally (see ProfRob's answer) is that the two forces you are exerting don't act at every point in the paper simultaneously; roughly, you're pulling half of the paper one way, half the other. $\endgroup$ Commented Aug 26, 2023 at 10:41
  • $\begingroup$ Yes I meant sphere. You are also right ther rotation doesn't matter, but for the motion - isn't it like the inertia would play some role? $\endgroup$
    – matej
    Commented Aug 26, 2023 at 12:41

5 Answers 5

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The Earth is roughly a sphere. If you excavated a hole in the centre and somehow transport yourself there, there would be no net gravitational force (due to the Earth) acting upon you, providing that the Earth has a spherically symmetric density distribution$^1$.

I think what you are getting at is - is this because there really is no gravitational field or because there are forces pulling equally in all directions?

It is the latter. The gravitational force between two bits of matter is universal. You would be attracted by gravity to every piece of the Earth. Because that external matter is distributed in a spherically symmetric way, these forces cancel out at every point that is inside a spherically symmetric shell - this is Newton's shell theorem.

$^1$ And if it doesn't, how are you defining the centre?

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    $\begingroup$ Ok, so if i understand your corretly (and the comment by Filip Milovanovic), even if i'm in the center of the sun it won't tear me apart, since there are no forces dragging me to opposite directions (like the example with the sheet of paper). Instead every single point in my body is affected so the result for every single point is to stay on the spot. Is that correct? $\endgroup$
    – matej
    Commented Aug 26, 2023 at 12:45
  • $\begingroup$ @matej the net gravitational force inside a spherical mass distribution is zero. That applies whether you are at the centre of that shell or not. $\endgroup$
    – ProfRob
    Commented Aug 26, 2023 at 12:56
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Gravity causes a mutual attraction of masses. If you want to move away from a mass, you have to use energy. So when you lift your arm, you feel that this is only possible by using energy.

If you are in the centre of the (ideally round) earth, the earth's gravity acts symmetrically on your arm and you do not need energy to overcome a gravitational attraction.

Now to a limit value consideration. In neutron stars, the gravitational potential is so great that the atomic structure as we know it no longer exists. Ergo, not only the distribution of masses is decisive, but also their quantity.

If I would be in the center of the sun, would I be in a weightlessness state or would I be torn to a pieces?

You are - if you make it to the centre of a celestial body - exposed to a symmetrical gravitational potential and will not be torn apart. However, as the potential increases, the structure of your atoms changes, and eventually they all become neutrons.

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The gravitational field $\textbf{g}$ is governed by Poisson's equation for Newtonian gravity $$\nabla\cdot\textbf{g}=-4\pi G\rho \tag{1}$$ where $G$ is the gravitational constant and $\rho$ is the density.

Since we are only interested in the gravitational field near the center at $\textbf{r}=\textbf{0}$, we can treat the density $\rho$ as constant and use its central value $\rho_\text{center}=13\text{ g/cm}^3$. Then the approximate solution of (1) near the center is $$\textbf{g}(\textbf{r}) \approx -\frac{4}{3}\pi G\rho_\text{center}\textbf{r}$$

Inserting the numbers we get $$\textbf{g}(\textbf{r}) \approx -3.6\cdot 10^{-6}\text{s}^{-2}\ \textbf{r}$$

So the gravitational force at the exact center of the earth is zero. And outside of the center it is still very small (compared to $\textbf{g}=9.8\text{ m/s}^2$ at the surface of the earth) and pointing to the center. Hence (supposed you are in a safe submarine-like chamber) you are not ripped apart, but essentially do not feel any force.

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Or in other words - If I would be in the center of the sun, would I be in a weightlessness state or would I be torn to pieces?

From Newton's shell theorem, there are no net forces acting on a particle, wherever it is in the cavity. If you extend your arms out, there would be no net force acting on a given particle in your left hand or on a given particle in your right hand, so there are no forces tending to move your left hand away from your right hand.

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The gravitational potential itself has zero gradient at the centre of the Earth or the Sun. However there is a very strong effect buoyancy, as your density will be inferior to that of your environment. I assume you are not carbonised and turned into diamond.

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