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Why does gravity act at the centre of earth and how does that happen?

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    $\begingroup$ Are you asking about the force of gravity at the centre of the earth ? Or are you asking why is the force of gravity at the surface of the earth directed towards its centre ? $\endgroup$ – gandalf61 Sep 5 '20 at 9:36
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    $\begingroup$ At what level are you asking this question? At the lowest level, it's something you'll probably solve in a 1st year college physics course. The answer involves some integration, but the answer is that in a uniform sphere, the forces from either side cancel out. (And interestingly, as you go towards the center, everything outside also cancels out, so at the center of the Earth you'd be weightless.) At the highest level, it's because that's the way the universe works :-) $\endgroup$ – jamesqf Sep 5 '20 at 21:32
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    $\begingroup$ @jamesqf - in the center of the earth: don't you have some momentum resulting from the elliptic track which the earth undergoes around the sun? $\endgroup$ – Gottfried Helms Sep 6 '20 at 7:10
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    $\begingroup$ @Gottfried Helm: OK, to a first approximation, and assuming a spherical Earth :-) But I think you wouldn't feel any effect from the sun, precisely because you are in orbit around it. Just as astronauts in orbit don't feel the effect of Earth's gravity. $\endgroup$ – jamesqf Sep 6 '20 at 19:35
  • $\begingroup$ @jamesqf - ah, that might be right, good example... $\endgroup$ – Gottfried Helms Sep 7 '20 at 0:58
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All parts of the Earth exert a pull on another body (such as an apple) and all parts of the Earth are pulled by external bodies (such as apples).

Let's consider the pull that the Earth exerts on the apple. Imagine the Earth divided up into kilogram portions by imaginary surfaces. The portion nearest the apple will exert the greatest pull, straight downwards on the apple. The (antipodean) portion furthest away will exert the least pull, though it will still be downwards, that is towards the centre of the Earth. Portions of the Earth to the North, South, East and West of the apple and at various depths inside the Earth will also pull the apple, but not straight downwards. The vector sum of these forces is the resultant pull.

Newton famously proposed the Law of Gravitation that every particle attracted every other particle with a force proportional to the mass of each particle and inversely proportional to the square of their separation. He also showed that if the Earth is a sphere, with its mass distributed with spherical symmetry about its centre, then, according to his Law, the resultant pull on the apple was the same as if all the Earth's mass were concentrated at its centre.

So gravity doesn't really "act at the centre of the Earth" but it behaves more of less as if it did. Only "more or less" because the Earth is not a perfect sphere and its mass is not distributed with perfect spherical symmetry.

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    $\begingroup$ And this "more or less" becomes important when you're trying to keep satellites precisely where you want them to be (or want to know precisely where they will be). $\endgroup$ – JiK Sep 6 '20 at 0:32
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Because the Earth is spherical (to the same level of approximation, to which we consider that the force of gravity is directed towards its center). In fact, there is a gravity force exerted by every element of the Earth, but when all these forces are added up, the net force is directed towards the center of the Earth.

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Given a spherically symmetric Earth and a point $P$, the gravitational field vector at $P$ can be determined in two steps:

  1. Why does the field point toward the Earth's center $C$? Consider rotational invariance about the axis passing through $C$ and $P$. The mass distribution of the Earth is unchanged by this rotation, and so must be the field -- i.e., the field is directed along the axis.

  2. Why is the field magnitude equivalent to concentrating the Earth's mass at $C$? This is a special property of the inverse-square law that is more easily understood via Gauss's law. Consider the spherical surface centered at $C$ and passing through $P$. The gravitational field on this surface is everywhere perpendicular to it, as argued above. Gauss's law says that the integral of the perpendicular field component (i.e., in this case, the field magnitude) over the surface is proportional to the mass $M$ enclosed by the surface. Thus, given spherical symmetry, the field magnitude depends only on $M$ and is the same as if it were a point mass at $C$. Note that if $P$ is at or above the Earth's surface, then $M$ is the Earth's total mass; if $P$ is below the Earth's surface, then $M$ is the amount of mass located deeper than $P$.

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Every single particle in Earth exerts gravitational attraction to an object in your example, and, all these effects add up/cancel, and the net of these is pointing towards the center.

Actually, at the center, all these effects cancel out, and you feel weightless.

Correct. If you split the earth up into spherical shells, then the gravity from the shells "above" you cancels out, and you only feel the shells "below" you. When you are in the middle there is nothing "below" you.

Would you be weightless at the center of the Earth?

This can be shown at the level of the Christoffel symbols and the Radial four acceleration, where the vector would always point towards the center of Earth, and its magnitude would only be zero at the center of Earth exactly.

When $r = 0$ the Christoffel symbol $\Gamma_{tt}^r$ is zero and that means the radial four-acceleration is zero and that means you're weightless.

What is the general relativity explanation for why objects at the center of the Earth are weightless?

So the answer to your question is that for Earth, the constituents exert gravitational pull from all directions, and the net of these points towards the center.

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