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At the centre of the Earth, if there is zero gravitational effect due to the Earth's mass, would the Sun's and the Moon's gravitational effect still be felt? So does the liquid centre get pulled towards the the surface facing the Sun or Moon as the Earth rotates?

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    $\begingroup$ Possible duplicate: Does the Earth's core have tides?. And related: How much of an effect does the moon have on Earth's liquid mantle? $\endgroup$ – lemon Oct 3 '16 at 8:55
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    $\begingroup$ What do you mean by "feeling gravitational effect"? Right now I don't feel any gravitational effect by Earth. I only feel the floor pushing my legs upwards and the chair pushing the core of my body upwards and my neck pushing my head upwards etc. To illustrate this, if the chair and the floor (and everything below me for a hundred meters) would magically disappear, I would be in free fall, where I couldn't distinguish up from down without looking or feeling the air moving relative to me. ... $\endgroup$ – JiK Oct 3 '16 at 11:26
  • $\begingroup$ ... But there would be a small tidal force between my feet and my head, which is too tiny for me to feel but is still physically there and observable in theory. And the third possibility is that you mean that I can observe the effects of Earth's gravity on me when looking at a bigger picture. Which one of these do you mean by feeling? $\endgroup$ – JiK Oct 3 '16 at 11:28
  • $\begingroup$ I don't quite understand your question, it seems obvious to me yes, given superposition of forces. $\endgroup$ – Shing Oct 5 '16 at 3:17
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The earth is in orbit around the sun, so the sun's gravitational pull isn't felt anywhere on the earth, because in effect we're continuously falling towards the sun.

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  • $\begingroup$ What about the moon then , in the same way the tides are affected? $\endgroup$ – Neil U Oct 3 '16 at 9:14
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    $\begingroup$ Good answer, but I don't quite agree that the the Sun's gravitation isn't felt anywhere on the Earth because there is a very small tidal effect owing to the Earth's nonzero size. This is a pedantic point, though, because it doesn't really change the spirit of your good answer, which certainly tells the full story as far as which way the liquid would go in the OP's question. $\endgroup$ – WetSavannaAnimal Oct 3 '16 at 9:29
  • $\begingroup$ @NeilU The same principle as in SuzuHirose's answer applies - or at least approximately: the Earth is also freefalling towards the Moon. Tidal effects are owing to a body's nonzero extent in a gravitational field, such that the body straddles regions of different gravitational field and, as such, not all points on them can be freefalling at once. Therefore, internal stresses can thus arise. $\endgroup$ – WetSavannaAnimal Oct 3 '16 at 9:34
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    $\begingroup$ @NeilU Actually, I've just noticed you're asking about the whole liquid core. This being a very substantial size, I retract my last comment, which applies for small bodies of liquid. I think you should probably read through the questions marked by Lemon and see whether they shed any light - I'm about to do the same! $\endgroup$ – WetSavannaAnimal Oct 3 '16 at 9:38
  • $\begingroup$ It's a weird way to express it. Isn't it true to say that we're continuously falling towards the sun because the sun's gravitational pull is felt everywhere on (and in) the earth? $\endgroup$ – wim Oct 3 '16 at 16:44
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As the gravity due to Earth is zero at the center, the mass at the center will continue moving along with the Earth's center around the sun, i.e., it will be in orbit. That is the effect of Sun's gravity.

As to the moon's effect, just like tides, the mass will suffer small fluctuations in position, if left totally free. In which case it will oscillate, and as soon as it deviates from the mean position, SHM motion due to Earth's effect will be added to it's motion.

Taking into account all the forces in the question, this is what I think of the final motion: the object will revolve around the sun in an orbit (alongwith the Earth) and execute SHM with mean position varying due to the effect of the moon.

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If Earth is really composed of concentric homogeneous spheroidal or ellipsoidal shells, then, the total Earth's gravitational field at its center is null. It is know in stellar dynamics as 1st and 2nd Newton's theorems. It is also valid for stars, spherical galaxies, or any other object with this type of mass distribution. 1st and 2nd Newton's theorem can be seen in Binney & Tremaine "Galactic Dynamics".

Sun gravitation affects Earth movement, of course. Earth acceleration is a = GM/r², where G= Gravitational constant, M= Sun Mass and r = distance Sun-Earth. all the objects on Earth feel the Sun gravitational field as well. Nevertheless, Earth different parts are pulled differently due to the tiny difference in distances between each one of these chunks and the Sun. These differences produce the Sun tidal force on Earth. Similar effect produce the Moon. The tidal acceleration is proportional to the Earth diameter D, an inverselly proportional to the distance Earth-Sun to the 3rd potence, D/R³.

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protected by Qmechanic Oct 6 '16 at 10:39

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