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?
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.
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.
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.
protected by Qmechanic♦ Oct 6 '16 at 10:39
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