Gravity at the centre of the Earth?

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Gravity at the center of the Earth is effectively zero due to the symmetrical distribution of mass surrounding that point. As one moves inward, the gravitational force is influenced only by the mass within the radius, while the outer shell's mass cancels out. At the center, this results in a weightless condition, as equal mass pulls in all directions. Calculations indicate that gravity increases until reaching the outer core boundary, where it peaks at approximately 10.88 m/s² before declining to zero at the center. Therefore, the concept of gravity changes significantly from the surface to the Earth's core.
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I know gravity on the surface of the Earth is 9.8m/s/s from

g=GM/d^2
(6.7x10^-11)(6x10^24) / (6.4x10^6)^2 =9.81

But if you go to the center of the Earth will there be no gravity?
As the mass at the center will be nothing and the distance will be nothing?
 
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As you proceed inwards towards the center, the net force of gravity is solely provided by the INNER massive ball (with center at the Earth's center); that is the outer shell of mass contributes nothing to the net force of gravity (we may prove it cancels out).

Thus, indeed, at the center of the Earth, there won't be any net force of gravity "felt".
 
Beaujolais said:
But if you go to the center of the Earth will there be no gravity?
Correct. If you imagine a hollowed out space at the center of the earth, and if you further treat the Earth as a spherically symmetric distribution of mass, then the gravitational field within that hollowed out space will be zero.
 
Beaujolais said:
I know gravity on the surface of the Earth is 9.8m/s/s from

g=GM/d^2
(6.7x10^-11)(6x10^24) / (6.4x10^6)^2 =9.81

But if you go to the center of the Earth will there be no gravity?
As the mass at the center will be nothing and the distance will be nothing?

Assuming you could actually exist in the center of the Earth, you would essentially be surrounded by equal mass therefore, not pulled in anyone direction greater than another. In essence, you would be weightless.
 
When you calculate the gravity from somewhere inside a sphere you can as has already been stated ignore the mass that is at a higher elevation than the point where we are measuring from. That being said the gravity can be calculated by the formula g=4pG(pi)r/3 where p is the density and r is the radius from the center if the body is of uniform density. This is not quite the case with the Earth. The inner core and outer core are much denser than the mantle so the gravity of the Earth actually goes up until you reach the boundary of the outer core, from then on it declines steadily to zero at the center.
 
what (aprox) would be the value of the FoG at its peak?
 
MoonKnight said:
what (aprox) would be the value of the FoG at its peak?

Using values I found on the internet I calculated an acceleration of 10.88 m/s^2 at the mantle/outer core boundary.
 
awesome, thanks
 

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