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It is said that value of $g=0$ at the centre of the Earth. $g$ is calculated using the formula

$$ g=G*M/R^2.$$

When $R = 0$, $g$ approaches infinity. I have two questions.

  1. Why is the $g= G*M/R^2$ not working?

  2. How to calculate the value of $g$ at centre of earth?

Qmechanic
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user213892
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    Because equation is only valid above the surface of the earth not below it. http://m.meritnation.com/ask-answer/question/why-accceleration-due-to-gravity-decreases-with-depth/physics/512175 ā€“ Paul Feb 08 '15 at 13:06
  • Possible duplicates: https://physics.stackexchange.com/q/2481/2451 , https://physics.stackexchange.com/q/18446/2451 and links therein. ā€“ Qmechanic Feb 08 '15 at 13:14
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    Consider only the mass in the sphere between you and the center of the earth. This scales with $r^3$ so the force scales with $r^3/r^2=r$ ā€“ Floris Feb 08 '15 at 13:16

2 Answers2

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Why is the g=Gāˆ—M/R2 not working

There are two regions to consider; the region outside the Earth and the region inside.

The solution inside (non-zero mass density) and the solution outside (zero mass density) are different but must give the same value at the radius of the surface.

Assuming a uniform mass density inside, the gravitational acceleration inside is proportional to the radius and, thus, goes to zero at the center.

enter image description here

Image credit

Alfred Centauri
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To expand on @Alfred's answer,

The gravitational attraction from the Earth as you dig down into the Earth is due only to the mass of the Earth enclosed by the sphere defined by the radius that you are at (see the derivation of Gauss's theorem for a proof of this). Therefore, as you go further down into the Earth, the amount of mass that is pulling you towards the center decreases, and the form of the equation that you cited changes.

Outside the earth (such that $R > R_E$), as you said, the gravitational acceleration is (classically),

$$g = \frac{G M}{R^2}$$

But once you enter the earth (at a radius such that $R < R_E$), the mass that is gravitationally attracting you gets smaller, according to the ration of the volume of the sphere inside your radius, and the sphere inside the Earth's radius$^1$.

$$M(R) = M \frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi R_E^3}$$ $$M(R) = M \left(\frac{R}{R_E}\right)^3$$

Inserting that into the equation above, we find that the acceleration for $R < R_E$ is

$$g = \frac{G}{R^2} M \left(\frac{R}{R_E}\right)^3$$

Simplifying that gives us

$$g = \frac{G M}{R_E^3} R$$

Since the first part of that term is a constant, we can say $\alpha = \frac{G M}{R_E^3}$, and

$$g = \alpha R$$

This is a simple linear equation (see the line part of the graph in @Alfred's answer), which goes to zero, not infinity, as $R$ goes to zero.

$^1$*Note that this is a simplification based on the Earth having uniform density. That is definitely not the case, but since the amount of mass inside a smaller radius is smaller regardless, it will serve our purpose well of determining the limiting behavior of gravitational acceleration as R goes to zero.*

Brionius
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