Why is the acceleration due to gravity at the center of the earth $0$?
My Attempt
actually at the center distance from the surface is equal to the radius so the formula $$g'=(1-\frac{d}{R})\times g$$ yields $0$.
But can anyone give me the other reason for this?
Intuitively, of course, it makes sense that the force in all directions at the center of a massive body would be equal. There are a few ways of formalizing this.
The most rigorous would be an integral:
$$G\rho\int_{0}^{r}\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\vec{r}}{|r|^3}|r|^2sin(\phi)d\theta d\phi dr$$
where $\rho$ is the mass density. You'll need to decompose the integral into components to account for the sign, but that's the idea.
Alternatively, you can prove Gauss' Law, which states that the gravitational force on a surface is proportional to the mass enclosed. If there is no mass enclosed by an infinitesimal surface at the center, there is no force on the surface. Notice that the integral above doesn't actually depend on the distance from the origin. That is the basic observation of Gauss' Law. I'd be happy to edit this in a few minutes with the full integral and derivation.
Gravitational acceleration is a vector and thus has a direction.
But, at the center of a spherically symmetric mass distribution, there is no preferred direction while, if there were a non-zero gravitational acceleration vector, it would point in a direction.
Now, if one rotated the mass distribution through some angle, the acceleration vector should rotate along with it. Yet, the mass distribution would be indistinguishable from the non-rotated distribution and thus, there is no physical reason for the acceleration vector to have changed direction.
And so, from just a symmetry argument, we conclude the gravitational acceleration vector at the center of a spherical mass distribution must be the zero vector.
