# Why is the gravity in a mine shaft independent of depth if the local density is 2/3 of the average density? [closed]

While searching for something, I have come across the fact that the gravitational acceleration in a mine shaft is independent of depth if the local density is $$2/3$$ of the average density (of the Earth).

Also, the gravitational acceleration increases only if the local density is greater than two-thirds of the average density. Note that everywhere a spherically symmetric model for the Earth was used.

Where does this figure come from?

This Question in Mentioned in Resnick Halliday Krane book as P-14-10.The reference is "Gravity in a Mine Shaft" by Peter M.Hall and David J.Hall,The Physics Teacer,November 1995 p.525.)

• Can you cite a source? Assuming homogenous earth, gravitational acceleration decreases linearly with depth, with zero at the center of earth. I don't know what you call a "local density", but any density disturbances close to the shaft should not have a significant effect. Nov 26 '15 at 16:58
• You'll have to be more specific than this. It isn't clear what ypu mean by local density. Nov 26 '15 at 17:30

If the earth is spherically symmetric, then its density $\rho$ only depends on the radial coordinate, and the gravitational field at any radial coordinate is given by

$$g(r)=\frac{G}{r^2}\int_0^r4\pi r'^2dr'\rho(r')$$

where the integral gives the mass inside the sphere of radius $r$. The average density within the sphere is just the mass divided by the volume, so

$$<\rho>=\frac{\int_0^r4\pi r'^2dr'\rho(r')}{\frac{4\pi r^3}{3}}$$

We then take the derivative of the first equation:

$$g'(r)=4\pi G\rho(r)-\frac{2G}{r^3}\int_0^r4\pi r'^2dr'\rho(r')$$

Setting this to zero and doing a little bit of algebra yields the result, $\rho(r)=\frac{2}{3}<\rho>$.

Near the surface of the earth the local density is actually a little less than that, so the gravitational field actually gets a little stronger as you go down until you're really quite deep.