The value of g measured at the bottom of a mineshaft is higher than that measured at the surface. Show that this implies the density of the Earth’s crust is less than 2/3 that of the mean density of the Earth.
g(r) = -GM(r)/r^2
G = 6.67384 × 10-11 m3 kg-1 s-2
The Attempt at a Solution
I started by showing that at a mineshaft depth of around 3500m, if the density of the earth was constant from surface to center, then the formula for g would be g(r) = 4/3 Pi G ρ r
where ρ is the density of the earth, and r is the distance from the center to the point of measurement.
Using a measurement of the radius of the earth of 6400km, g at the bottom of the shaft would be 0.055% smaller than on the surface. We know, however, that it is greater.
I'm unsure as to how it would be shown that the density of the crust is less 2/3 than that of the mean. Assumptions may have to be made.
Thanks in advance for any help! :)