DaveC426913 said:
But if you do want to be picky about reality, it's still perfectly true. It is still Newton's Shell Theorem. It's just that you need to factor in the varying density.
Additonally, the poster's hole is only a few metres deep. While your factoring of the varying density will indeed change the numerical answer (which was not asked for), it will not change the principle of the answer. The person in the hole will still weigh less than the person standing above the hole.
So, really, I'm not sure where you're coming from.
Where I am coming from is that you are wrong. The person will weigh more at the bottom of the hole.
You are right that the shell theorem still applies. Ignoring the rotation of the Earth and assuming the Earth has a spherical mass distribution (density depends only on distance from the center of the Earth), the gravitational acceleration experienced at some distance
r from the center of the Earth is exactly that given by Newton's law of gravitation,
a(r) = \frac {Gm(r)}{r^2}
where
m(r) is the mass of that portion of the Earth inside a sphere of radius
r centered at the center of the Earth.
Suppose the Earth comprises an inner shell of mass surrounded by an outer shell. The inner shell has spherical mass distribution, mass
m0, and radius
r0. The outer shell is a spherical shell of constant density ρ
s. The mass of the part of the Earth within a sphere of radius
r,
Re>r>r0 is
m(r) = m_0 + \frac 4 3 \pi \rho_s \left(r^3-r_0^{\,3}\right)
Denote the difference between the mass of the inner shell and a mass of the same volume but the density of the outer shell as Δ
m,
\Delta m \equiv m_0 - \frac 4 3 \pi \rho_s r_0^{\,3}
With this, the mass
m(r) becomes
m(r) = \frac 4 3 \pi \rho_s r^3 + \Delta m
The gravitational acceleration at some point
r within this outer shell is thus
a(r) = \frac {Gm(r)}{r^2} = \frac 4 3 \pi G \rho_s r + \frac {G\Delta m}{r^2}
The gravitational acceleration will decrease/increase with depth if the derivative of the acceleration with respect to
r is positive/negative. Differentiating,
\frac{\partial a(r)}{\partial r} = \frac 4 3 \pi G \rho_s - 2 \frac {G\Delta m}{r^3}
The conditions for an decrease gravitational acceleration with respect to depth is
\rho_s > 2 \frac {\Delta m}{4/3 \pi r^3}
Substituting \Delta m = \left(m(r) - 4/3 \pi \rho_s r^3 and simplifying,
\rho_s > 2 \frac {\left(m(r)}{4/3 \pi r^3} - 2 \rho_s
The first term on the right hand side, m(r)/(4/3\pi r^3) is the mean density of of that portion of the Earth inside a sphere of radius
r. With this,
\rho_s > \frac 2 3 \bar{\rho}
Two thirds of the mean density of the Earth is significantly greater than the density of granity (2.7 grams/cc) or even basalt (3.0 grams/cc). Gravitational acceleration increases with depth near the surface of the Earth.
In fact, gravitational acceleration increases all the way down to the mantle/crust boundary. It then decreases a bit with increasing depth but then starts increasing again, reaching a maximum at the mantle/outer core boundary. The gravitational acceleration at the mantle/outer core boundary, 2,890 km below the surface, is greater than the gravitational acceleration at the Earth's surface.
