Gravity at the bottom of a mineshaft

[THarper]

Homework Statement

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.

Homework Equations

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! :)

 

[D H]

The only assumption you need to make is that density is a function of distance from the center of the Earth.

 

[THarper]

Thanks DH :), I don't understand how I would get to an answer of 2/3 using that assumption, however?

 

[D H]

Find the gradient of gravitation acceleration with respect to radial distance, $\frac{dg(r)}{dr}$.

Newton's shell theorem may be of assistance. Note: Does this theorem apply with a non-constant density? Under what conditions does this theorem apply?

 

[THarper]

I've been given this question as an example of what I'll be facing next year. It's really daunting not having questions with specific steps for the first time!

I was thinking that I would have to somehow show that at a limit near to 2/3, there is a point where the Earth's crust above is much lighter than the Earth's magma below, and so the weight increases, because you're getting closer to "more mass".

I think I'm out of my depth, because I have no idea how to quantify any of this, or what finding the gravitation acceleration with respect to radial distance will do to help! Really sorry for my stupidity :/

 

[D H]

Let's look at this problem from the perspective of Newton's shell theorem. A spherical shell of constant density looks just like a point mass to points outside the shell. To points inside the shell, the shell of matter results in zero gravitational acceleration. What this means is that for a point inside a planet whose density is a function of distance only, it's only the mass closer to the center than the point in question that counts. All of the mass outside: No effect. In other words,
g(r) = \frac {G M(r)}{r^2}
where $M(r)$ is the mass of that part of the planet that is at a distance $r$ from the center of the planet or less.

Differentiating with respect to $r$ yields
\frac{dg(r)}{dr} = \frac{G}{r^2}\frac {dM(r)}{dr} - 2\frac {G M(r)}{r^3}
Let's look at two ways to express this inner mass M(r). One is via the average density $\bar{\rho}(r)$: $M(r) = V(r)\bar{\rho}(r) = \frac 4 3 \pi r^3 \bar{\rho}(r)$. The other is to integrate density from the center on outward, $M(r) = \int_0^r 4\pi \xi^2 \rho(\xi) d\xi$.

What does the fundamental theorem of calculus say about the second form and $\frac {dM(r)}{dr}$?
What does this mean with regard to $\frac{dg(r)}{dr}$?