Gravity at the bottom of a mineshaft

THarper · Sep 24, 2013

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
g(r) = -GM(r)/r^2
G = 6.67384 × 10-11 m3 kg-1 s-2

3. 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 · Sep 24, 2013

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

THarper · Sep 24, 2013

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

D H · Sep 24, 2013

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 · Sep 24, 2013

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 · Sep 24, 2013

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,
[tex]g(r) = \frac {G M(r)}{r^2}[/tex]
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
[tex]\frac{dg(r)}{dr} = \frac{G}{r^2}\frac {dM(r)}{dr} - 2\frac {G M(r)}{r^3}[/tex]
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}##?

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