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Gravity at the bottom of a mineshaft

  1. Sep 24, 2013 #1
    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! :)
     
  2. jcsd
  3. Sep 24, 2013 #2

    D H

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    The only assumption you need to make is that density is a function of distance from the center of the Earth.
     
  4. Sep 24, 2013 #3
    Thanks DH :), I don't understand how I would get to an answer of 2/3 using that assumption, however?
     
  5. Sep 24, 2013 #4

    D H

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    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?
     
  6. Sep 24, 2013 #5
    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 :/
     
  7. Sep 24, 2013 #6

    D H

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    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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