Gravity at the bottom of a mineshaft

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In summary: The fundamental theorem of calculus says that if you have a function of several variables and you want to find the derivative of that function at a particular point, you need to integrate over the domain of that function. In this case, we're looking to find the derivative of the gravitational acceleration function with respect to radial distance. The derivative of this function is given by\frac{dg(r)}{dr} = \frac{GM}{r^2}\frac{dM}{dr} - 2\frac{G M}{r^3}So, if we take the inner mass M(r) from before and integrate it from the center outwards, we get\frac{dM}{dr
  • #1
THarper
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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! :)
 
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  • #2
The only assumption you need to make is that density is a function of distance from the center of the Earth.
 
  • #3
Thanks DH :), I don't understand how I would get to an answer of 2/3 using that assumption, however?
 
  • #4
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?
 
  • #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 :/
 
  • #6
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}##?
 

1. What is the effect of gravity at the bottom of a mineshaft?

The effect of gravity at the bottom of a mineshaft is the same as it is at the surface of the Earth. Gravity is a force that pulls objects towards the center of the Earth, and this force remains constant regardless of depth or location.

2. Does gravity change as you descend deeper into a mineshaft?

No, gravity does not change as you descend deeper into a mineshaft. As mentioned, gravity is a constant force that remains the same at all points on Earth.

3. Is the force of gravity stronger or weaker at the bottom of a mineshaft?

The force of gravity is the same at the bottom of a mineshaft as it is at the surface of the Earth. However, other factors such as air resistance and buoyancy may affect the movement of objects in a mineshaft.

4. Can gravity be completely eliminated at the bottom of a mineshaft?

No, it is not possible to completely eliminate gravity at the bottom of a mineshaft. The force of gravity is always present and cannot be turned off or removed. However, there may be instances where the effects of gravity are reduced or counteracted by other forces.

5. How does gravity affect the shape of a mineshaft?

Gravity does not have a direct effect on the shape of a mineshaft. The shape of a mineshaft is determined by the engineering and construction methods used to create it. However, gravity does play a role in keeping the mineshaft stable and preventing it from collapsing.

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