What would the weight of a column of water be at the center of the Earth?

[Donald E Sawyer]

In reading Jules Verne's book A Journey to the Center of the Earth I came to a section where one of the characters questions if the massive weight of the air would be fatal at the center of the Earth. I lack the mathematical skills to derive an equation to graphically represent the results. My idea was to represent a column of water as it descended into a uniform sphere of significantly larger dimension.

 

[Simon Bridge]

You know that pressure increases with depth... that is often described as being due to the weight of the water above you... which is accurate.
This is what Verne was talking about (through a character who may or may not be wrong...)

Anyway ... gravity decreases linearly to zero as you approach the center of a uniform sphere.

You know the force at the surface (r=R) R^2 where M and R are the mass and radius of the sphere in question.
You know it is zero at r=0 ... and you know the graph is linear... and you know the equation for a straight line...

acceleration of gravity comes out as: g(r)= (GM/R^3)r = g(R)r/R

So the mass dm of water between r and r+dr experiences force g(r).dm ... add up all the "dm"'s.

 

[andrewkirk]

The calculation is easier for a liquid (water) than it is for a gas (air) because the latter compresses under pressure whereas the former hardly does at all. To do the calculation for air requires using the Ideal Gas Law $PV=nRT$. Further, to do the calculation one needs to know the temperature at all points in the air shaft, which will be mostly determined by the temperature of the walls.

Prof Lidenbrock in Verne's book believed that, contrary to mainstream scientific opinion at the time, the temperature never strayed very far from surface temperature as one traveled to the centre of the Earth. Indeed, if that were not the case, Arne Saknussem would never have survived the journey (unless he had lied about getting there - a question that is never resolved in the novel, as Lidenbrock never gets deeper than about 50km).

In contrast, now scientific opinion is that the temperature at the centre of the Earth is more than 5000 degrees C. That makes for much less dense, and hence lighter, air, giving a much lower pressure at the centre than Lidenbrock would have calculated.

Assume the column is straight and has cross-sectional area of one square metre. The actual area or direction of the hole makes no difference, but the calculations are easier that way. We can calculate the contribution $dP$ to pressure (weight per unit area) from a section of the column of air of height $dx$ at depth $x$ below the Earth's surface as:
$$dP= m\,g(x)n(x) dx
=m(D-x)g_s\frac{ P(x)dx}{RT(x)}$$
where $D$ is the Earth's radius, $g_s$ is the gravity at Earth's surface (about 9.8 $ms^{-2}$) and $m$ is the average molecular mass of air.

Using separation of variables, we can re-write this differential equation as:

$$\frac{dP}P = \frac{mg_s}R \frac{(D-x)dx}{T(x)}$$

so that
$$\log \frac{P_c}{P_s} = \frac{mg_s}R \int_0^D \frac{(D-x)dx}{T(x)}$$
where $P_c,P_s$ are the air pressure at the centre and the surface of the Earth respectively, so that
$$\frac{P_c}{P_s} = \exp\left(\frac{mg_s}R \int_0^D \frac{(D-x)dx}{T(x)}\right)$$

Under Lidenbrock's (erroneous) assumption of approximately constant temperature $T$ along the shaft, this is
$$\exp\left(\frac{mg_s}{RT} \int_0^D (D-x)dx\right)
=\exp\left(\frac{mg_s}{RT} [Dx-x^2/2]_0^D\right)
=\exp\left(\frac{mg_sD^2}{2RT} \right)
$$
Alternatively, assuming the temperature increases linearly from say 20 degrees to 6020 degrees C along the shaft (293 to 6293 Kelvin) the calculation is
$$\frac{P_c}{P_s} = \exp\left(\frac{mg_s}R \int_0^D \frac{(D-x)dx}{293+6000x/D}\right)$$
for which the integral is evaluated easily on Wolfram Alpha if you're lazy like me.
Plug in the values of $D, m, g_s, R$ and we'll have the answer. I expect Lidenbrock's number to be much bigger.