What is the pressure at the centre of the Earth?

Question

If the (net) gravity (at the centre of the Earth) is $0$, then what is the pressure? Some say it's astronomically big, but how can that be? Is it because of the (high) temperature? I can't think of anything else. And how about a neutron star and a black hole then? What is the temperature at the centre of those?

Answer 1

According to this Wikipedia article, the pressure in the center of the earth ranges from about $330$ to $360$ gigapascals ($10^9 \, \rm{Pa}$). Using Newtonian mechanics, the order of magnitude of this value can be obtained.

The pressure due to the gravitational force of the Earth in the center of the earth can be calculated through the definition of the pressure as $$dP = \frac{{dF}}{{dA}}\,\, \Rightarrow \,\,\,P = \int {dP} .$$ To do this calculation, it is necessary to consider the earth as an infinitely many spherical shells with the radius $0 \le r \le R$. So, for simplicity, we can assume that the earth is spherically symmetric and also has a uniform density ($\rho = \frac{M}{V} = \frac{M}{{\frac{4}{3}\pi {R^3}}} = \frac{{dm}}{{dV}}$). The gravitational force between each shell and the mass inside that shell can also be written as $dF = G\frac{{m\,dm}}{{{r^2}}}$. Therefore, the pressure at the center of the Earth due to the gravitational force is computed as
$$P = \frac{3}{8 \pi}\frac{{G{M^2}}}{{{R^4}}} \simeq 1.7 \times {10^{11}}\,{\rm{Pa}},$$ which is close to the result of that article. Although it is a little less than the more realistic estimate, i.e., $3.3-3.6 \times {10^{11}}$ $\rm{Pa}$. Obviously, the gravitational pressure component has the major role (I think that the thermal pressure due to the Earth's hot inner core has a decreasing contribution to this estimation, but seemingly very small). However, in order to have an accurate answer, the exact details of the density of different layers of the earth must be specified.

Neutron stars are so dense, compact astrophysical objects. The surface temperature of neutron stars ($\sim 6 \times 10^5$ $\rm{K}$) is much greater than the Earth's core temperature. Naturally, the pressure inside them is very big. By repeating the calculations we can show that the order of magnitude of the pressure inside the neutron star is equal to $\sim 4 \times 10^{45}$ $\rm{Pa}$ which is very big and perhaps far away from reality. This estimate is very naïve since on has to study these objects more carefully using quantum mechanics (see Neutron degeneracy and also Degenerate matter for more information). In fact, in neutron stars, a third pressure component is present due to the Pauli exclusion principle in quantum mechanic, meaning that identical neutrons cannot occupy the same quantum state within a neutron star. Assuming the neuron stars as relativistic ideal Fermi systems, it can be shown that the maximum pressure is of order of magnitude of $\sim 10^{33}-10^{35}$ $\rm{Pa}$.Therefore, in neutron stars, this pressure (degenerate matter component) plays a dominant role.

The story for black hole is totally different because of appearance of new concepts such as spacetime singularity, event horizon, Hawking temperature etc. For black hole temperature, we can only have a definition of temperature for the black hole's horizon. From quantum field theory in the curved black hole background, it can be found that the temperature is given by Hawking radiation relation as (for static black holes) $${T_{BH}} \approx 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{\rm{K} },$$ where $M_\odot \approx 2×10^{30} \rm{kg}$ is the solar mass. As seen, for solar mass black holes it is near the absolute zero. On the other hand, we can have no reliable estimate of the pressure inside black holes due to the singularity at the origin (a place where matter is compressed down to an infinitely tiny point). Therefore, the theory of general relativity does not work well in this area.

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Thanks to @Bowl Of Red for pointing out a mistake in my answer.

Answer 2

The pressure at a point near the centre of the earth depends on the weight of matter above, and that depends on the mass and radius of the earth.

The best way to get an estimate is from dimensional analysis

$P=\frac{F}{A}$, where $F$ is a force and $A$ is an area,

The only sensible formula that has the right dimensions is

$$\frac{GM^2}{R^4}$$

To check it's got the right dimensions, think of it as force (Newtons gravity), divided by an area.

Using $M=6\times 10^{24}$kg and $R=6.37\times 10^6$m, gives about $1.5\times 10^{12}$ Pa

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The above is a way to get a quick estimate. For constant density $\rho$ integration (as requested in the comments) gives a similar result like this:

We can use $P=\frac{F}{A}$, where $F$, the force pressing down on the area, is due to the total weight of matter above and will have to be calculated by integration.

Imagine an area $A$, near the centre of the earth. The weight of a slice of matter above, of the same area at radius $r$ is $W=mg(r) = (A\delta r \rho) \times (G {\frac{(4/3)\pi r^3\rho}{r^2}})$

Integrating between $r$ and the radius of the earth $R$

$\int_r^R \rho^2 A G \frac{4\pi}{3}r dr = \rho^2 A G\frac{2\pi}{3}(R^2-r^2)$

letting $r$ tend to zero and dividing by the area $A$ to get the pressure gives $P = \rho^2 G \frac{2\pi}{3}R^2$

substituting for $\rho = M/(\frac{4 \pi R^3}{3})$ where M is the mass of the earth, gives the pressure

$$P=\frac{3GM^2}{8\pi R^4}$$