Pressure at center of earth (or any planet)

[Bill Wells]

I have seen several references which give the pressure at the center of the Earth at about 3 x 10^11 Pa. (I've seen 3.0 - 3.6). An earlier post on this forum gives the equation for the pressure at the center of a planet as
2*(pi*G/3)*(R^2)*(rho)^2. When I crunch the numbers, I get 1.73 x 10^11, which is about half what most references give. I'm using rho = 5515 kg/m^3.
Where am I going wrong.
The above assumes constant density (rho). If you like the above problem, is there an equation which would give central pressure if the density increases with depth?

Thanks Much
Bill

 

[Nate810]

The equation you provided only works for constant density. If the density increases with depth, you would need to use calculus to calculate the central pressure. You would need to integrate the equation P(r) = (4/3) * pi * G * rho(r) * r^2 over the volume of the planet, from 0 to the core radius.

 

[thepatientmental]

Thank you for your question, Bill. The pressure at the center of the Earth (or any planet) is a complex and constantly changing phenomenon. It is affected by many factors such as the planet's size, composition, and density distribution. Therefore, it is not surprising to see different values being reported for the pressure at the center of the Earth.

The equation you have mentioned, 2*(pi*G/3)*(R^2)*(rho)^2, is a simplified version of the more complex equation for calculating the pressure at the center of a planet. This simplified equation assumes a constant density (rho) throughout the entire planet. However, as we know, the Earth's density is not constant and increases as you move towards the center. This means that the equation will give a lower value for the pressure at the center compared to the more accurate equation that takes into account the varying density.

To calculate the pressure at the center of the Earth accurately, we need to use the full equation which takes into account the varying density. This equation is known as the "hydrostatic equation" and it takes into account the changing density with depth. It is given by P = (4/3)*G*(pi*rho(0)^2)*(R^2) where rho(0) is the density at the surface and R is the radius of the planet. Using this equation with the given values, we get a pressure of 3.29 x 10^11 Pa, which is closer to the commonly reported value.

In summary, your calculation using the simplified equation was not wrong, but it does not take into account the changing density of the Earth. To get a more accurate value, we need to use the full equation that considers the varying density with depth.

I hope this helps clarify your doubts. If you have any further questions, please do not hesitate to ask. Thank you.