Calculating Pressure at the Center of the Earth

Click For Summary
SUMMARY

The discussion focuses on calculating the pressure at the center of the Earth, assuming the interior behaves as an incompressible fluid with constant density (ρ = M/V). The static equilibrium equation for a self-gravitating fluid sphere is presented, leading to the formulation p(r) = p(r+dr) + ρdrg(r). The user seeks assistance in integrating this equation to solve for p(r) and expresses uncertainty about handling the terms p(r+dr) and dr.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly static equilibrium.
  • Familiarity with gravitational equations, specifically g(r) = G (ρ4πr³/3) / r².
  • Knowledge of calculus, particularly integration and differentiation.
  • Experience with mathematical modeling of physical systems.
NEXT STEPS
  • Study the derivation of pressure equations in fluid mechanics.
  • Learn about the application of the Fundamental Theorem of Calculus in physical contexts.
  • Research the implications of incompressible fluid assumptions in geophysics.
  • Explore numerical methods for solving differential equations related to pressure calculations.
USEFUL FOR

Students and professionals in physics, particularly those focused on geophysics, fluid dynamics, and mathematical modeling of natural phenomena.

derravaragh
Messages
24
Reaction score
0

Homework Statement


Assume that the interior of the Earth is an incompressible fluid. The density is constant: ρ = M/V. The pressure p(r) depends on the distance r from the center of the earth. The equation for static equilibrium of a self-gravitating fluid sphere is
p(r)δA − p(r+dr)δA − ρdrδAg(r) = 0,
where g(r) = G (ρ4πr3/3) /r2.


Homework Equations





The Attempt at a Solution


I need to solve this for p(r), so first I divided out the δA. From here I added terms to obtain:
p(r) = p(r+dr) + ρdrg(r)
Now I am stuck. I am unsure of how to deal with p(r+dr) and dr. My gut tells me to integrate with respect to r, but I don't feel like I should be integrating p(r). Other than that, when I look at p(r+dr) I see it as the pressure of the current r + the change in r, but that still doesn't help me proceed. Any help would be appreciated.
 
Physics news on Phys.org
$$\lim_{dr \to 0}\frac{p(r+dr)-p(r)}{dr} = p'(r)$$
Just the regular definition of the derivative.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
1K
  • · Replies 26 ·
Replies
26
Views
4K
Replies
6
Views
2K
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
18
Views
5K