Pressure in Hydrostatic Equlibrium

In summary, the conversation discusses finding the pressure as a function of r for the region r<< R, where R is the radius of the object. The density follows a specific equation, but there is an issue with dealing with the infinite density at r = 0. Several models are suggested, but it is concluded that the given formula is not sufficient for calculating pressure near r=0.
  • #1
zachzach
258
1

Homework Statement



Find the pressure as a function of r for the region r<< R where R is the radius of the object. The density goes as [tex]\rho = \rho_o {(\frac{r_o}{r})}^2[/tex].

Homework Equations


I know how to get to the answer my problem is dealing with the infinite density at r = 0.

The Attempt at a Solution



My pressure integrates to an equation that is inversely proportional to the square of r. But at r = 0 the central pressure, will be infinite. How can you deal with this infinite. It makes sense since the density is infinite at r = 0, but there must be an answer. If the [tex]r_o[/tex] was an R you could do a taylor expansion but it is not.
 
Physics news on Phys.org
  • #2
Then just take it as an approximate model :wink: A better model may look somewhat like this: [tex]\rho = \rho _0 (\frac{r_o}{r+\epsilon})^2[/tex] where [tex]\epsilon << R[/tex] (I devise it, so don't take it for real :biggrin:). If you're interested in the pressure at r=R, r=R/2, etc, then the model given by the problem might be sufficient. If you're interested in the pressure near the center of the object, then there is a need for another model.
In short, don't take the formula given in the problem too seriously :smile:
 
  • #3
Hi,
Sorry, I overlooked the r<<R part, which made my reply above rather stupid :biggrin: But then, my conclusion is still the same: this model is not sufficient for calculating pressure near r=0. It's very non-intuitive to have the density to go to infinity, and thus, this non-intuitive model will probably lead to non-intuitive result.
 

What is pressure in hydrostatic equilibrium?

Pressure in hydrostatic equilibrium refers to the balance between the downward force of gravity and the upward force of pressure in a fluid at rest. This equilibrium exists when the pressure at any given point in the fluid is equal in all directions.

How is pressure calculated in hydrostatic equilibrium?

The pressure in hydrostatic equilibrium can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column. This formula is also known as the hydrostatic pressure equation.

What does it mean for a fluid to be in hydrostatic equilibrium?

A fluid is in hydrostatic equilibrium when there is no net force acting on it and it is at rest. This means that the fluid is not accelerating and the pressure is consistent throughout the fluid.

How does pressure change in hydrostatic equilibrium with depth?

In hydrostatic equilibrium, pressure increases with depth in a fluid. This is because the weight of the fluid above exerts a greater downward force on the layers below, resulting in an increase in pressure.

What are the real-life applications of hydrostatic equilibrium?

Hydrostatic equilibrium is an important principle in various fields, such as meteorology, oceanography, and engineering. It helps us understand and predict the behavior of fluids in various systems, such as weather patterns, ocean currents, and hydraulic systems.

Similar threads

Replies
1
Views
870
  • Advanced Physics Homework Help
Replies
15
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
16
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Back
Top