Pressure at the center of a planet

In summary, the conversation discusses finding the pressure at the center of a planet using the equation for hydrostatic equilibrium and the acceleration due to gravity. It also introduces the concept of planetary differentiation and compressibility, which make the assumption of constant density invalid in most cases. The final equation for pressure as a function of radius inside a planet is dependent on the density and radius of the planet. However, this is based on the assumption of constant density, which is not realistic for most planets.
  • #1
Psycopathak
13
0

Homework Statement



Find the pressure at the center of a planet

Homework Equations



dP/dr = -ρg (Hydrostatic Equilibrium)

g = GM/r2 (acceleration due to gravity)

The Attempt at a Solution



dP/dr = -ρg

Assume that density is constant.

subsitute GM/r2 for g in the pressure gradient formula

dP/dr = -ρGM/r2


Now we need Mass as a function of radius. Divide the planet up into differential concentric rings

then the differential mass element is related to the surface area.

dM = 4πr2ρdr or dM/dr = 4πr2ρ (conservation of mass)

Then integrating both sides of the the above equation gives.

∫dM = ∫4πr2ρdr

M(r) = ∫4πr2ρdr With limits of integration from 0 to r

and since density is constant

M(r) = 4/3πr3ρ

so the pressure gradient with a constant density gives

dP/dr = (-ρg/r2)(4/3πr3ρ)

Solving the differential equation for P gives

∫dP = ∫-(4πGρ2rdr)/3

with limits of integration from r1 to r2 (radius) and P1 to P2 (pressure)

gives:

P2-P1 = (-4Gπρ2/3)(r22-r12/2)

Solving for P1 gives



P1 = P2+(-2Gπρ2/3)(r22-r12/2)

Setting the boundry condition as

r2 = R1 and P2=0

gives the final equation for Pressure as a function of radius inside the planet with density ρ, planetary radius R and varying radius r as

P(r) = (2πGρ2R2/3)(1-r2/R2)

Is this correct?

Does this mean that inserting 0 for r will give you (1-0) or just 1 and then the pressure at the center of the planet is dependent on just the density of the planet and the radius of the planet?
 
Physics news on Phys.org
  • #2
Psycopathak said:
P(r) = (2πGρ2R2/3)(1-r2/R2)

Is this correct?
Very good.

Does this mean that inserting 0 for r will give you (1-0) or just 1 and then the pressure at the center of the planet is dependent on just the density of the planet and the radius of the planet?
Also correct, in terms of the problem at hand. However, you had to make a very big and rather invalid assumption to get to this simple result:
Assume that density is constant.
A planet does not have a constant density. Far from it! The Earth's inner core is more than 4 times as dense as the Earth's crust. Two things are going on here: planetary differentiation (the crust, mantle, inner core, and outer core are composed of different materials) and compressibility (solids aren't quite as solid as you think; they become smaller under pressure).
 
  • #3
D H said:
Very good.


Also correct, in terms of the problem at hand. However, you had to make a very big and rather invalid assumption to get to this simple result:

A planet does not have a constant density. Far from it! The Earth's inner core is more than 4 times as dense as the Earth's crust. Two things are going on here: planetary differentiation (the crust, mantle, inner core, and outer core are composed of different materials) and compressibility (solids aren't quite as solid as you think; they become smaller under pressure).

Thank you sir!

Yes, I am aware that it is a highly unrealistic assumption, but on my homework problem It gives me the question to simply find the pressure at the center of Saturn and Uranus. It gives no density function to integrate.

The more difficult problems give you density functions to integrate.

I just wanted to see if I had carried out the derivation correctly.
 

1. What is the pressure like at the center of a planet?

The pressure at the center of a planet is extremely high. It is caused by the weight of the layers of the planet above pushing down on the core.

2. How does the pressure at the center of a planet compare to the pressure at the surface?

The pressure at the center of a planet is much greater than the pressure at the surface. As you move closer to the core, the weight of the layers above increases, causing the pressure to increase as well.

3. What factors affect the pressure at the center of a planet?

The pressure at the center of a planet is affected by the size and mass of the planet, as well as the density and composition of the layers above the core. Gravity also plays a major role in determining the pressure at the center.

4. Can the pressure at the center of a planet ever be relieved?

It is possible for the pressure at the center of a planet to be relieved through processes such as volcanic activity or plate tectonics. These processes can release built-up pressure and heat from the core.

5. How is the pressure at the center of a planet measured?

The pressure at the center of a planet cannot be directly measured, as humans are unable to reach the center. However, scientists use mathematical models and data from seismic waves to estimate the pressure at the core of a planet.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
77
  • Introductory Physics Homework Help
Replies
4
Views
570
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
442
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
823
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
865
Back
Top