Pressure at the center of a planet

[Psycopathak]

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 boundary 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?

 

[D H]

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).

 

[Psycopathak]

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.