# Pressure at the center of a planet

1. Dec 9, 2008

### Psycopathak

1. The problem statement, all variables and given/known data

Find the pressure at the center of a planet

2. Relevant equations

dP/dr = -ρg (Hydrostatic Equilibrium)

g = GM/r2 (acceleration due to gravity)

3. 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?

2. Dec 10, 2008

### D H

Staff Emeritus
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).

3. Dec 10, 2008

### Psycopathak

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.