(adsbygoogle = window.adsbygoogle || []).push({}); 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/r^{2}(acceleration due to gravity)

3. The attempt at a solution

dP/dr = -ρg

Assume that density is constant.

subsitute GM/r^{2}for g in the pressure gradient formula

dP/dr = -ρGM/r^{2}

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 = 4πr^{2}ρdr or dM/dr = 4πr^{2}ρ (conservation of mass)

Then integrating both sides of the the above equation gives.

∫dM = ∫4πr^{2}ρdr

M(r) = ∫4πr^{2}ρdr With limits of integration from 0 to r

and since density is constant

M(r) = 4/3πr^{3}ρ

so the pressure gradient with a constant density gives

dP/dr = (-ρg/r2)(4/3πr^{3}ρ)

Solving the differential equation for P gives

∫dP = ∫-(4πGρ^{2}rdr)/3

with limits of integration from r_{1}to r_{2}(radius) and P_{1}to P_{2}(pressure)

gives:

P_{2}-P_{1}= (-4Gπρ^{2}/3)(r_{2}^{2}-r_{1}^{2}/2)

Solving for P1 gives

P_{1}= P_{2}+(-2Gπρ^{2}/3)(r_{2}^{2}-r_{1}^{2}/2)

Setting the boundry condition as

r_{2}= R_{1}and P_{2}=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) = (2πGρ^{2}R^{2}/3)(1-r^{2}/R^{2})

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 Forums - The Fusion of Science and Community**

# Pressure at the center of a planet

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Pressure at the center of a planet

Loading...

**Physics Forums - The Fusion of Science and Community**