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Pressure at the center of a planet 
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#1
Dec908, 10:57 PM

P: 13

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)(1r^{2}/R^{2}) Is this correct? Does this mean that inserting 0 for r will give you (10) 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
Dec1008, 04:49 AM

Mentor
P: 15,067




#3
Dec1008, 10:59 PM

P: 13

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. 


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