What is the pressure gradient towards the centre of a large planet?

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SUMMARY

The pressure gradient towards the center of a large planet, such as Earth, reaches its maximum due to the weight of the overlying material, despite the gravitational field falling to zero at the center. The pressure can be mathematically represented by the equation P=∫ρgdr, where g is defined as g=Gm/r² and m=∫4πr²ρdr. The radial distribution of density (ρ) is crucial in determining the pressure curve, which eases off as one approaches the center. High temperatures do not significantly affect this distribution in this context.

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kevindin
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Given that the gravitational field falls to zero at the centre of a large body (e.g. the earth), what happens to the pressure curve? (Assuming no effects due to high temperature.) Does it ease off too? What would the curve look like and what would the formula be?
 
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kevindin said:
Given that the gravitational field falls to zero at the centre of a large body (e.g. the earth), what happens to the pressure curve?

It reaches its maximum, as the pressure is the result of the weight of the overlying material.

kevindin said:
What would the curve look like and what would the formula be?

That I can't answer. Perhaps someone here can provide the math.
 
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Simple rules
Pressure is
P=∫ρgdr
Where g is
g=Gm/r2
and
m=∫4πr2ρdr
Hm, cannot figure out an easy way to place the bounds on your ∫.
There would be no effects due to high temperature. What matters is the radial distribution of ρ.
 
snorkack said:
There would be no effects due to high temperature. What matters is the radial distribution of ρ.

High temperatures have an effect on the radial distribution of ρ.
 
Drakkith said:
It reaches its maximum, as the pressure is the result of the weight of the overlying material.

As you approach the center, less and less mass is added to the already overlying material. Hence the pressure gradient will be at its minimum.
 

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