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For example the core of the planet Jupiter, how to find out the magnitude of the force that compresses its core? and can that be thought of as the total force that holds the planet together?

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Model the planet as an incompressible material with constant density $\rho$ and radial pressure $p(r)$ for $0\leq r\leq R$, where $R$ is the radius of the planet. The force per unit volume due to pressure is given by $$F_p=-\nabla p(r).$$

Meanwhile, the force per unit volume due to gravity at a depth $r$ is given by Newton's law: $$F_g=\frac{G\rho\left(\frac{4}{3}\pi r^3\rho\right)}{r^2}=\frac{4}{3} \pi G \rho ^2 r.$$ Setting $F_p=F_g$ and integrating to find $p(r)$ along with the condition $p(R)=0$ yields $$p(r)=\frac{2}{3} \pi G \rho ^2 \left(R^2-r^2\right).$$ Note that the pressure is actually highest at the core, and that as you travel towards the surface, it decreases quadratically to zero. The core pressure is $p(0)=\frac{2}{3}\pi G \rho ^2 R^2$.

For nonconstant $\rho$ the answer will be slightly different, but the concept is the same.

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  • $\begingroup$ Hi, I just tried to use your formula (which also is consistent with other formulae online) to calculate the maximum diameter of a planet consisting of pure water all the way through. It came up with an absurd answer. Can you tell me what I'm doing wrong? $\endgroup$ – Robert Walker Nov 3 '16 at 20:07
  • $\begingroup$ My calculation is P = (2/3) * π * G * ρ^2 * R^2 There using SI units, the density of water, ρ = 1000 kg / m3, Pascal is the SI unit for pressure, and meter is the SI unit for length. I'm using the triple point of Ice VI with liquid water and Ice V at -0.16 °C, 632.4 MPa So that makes R = 632.4*10^6/ sqrt((2/3) * π * G * 10^6) meters. But I make that a rather absurd (2/3) * π * G * 10^6 = 0.000139781605 in SI units R = 632.4*10^6/ sqrt(0.000139781605) which works out at around 54 million kilometers. That can’t be right. Anyone know what I’ve done wrong here? $\endgroup$ – Robert Walker Nov 3 '16 at 20:07
  • $\begingroup$ It's for my quora answer here, so you can read it better formatted than I can in a comment quora.com/Do-water-planets-exist/answer/Robert-Walker-5 $\endgroup$ – Robert Walker Nov 3 '16 at 20:10
  • $\begingroup$ Answered my own question, I was missing a square root. $\endgroup$ – Robert Walker Jan 11 '17 at 2:42
  • $\begingroup$ P = (2/3) * π * G * ρ^2 * R^2. There using SI units, the density of water, ρ = 1000 kg / m3, Pascal is the SI unit for pressure, and meter is the SI unit for length. There P for Ice V at -0.16 °C, is 632.4 MP = 632.4*10^6 Pascals G = 6.674×10^−11 N⋅m² / kg² Want to solve for R.<br> So R = sqrt ( 632.4*10^6 / ((2/3) * π * 6.674×10^−11*10^6 )) meters. = 2,127,029 meters or around 2,127 km $\endgroup$ – Robert Walker Jan 11 '17 at 2:42

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