*Originally posted by joe shmo*

**As the pressure inceases, the temperature that solid H20 (saturated) can exist at also increases. So thats promising for the argument. However, the data tables out of my text don't cover pressures as extream as would be experienced by a planetary sized body of water.**

As the pressure inceases, the temperature that solid H20 (saturated) can exist at also increases. So thats promising for the argument. However, the data tables out of my text don't cover pressures as extream as would be experienced by a planetary sized body of water.

If you want a place to start looking in the realm of pressure its going to be a function of the radius of the planet. I say this is a ball park estimate, but you probably searching for something the size of a quarter in the ball park.

Start with the hydrostatic differential equation

dP/dz = -pg (eq1)

P=pressure

z= elevation

p=density of water

g=acceleration due to gravity

we are going to assume that the density of water is constant ( this is in direct conflict with the objective (ie. the solid core of ice has differing density), and depending on the size of your planet(ie. if really large) compressibility of the fluid could come into play).

To get "g" the acceleration due to gravity, use Newtons Law of Gravitation:

F=G*m1*m2/(r^2)

arbitrarily divide through by one of the masses, and make the mass and the force differential.

dF/m1 = G*dm2/r^2

dm2= p*dV (assuming p is constant)

dV= 4*pi*r^2*dr (assuming the planet is spherical, r being it radius)

combine the above equations by substitution and integrate

1/m1*Int(dF)= 4*pi*G*p*Int(dr)

g= 4*pi*G*p*r (with initial conditions r=0,g=0) (eq2)

substitute eq2 into eq1

dP/dz= -p^2*4*pi*G*r

dz=dr

separate variables and integrate with conditions(r=r_planet,P=0)

P=(p^2*4*pi*G*(r_planet)^2)/2 -(p^2*4*pi*G*(r_core)^2)/2 (eq3)

G=Gravitational constant 6.67428*10^(-11) m^3/(kg*s^2)

everything else you pick( just make sure units are metric)

This would be the(ball park) pressure at the liquid, solid interface.

Using the pressure, and a suitable temp(?) you can define the state the water is in(that is assuming the thermo data can be found in the vicinity of the result).

edit: I remember my Thermo Prof say that ice can be compressed isothermally back to liquid quite easily, this now makes me think that the core temp would have to be really cold(this opposes my first post, and must be because the tables were for solid to vapor).