# LARGE Water Body

1. Jan 3, 2012

### Algren

We have heard of celestial bodies, but who heard of a large Water Body(made only of water)?

Considering that the water will not convert into other chemicals under pressure and enclosing it in a box full of 1 atm air and 300 K, How large/massful/volume should the water body be to get 9.81 m/s^2 at its surface?

Consider:
- Surface Tension
- Gravitation Force
- Atmospheric Pressure
- Any other cohesive forces
- Made of H2O, for H2O, and by H2O
- If there is a variation in density, it is uniformally distributed through the body.

I have tried to derive the formula for a given 'g', but the gravitational force and surface tension has to be considered for a pirticular 'area' and not a 'mass' to get the contraction/expansion of the water body.

Last edited: Jan 3, 2012
2. Jan 4, 2012

### xts

You may neglect surface tension and similar forces. It counts for few milimeter droplets, but not for a large body.
If your body is not rotating (or the rotation is slow), you may assume spherical symmetry.
You may probably assume that the density $\rho$ is uniform.
So now the gravity on the surface is given as
$$g(r) = \frac{4}{3}\pi \,r^3 \rho\cdot \frac{G}{r^2} = \frac{4\pi}{3}\,G\,\rho\, r$$
If you want to include the change of density with pressure you must write appropriate integrational equation and solve it, but as the $\rho(P)$ is nonlinear, it might be pretty difficult - you must use probably some iterative method or compute it numerically.

3. Jan 4, 2012

### D H

Staff Emeritus
That's not a good assumption for large solid bodies, let alone a large liquid one. Iron is a lot less compressible than is water, yet the density of the Earth's (mostly) iron core is 13.088 gm/cm3. Compare that to the density of iron at STP, 7.874 gm/cm3.

4. Jan 11, 2012

### Algren

That is exactly why i came here to ask =D, I was unable to make a proper integral equation, =( .

And, if we do not consider the ρ to be uniform throughout the body at any instant, then the way to get to the answer will become much more complicated.

And, density of the Earth varies with depth around many conditions and effects, as its not made of the same material. Here, we are considering that excessive pressure on water molecules down deep wont make different compounds.

5. Jan 11, 2012

### D H

Staff Emeritus
You cannot with any validity assume that density for a "LARGE water body" (your words) will be uniform. Water is much more compressible than is iron, and a ball of solid iron at the center of the Earth is compressed to 3/5 its vacuum density due to the weight of the mass above it.

6. Jan 12, 2012

### D H

Staff Emeritus
Sorry about the bad wording. I meant iron is compressed to 3/5 of its uncompressed volume (or 5/3 its uncompressed density).

7. Jan 12, 2012

### Neandethal00

This formula is right.
It makes radius of earth would be around 35 Km with all water mass.

Almost 6 times larger than the current radius.

Is it physically possible to have all liquid mass no rocky core?

8. Jan 12, 2012

### sophiecentaur

Could you be sure that there would be no drastic molecular change near the centre under that pressure? (Or perhaps that's what's implied at the end of the previous post.)

9. Jan 12, 2012

### rollcast

I remembered something about water at very high pressures but can't find a source other than wiki, the new scientist link doesn't work unless you're a subscriber.

I can't find any reference to how high a pressure if needed though.

10. Jan 13, 2012

### Algren

Uhh, if we consider the 'practicality' of stuff, such as dissoctiation of h2o at high pressures, variation in density, etc. etc. etc. then please do derive the answer using the same, =D

Cos, in the short sight, if we consider too many stuff, we will get too complicated of an answer.

11. Jan 13, 2012

### sophiecentaur

This thread needs some ground rules. Are we dealing with an incompressibke fluid (I.e. total theory) or a half way house involving the 'linear' modulus of water OR are we attempting to go further? We need to decide. There's no point arguing between different standpoints.

12. Jan 13, 2012

### Algren

For a compressible fluid with uniform density we dont, that is what this thread is about.

Once we get the above answer, only then we can talk about density variation and chemical changes. Its better to go step by step from rudimentary to advanced.

So, can somebody give me an idea who/what to integrate for compressible fluid with uniform density.

And we are considering bulk modulus. So, there will be a change in radius.

13. Jan 13, 2012

### D H

Staff Emeritus
Hydrostatic equilibrium, $\partial \rho/\partial r = -\rho g$.

Last edited: Jan 13, 2012
14. Jan 13, 2012

### sophiecentaur

If it compressible then its density won't be uniform. You probably know that but what do you mean here?
It strikes me that you could look at the equations for a gas atmosphere (and the derivations used) and find a way to include modulus in the formula.

15. Jan 13, 2012

### Algren

There is a difference between uniform density and constant density.

Uniform, i.e. if the liquid collapses, its net density increase uniformly everywhere, hence density is uniform. Its not like earth, where we have varying densities as we go deep.

Isnt that for a liquid column? its straight from ρgh=P. I thought PA= - B x ΔV/V will be used.

16. Jan 13, 2012

### sophiecentaur

"its net density increase uniformly everywhere"??

I think you mean its modulus is uniform or the substance is the same throughout. Uniform means the same everywhere and its density wouldn't be. It would increase in density towards the centre - which is not what 'uniform' means'.

17. Jan 13, 2012

### IsometricPion

The bulk modulus of water is needed to determine the variation in density with pressure, but it says nothing about how that pressure comes about. Think about a ball of water, consider a cylindrical section of it that passes through the center and has an arbitrarily small radius. The pressure must be distributed such that it is equal at every distance from the center (but will vary with said distance), this is Pascal's law (it also follows from the fact that we are considering fluid in equilibrium). Therefore, there are no forces on the column other than its own weight. So, the pressure at any given point in the ball must be the sum of the external pressure and the integral version for the pressure due to a column of water (since both g and ρ vary with distance from the center).
Wikipedia links to a page with lots of information and diagrams on the properties of water. I also found a NIST webpage that gives lots of information (link to page where one can get the variation of density with pressure up to 10000bar (1GPa) at a given temperature).

It turns out that the water in the central regions of the ball turns into a high pressure form of ice (ice IV assuming the ball is isothermal at 300K) long before the ball is massive enough to have a surface gravity of 9.81m/s2 (at least according to my preliminary calculations and what I have heard on TV). I am getting values of between ~2229Km and ~2360Km for the minimum radius to produce ice (assuming it forms at about 9500bar, the corresponding surface gravities are: 0.697 to 0.709m/s2), depending on the value of the compressibility I use (assuming it is constant, the larger value corresponding to one that interpolates between the minimum and maximum densities, the smaller to that which obtains at everyday pressures). I might be able to extend this analysis to the necessary ball masses if I knew how we wanted to treat the density above 1GPa, for which I don't have any quantitative data (though I expect it exists somewhere on the internet).

Last edited: Jan 13, 2012
18. Jan 14, 2012

### Algren

Hmm, so how will you define the distribution? Im not familiar with that topic yet..

19. Jan 14, 2012

### sophiecentaur

The density distribution in an ideal gaseous atmosphere is along the same lines.

20. Jan 16, 2012

### Algren

Ok, So considering density variation too, and not chemical changes, we first compute the pressure on a spherical shell of water with ∂x thickness and x radius.

So, Gravitational Pressure comes to: (G*4*∏*ρ'*ρ'*x*∂x) / 3;

Where ρ' is the new net pressure of the contained liquid.

There will also be Atmospheric Pressure, Patm

And, there will be (R-x)*ρ''*g'

Where, ρ'' is the net density of the outer spherical shell and g' will be the net gravitational acceleration integrated over the outer shell.

So, g', ρ', and ρ'' and x are variables. Its coming to be pretty hard to make a integration equation.

21. Jan 16, 2012

### sophiecentaur

Why would it be any harder than working our the pressure at the bottom of a column of air?
This link shows you how to do that.

22. Jan 16, 2012

### IsometricPion

In the second section of the link sophiecentaur posted there is an equation for the change in pressure with height: ΔP=-ρgΔh. Dividing both sides by Δh and taking the limit as Δh→0 and replacing h by r, yields the equation D H provided in post #13. Taking ρ and g to be functions only of r, the radius of the ball, the equation is an ordinary differential equation. Integrating both sides of this equation yields an integral expression for the pressure as a function of radius in the ball. g(r) is the acceleration acting on the material at a given radius, r, in this case it is just the gravitational acceleration due to the mass of water at a radius less than r from the center of the ball. ρ(r) is the density of the water as a function of distance from the center of the ball, this can be obtained from the equation of state which gives ρ as a function of temperature and pressure. One easy way to deal with the variation in ρ at constant temperature (the case of interest here) is to take it to be a linear function of pressure: ρ=ρ0+βP for some constant β (which is the bulk modulus or isothermal compressability at the pressure at which ρ=ρ0). Note, however, that this linear function with constant β is only an approximation, it may not be a good model if the pressure is too far from that at which one obtains β (e.g., for water beyond ~2000atm. for β measured at 1atm.). At this point one has a coupled set of algebro-integral equations (an integral equation for P in terms of ρ, and an algebraic equation for ρ in terms of P).

There are several ways of solving such equations, one could substitute the algebraic equation for ρ into the P equation and then successively differentiate and rearrange the resulting equation until one obtained an ordinary differential equation, and then numerically integrate. I prefer the method of successive approximation wherein one starts with a naive guess for P and ρ (such as P=0, ρ=ρ0) and by plugging these guesses into the equations obtains a better estimate of each (which one plugs back into the equations again, and so on until the results converge within the desired accuracy).