# LARGE Water Body

 P: 47 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.
 P: 882 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.
Mentor
P: 15,201
 Quote by xts You may probably assume that the density $\rho$ is uniform.
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.

P: 47
LARGE Water Body

 Quote by xts 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.
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.
 Mentor P: 15,201 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.
Mentor
P: 15,201
 Quote by D H 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.
Sorry about the bad wording. I meant iron is compressed to 3/5 of its uncompressed volume (or 5/3 its uncompressed density).
P: 168
 Quote by 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.
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?
 Sci Advisor Thanks PF Gold P: 12,253 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.)
P: 418
 Quote by 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.)
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.

 At high temperatures and pressures, such as in the interior of giant planets, it is argued that water exists as ionic water in which the molecules break down into a soup of hydrogen and oxygen ions, and at even higher pressures as superionic water in which the oxygen crystallises but the hydrogen ions float around freely within the oxygen lattice.
I can't find any reference to how high a pressure if needed though.
 P: 47 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.
 Sci Advisor Thanks PF Gold P: 12,253 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.
P: 47

 Quote by xts $$g(r) = \frac{4}{3}\pi \,r^3 \rho\cdot \frac{G}{r^2} = \frac{4\pi}{3}\,G\,\rho\, r$$
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.
 Mentor P: 15,201 Hydrostatic equilibrium, $\partial \rho/\partial r = -\rho g$.
Thanks
PF Gold
P: 12,253
 Quote by Algren So, can somebody give me an idea who/what to integrate for compressible fluid with uniform density.
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.
P: 47
 Quote by 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.
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.

 Quote by D H Hydrostatic equilibrium, $\partial \rho/\partial r = -\rho g$.
Isnt that for a liquid column? its straight from ρgh=P. I thought PA= - B x ΔV/V will be used.
 Sci Advisor Thanks PF Gold P: 12,253 "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'.
P: 177
 Quote by Algren Isnt that for a liquid column? its straight from ρgh=P. I thought PA= - B x ΔV/V will be used.
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).
 Quote by 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.
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).
P: 47
 Quote by sophiecentaur It would increase in density towards the centre - which is not what 'uniform' means'.
Hmm, so how will you define the distribution? Im not familiar with that topic yet..

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