Originally posted by googlefudgeI'm just considering hypotheticals. I've thought about it turning to ice, but the freezing point of water goes down with pressure, not up. Also ice floats, so you're really saying the whole ball of water would simply freeze solid. I'm not sure about that. You could be right, though...
I believe [need to check] that below a certain depth the water forms a solid
pressure ice... and below that you might get such exotics as metallic hydrogen...
But nature is never going to build a ball of water that big with no silicate rocks in it.
Originally posted by SoothfastI now gather after looking at it that you are correct.
Joe Shmo supplied such an equation, and it says pressure at the core is zero IF we assume the fluid is essentially incompressible. Of course, I'm just assuming Joe's equation is correct; but if it were wrong, then it's practically a miracle that it yields the tidy answer I derived.
Just sticking to that one column of water you speak of, you have to con ...[text shortened]... top of the column will have a gravity that pulls "up" on the water at the bottom of the column.
assuming incompressible fluid ( which i'm not sure is a valid assumption), and assuming my general equation is correct.
My final result is
P(h) = 4/3*G*pi*p*(h²-R*h)...(slightly different from your result)
Evaluating
P(0) = 0 (at the surface)
P(R) = 0 (at the center)
So, Cudos on the good intuition. I stand corrected
Originally posted by joe shmoActually I'm still thinking about this, and I think Google is right: there has to be pressure at the core, and it must maximize there. The equation you supplied is probably not applicable to a "global scale" problem like this. I've been thinking about my 200 km tube of water, and letting it freeze into ice in intergalactic space after all. If the ends of two 100 km cylinders of ice came together in space under their mutual gravitational attraction, and I was floating right at the point of contact, I'm damn tootin' going to feel the pressure! And that pressure is not going to magically disappear if the ice melts into water inside a containing tube.
I now gather after looking at it that you are correct.
assuming incompressible fluid ( which i'm not sure is a valid assumption), and assuming my general equation is correct.
My final result is
P(h) = 4/3*G*pi*p*(h²-R*h)...(slightly different from your result)
Evaluating
P(0) = 0 (at the surface)
P(R) = 0 (at the center)
So, Cudos on the good intuition. I stand corrected
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Originally posted by SoothfastYou just need to remember your Newton.
Actually I'm still thinking about this, and I think Google is right: there has to be pressure at the core, and it must maximize there. The equation you supplied is probably not applicable to a "global scale" problem like this. I've been thinking about my 200 km tube of water, and letting it freeze into ice in intergalactic space after all. If the ends o ...[text shortened]... essure is not going to magically disappear if the ice melts into water inside a containing tube.
A planet (water or otherwise) is always trying to collapse under it's own gravity.
The force resisting that is pressure.
The pressure on the surface of a mass centred sphere inside of the planet is going
to equal the weight of the mass in the volume above it divided by the surface
area of the surface.
And while gravity will linearly reduce (given uniform density and symmetry) from
the surface to the core, it is always acting inwards. And so the pressure will always
max out at the very centre.... Even though at the centre you would be weightless.
Originally posted by googlefudgeLets suppose you have a dude in space with two cylinders of water, say 1Km wide and 100Km long. His job is to keep them separated like Sampson shoving apart the columns. Doesn't it stand to reason the gravitational attraction of the cylinders will try to force the ends together where our super hero is trying to keep them apart? Assuming the cylinders of water have closed ends.
You just need to remember your Newton.
A planet (water or otherwise) is always trying to collapse under it's own gravity.
The force resisting that is pressure.
The pressure on the surface of a mass centred sphere inside of the planet is going
to equal the weight of the mass in the volume above it divided by the surface
area of the surface.
...[text shortened]... will always
max out at the very centre.... Even though at the centre you would be weightless.
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Originally posted by sonhouseYes, the two cylinders (of whatever uniform density material) will be
Lets suppose you have a dude in space with two cylinders of water, say 1Km wide and 100Km long. His job is to keep them separated like Sampson shoving apart the columns. Doesn't it stand to reason the gravitational attraction of the cylinders will try to force the ends together where our super hero is trying to keep them apart? Assuming the cylinders of water have closed ends.
gravitationally attracted to one another.
And the guy in the middle will get crushed like a bug.
You have however just made the equations harder by not using spherical
masses that can be modelled as point masses.
So, my original intuition was correct, however the differential equation I derived didn't support it...Yikes, that's what you get when an engineer attempts physics on a planetary scale. However, i'd really like to get some mathematical closure on the subject (is anyone working it out). CALLING ALL RHP PHYSICIST'S!
Originally posted by joe shmoYour original equation leads to two integrals that cancel out if the fluid is incompressible. Perhaps there's a sign error. The negatively valued integral may need to be positive.
So, my original intuition was correct, however the differential equation I derived didn't support it...Yikes, that's what you get when an engineer attempts physics on a planetary scale. However, i'd really like to get some mathematical closure on the subject (is anyone working it out). CALLING ALL RHP PHYSICIST'S!
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Ok here goes...
P is Pressure
Ps is Pressure at the surface
r is Radius
R is Radius at surface
D is density = 1,000 [kg m-3]
G is the gravitational constant = 6.67E-11 m3 [kg-1 s-2]
Mr is Mass inside radius r
Vr is volume at radius r
Pi is 3.14159...
h is depth which is = R - r
gr is gravity at radius r
gs is gravity at radius R [surface]
Pa is Atmospheric Pressure
Ph is Pressure at depth h
Pc is pressure at core.
Gravitational acceleration at radius r = gr = G * Mr / r^2
Mr = Vr * D
Vr = 4/3 * Pi * r^3
Giving:
Mr = 4/3 * Pi * D * r^3
Therefore: [internally to the Water World where density is a constant]
gr = G * (Vr * D) / r^2 = G * (4/3 * Pi * r^3 * D) / r^2 = 4/3 * G * pi * D * r
The hydrostatic equation for hydrostatic equilibrium gives the rate of change of pressure with respect to radius [dP/dr] as...
dP/dr = - (G * Mr * D) / r^2
Substituting Mr into dP/dr gives...
dP/dr = -(4/3 * G * Pi * D * D * r^3) / r^2
Which simplifies to...
dP/dr = -4/3 * G * Pi * D^2 * r
Which means physically that the change of pressure is linearly proportional to radius, and reduces as r increases.
This makes sense, it means that pressure drops the less stuff there is over your head and that pressure is at a maximum
at r=0
However, we want dP/dh ... The change of pressure with depth [h]... Which is the inverse.
Which gives us dP/dh = 4/3 * G * Pi * D^2 * h = 4/3 * G * Pi * D^2 * (R-r)
The pressure at the surface [Ps] cannot be zero.
The reasons being two fold.
One this is a liquid and will boil at zero pressure, this will produce an atmosphere if there was not one already, and
this atmosphere will exert a pressure on the liquid surface. [Pa]
Two we can't have unbalanced forces and gravity is pulling down at the surface and this pressure must be pushing up
to counter balance.
Thus the surface pressure will be the sum of the gravitational force at the surface plus the pressure exerted by the atmosphere
above.
Ps = gs + Pa
gs = 4/3 * G * pi * D * R
Ps = (4/3 * G * Pi * D * R) + Pa
(approximates to Pa for low surface gravities)
Given all that...
The pressure at depth h is given by...
Integral dP/dh = (4/6 * G * Pi * D^2 * h^2) + C
Where C = Ps
So. Ph = (4/6 * G * Pi * D^2 * h^2) + (4/3 * G * Pi * D * R) + Pa
h = R - r
So Ph = (4/6 * G * Pi * D^2 * (R-r)^2) + (4/3 * G * Pi * D * R) + Pa
For r=0 h=R
Pc = (4/6 * G * Pi * D^2 * R^2) + (4/3 * G * Pi * D * R) + Pa
Assuming a radius of 100 km and an atmospheric pressure of 0.35 atmospheres [atm] (3.54 KPa) (boiling point of water 27 Celsius)
This gives a core pressure of 14.15 atm or 1.4 MPa
Assuming a radius of 6400 km [earth radius] and 1 atm atmospheric pressure (101.3KPa)
This gives a core pressure of appx 56,519 atm or 5.725 GPa
However a water body of such size wouldn't have constant density due to water compression and then state change.
EDIT: Hoping I got this all right. It looks physical, but I've gotten a bit rusty... Too much time mending PCs and not
enough doing math. I have a spreadsheet calculating this so I can calculate the pressure at any depth for a water world
of any size with any given atmospheric pressure. (for any liquid density)
Entertainingly, the Neutronium inside a Neutron Star is a liquid... (albeit not one of uniform density, but hey)
If you input a density of 5E17 kg/m3 and a radius of 12 km.
You get a surface pressure of 1.677E12 Pa.
You get a core pressure of ~ 5E33 Pa... or ~5E28 atm.
The actual density varies from ~ 1e9 kg/m3 in the crust, to 6~8E17 kg/m3 in the core.
Which is why if you got stuck on a Neutron star is the ultimate in thinning... you would be lucky to still be a few microns thick.
EDIT2: Nope. Some thing's not right. the pressures I'm getting are nonsense. Don't know if it's cos I balls'd the math or
the spreadsheet... I need to look at it again when it's not 1:30 am on a work day.
Originally posted by SoothfastOk, i'll show how I arrived at the equation
Your original equation leads to two integrals that cancel out if the fluid is incompressible. Perhaps there's a sign error. The negatively valued integral may need to be positive.
Start with a fluid element submerged in a liquid at a depth "h"
The element has a differential thickness "dh"
The force on the elements surface at depth "h" is:
F_h = -P*A
The force on the elements surface at depth h + dh is:
F_h+dh = ( P + dP)*A
The only other force acting on the element is it own differential force of weight:
F_w = -m*g
F_w = -p*V*g
F_w = -p*A*h*g
Given all parameters except for "A" vary with depth "h":
dF_w = -A(dp*g*h + p*dg*h + p*g*dh)
Then the sum of the forces = 0 (static equilibrium)
(P + dP)*A - P*A - A(dp*g*h + p*dg*h + p*g*dh) = 0
Simplify ( Divide by A, and collect terms)
dP = dp*g*h + p*dg*h + p*g*dh
Can anyone see what may not be valid about that part?
Originally posted by joe shmoI understand your development of dF_w based on F_w. I'm not sure why your F_h=-P*A is not similarly developed as dF_h=-A*dP. Instead you have
Ok, i'll show how I arrived at the equation
Start with a fluid element submerged in a liquid at a depth "h"
The element has a differential thickness "dh"
The force on the elements surface at depth "h" is:
F_h = -P*A
The force on the elements surface at depth h + dh is:
F_h+dh = ( P + dP)*A
The only other force acting on the element is ...[text shortened]...
[b]dP = dp*g*h + p*dg*h + p*g*dh
Can anyone see what may not be valid about that part?[/b]
F_{h+dh} = (P+dP)*A,
but isn't that missing a negative sign? Shouldn't it be
F_{h+dh} = -(P+dP)*A …?
Then your equation
(P + dP)*A - P*A - A(dp*g*h + p*dg*h + p*g*dh) = 0
would become
-(P + dP)*A - P*A - A(dp*g*h + p*dg*h + p*g*dh) = 0,……(Eq1)
which still seems wrong. It's been a long time since I've dabbled seriously in physics and its handling of so-called "differentials," but it seems to me that non-differential quantities such as P*A are incommensurate with differential quantities. They're being mixed together here. In your version of the equation the P*A terms just happen to cancel out, and in my version they don't. But anyway, if we just replace -(P + dP)*A - P*A with dF_h=-A*dP to get
-A*dP - A(dp*g*h + p*dg*h + p*g*dh) = 0,
we seem to get crap:
dP = -(dp*g*h + p*dg*h + p*g*dh).
That negative sign just isn't going to help! But maybe (Eq1) is salvageable?
But what's the meaning of F_w = -p*V*g = -p*A*h*g? You're writing V=A*h, but are you saying the volume of the fluid element is its cross-sectional area times its depth? Is this truly a submerged fluid element, or the column of fluid above the element? Are you maybe mixing up the calculation of P at depth h with the calculation of F_w?
Finally, how can F_h and F_w cancel out, given they are two force vectors that point downward? In fact, how can F_h=-P*A be a vector at all, considering that pressure and area are scalar quantities? Are we dealing with magnitudes instead? But the magnitudes can't cancel out either.
I'm discombooberated and conflusterflated. 😉
Originally posted by googlefudgeBummer. This isn't a trivial problem, seemingly. I'll have to flip through a physics text to remind myself of the basic concepts when I get the time.
Ok here goes...
P is Pressure
Ps is Pressure at the surface
r is Radius
R is Radius at surface
D is density = 1,000 [kg m-3]
G is the gravitational constant = 6.67E-11 m3 [kg-1 s-2]
Mr is Mass inside radius r
Vr is volume at radius r
Pi is 3.14159...
h is depth which is = R - r
gr is gravity at radius r
gs is gravity at radius R [surfa ...[text shortened]... the math or
the spreadsheet... I need to look at it again when it's not 1:30 am on a work day.
Originally posted by Soothfast"I understand your development of dF_w based on F_w. I'm not sure why your F_h=-P*A is not similarly developed as dF_h=-A*dP."
I understand your development of dF_w based on F_w. I'm not sure why your F_h=-P*A is not similarly developed as dF_h=-A*dP. Instead you have
F_{h+dh} = (P+dP)*A,
but isn't that missing a negative sign? Shouldn't it be
F_{h+dh} = -(P+dP)*A …?
Then your equation
(P + dP)*A - P*A - A(dp*g*h + p*dg*h + p*g*dh) = 0
would become
-(P + dP ...[text shortened]... d? But the magnitudes can't cancel out either.
I'm discombooberated and conflusterflated. 😉
Because Its not a differential pressure. At depth h the pressure is not changing, thus the force is not changing and its magnitude and direction are respectively
P*A <-j> (because the pressure from the liquid above is acting perpendicular to the top surface of the element)
At the bottom surface {h + dh} of the element the pressure is what it was at the top surface "P" plus a small change in pressure "dP". The force on the bottom surfaces magnitude and direction is respectively:
(P + dP)*A , <j>
It is positive because the pressure in this scenario is trying to compress the body.
The only remaining force acting on the element is the elements weight.
Which incidentally I now think I see a problem with my thought process ( pointed out by you incidentally)
F_w = A*p*g*dh, <-j> (note: the elements has a differential thickness already, but we also have to allow p and g to vary over the thickness dh
so
dF_w = A*(dp*g*dh + p*dg*dh + p*g *d²h), <-j>
Which if we allow p to be constant (which probably shouldn't be the case since the pressures at the core would be immense) after simplification,
dP = ( p*dg*dh + p*g *d²h)
Which looks like it would be a second order, non-linear differential equation...
yikes
Could be wrong of course if my method on dF_w is wrong.
Originally posted by googlefudgeYo, I want to be sure: is your r, which you call the radius, really the distance from the center of the planet? Here's what you got:
Ok here goes...
P is Pressure
Ps is Pressure at the surface
r is Radius
R is Radius at surface
D is density = 1,000 [kg m-3]
G is the gravitational constant = 6.67E-11 m3 [kg-1 s-2]
Mr is Mass inside radius r
Vr is volume at radius r
Pi is 3.14159...
h is depth which is = R - r
gr is gravity at radius r
gs is gravity at radius R [surfa ...[text shortened]... he inverse.
Which gives us [b]dP/dh = 4/3 * G * Pi * D^2 * h = 4/3 * G * Pi * D^2 * (R-r)[/b]
dP/dh = 4/3 * G * Pi * D^2 * h = 4/3 * G * Pi * D^2 * (R-r).
We should expect dP/dh to tend to zero as we approach the center of the planet, since gravity gets weaker. If r represents the distance to the center, then it's going the opposite way. If r represents the depth (i.e. distance from surface), however, things look good. I just worked out the problem by completely different means (setting up an integral as a limit of Riemann sums) and I get
P'(d) = 4/3 * pi * G * D^2 * (R - d),
where d represents the distance from the surface (i.e. the depth), and thus
P(d) = 4/3 * pi * G * D^2 * d * (R - d/2).
So anyway, the pressure at the center I work out to be
P(R) = 2/3 * pi * G * D^2 * R^2.