# Question about water pressure in oceans:

sonhouse
Science 16 Dec '13 16:55
1. sonhouse
Fast and Curious
16 Dec '13 16:55
So just saw a tweet about a dude in a sub at the bottom of the Mariana trench.

So that is a lot of pressure, not sure how many bars but the moons of Saturn and Jupiter may have deep, ultradeep oceans.

Anyone here have any idea of what the pressures would be like compared to the same depth in an Earth ocean?

For instance, what if you had a total water world. Would the pressure be the same assuming the same radius as Earth?

I assume the pressure would go down at X depth depending on the size of the world, say assume the total water world was 1000 km radius, say 1/6th of Earth radius. Would it be safe to say at X depth the pressure would also be 1/6th of Earth at the same depth?
2. 16 Dec '13 20:50
Originally posted by sonhouse
So just saw a tweet about a dude in a sub at the bottom of the Mariana trench.

So that is a lot of pressure, not sure how many bars but the moons of Saturn and Jupiter may have deep, ultradeep oceans.

Anyone here have any idea of what the pressures would be like compared to the same depth in an Earth ocean?

For instance, what if you had a total wa ...[text shortened]... Would it be safe to say at X depth the pressure would also be 1/6th of Earth at the same depth?
pressure is a function of the weight of the water above you.

And weight is dependent on gravity.

The higher the gravity, the heavier the water, the higher the pressure for any given depth.
3. sonhouse
Fast and Curious
16 Dec '13 22:44
pressure is a function of the weight of the water above you.

And weight is dependent on gravity.

The higher the gravity, the heavier the water, the higher the pressure for any given depth.
That's what I thought. So they should be able to go really deep into the ocean on Europa should they prove it has oceans.
4. Soothfast
0,1,1,2,3,5,8,13,21,
16 Dec '13 23:50
pressure is a function of the weight of the water above you.

And weight is dependent on gravity.

The higher the gravity, the heavier the water, the higher the pressure for any given depth.
So, at the center of a world made completely of water there should be (virtually) no pressure because there is no gravity, yes? Starting from the surface and making toward the center, then, the pressure should increase up to some maximum, and then with additional depth should begin to decrease.
5. joe shmo
Strange Egg
17 Dec '13 00:47
Originally posted by Soothfast
So, at the center of a world made completely of water there should be (virtually) no pressure because there is no gravity, yes? Starting from the surface and making toward the center, then, the pressure should increase up to some maximum, and then with additional depth should begin to decrease.
I think the pressure would be at its maximum at the center of the globe, then decrease as you approached the surface.
6. joe shmo
Strange Egg
17 Dec '13 00:591 edit
The Hydrostatic Differential Equation in its general form:

dP = -[dp*h*g + p*dh*g + p*h*dg] Eq(1)

where P = pressure
p = density of liquid (a function of "h"ðŸ˜‰
h = height from surface
g = accel. due to gravity (a function of "h"ðŸ˜‰

let ƒ(h) = F'(h)
if: p = F(h)
then: dp = ƒ(h)*dh

and similarly
g = F_1(h)
dg = ƒ_1(h)*dh

Substitute all that in to Eq(1)

dP = -[ƒ(h)*dh*h*F_1(h) + F(h)*dh*F_1(h) + F(h)*h* ƒ_1(h)*dh] ...Eq(2)

Then

P(h) = Int(Eq(2)) + C

Anyhow, the point I was trying to make is that it stands to reason that through all of this there could be a term independent of "g" in this equation, meaning that simply because g = 0 the pressure does not equal 0.
7. sonhouse
Fast and Curious
17 Dec '13 11:511 edit
Originally posted by joe shmo
The Hydrostatic Differential Equation in its general form:

dP = -[dp*h*g + p*dh*g + p*h*dg] Eq(1)

where P = pressure
p = density of liquid (a function of "h"ðŸ˜‰
h = height from surface
g = accel. due to gravity (a function of "h"ðŸ˜‰

let ƒ(h) = F'(h)
if: p = F(h)
then: dp = ƒ(h)*dh

and similarly
g = F_1(h)
dg = ƒ_1(h)*dh

Substitute all t ...[text shortened]... endent of "g" in this equation, meaning that simply because g = 0 the pressure does not equal 0.
I was thinking about places like Europa where there are supposedly 40 mile deep oceans. If it is true the depth you can visit is related to the gravity field of the planet in question, then on Earth we know we can visit down to some 30,000 feet deep with current technology. So it seems on Europa we should be able to get to at least 5 times that depth with the same pressure load so that would put such machines able to explore a rather large percentage of said ocean.

I would assume such machines of the future would be powered by some kind of nuclear reaction since there would be a need for long term exploration, whether manned or robot.
8. Soothfast
0,1,1,2,3,5,8,13,21,
17 Dec '13 12:096 edits
Originally posted by joe shmo
The Hydrostatic Differential Equation in its general form:

dP = -[dp*h*g + p*dh*g + p*h*dg] Eq(1)

where P = pressure
p = density of liquid (a function of "h" )
h = height from surface
g = accel. due to gravity (a function of "h" )

let ƒ(h) = F'(h)
if: p = F(h)
then: dp = ƒ(h)*dh

and similarly
g = F_1(h)
dg = ƒ_1(h)*dh

Substitute all ...[text shortened]... endent of "g" in this equation, meaning that simply because g = 0 the pressure does not equal 0.
M'kay, so, to find the pressure at some specific depth (i.e height from surface) d, presumably the limits of integration for

P(h) = Int(Eq(2)) + C, … Eq(3)

are 0 and d, as in:

P(d) = Int_0^d(Eq(2))

For the sake of simplicity we can assume that water is utterly incompressible, so f(h)=0 for all h, and F(h) is a constant C. Now Eq(3) becomes

P(d) = C*Int_0^d F_1(h)dh + C*Int_0^d h*f_1(h)dh, … Eq(4)

We need F_1(h), the gravity at h. Fortunately this is relatively easy if we assume a water world (i.e. water all the way to the center) with constant density. Let's say it has radius 100 km, assuming that's large enough to "keep it together" as a sphere. At "height from surface" h (in kilometers) the distance to the core is 100-h. The mass M(h) enclosed within a sphere of radius 100-h centered at the core is

M(h) = (density)(volume) = C(4/3)(pi*(100-h)^3),

assuming the constant C to be in the proper units. Gravitational acceleration is then given as

F_1(h) = -G*M(h)/(100-h)^2 = -G*C*4/3*pi*(100-h)^3*(100-h)^-2 = -4GC*pi/3*(100-h),

a linear function! (G is the gravitational constant.) Now

f_1(h) = F'_1(h) = 4GC*pi/3.

Putting this all into Eq(4):

P(d) = C*Int_0^d -4GC*pi/3*(100-h)dh + C*Int_0^d 4GC*pi/3*h dh.

Letting d=100, which puts us at the core, and pulling out the constant factor 4GC^2*pi/3, we get

P(100) = 4GC^2*pi/3[ Int_0^100 (h-100)dh + Int_0^100 (h)dh ] = 4GC^2*pi/3*Int_0^100 (2h-100)dh

The last integral, you can see, works out to zero. Thus:

P(100) = 0.

Unless I screwed something up real nice, it looks like there's no pressure at the core of a world made entirely of an incompressible fluid!
9. sonhouse
Fast and Curious
17 Dec '13 12:30
Originally posted by Soothfast
M'kay, so, to find the pressure at some specific depth (i.e height from surface) d, presumably the limits of integration for

P(h) = Int(Eq(2)) + C, … Eq(3)

are 0 and d, as in:

P(d) = Int_0^d(Eq(2))

For the sake of simplicity we can assume that water is utterly incompressible, so f(h)=0 for all h, and F(h) is a constant C. Now Eq(3) becomes
...[text shortened]... looks like there's no pressure at the core of a world made entirely of an incompressible fluid!
So a curve of the pressure vs depth would be an upswing in pressure as you go down but then reaching a minimum and decreasing to zero at dead center?

With max pressure at say, 0.7 radius?
10. Soothfast
0,1,1,2,3,5,8,13,21,
17 Dec '13 12:491 edit
Originally posted by sonhouse
So a curve of the pressure vs depth would be an upswing in pressure as you go down but then reaching a minimum and decreasing to zero at dead center?

With max pressure at say, 0.7 radius?
Letting constant=4GC^2*pi/3, we have

P(d) = (constant)*Int_0^d (2h-100)dh.

So

P'(d) = (constant)(2d-100)

by Ye Olde Fundamental Theorem of Calculus. To find the maximum for P(d), set P'(d)=0 and solve for d:

(constant)(2d-100)=0, or 2d-100=0.

So, according to this (unless again I screwed somethin' up re-e-eaal nice), it appears that fluid pressure is maximum when d=50 kilometers. That would be halfway down, which I guess makes some sense. If the fluid is actually compressible, my guess is that it would maximize somewhere else, like maybe 60 km or 40 km. EDIT: I'll bet on something less than 50 km.

At http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth there's an interesting chart that shows how gravity actually increases as you descend into the Earth until you're nearly halfway to the core.
11. sonhouse
Fast and Curious
17 Dec '13 13:013 edits
Originally posted by Soothfast
We have, where constant=4GC^2*pi/3,

P(d) = (constant)*Int_0^d (2h-100)dh.

So

P'(d) = (constant)(2d-100)

by Ye Olde Fundamental Theorem of Calculus. To find the maximum for P(d), set P'(d)=0 and solve for d:

(constant)(2d-100)=0, or 2d-100=0.

So, according to this (unless again I screwed somethin' up re-e-eaal nice), it appears that flu ...[text shortened]... ravity actually increases as you descend into the Earth until you're nearly halfway to the core.
Of course in the case of Europa, it's a pretty big moon, something like our own moon in size(4% bigger) and if it has a 40 mile deep ocean, it would turn to rock most likely underneath that so 40 miles deep would still be relatively close to the surface compared to the total radius.

Which would suggest a higher density volume under said ocean which would upset the nice applecart you just builtðŸ™‚

Well, to a certain extent anyway, I am sure gravity would still be zero at dead center, even if it was 1000 degrees CðŸ™‚
12. Soothfast
0,1,1,2,3,5,8,13,21,
17 Dec '13 13:182 edits
Originally posted by sonhouse
Of course in the case of Europa, it's a pretty big moon, something like our own moon in size and if it has a 40 mile deep ocean, it would turn to rock most likely underneath that so 40 miles deep would still be relatively close to the surface compared to the total radius.

Which would suggest a higher density volume under said ocean which would upset the ...[text shortened]... nt anyway, I am sure gravity would still be zero at dead center, even if it was 1000 degrees CðŸ™‚
Gravity is always zero at the center. But the interplay between gravity and pressure is tricky. It depends on how density varies, and whether you're going into a gas, liquid, or solid. The deeper you go, gravity tends to decrease while the pressure tends to increase. Unless the planet is much denser near the core than at the surface, in which case both may increase for a fraction of a radius down. In the case of Jupiter or the sun, I think pressure increases just about all the way down on account of continuously increasing density. Gases and fluids are quite different in the compressibility department!

How about a planet made entirely of water the size of the Earth? I should think the water molecules would break down at some point because of the extreme pressure half a radius down.
13. 17 Dec '13 13:21
Originally posted by Soothfast
So, at the center of a world made completely of water there should be (virtually) no pressure because there is no gravity, yes? Starting from the surface and making toward the center, then, the pressure should increase up to some maximum, and then with additional depth should begin to decrease.
I am going to have to look carefully at the maths later, but pressure is at
maximum in the core of a water world. (or any other world).

It is true that g in the centre of a water world would be 0.

But the water column 'above' you has gravity acting on it to crush it
down towards the centre of the planet.

The force acting against this crushing is pressure, and that must maximise at
the centre for a stable (non-collapsing or expanding) planet.

What we are looking for here is an equation that integrates over the weight
of the column above a specific location with gravity varying as a function of radius.
14. 17 Dec '13 13:24
Originally posted by Soothfast
Gravity is always zero at the center. But the interplay between gravity and pressure is tricky: the deeper you go, gravity decreases while (at first) the pressure increases. Unless the planet is much denser near the core than at the surface, in which case both may increase for a fraction of a radius down. In the case of Jupiter or the sun, I think pres ...[text shortened]... ter molecules would break down at some point because of the extreme pressure half a radius down.
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.
15. Soothfast
0,1,1,2,3,5,8,13,21,
17 Dec '13 13:46
I am going to have to look carefully at the maths later, but pressure is at
maximum in the core of a water world. (or any other world).

It is true that g in the centre of a water world would be 0.

But the water column 'above' you has gravity acting on it to crush it
down towards the centre of the planet.

The force acting against this crushin ...[text shortened]... e weight
of the column above a specific location with gravity varying as a function of radius.
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 consider what it means for there to be an "up" and a "down". Take a column of water and bottle it in a cylindrical tube 200 km long. Put it in intergalactic space (but don't let it freeze!) Put yourself inside the tube at one end. What's the pressure? Gotta be zip, although there will be a gravitational tug on you from the sheer mass of the water. Now put yourself at the center of the tube, 100 km from either end. What's the pressure? Surely zip! And there will be no net gravitational tug, since the 100 km of water on either side of you has equal mass and opposite gravitational acceleration vectors. This 200 km tube of water represents a "slice" of my hypothetical water world with radius 100 km.

Now, run another 200 km tube of water through the midpoint of the first one, at a right angle, say. Put yourself at the intersection of this gigantic tire-iron of two tubes of water. Pressure? Still zero. Run more and more tubes of water through the intersection point until you've essentially built a water world with radius 100 km. The pressure at the center is not going to deviate from zero. All you're doing is adding zeros. "Down" is the point of intersection of all the tubes, though.

Finally, if you remove the tubing so that the hundreds of separate columns of water come together to form a sphere, I don't see anything changing. You have a water world with no hydraulic pressure at the center.

As you descend "down" a column of water in a water world, the inverse-square law causes all kinds of crazy cancellations of gravitational force. In fact, water at the top of the column will have a gravity that pulls "up" on the water at the bottom of the column.