Originally posted by Soothfast
Yo, 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:

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 ...[text shortened]... ).

So anyway, the pressure at the center I work out to be

P(R) = 2/3 * pi * G * D^2 * R^2.

Ok, something is not kosher with the way I did the differential F_w

I just decided to use deltas and take the limit
Let delta = D

So the equation is



-P*A + (P + DP)A = A*(p + Dp)*(g + Dg)*Dh

DP = p*g*Dh + p*Dg*Dh + g*Dp*Dh + Dp*Dg*Dh

Divide through by Dh

DP/Dh = p*g + p*Dg + g*Dp + Dp*Dg

Take the limit as Dh approaches zero and the deltas become differentials

dP/dh = p*g + p*dg + g*dp + dp*dg

Since the density is constant the last two terms drop out.

dP/dh = p*g + p*dg Equation 2

g = 4/3*pi*G*p*(R - h)

dg = -4/3*pi*G*p*dh

dP/dh = 4/3*pi*G*p²*(R - h) - 4/3*pi*G*p²*dh

Factor out the constants

dP/dh = 4/3*pi*G*p² [ (R - h) - dh ]

assume the term dh is appoximately zero

dP/dh = 4/3*pi*G*p² (R - h)

Separate and Integrate assuming P(0) = 0

P(h) = 4/3*pi*G*p²*[ R*h - ½*h² ]

Then Evaluate h = R

P(R) = 2/3*pi*G*p²*R²

Just like you guys arrived at. So I hope we are finally in agreement now!!!

Originally posted by joe shmo
Ok, something is not kosher with the way I did the differential F_w

I just decided to use deltas and take the limit
Let delta = D

So the equation is



-P*A + (P + DP)A = A*(p + Dp)*(g + Dg)*Dh

DP = p*g*Dh + p*Dg*Dh + g*Dp*Dh + Dp*Dg*Dh

Divide through by Dh

DP/Dh = p*g + p*Dg + g*Dp + Dp*Dg

Take the limit as Dh approaches zero and the ...[text shortened]... 2/3*pi*G*p²*R²

Just like you guys arrived at. So I hope we are finally in agreement now!!!

Tah-dah! I guess there is little doubt that this is the correct answer. And it wouldn't be difficult to adjust it for fluid that is compressible in some way.

I guess I shouldn't say the limit of Riemann sums approach (i.e. using the definition of the Riemann integral) is "completely different." It's really just the grandfather of the approach using differentials (or infinitesimals). Generally, though, I've had better luck working with deltas (increments) and taking a limit. I don't actually hassle with the limit directly, I just set it up and then convert to an integral. If I did more of this and paid closer attention to the moving parts, I guess I'd see how it all connects naturally with differentials.

Originally posted by Soothfast
Tah-dah! I guess there is little doubt that this is the correct answer. And it wouldn't be difficult to adjust it for fluid that is compressible in some way.

I guess I shouldn't say the limit of Riemann sums approach (i.e. using the definition of the Riemann integral) is "completely different." It's really just the grandfather of the approach using d ...[text shortened]... attention to the moving parts, I guess I'd see how it all connects naturally with differentials.

Yeah, I got all confused for a while on what was going on with the differentials. To be honest I have never had much experience with them in this context before this problem. I remembered just a little while ago that if:

F_w = A*p*g*dh

then the Total Differential is the sum of the Partial Derivatives times the incremental

dF_w =A [ƒ_p(p,g,dh)*dp + ƒ_g(p,g,dh)*dg + ƒ_dh(p,g,dh)*dh]

Where;

ƒ_p(p,g,dh)*dp = g*dh*dp

ƒ_g(p,g,dh)*dg = p*dh*dg

ƒ_dh(p,g,dh)*dh = g*p*dh

After that, it all comes together in the same way. A nice obsession for a few days, but I'm glad I pushed through it, just for the gain of understanding the math of differentials a bit more.

I have 12 sheets of paper (front and back) scribbled with failed attempts!
🙄 haha, good times!

Originally posted by joe shmo
Yeah, I got all confused for a while on what was going on with the differentials. To be honest I have never had much experience with them in this context before this problem. I remembered just a little while ago that if:

F_w = A*p*g*dh

then the Total Differential is the sum of the Partial Derivatives times the incremental

dF_w =A [ƒ_p(p,g,dh)*dp ...[text shortened]... have 12 sheets of paper (front and back) scribbled with failed attempts!
🙄 haha, good times!

Nice work guys! One question, in the final equation, what is little p? the p^2 bit?

So I guess the pressure does not drop to zero at dead center then.

p^2 is pressure at the surface? Like atmospheric pressure?

If it is, and that pressure is zero, doesn't that bugger up the equation?

Originally posted by sonhouse
Nice work guys! One question, in the final equation, what is little p? the p^2 bit?

So I guess the pressure does not drop to zero at dead center then.

p^2 is pressure at the surface? Like atmospheric pressure?

If it is, and that pressure is zero, doesn't that bugger up the equation?

Surface pressure cannot ever be zero

Both because gravity will always be exerting a force (pressure) on the surface,
and because the liquid will have a gas [atmosphere] above it which will have a pressure.

Originally posted by googlefudge
Surface pressure cannot ever be zero

Both because gravity will always be exerting a force (pressure) on the surface,
and because the liquid will have a gas [atmosphere] above it which will have a pressure.

Will the units come out in Newtons? But, looking at that equation, the lower the surface pressure the lower the center pressure, right? For instance, let's suppose the liquid is mineral oil or a good vacuum oil. Low vapor pressure at the surface, say the pressure is 0.001 units, wouldn't that make the center pressure way lower than it would be if it was water with some kind of normal vapor pressure at the surface?

Originally posted by sonhouse
Nice work guys! One question, in the final equation, what is little p? the p^2 bit?

So I guess the pressure does not drop to zero at dead center then.

p^2 is pressure at the surface? Like atmospheric pressure?

If it is, and that pressure is zero, doesn't that bugger up the equation?

p represents the density of the material with Joe (ideally the Greek letter rho is used, which looks like a p). I used D like Googlefudge does. But Joe and I assume the surface pressure to be ZERO. If it isn't zero, all you have to do is add whatever the surface pressure is to the final result.

Originally posted by googlefudge
Surface pressure cannot ever be zero

Both because gravity will always be exerting a force (pressure) on the surface...

You speak of "pressure" here when you're really talking about the weight an object on the surface has on account of its own mass. That usually isn't counted as pressure. The atmospheric pressure being exerted on me is not more or less than the next guy in line depending on whether he weighs more or less than me. Since pressure is |F|/A (magnitude of force per unit area), we calculate it as a scalar quantity that prevails at a locality, without thinking about what kind of mass a physical object at that locality may possess.

Originally posted by Soothfast
p represents the density of the material with Joe (ideally the Greek letter rho is used, which looks like a p). I used D like Googlefudge does. But Joe and I assume the surface pressure to be ZERO. If it isn't zero, all you have to do is add whatever the surface pressure is to the final result.

Ah, thanks. So if we are talking about water, p and p^2 are both 1, right?

Originally posted by sonhouse
Ah, thanks. So if we are talking about water, p and p^2 are both 1, right?

for water around 0°F, and atmospheric pressure

p = 1000[ kg/m^3]

p^2 = 10^6* [kg²/m^6]

units matter

Edit: To answer your earlier question a Newton is a measure of force(not pressure). If all units are standard metric it comes out to [N/m²] otherwise known as the Pascal [Pa].

uOriginally posted by joe shmo
for water around 0°F, and atmospheric pressure

p = 1000[ kg/m^3]

p^2 = 10^6* [kg²/m^6]

units matter

Edit: To answer your earlier question a Newton is a measure of force(not pressure). If all units are standard metric it comes out to [N/m²] otherwise known as the Pascal [Pa].

Can't you say that a certain mass, say 1000 Kg and the bottom is 1 meter in diameter, wouldn't that equate to a certain amount of newtons AND pascals on the bottom of the piece? On Earth against gravity?

Originally posted by sonhouse
Can't you say that a certain mass, say 1000 Kg and the bottom is 1 meter in diameter, wouldn't that equate to a certain amount of newtons AND pascals on the bottom of the piece? On Earth against gravity?

Yeah, the volume would have a weight (N), and there would be a pressure at the bottom (Pascals), but they are not the same quantity.

If the volume of water had a mass of 1000 kg, the Force of weight would be

W = 1000[kg]*9.81[m/s²] = 9810[kg*m/s²] = 9810[Newtons]

The pressure is the Weight/Area which varies with the Area

So the pressure that mass of water exerts on a circular plate of diameter 1 meter is

A = Pi*D^2/4 = Pi/4[m²]
P = W/A = 9810[N]/(Pi/4[m²]) = 12490 [Pascals]

While if the plate were a 1 meter by 1 meter square the pressure would be

P = W/A = 9810[N]/(1[m²]) = 9810[Pascals]

As you can see we have the same mass, but different pressures. How do you explain this?

Originally posted by joe shmo
Yeah, the volume would have a weight (N), and there would be a pressure at the bottom (Pascals), but they are not the same quantity.

If the volume of water had a mass of 1000 kg, the Force of weight would be

W = 1000[kg]*9.81[m/s²] = 9810[kg*m/s²] = 9810[Newtons]

The pressure is the Weight/Area which varies with the Area

So the pressure that mas ...[text shortened]... cals]

As you can see we have the same mass, but different pressures. How do you explain this?

Ok, got it now. I remember the old vinyl record days where a single gram of pressure was some ungodly amount of pressure per square inch since the area on the record touched by the needle was measured in square microns, so if the area of contact was 10 square microns (just a guess) then the pressure of one gram is what? Not sure how do do the math here. Let's see, if the area of contact was one meter square, than the pressure would be one gram per square meter. If the area of contact was one millimeter^2 then the pressure would be 1,000,000 grams per mm^2? No, It looks like still 1 gram per square mm. Ok, how do I get out of this mess?🙂

Originally posted by sonhouse
Ok, got it now. I remember the old vinyl record days where a single gram of pressure was some ungodly amount of pressure per square inch since the area on the record touched by the needle was measured in square microns, so if the area of contact was 10 square microns (just a guess) then the pressure of one gram is what? Not sure how do do the math here. Let ...[text shortened]... ams per mm^2? No, It looks like still 1 gram per square mm. Ok, how do I get out of this mess?🙂

If you have 1 gram of mass per square meter the pressure isn't 1 gram per square meter.

gram is a measurement of mass.
pressure is a measurement of Force per unit Area.

You need to turn that mass into a force(the force of the masses weight here on earth) by multiplying it by the acceleration due to gravity on earth (9.81 m/s² ), and change the gram into kg, because;

1N = 1kg*m/s²

The pressure would be

P = [1gram]*[1kg/1000gram]*[9.81 m/s²]/1[m²] = 0.00981[N/m²]

Now, that same mass (on earth) distributed over an area of 1[mm²] has a pressure equal to

P = [1gram]*[1kg/1000gram]*[9.81 m/s²]/1[mm²] = 0.00981[N/mm²]

Now, which is the greater pressure?

Originally posted by joe shmo
If you have 1 gram of mass per square meter the pressure isn't 1 gram per square meter.

gram is a measurement of mass.
pressure is a measurement of Force per unit Area.

You need to turn that mass into a force(the force of the masses weight here on earth) by multiplying it by the acceleration due to gravity on earth (9.81 m/s² ), and change the gram i ...[text shortened]... 1gram]*[1kg/1000gram]*[9.81 m/s²]/1[mm²] = 0.00981[N/mm²]

Now, which is the greater pressure?

The latter, by what, a thousand times. I guess you could rate pressure in its ability to punch holes in the object being touched.