- #1

- 17

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter uros
- Start date

- #1

- 17

- 0

- #2

Delphi51

Homework Helper

- 3,407

- 10

- #3

- 17

- 0

Also... at the distance x from the center due its own gravity.

x<r

- #4

- 17

- 0

V'=4πx³/3

m'=4πx³ρ/3

The pressure at the distance x:

P=ρgh

P=ρGm'(r-x)/x²

P=ρ²G4πx(r-x)/3

- #5

Delphi51

Homework Helper

- 3,407

- 10

We are on different wavelengths, here. I'm picturing a sphere of liquid in space where g = 0 and P = ρgh is zero. Looking for the pressure inside the sphere "due its OWN gravity". Remember it "floats in vacuum" whereas if it were at the surface of the Earth, g would pull it down to smash on the bottom of its container.P=ρgh

- #6

gneill

Mentor

- 20,925

- 2,866

We are on different wavelengths, here. I'm picturing a sphere of liquid in space where g = 0 and P = ρgh is zero. Looking for the pressure inside the sphere "due its OWN gravity". Remember it "floats in vacuum" whereas if it were at the surface of the Earth, g would pull it down to smash on the bottom of its container.

I think he means that g is the 'local' value of gravitational acceleration due to the mass below x. The problem though, is that 'g' will not remain constant from x to the top of the surface of the sphere.

- #7

- 17

- 0

V'=4πx³/3

m'=4πx³ρ/3

The pressure at the distance x:

P=ρgh

P=ρGm'(r-x)/x²

P=ρ²G4πx(r-x)/3

V'→ spherical volume with radius x

V'=4πx³/3

m'→ mass of the spherical volume with radius x

ρ → density of the fluid

m'=ρV'

m'=ρ4πx³/3

x → distance from the center (x<r)

r → radius of the whole fluid sphere

h → height of the fluid shell from x to r

h=r-x

g → gravitational field in function of x and m'

g=Gm'/x²

g=Gρ4πx/3

P → hydrostatic pressure

P=ρgh

P=ρ²G4πx(r-x)/3

- #8

gneill

Mentor

- 20,925

- 2,866

P → hydrostatic pressure

P=ρgh

P=ρ²G4πx(r-x)/3

The 'g' is not constant over the distance (r - x).

Perhaps you might consider calculating the total weight of a column of arbitrary base area A. The pressure will be equal to the weight divided by the area. (so you'll find that the A will eventually cancel out in the workings). Each differential element dV of the the column will have cross sectional area A and height dr. It will be located at distance r from the center. Sum up the weights of all the dV's.

- #9

Delphi51

Homework Helper

- 3,407

- 10

Maybe use R for positions between x and r?

Then g = Gρ4πR/3, dF = g*dm

Then g = Gρ4πR/3, dF = g*dm

- #10

- 17

- 0

wight of the spherical shell between x and r, F

area of the sphere with radius x, A

mass of the spherical shell, m

P=F/A

A=4πx²

F=mg

m=ρ4π(r³-x³)/3

g=Gρ4πx/3

F=Gρ²16π²x(r³-x³)/9

P=Gρ²4π(r³-x³)/9x

- #11

gneill

Mentor

- 20,925

- 2,866

Once again, you're assuming that the value of g remains constant over the whole height of the shell above point x. This is not the case. You can only assume that it is constant over a very small change in height. So you need to take very thin shells of height dr.

But you don't need to use whole spherical shells. It would be fine to assume a thin column with small base area A (any shape -- the A will cancel out later).

- #12

Delphi51

Homework Helper

- 3,407

- 10

A thin "column" would have to be wider at the top than at the bottom, wouldn't it?

- #13

gneill

Mentor

- 20,925

- 2,866

A thin "column" would have to be wider at the top than at the bottom, wouldn't it?

Nope. Think of a thin straw sticking up from the bottom to the top. It'll work fine, and you won't have to deal with the surface area at each radial distance.

- #14

Delphi51

Homework Helper

- 3,407

- 10

I attempted to work it out with the spherical shell and with the straw; got different answers. With the shell method, for the weight or force I get an integral from x to r of R^3. One R comes from the g, two more from the surface area of the shell. After dividing by the base area 4πx² the final answer for the pressure has (r⁴ - x⁴)/x² multiplied by some constants.

For the straw method, I have only one R from the expression for g. Integrating that and dividing away the base area I get pressure proportional to r² - x², extra factor of 2 in the constants.

I see that my spherical surface result is infinite when x -> 0, reflecting the fact that I divided the force by an area of zero. The straw answer is finite and looks reasonable but I'm uncomfortable with the assumption of constant area in the derivation in this extreme case.

When x -> r, the two solutions converge. As expected - the effect of the straw being too narrow at the top is gone.

Very interesting! I'm doing physics for pleasure and entertainment in my retirement.

For the straw method, I have only one R from the expression for g. Integrating that and dividing away the base area I get pressure proportional to r² - x², extra factor of 2 in the constants.

I see that my spherical surface result is infinite when x -> 0, reflecting the fact that I divided the force by an area of zero. The straw answer is finite and looks reasonable but I'm uncomfortable with the assumption of constant area in the derivation in this extreme case.

When x -> r, the two solutions converge. As expected - the effect of the straw being too narrow at the top is gone.

Very interesting! I'm doing physics for pleasure and entertainment in my retirement.

Last edited:

- #15

gneill

Mentor

- 20,925

- 2,866

The "thin straw" method avoids dealing with radially divergent forces over the patch by allowing the area to be the same, arbitrarily small flat patch at all heights. In the limit as dA goes to zero, the "column" becomes a radial line with a linear mass density.

- #16

Delphi51

Homework Helper

- 3,407

- 10

Using the usual g = 4/3*πρGR and integrating we get the same answer as you would with the straw model.

- #17

gneill

Mentor

- 20,925

- 2,866

Nifty. I like it.

Share: