Calculate the pressure inside a fluid sphere due its own gravity.

In summary, we discussed the problem of calculating the pressure at a distance x from the center of a spherical mass m with radius r of a known fluid with density ρ that floats in the vacuum. It was determined that the pressure is due to the gravitational force of the mass of fluid inside radius x attracting the mass of the fluid outside x. The spherical symmetry of the problem was noted, as well as the fact that the whole mass inside radius R acts as if it was a point mass at the center. The equation dF = GM/R²*dm was suggested for calculating the force acting on a dm in the outside layer, with M being a function of R. The concept of using a layer of thickness dR and integrating from x to
  • #1
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A spherical mass m with radius r of a known fluid with density ρ floats in the vacuum. Calculate the pressure P at the distance x from the center due its own gravity.
 
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  • #2
Does that mean due to the gravitational force of the mass of fluid inside radius x attracting the mass of the fluid outside x? If so, you can make good use of the spherical symmetry. But you must keep in mind that at radius R > x, the whole mass inside radius R will act. The mass inside radius R acts as if it was a point mass at the center. Use dF = GM/R²*dm for the force acting on a dm in the outside layer. Remember M is a function of R. I would consider dm to be the mass of a layer on the sphere at radius r of thickness dR. Integrate from x to r.
 
  • #3
uros said:
... at the distance x from the center due its own gravity.
Also
x<r
 
  • #4
The mass from the radius x:
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
P=ρgh
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.
 
  • #6
Delphi51 said:
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
uros said:
The mass from the radius x:
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
uros said:
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
Maybe use R for positions between x and r?
Then g = Gρ4πR/3, dF = g*dm
 
  • #10
pressure, P
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
Are you trying to avoid doing the necessary integration?

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
Wouldn't the spherical shell with thickness dR be easier to work with?
A thin "column" would have to be wider at the top than at the bottom, wouldn't it?
 
  • #13
Delphi51 said:
Wouldn't the spherical shell with thickness dR be easier to work with?
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
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.
 
Last edited:
  • #15
I'm sure there's a perfectly good reason why the spherical shell method doesn't work, correctly, but I can't say what that is (definitively) at the moment. I suspect is has something to do with the fact that the contributions to the weight vector at x should be the component of the forces acting that are parallel to the normal vector of differential area patch dA at distance x. The normal vector is is perpendicular to the patch at its geometrical center.

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
I found a way that avoids the problems of the spherical shell AND the straw. One thinks of the change in pressure at each layer, not the change in force or weight: dP = ρg*dR
Using the usual g = 4/3*πρGR and integrating we get the same answer as you would with the straw model.
 
  • #17
Nifty. I like it.
 

1. How do you calculate the pressure inside a fluid sphere due to its own gravity?

In order to calculate the pressure inside a fluid sphere due to its own gravity, you will need to use the formula P = (2/3) * (G * rho^2) * (M^2 / R^4), where P is the pressure, G is the gravitational constant, rho is the density of the sphere, M is the mass of the sphere, and R is the radius of the sphere.

2. What factors affect the pressure inside a fluid sphere due to its own gravity?

The pressure inside a fluid sphere due to its own gravity is affected by the density and mass of the sphere, as well as the gravitational constant and the radius of the sphere.

3. Is the pressure inside a fluid sphere due to its own gravity constant throughout the entire sphere?

No, the pressure inside a fluid sphere due to its own gravity is not constant throughout the entire sphere. It decreases towards the outer edges of the sphere due to the decreasing strength of the gravitational force.

4. How does the pressure inside a fluid sphere due to its own gravity compare to the pressure at the surface of the sphere?

The pressure inside a fluid sphere due to its own gravity is higher than the pressure at the surface of the sphere. This is because the weight of the fluid above the surface exerts a downward force, increasing the pressure at the center of the sphere.

5. Can the pressure inside a fluid sphere due to its own gravity be negative?

No, the pressure inside a fluid sphere due to its own gravity cannot be negative. This is because the gravitational force always acts inwards, causing the pressure to be positive at all points within the sphere.

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