
#1
Jan2911, 02:20 PM

P: 17

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.




#2
Jan2911, 02:30 PM

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P: 3,394

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
Jan2911, 02:37 PM

P: 17

x<r 



#4
Jan2911, 03:12 PM

P: 17

Calculate the pressure inside a fluid sphere due its own gravity.
The mass from the radius x:
V'=4πx³/3 m'=4πx³ρ/3 The pressure at the distance x: P=ρgh P=ρGm'(rx)/x² P=ρ²G4πx(rx)/3 



#5
Jan2911, 03:20 PM

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P: 3,394





#6
Jan2911, 03:33 PM

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#7
Jan2911, 04:37 PM

P: 17

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=rx g → gravitational field in function of x and m' g=Gm'/x² g=Gρ4πx/3 P → hydrostatic pressure P=ρgh P=ρ²G4πx(rx)/3 



#8
Jan2911, 04:46 PM

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P: 11,453

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
Jan2911, 04:56 PM

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P: 3,394

Maybe use R for positions between x and r?
Then g = Gρ4πR/3, dF = g*dm 



#10
Jan2911, 05:18 PM

P: 17

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(r³x³)/3 g=Gρ4πx/3 F=Gρ²16π²x(r³x³)/9 P=Gρ²4π(r³x³)/9x 



#11
Jan2911, 05:26 PM

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P: 11,453

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
Jan2911, 06:38 PM

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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
Jan2911, 06:43 PM

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#14
Jan2911, 07:32 PM

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P: 3,394

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. 



#15
Jan2911, 08:07 PM

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P: 11,453

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
Jan2911, 09:07 PM

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P: 3,394

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
Jan2911, 09:31 PM

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P: 11,453

Nifty. I like it.



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