Static sphere with gravitating fluid

  • #1
VincentIsoz78
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1
Hi

The rotating bucket problem with a fluid is well known as a homework. For the fun i wanted to adapt it to the case of a massive non-rotating sphere surrounded by a fluid. However i don't know if the calculations i made are correct or don't make sense at all (even if the result lead to an intuitive conclusion).

Here is how i did it:

Let us consider a volume of fluid of uniform density put beside a material sphere of mass ##M## and radius ##R## and such that the volume of the fluid is very small in comparison of the material sphere (i.e. the sphere is surrounded by the liquid). We assume to be under the incompressibility assumption and steady flow and we neglect the self-gravity of the fluid!

Based on our assumption, the Euler equation :

[tex]\dfrac{\partial \vec{v}}{\partial t}+(\vec{v} \circ \vec{\nabla}) \vec{v}=-\vec{\nabla} \dfrac{p}{\rho}+\vec{g}[/tex]

reduces to (as we are interested of what happens when the fluid is static !) :

[tex]\vec{0}=-\vec{\nabla} \dfrac{p}{\rho}+\vec{g} \Leftrightarrow \vec{\nabla} p=\rho \vec{g}[/tex]

Now we know very well that we have outside of the material sphere:

[tex]g(r)=\dfrac{GM}{r^2}[/tex]

Then the Euler equation can be expressed as (using the gradient in spherical coordinates) :

[tex]\dfrac{\partial p}{\partial r}=\rho \dfrac{G M}{r^2}[/tex]

Integrating we get :

[tex]p=-\rho \dfrac{G M}{r}+C^{t e}[/tex]

Given ##p=c^{\text {te }}## at a free surface, we have :

[tex]r=\dfrac{C^{t e}-c^{t e}}{\rho G M}=\dfrac{k}{\rho G M}[/tex]

Let us put ##c:=k /(\rho G M)##. Therefore we get :

And this is the equation of a sphere in spherical coordinates! Hence the final shape of a fluid surrounding a material sphere is itself a sphere.

What are you thoughts? If this is the wrong place to ask for feedbacks let me know where i could post the above development instead.

Thanks

MENTOR NOTE to POSTER:

We use Mathjax for rendering Latex expressions:
- mathjax keys on double $ quotes for expressions on their own line
- mathjax keys on double # quotes for expressions embedded in text

Adjusted some of your expressions from single $ quoted to double # quoted.
 
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  • #2
VincentIsoz78 said:
Hence the final shape of a fluid surrounding a material sphere is itself a sphere.
Wow, I expected a dodecahedron!
 
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  • #3
A.T. said:
Wow, I expected a dodecahedron!
LoL The idea was the prove it using maths for fun and not just claim it.
 
  • #4
VincentIsoz78 said:
The idea was the prove it using maths for fun and not just claim it.
Since the setup has spherical symmetry, a non-spherical result would violate rotational invariance of space.

I also don't see the final equation after: "Therefore we get :" Maybe the Mod adjusting the Mathjax quotes dropped it?
 
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  • #5
A.T. said:
Since the setup has spherical symmetry, a non-spherical result would violate rotational invariance of space.
Actually, non-spherical solutions are permitted, so long as the set of all solutions has spherical symmetry. But I agree with your point (which can also be shown to be correct in this case.)
 
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  • #6
A.T. said:
Since the setup has spherical symmetry, a non-spherical result would violate rotational invariance of space.

I also don't see the final equation after: "Therefore we get :" Maybe the Mod adjusting the Mathjax quotes dropped it?
The missing part removed by the Mod is

$$r=c$$.

I get your point about rotation invariance. But that's the same for the rotating bucket that you deal in cylindric coordinates and therefore you get obviously a parabola as a result and not a hyperbola.

However my question remains. Does the proof makes sense or is it just good for the garbage?
 
  • #7
VincentIsoz78 said:
I get your point about rotation invariance. But that's the same for the rotating bucket that you deal in cylindric coordinates and therefore you get obviously a parabola as a result and not a hyperbola.
The rotating bucket has axial symmetry, and there are infinitely many axially symmetric surfaces, so you have to derive the right one. Your setup has spherical symmetry, so there is only one possible surface.
 

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