# Deformation of a self-gravitating sphere from two forces

I have a fluid sphere (say a gas or a liquid of uniform density, under its own gravity) on which forces is applied to its surface. I would like to find its approximate shape (most probably an oblate ellipsoid), from the forces applied on its (initialy) spherical surface. Using spherical coordinates, the radial (pressure) and tangential (shear) forces are these : \begin{align}\tag{1} F_r(\vartheta) &= C \sin^3 \vartheta, \\[12pt] F_{\vartheta}(\vartheta) &= 4C \, \sin^2 \vartheta \, \cos{\vartheta}, \tag{2} \end{align} where $$C$$ is an arbitrary positive constant. In vectorial form : $$\begin{equation}\tag{3} \vec{\boldsymbol{\mathrm{F}}} = C \sin^3 \vartheta \, \vec{\boldsymbol{\mathrm{u}}}_r + 4C \, \sin^2 \vartheta \, \cos{\vartheta} \, \vec{\boldsymbol{\mathrm{u}}}_{\vartheta}. \end{equation}$$ There's an axial symetry around the $$z$$ axis. The deformation may be considered "weak", as a first approximation ($$C$$ may be "small", compared to the gravitationnal force on the surface : $$C \ll G M^2/R^2$$).

Note that the pressure force is 0 at the poles, and maximal at the equator, so it tends to "squash" the sphere to an oblate ellipsoid (of unknown ellipticity). The shear force is 0 at the poles and at the equator.

Any idea about how to find the deformation's ellipticity?

• Makes a difference what the fluid sphere is made of, and the pressure it would exert back. – Bob Bee Jul 17 '17 at 1:33
• Well, lets consider an uncompressible liquid. – Cham Jul 17 '17 at 1:55
• If incompressible it won't change shape – Bob Bee Jul 17 '17 at 2:06
• It should change shape ! It's liquid ! The volume would stay the same, but you could have an ellipsoid of the same volume as the initial sphere. The pressure is variable with $\vartheta$, and it's squashing the sphere at the poles. – Cham Jul 17 '17 at 12:03
• I agree. Forgot about changing shape. – Bob Bee Jul 18 '17 at 0:30

The problem is to find $r(\theta, \phi) = r(\theta)$ with r the radius of the ellipsoide.

I think we can assume that the gravitationnal field created by the fluid is the same as the one from a sphere (weak deformation).

$$\mathbf{G}(r) = - \frac{\mathcal{G}M}{R^3}{r} \mathbf{e}_r$$ Where M is the mass of fluid, and R the radius of the sphere

Fluid statics gives $$\boldsymbol{\nabla} P = \mathbf{f}_v$$ With P the pressure and $\mathbf{f}_v$ the gravitationnal force.

So we reach $$P(r, \theta, \phi) = P(r) = P(0) - \frac{\rho GM}{2}\frac{r²}{R^2}$$

And we should try to make it equal the tangential part of the pressure F.

With $\alpha$ the angle between the normal vector to the surface for the sphere and the ellipsoid, we have $$\mathbf{F}_N = \mathbf{F}_R \cos(\alpha) + \mathbf{F}_{\theta} \sin(\alpha)$$

As we can assume $\alpha <<1$, $$\mathbf{F}_N = \mathbf{F}_R+ \alpha\mathbf{F}_{\theta}$$

And I think we can find $$\alpha = \frac{1}{r} \frac{\textrm{d}r}{\textrm{d}\theta}$$

So we have a differential equation on $r(\theta)$! The inital condition would be that the volume must remain the same.

If the deformation is minimal, I think you can approximate the problem by just increasing the radius of the sphere by the amount that it is deformed outwards (of course on the poles this gives the wrong results then).

Then you can require the increased gravitational force to be equal to the force acting on the equator due to rotation.

This approximation should for example work on earth.