# Show that a rotating (fluid) planet is an oblate spheroid

• I
The idea is that, if we take the rotating frame of the fluid-planet so that it is at rest, then the sum of gravity and the centrifugal force must be perpendicular to the surface.

Take the planet to be rotating about the y axis at a rate ω. By symmetry we only need to work in 2 dimensions. So what we get is that the surface’s normal is directed along N = <Nx, Ny> = <ω2x - GMx/(x2+y2)3/2, -GMy/(x2+y2)3/2>

So the goal is to find the surfaces which are everywhere perpendicular to N. Well N is curl-less and so can be written as the gradient of a scalar potential, and then the surfaces we wish to find will simply be equipotentials. The scalar field with gradient equal N is GM/√(x2+y2) + 0.5(ωx)2 but if we take this expression equal to a constant, we do not get the equation of an ellipse. (The 2D cross section of the surface should be an ellipse.) What went wrong?

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Actually I just realized that my expression for gravitational force comes from assuming the planet is spherical. That’s one mistake.

If we properly accounted for the gravity would the answer actually come out as an exact ellipse? Or is the oblate spheroid just an approximation?

If it is exact (for fluid planets in the Newtonian analysis) then is there any simple derivation? (Integrating gravity of a spheroid doesn’t sound simple.)

For small ω the spherical gravity should be appoximately correct. If you solve for y in the equipotential equation then take the lowest order Taylor-approximation about ω ≈ 0 then you actually do get an equation for an ellipse.

With that, I guess it is just an approximation for low rotational speeds. (I suppose it’s still possible that taking the exact gravity expression will make it an ellipse for any ω, but that seems quite unlikely.)

Guess I jumped the gun on creating this thread.

sophiecentaur
Gold Member
The idea is that, if we take the rotating frame of the fluid-planet so that it is at rest, then the sum of gravity and the centrifugal force must be perpendicular to the surface.
Is that 'obvious'? I think it needs justifying: There is another (hydrostatic) force, due to the pressure of the rest of the fluid? The hydrostatic force is always normal to the surface and, if there is equilibrium, that must mean the sum of the other two forces must also be normal to the surface and of same magnitude as the hydrostatic force.
I don't think there's another force involved???

@sophiecentaur Right, I could’ve explained better. The real idea is that a fluid can’t withstand any shear stress, so that any component of force along the surface would just cause the fluid to redistribute until the perpendicular-condition is satisfied.

There are probably some other subtleties of a rotating fluid planet that I’m overlooking, but my concern was more about the math than the physics.

• sophiecentaur