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The Shape of a Rotating Gas Cloud

The purpose of this material is to derive the shape of the equipotential surfaces
of a rotating gas cloud held together by gravitational attraction to a central mass. The equipotential
surfaces describe the shape of a body because the gradient of these surfaces gives the
force which counterbalances the gradient of the pressure in the body.

Let the y-axis be the axis of rotation. For a particle at the point (x,y) the components
of the forces on it are:

F_{x} = −(GM/r²)(x/r) + ω²r
F_{y} = −(GM/r²)(y/r)

where r²=x²+y², G is the gravitational constant and M is the central mass.

The equation of the equipotential surface is given by:

dy/dx = −(F_{y}/F_{x})

For the components of force given above this equation reduces to:

dy/dx = y/[x - α(x²+y²)²

where α=ω²/GM.

A numerical solution of this equation for arbitrary values of the variables and the parameter
is shown below. This shows the general shape of the solutions to the equation.

For x<<y this is approximately

dy/dx = y/[−αy^{4}] = −(1/α)y^{-3} and hence equivalent to
y³dy = −(1/α)dx
which has the solution
y^{4} = C −(4/α)x

Let y_{0} be the value of y at x=0. Then

y^{4} = y_{0}^{4} −(4/α)x
= y_{0}^{4}(1 −(4/α)(x/y_{0}^{4})
and hence
y = y_{0}(1 −(4/α)(x/y_{0}^{4})^{1/4}

On the other hand, where y<<x the equation for the equipotential surfaces reduces to

dy/dx = −y/[x - αx^{4}] which is equivalent to
dy/y = −(1/[x - αx^{4}])dx
which has the solution
ln(y) = −(1/3)ln(x³/(1-αx³))
or
y = (1-αx³)^{1/3}/x

The equation for the profile of an equipotential surface has a far simpler form in polar
coordinates; i.e., r=r(θ).

y = r(θ)sin(θ)
x = r(θ)cos(θ)
therefore
dy/dθ = r'(θ)sin(θ) + r(θ)cos(θ)
dx/dθ = r'(θ)cos(θ) − r(θ)sin(θ)
and hence
dy/dx = (dy/dθ)/(dx/dθ)
= [r'(θ)sin(θ) + r(θ)cos(θ)]/[r'(θ)cos(θ) − r(θ)sin(θ)]

With the elimination of the common term r'sin(θ)cos(θ) and the combining
of rsin²(θ) and rcos²(θ) the result is

r - αr'r²sin(θ) = 0 which reduces to
αr²r'(θ) = 1/sin(θ)

The quadrature (integration) of this last equation gives

(α/3)r² = ln(tan(θ)) + C

Imposing a boundary condition at θ=0 is inappropriate. A more convenient point is
θ=π/4. If r(π/4) = r_{0} then C=(α/3)r_{0}² so
the polar form of the equation for the profile of the equipotential surface is