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/[−αy⁴] = −(1/α)y^-3
and hence equivalent to
y³dy = −(1/α)dx
which has the solution
y⁴ = C −(4/α)x
 

Let y₀ be the value of y at x=0. Then

y⁴ = y₀⁴ −(4/α)x
= y₀⁴(1 −(4/α)(x/y₀⁴)
and hence
y = y₀(1 −(4/α)(x/y₀⁴)^1/4
 

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

dy/dx = −y/[x - αx⁴]
which is equivalent to
dy/y = −(1/[x - αx⁴])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)]
 

But

dy/dx = y/[x - αr⁴]
= (y/r)/[(x/r) - αr³]
= sin(θ)/[cos(θ) - αr³]
 

Equating these two expressions for dy/dx and cross multiplying to clear fractions gives

r'sin(θ)cos(θ) + rcos²(θ) - αr'r³sin(θ)
= r'sin(θ)cos(θ) - rsin²(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) = r₀ then C=(α/3)r₀² so the polar form of the equation for the profile of the equipotential surface is

(α/3)[r² − r₀²] = ln(tan(θ))
 

(To be continued.)