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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:

#### Fx = −(GM/r²)(x/r) + ω²r Fy = −(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 = −(Fy/Fx)

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-3and 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).

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) = 0which 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.)