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 The Dynamics of the Gravitational Collapse of a Gas Cloud

This is an investigation of the collapse of a gas cloud under gravitation. As the collapse of the cloud procedes a radial gradient in pressure builds up which interferes with the further collapse. Hydrostatic equilibrium is not immediately achieved because the dynamics of flow causes it to overshot the equilibrium. p>The analysis requires a Lagrangian perspective for the interior of the cloud, but an Eulerian for the surface elements. The analysis involves the assumption of a thin layer on the surface of thickness δ. In effect, the analysis assumes the gas cloud is a gas bubble. This is largely a mathematical convenience in order to have a bounded region of analysis, but there is a tendency of unbalanced intermolecular forces at a boundary to form such a layer.

## The Case of a Nonspinning Cloud

In this case everything is spherically symmetric. Let p(r,t), ρ(r,t), v(r,t) and T(r,at) be the radial profiles of pressure, mass density, radial velocity and temperature, respectively, at time t. These profile run from 0 to R(0). At time zero the variables are all uniform. The radial velocity, in particular, at time zero is uniformly zero. The gas is presumed to be hydrogen obeying the ideal gas law.

The other key variable of the analysis is the cumulative mass, M(r). This is the mass enclosed within a radius r; i.e.,

#### M(r) = ∫0r(4πs²)ρ(s)ds

The gravitation attraction at a distance r from the center of a spherical distribution of mass is equal to that of the mass M(r) concentrated at the center. The attraction of the mass beyond r simply cancels to zero.

## The Surface of the Cloud

The out limit of the gas cloud, R(0), is such that the gravitational attraction exactly counterbalances the pressure force; i.e.,

#### p0dA = (GMρ0/R(0)²)δdA and M = ρ0[(4/3)πR(0)³]

where dA is an infinitesimal spherical area and δ is a small but not infinitesmial radial thickness. M is the mass of the spherical cloud and G is the gravitational constant.

The above two equations reduce to

#### p0 = (Gρ0²(4/3)πR(0))δ which means that R(0) = (3/4)p0/(Gρ0²πδ)

The dynamics of a surface element at times after t=0 is given by:

## The Interior of the Cloud

### The Momentum Equation

In the interior of the cloud

#### ρ(dv(r)/Δt)(dAdr) = (∂p/∂r)dAdr - GM(r)ρ(dAdr)/r² which reduces to dv(r)/Δt = (1/ρ) (∂p/∂r) - GM(r)/r²

The parcel-following Eulerian derivative dv/Δt can be expressed as

#### dv/Δt = ∂v/∂t + v∂v/∂r

Therefore the momentum equation in Lagrangian form is:

### The Continuity Equation

The continuity equation in general is

#### ∂ρ/∂t = −∇(ρv).

In spherical coordinates for this spherically symmetric case the continuity equation reduces to:

#### ∂ρ/∂t = −(2ρv/r + ∂(ρv)/∂r)

The center of the gas cloud is a special case. For a sphere of radius Δr the mass flow through its surface is 4π(Δr)²(ρv). The mass contained in this sphere is (4/3)π(Δr)³ρ. The velocity at r=0 must be zero so the mass flow at Δr is, to the first approximation, (∂(ρv)/∂r)(Δr). Therefore the rate of increase of mass within the sphere is

#### (4/3)π(Δr)³(∂ρ/∂t) = 4π(Δr)²(∂(ρv)/∂r)Δr which reduces to ∂ρ/∂t = 3(∂(ρv)/∂r)

Since ρv/r → (∂(ρv)/∂r) as r → 0 this special case of the center fits in with the general case.

### The Gas Equations

The equation of state for an ideal gas is

#### p = RgρT

where Rg is the gas constant.

For adiabatic processes the Poisson relation

###### p = Cργ

holds, where γ and C are constants. The parameter γ is equal to 1.4 for an ideal gas. The value of the constant C is whatever value p/ργ has for the initial conditions. For such processes the temperature T is also a function of ρ; i.e.,

Since

### The Dynamic Equations

The dynamic equations for the interior of the cloud are therefore

###### (I) ∂v/∂t = −v∂v/∂r + (C/γ)ργ-2(∂ρ/∂r) - GM(r)/r² (II) ∂ρ/∂t = −[2ρv/r + ∂(ρv)/∂r] (III) ∂M/∂t = −(4πr²)ρv

This is a system of three nonlinear partial differential equations in three dependent variables (ρ, v and M) and two independent variables, r and t.

## Initial Conditions

### Density

This is an attempt to get realistic values as the initial conditions of a gas cloud. Consider the mass of the solar system distributed over a sphere of radius twice that of Pluto now.

The mass of all the planets is 449 times the mass of the Earth; the Sun has a mass 332,800 times that of Earth. The mass of all the asteroids, planetary satellites and comets amounts to an insignificant figure compared to the sun. Take 2×1030 kg as the mass of the solar system.

The radius of Pluto's orbit is roughly 6 billion km. A sphere of 10 billion km or 1.0×1010 km has a volume of 4.2×1030 km³. The mass of the solar system spread over this sphere would give a density of 0.48 kg/km³. Rounding that would be 0.5 kg/km³ or 5×10-10 kg/m³. This will be the initial density of the gas cloud.

The initial temperature of the gas cloud will be taken to be about the boiling point of liquid hydrogen; i.e., 20 K.

The gravitational constant is 6.67300ū10-11 m3 kg-1 s-2

The gravitational acceleration GM/r² is, under conditions of uniform density, just proportional to distance; i.e.,

#### GM/r² = G((4/3)πr³ρ0/r² = (4/3)πρ0Gr = (1.4×10-19)r

At the distance of Earth's orbit the gravitational acceleration would be

#### (1.4×10-19)(1.5×1011 m) = 2.1×10-8 ms-2.

At the edge of the gas cloud the gravitational acceleration would be

#### (1.4×10-19)(1.0×1013 m) = 1.4×10-6 ms-2

If a particle continued to accelerated at this rate it would take a time equal to

#### tc = (2×1010/1.4×10-6)1/2 = 1.2×108 seconds = 3.8 years

to reach the center. This figure gives the order of the time scale of the collapse of the cloud.

### Initial Development of the Collapse

The initial radial profile of the acceleration was found in the preceding to be

#### GM/r² = −βr where β= (1.4×10-19).

After an increment of time Δt the radial velocity profile is

#### v = −β(Δt)r

The rate of change of density is then

#### ∂ρ/∂t = −[2ρv/r + ∂(ρv)/∂r] = −[−2ρ(βΔt)r/r + ρ(−βΔt)] = 3ρβ(Δt)

Since initially ρ is independent of distance the density increases uniformly at all radii. Density continues to be radially uniform.

Since the gradient of density is still zero, the momentum equation then give the rate of increase of velocity as

#### ∂v/∂t = −(β(Δt))²r −βr = −[(β(Δt))² + β]r

The velocity gradient continues to be uniform over all radii. This is significant in that it means that the matter moves toward the center uniformly without some parts overtaking or lagging behind other parts. At the center however the matter must build up creating a gradient in density that interferes with the further internal flow near the center.

The cumulative mass M(r) increases in an exponential manner when the density is uniform and the velocity gradient is constant; i.e.,

## Numerical Simulation

(To be continued.)