Gas Cloud

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applet-magic.comThayer Watkins

Silicon Valley

& Tornado Alley

USA

The Dynamics of

the Gravitational Collapseof 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 . In effect, the analysis assumes

the gas cloud is a gas bubble. This is largely a

mathematical convenience in order to have a boundedregion 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

Dynamics of the Gravitational Collapse of a Gas Cloud http://www.sjsu.edu/faculty/watkins/gravcollap

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In this case everything is spherically symmetric. Let

p(r,t), (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(4s)(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 = (GM0/R(0))dA

andM = 0[(4/3)R(0)]

where dA is an infinitesimal spherical area and is a

small but not infinitesmial radial thickness. M is the

mass of the spherical cloud and G is the gravitational

constant.


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The above two equations reduce to

p0 = (G0(4/3)R(0))

which means thatR(0) = (3/4)p0/(G0)

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

given by:

(R)(dv(R)/t) = p(R) - GM(R)(R)/Ror, equivalently

dv(R)/t = (1/(R))p(R)/ - GM(R)/Rand

dR/t = v(R).

The Interior of the Cloud

The Momentum Equation

In the interior of the cloud

(dv(r)/t)(dAdr) = (p/r)dAdr -GM(r)(dAdr)/r

which reduces todv(r)/t = (1/) (p/r) - GM(r)/r

The parcel-following Eulerian derivative dv/t can be

expressed as

dv/t = v/t + vv/r

Therefore the momentum equation in Lagrangian form


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is:

v/t = vv/r + (1/) (p/r) - GM(r)/r

The Continuity Equation

The continuity equation in general is

/t = (v).

In spherical coordinates for this spherically symmetric

case the continuity equation reduces to:

/t = (2v/r + (v)/r)

The center of the gas cloud is a special case. For a

sphere of radius r the mass flow through its surface is4(r)(v). The mass contained in this sphere is

(4/3)(r). The velocity at r=0 must be zero so the

mass flow at r is, to the first approximation,

((v)/r)(r). Therefore the rate of increase of mass

within the sphere is

(4/3)(r)(/t) = 4(r)((v)/r)r

which reduces to/t = 3((v)/r)

Since v/r ((v)/r) as r 0 this special case of the

center fits in with the general case.


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The Gas Equations

The equation of state for an ideal gas is

p = RgT

where Rg is the gas constant.

For adiabatic processes the Poisson relation

p = C

holds, where and C are constants. The parameter is

equal to 1.4 for an ideal gas. The value of the constant C

is whatever value p/

has for the initial conditions. For

such processes the temperature T is also a function of ;

i.e.,

T = p/(Rg) = (C-1

/Rg)

The Cumulative Mass Function

Since

M(r) = 0r(4s)(s)ds

M/t = (4r)v(r)

The Dynamic Equations


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The dynamic equations for the interior of the cloud are

therefore

(I) v/t = vv/r + (C/)

-2

(/r) -GM(r)/r

(II) /t = [2v/r + (v)/r]

(III) M/t = (4r)v

This is a system of three nonlinear partial differential

equations in three dependent variables (, 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 21030

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 1.01010

km has a volume of


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4.21030

km. The mass of the solar system spread over

this sphere would give a density of 0.48 kg/km.

Rounding that would be 0.5 kg/km or 510-10

kg/m.

This will be the initial density of the gas cloud.

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

The gravitational constant is 6.6730010-11

m3

kg-1

s-2

The gravitational acceleration GM/r is, under

conditions of uniform density, just proportional to

distance; i.e.,

GM/r = G((4/3)r0/r = (4/3)0Gr =

(1.410-19

)r

At the distance of Earth's orbit the gravitational

acceleration would be

(1.410-19

)(1.51011

m) = 2.110-8

ms-2

.

At the edge of the gas cloud the gravitational

acceleration would be

(1.410-19

)(1.01013

m) = 1.410-6

ms-2

If a particle continued to accelerated at this rate it

would take a time equal to

tc = (21010

/1.410-6

)1/2

= 1.2108

seconds =

3.8 years


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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 inthe preceding to be

GM/r = r

where = (1.410-19).

After an increment of time t the radial velocity profile

is

v = (t)r

The rate of change of density is then

/t = [2v/r + (v)/r]

= [2(t)r/r + (t)]= 3(t)

Since initially 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, themomentum equation then give the rate of increase of

velocity as

v/t = ((t))r r = [((t)) + ]r

The velocity gradient continues to be uniform over all


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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.,

M/t = (4r)v = (4r)((t))r= 4r)(t) = (1/3)M(t)

and thus

M(r,t) = M(r,0)e((t/3)t

Numerical Simulation

(To be continued.)

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