Lecture 3: Minimum Mass Model

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Lecture 3: Minimum Mass Model
Lecture 3 - Minimum mass model of the solar nebula
o Topics to be covered:
o Minimum mass nebula (continued)
o Pressure distribution
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o Temperature distribution
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Gas pressure distribution
o
Know surface mass and density of the early nebula ( (r) = 0 r -3/2), but how is
gas pressure and temperature distributed?
o
Nebula can be considered a flattened disk, each element subject to three forces:
1. Gravitational, directed inward.
2. Inertial, directed outward.
3. Gas pressure, directed upward and outward.
inertial
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Gas pressure distribution
o
o
o
Assume pressure gradient balances z-component of gravity. Hydrostatic equilibrium therefore
applies:
dP = -gzdz
Eqn. 1
z

 GM  z
But, gz  gsin    2 2  2 2 
(r  z ) r  z 
GMz
As (r2 + z2) ~ r2 => gz  3
Eqn. 2
r
g

r

o Substituting Eqn. 2 and Perfect Gas Law (P = RT/) into Eqn. 1:
o
gz
P GMz 
dP    3 dz
RT  r 
Pz 1

Rearrange:
Pc
GM  z
dP  
zdz
3  0


P
RTr

where Pc is the pressure in the central plane, and Pz is the pressure at a height z.
 GMz 2 
Pz  Pc exp
3 
 2RTr 
o

Integrating,
o
The isothermal pressure distribution is flat for small z, but drops off rapidly for larger z.

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Gas pressure distribution
o
Pz drops to 0.5 central plane value, when
 GMz 2 
Pc /2  Pc exp
3 
 2RTr 
o
Rearranging,

ln( 0.5)  0.7  
GMz 2
2RTr 3
1.4RTr 3
 z 
GM
o
Using typical value => z ~ 0.2 AU
(1 AU 
= 1.49 x 1013 cm)
o
Thus z/r is extremely small for bulk of disk gas.
o
Effective thickness of nebular disk as seen from the
Sun is only a few degrees.
PY4A01 Solar System Science
Gas pressure distribution
o
How does the central gas pressure vary with r? Know that the surface density is

 






Pc
RT
dz
Pz / RTdz

 GM  2
exp

z dz

 2RTr 3 

which is a standard definite integral

 = (Pc/RT)(2RTr3/GM)1/2
or
 /Pc = (2r3/RTGM)1/2 = 2160r3/2T1/2
o
Will be given if examined
Using  =3300 r-3/2, P=T1/2/(2160r3/2)=3300T1/2r1.5/2160r1.5
 Pc = 1.5T1/2r -3 Eqn. 3
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Gas temperature distribution
o
In z-direction, temperature gradients are very large (dT/dz>>0), due to the large temperature
difference between the interior and the edge of the disk and the very thin disk in the vertical
direction. This drives convection.
o
We know that P = RT/ and R ~ Cp for the solar nebula.
o
Therefore,
o
dT/dz>>0
As dP/dz = -g;

o
dP C P dT

dz
 dz
dT
 dP


dz C p dz
dT


(g)  g /CP
dz C p
dT/dr<<0
And using g = GMz / r3 (Eqn. 2):
dT
GMz

 1.5 108 zr3
3
dz
CP r

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
Gas temperature distribution
o
Horizontally (in r-direction), temperature gradient is not significant. As little heat
enters or leaves the disk in this direction, the gas behaves adiabatically.
o
The temperature and pressure changes can therefore be related by:
P/P0 = (T/T0)Cp/R
o
For 150 <T<2000K, Cp/R ~7/2 for the nebula, therefore:
Pc ~ Tc7/2 or Tc ~ Pc2/7
o
From Eqn. 3, we know that Pc ~ r -3, thus: Tc ~ (r -3)2/7 ~ r -6/7
o
The temperature therefore falls off relatively slowly with r.
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Minimum mass model of solar nebula
o
Based on current distribution of planet positions and masses, now have a crude model of the
nebula that collapsed, condensed and accreted to form planets.
o
Surface density:
~ r -3/2
o
Force due to gravity:
g(z) = GMz / r3
o
Pressure in z:
 GMz 2 
Pz  Pc exp
3 
 2RTr 
o
Pressure with r:
Pc = 1.5T1/2r-3
o
Temperature with
 z:
dT
GMz

dz
CP r 3
o
Temperature with r:
Tc ~ r-6/7
o

See Chapter IV of Physics
and Chemistry of the Solar System by Lewis.
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