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Lecture 4 – Protoplanetary disk structure
o  Protoplanetary disk
o  Pressure distribution
o  Density distribution
o  Temperature distribution
o  Cloud collapse & Disks
The Protoplanetary Disk of HL Tauri from ALMA
PY4A01 Solar System Science
Gas pressure distribution
o 
Know surface mass and density of the early nebula (σ (r) = σ0 r -3/2), but how is
gas pressure, density 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
Gas pressure distribution
o  Assume pressure gradient balances z-component of gravity. Hydrostatic equilibrium
therefore applies:
dP = -ρgzdz
o  But,
Eqn. 1
z
gz = gsin φ
,
# GM &)
z
= % 2 2 (+
.
$(r + z )'* r 2 + z2 -
o  As (r2 + z2) ~ r2 =>
GMz
€
gz = 3
g
ϕ
Eqn. 2
r
o  Substituting Eqn. 2 and Perfect Gas Law (P = ρRT/µ) into Eqn. 1:
€
# Pµ &# GMz &
dP = −% (% 3 ( dz
$ RT '$ r '
€
gz
r
Gas pressure distribution
o  Rearrange:
∫
Pz
Pc
$ µGM ' z
1
dP = −&
) ∫ zdz
% RTr 3 ( 0
P
where Pc is the pressure in the central plane, and Pz is the pressure at a
height z.€
o  Integrating,
# µGMz 2 &
Pz = Pc exp%−
3 (
$ 2RTr '
Eqn. 3
€
o  The isothermal
pressure distribution is flat for small z, but drops
off rapidly for larger z.
Gas pressure distribution
o  Pz drops to 0.5 central plane value, when
# µGMz 2 &
Pc /2 = Pc exp%−
3 (
$ 2RTr '
o  Rearranging,
€
µGMz 2
ln(0.5) = −0.7 = −
2RTr 3
1.4RTr 3
=> z =
µGM
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.
Figure from Lewis
Gas density distribution
o  For a thin disk, z << r, and gz ≅ Ω2 z where
Ω=
GM
r3
€
is the Kelperian angular
velocity.
€
o  Using Eqn. 3, the vertical density has simple form:
ρ = ρ0 e
−z2 /2h 2
o  where ρ0 is the mid-plane density ρ = 1 σ
0
2π h
€
and the vertical disk scale height is h ≡ cs => h /r = M −1
Ω
€
Gas pressure distribution
o  How does the central gas pressure vary with r? Know that the surface density is
∫
= ∫
σ=
=
+∞
−∞
+∞
−∞
µPc
RT
ρdz
Pzµ /RTdz
∫
' µGM * 2
exp
z dz
)−
3,
−∞
( 2RTr +
+∞
which is a standard definite integral
€
σ = (µPc/RT)(2πRTr3/µGM)1/2
Will be given if examined
or
σ /Pc = (2πµr3/RTGM)1/2 = 2160r3/2T1/2
Pc = σ r -3/2T -1/2/2160
and using σ = 3300 r -3/2
=> Pc = 1.5T -1/2r -3
Eqn. 4
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 = ρRT/µ and R ~ Cp for the solar nebula.
o 
Therefore,
o 
dT/dz>>0
As dP/dz = -ρg;
€
o 
dP ρCP dT
=
dz
µ dz
dT
µ dP
=>
=
dz ρC p dz
dT
µ
=
(−ρg) = −µg /CP
dz ρC p
And using g = GMz / r3:
dT
µGMz
=−
= −1.5 ×10 8 zr−3
3
dz
CP r
€
€
dT/dr ~ 0
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. 4, 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.
Gas properties in protoplanetary disk
o  Surface density:
σ~ r -3/2
o  Force due to gravity:
g(z) = GMz / r3
# µGMz 2 &
Pz = Pc exp%−
3 (
$ 2RTr '
o  Pressure in z:
Pc = 1.5T-1/2r-3
o  Pressure with r:
o  Density with r
€
ρ = ρ0e −z
o  Temperature with z:
€
o  Temperature with r:
dT ~ −
2 /2h 2
µGMz
dz
CP r 3
Tc ~ r-6/7
€
o  See Chapter IV of Physics and Chemistry of the Solar System by Lewis.
Cloud Collapse and Star/
Planet Formation
o  Jeans cloud collapse equations
describe the conditions required for
an ISM cloud to collapse.
o  As a cloud collapses, central
temperature increases.
o  This is accompanied by spinning-up
of the central star (to conserve AM).
o  Disk also flattens into an oblate
spheroid.
Stars and disks in Yong Stellar Objects (YSOs)
o 
Class 0: Gravity causes cloud to fragment into
dense cores. One core further collapses, becoming
hot enough to start emitting in far IR.
o 
Class I: By Cloud will start to rotate faster and
dust will settle in disk. Inner part of cloud will
start to fuse hydrogen => star is born. Spectrum
does not look like a star yet as is embedded in
envelope of gas and dust.
o 
Class II: Envelope dissipated due to stellar wind.
Due to the dense disk, object will still radiate in
the IR.
o 
Class III: Consist of young star surrounded by a
transparent disk. Disk matter is discarded through
various mechanisms such as photo-evaporation
and accretion onto central star.
o 
Planet formation thought to occur during
transition from class II to class III.