PY4A01 Solar System Science Lecture 3 - Minimum mass model of the solar nebula oTopics to be...
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PY4A01 Solar System Science Lecture 3 - Minimum mass model of the solar nebula Lecture 3 - Minimum mass model of the solar nebula o Topics to be covered: o Minimum mass nebula (continued) o Pressure distribution o Temperature distribution QuickTime™ and a Video decompressor are needed to see this picture.
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Transcript of PY4A01 Solar System Science Lecture 3 - Minimum mass model of the solar nebula oTopics to be...
PY4A01 Solar System Science
Lecture 3 - Minimum mass model of the solar nebulaLecture 3 - Minimum mass model of the solar nebula
o Topics to be covered:
o Minimum mass nebula (continued)
o Pressure distribution
o Temperature distribution
QuickTime™ and aVideo decompressor
are needed to see this picture.
PY4A01 Solar System Science
PY4A01 Solar System Science
PY4A01 Solar System Science
Gas pressure distributionGas 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
PY4A01 Solar System Science
Gas pressure distributionGas pressure distribution
o Assume pressure gradient balances z-component of gravity. Hydrostatic equilibrium therefore applies: dP = -gzdz Eqn. 1
o But,
o As (r2 + z2) ~ r2 => Eqn. 2
o Substituting Eqn. 2 and Perfect Gas Law (P = RT/) into Eqn. 1:
o Rearrange:
where Pc is the pressure in the central plane, and Pz is the pressure at a height z.
o Integrating,
o The isothermal pressure distribution is flat for small z, but drops off rapidly for larger z.
€
gz = gsinφ =GM
(r2 + z2)
⎡
⎣ ⎢ ⎤
⎦ ⎥z
r2 + z2
⎛
⎝ ⎜
⎞
⎠ ⎟
€
gz =GMz
r3
€
dP = −Pμ
RT
⎛
⎝ ⎜
⎞
⎠ ⎟GMz
r3
⎛
⎝ ⎜
⎞
⎠ ⎟dz
€
1
PdP
Pc
Pz∫ = −μGM
RTr 3
⎛
⎝ ⎜
⎞
⎠ ⎟ zdz
0
z
∫
€
Pz = Pc exp −μGMz2
2RTr 3
⎛
⎝ ⎜
⎞
⎠ ⎟
gzg
r
z
PY4A01 Solar System Science
Gas pressure distributionGas pressure distribution
o Pz drops to 0.5 central plane value, when
o Rearranging,
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.
€
Pc /2 = Pc exp −μGMz2
2RTr 3
⎛
⎝ ⎜
⎞
⎠ ⎟
€
ln(0.5) = −0.7 = −μGMz2
2RTr 3
=> z =1.4RTr 3
μGM
PY4A01 Solar System Science
Gas pressure distributionGas pressure distribution
o How does the central gas pressure vary with r? Know that the surface density is
€
= dz−∞
+∞
∫= Pzμ /RTdz
−∞
+∞
∫
=μPc
RTexp −
μGM
2RTr 3
⎛
⎝ ⎜
⎞
⎠ ⎟z2dz
−∞
+∞
∫
Will be given if examined which is a standard definite integral
= (Pc/RT)(2RTr3/GM)1/2
or
/Pc = (2r3/RTGM)1/2 = 2160r3/2T1/2
o Using =3300 r-3/2, P=T1/2/(2160r3/2)=3300T1/2r1.5/2160r1.5
Pc = 1.5T1/2r -3 Eqn. 3
PY4A01 Solar System Science
Gas temperature distributionGas 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 As dP/dz = -g;
o And using g = GMz / r3 (Eqn. 2):
dT/dz>>0
dT/dr<<0€
dP
dz=
ρCP
μ
dT
dz
=>dT
dz=
μ
ρCp
dP
dz
€
dT
dz=
μ
ρCp
(−ρg) = −μg /CP
€
dT
dz= −
μGMz
CP r3= −1.5 ×108 zr−3
PY4A01 Solar System Science
Gas temperature distribution 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 ~ Pc
2/7
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.
PY4A01 Solar System Science
Minimum mass model of solar nebulaMinimum 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:
o Pressure with r: Pc = 1.5T1/2r-3
o Temperature with z:
o Temperature with r: Tc ~ r-6/7
o See Chapter IV of Physics and Chemistry of the Solar System by Lewis.
€
Pz = Pc exp −μGMz2
2RTr 3
⎛
⎝ ⎜
⎞
⎠ ⎟
€
dT
dz= −
μGMz
CP r3
