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Lecture 4 – Protoplanetary Disk Structure

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Lecture 4 – Protoplanetary Disk Structure 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 -3/2 o Know surface mass and density of the early nebula (σ (r) = σ0 r ), 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 Eqn. 1 z o But, g gz = gsinφ gz # GM &) z , ϕ = % 2 2 (+ 2 2 . r $ (r + z )'* r + z - o As (r2 + z2) ~ r2 => GMz Eqn. 2 € g = z r3 o Substituting Eqn. 2 and Perfect Gas Law (P = ρRT/µ) into Eqn. 1: € # Pµ&# GMz& dP = −% (% ( dz $ RT '$ r3 ' € Gas pressure distribution Pz 1 $ µGM ' z o Rearrange: dP zdz ∫ = −& 3 ) ∫ Pc P % RTr ( 0 where Pc is the pressure in the central plane, and Pz is the pressure at a height z.€ # µGMz2 & P P exp o Integrating, z = c % − 3 ( Eqn. 3 $ 2RTr ' 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 # µGMz2 & Pc /2 = Pc exp% − ( $ 2RTr3 ' µGMz2 o Rearranging, ln(0.5) = −0.7 = − 2RTr3 € 1.4RTr3 => z = µGM o Using typical value => z ~ 0.2 AU 13 (1 AU €= 1.49 x 10 cm) o Thus z/r is extremely small for bulk of disk gas. Figure from Lewis o Effective thickness of nebular disk as seen from the Sun is only a few degrees. Gas density distribution 2 o For a thin disk, z << r, and g z ≅ Ω z where GM Ω = r3 is the Kelperian angular€ velocity. o Using Eqn. 3,€ the vertical density has simple form: −z2 /2h2 ρ = ρ0e o where ρ0 is the mid-plane density 1 σ ρ0 = 2π h € c and the vertical disk scale height is h ≡ s => 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 = ρdz ∫−∞ +∞ = P µ /RTdz ∫−∞ z µP +∞ ' µGM * = c exp) − , z2dz RT ∫−∞ ( 2RTr3 + which is a standard definite integral € 3 1/2 σ = (µPc/RT)(2πRTr /µGM) Will be given if examined or 3 1/2 3/2 1/2 σ /Pc = (2πµr /RTGM) = 2160r T -3/2 -1/2 Pc = σ r T /2160 -3/2 and using σ = 3300 r -1/2 -3 => Pc = 1.5T r 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. dP ρC dT o Therefore, = P dz µ dz dT µ dP => = dz ρCp dz dT/dz>>0 o As dP/dz = -ρg; dT µ € = (−ρg) = −µg/C P dT/dr ~ 0 dz ρCp o And using g = GMz / r3: dT µGMz 8 −3 € = − 3 = −1.5 ×10 zr dz CP r € 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: Cp/R P/P0 = (T/T0) o For 150 <T<2000K, Cp/R ~7/2 for the nebula, therefore: 7/2 2/7 Pc ~ Tc or Tc ~ Pc -3 -3 2/7 -6/7 o From Eqn. 4, we know that Pc ~ r , thus: Tc ~ (r ) ~ r 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 # µGMz2 & P P exp o Pressure in z: z = c % − 3 ( $ 2RTr ' -1/2 -3 o Pressure with r: Pc = 1.5T r € −z2 /2h2 o Density with r ρ = ρ0e µGMz dT ~ − 3 dz o Temperature with z: CPr € -6/7 o Temperature with r: Tc ~ r € 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. .
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