PY4A01 Solar System Science Lecture 3 - Minimum mass model of the solar nebula oTopics to be covered: oMinimum mass nebula (continued) oPressure distribution.

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PY4A01 Solar System Science Lecture 3 - Minimum mass model of the solar nebula oTopics to be covered: oMinimum mass nebula (continued) oPressure distribution oTemperature distribution

PY4A01 Solar System Science

Gas pressure distribution oKnow surface mass and density of the early nebula (  (r) =  0 r -3/2 ), but how is gas pressure and temperature distributed? oNebula 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 distribution oAssume pressure gradient balances z-component of gravity. Hydrostatic equilibrium therefore applies: dP = -  g z dz Eqn. 1 oBut, oAs (r 2 + z 2 ) ~ r 2 => Eqn. 2 oSubstituting Eqn. 2 and Perfect Gas Law (P =  RT/  ) into Eqn. 1: oRearrange: where P c is the pressure in the central plane, and P z is the pressure at a height z. oIntegrating, oThe isothermal pressure distribution is flat for small z, but drops off rapidly for larger z. gzgz g  r z

PY4A01 Solar System Science Gas pressure distribution oP z drops to 0.5 central plane value, when oRearranging, oUsing typical value => z ~ 0.2 AU (1 AU = 1.49 x cm) oThus z/r is extremely small for bulk of disk gas. oEffective thickness of nebular disk as seen from the Sun is only a few degrees.

PY4A01 Solar System Science Gas pressure distribution oHow does the central gas pressure vary with r? Know that the surface density is Will be given if examined which is a standard definite integral  = (  P c /RT)(2  RTr 3 /  GM) 1/2 or  /P c = (2  r 3 /RTGM) 1/2 = 2160r 3/2 T 1/2 o Using  =3300 r -3/2, P=  T 1/2 /(2160r 3/2 )=3300T 1/2 r 1.5 /2160r 1.5  P c = 1.5T 1/2 r -3 Eqn. 3

PY4A01 Solar System Science Gas temperature distribution oIn 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. oWe know that P =  RT/  and R ~ C p for the solar nebula. oTherefore, oAs dP/dz = -  g; oAnd using g = GMz / r 3 (Eqn. 2): dT/dz>>0 dT/dr<<0

PY4A01 Solar System Science Gas temperature distribution oHorizontally (in r-direction), temperature gradient is not significant. As little heat enters or leaves the disk in this direction, the gas behaves adiabatically. oThe temperature and pressure changes can therefore be related by: P/P 0 = (T/T 0 ) Cp/R oFor 150 <T<2000K, C p /R ~7/2 for the nebula, therefore: P c ~ T c 7/2 or T c ~ P c 2/7 oFrom Eqn. 3, we know that P c ~ r -3, thus: T c ~ (r -3 ) 2/7 ~ r -6/7 oThe temperature therefore falls off relatively slowly with r.

PY4A01 Solar System Science Minimum mass model of solar nebula oBased on current distribution of planet positions and masses, now have a crude model of the nebula that collapsed, condensed and accreted to form planets. oSurface density:  ~ r -3/2 oForce due to gravity:g(z) = GMz / r 3 oPressure in z: oPressure with r: P c = 1.5T 1/2 r -3 oTemperature with z: oTemperature with r:T c ~ r -6/7 oSee Chapter IV of Physics and Chemistry of the Solar System by Lewis.