py4a01 solar system science lecture 2 - minimum mass model of solar nebula otopics to be covered:...
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PY4A01 Solar System Science
Lecture 2 - Minimum mass model of solar nebulaLecture 2 - Minimum mass model of solar nebula
o Topics to be covered:
o Composition and condensation
o Surface density profile
o Minimum mass of solar nebula
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PY4A01 Solar System Science
Terrestrial planets’ compositionTerrestrial planets’ composition
o Mercury has a very large iron core about 3,500 km in diameter that makes up 60% of its total mass, surrounded by a silicate layer ~700 km thick. Its core is probably partially molten.
o Mars has a solid Fe and/or iron-sulfide core ~2,600-4,000 km in diameter, surrounded by a silicate mantle and rocky crust that is probably several hundred km thick.
o Venus' interior is like the Earth's, except its Fe-Ni core probably makes up a smaller percentage of its interior.
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PY4A01 Solar System Science
Jovian planets’ interiorsJovian planets’ interiors
o Jupiter's H/He atmosphere is ~1,000 km thick and merges smoothly with the layer of liquid molecular H, which is ~20,000-21,000 km thick. Pressure near center is sufficient to create a liquid metallic H layer ~37,000-38,000 km thick. Probably has silicate/ice core twice diameter of Earth with ~14 times Earth's mass.
o Saturn is smaller version of Jupiter: silicate core ~26,000 km in diameter, ice layer about 3500 km thick, beneath a ~12,000 km thick layer of liquid metallic H. Then liquid molecular H layer around 28,000 kilometers thick, and atmosphere about 2,000 km thick.
o Compression on Uranus/Neptune probably not enough to liquefy H. Uranus/Neptune have silicate cores ~8,000-8,500 km in diameter surrounded by a slushy mantle of water mixed with ammonia and methane ~7,000-8,000 kilometers thick. At top is a 9000 -10000 km thick atmosphere of H and He.
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PY4A01 Solar System Science
Minimum mass of the solar nebulaMinimum mass of the solar nebula
o Can make approximation of minimum amount of solar nebula material that must have been present to form planets. Know:
1. Current masses, composition, location and radii of the planets.
2. Cosmic elemental abundances.
3. Condensation temperatures of material.
o Given % of material that condenses, can calculate minimum mass of original nebula from which the planets formed.
o Steps 1-8: metals & rock, steps 9-13: ices
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PY4A01 Solar System Science
Nebula compositionNebula composition
o Assume solar/cosmic abundances:
Representative
elements
Main nebular
Low-T material
Fraction of
nebular mass
H, He Gas
H2, He
98.4 %
C, N, O Volatiles (ices)
H2O, CH4, NH3
1.2 %
Si, Mg, Fe Refractories
(metals, silicates)
0.3 %
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PY4A01 Solar System Science
Minimum mass for terrestrial planetsMinimum mass for terrestrial planets
o Mercury:~5.43 gcm-3 => complete condensation of Fe (~0.285% Mnebula).
0.285% Mnebula = 100 % Mmercury
=> Mnebula = (100/ 0.285) Mmercury
= 350 Mmercury
o Venus: ~5.24 g cm-3 => condensation from Fe and silicates (~0.37% Mnebula).
=>(100% / 0.37% ) Mvenus = 270 Mvenus
o Earth/Mars: 0.43% of material condensed at cooler temperatures.
=> (100% / 0.43% ) Mearth = 235 Mearth
o Asteroids: Cooler temperatures produce more condensation ~ 0.5 %.
=> (100% / 0.5%) = 200 Masteroids
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PY4A01 Solar System Science
Minimum mass for terrestrial planetsMinimum mass for terrestrial planets
o What is the minimum mass required to make the Terrestrial planets?
o Total of the 4th column is 29881x1026 g. This is the minimum mass required to form the Terrestrial planets =>2.9881x1030 g ~ 500 Mearth
.
Planet Factor Mass
(x1026 g)
Min Mass
(x1026 g)
Mercury 350 3.3 1155
Venus 270 48.7 13149
Earth 235 59.8 14053
Mars 235 6.4 1504
Asteroids 200 0.1 20
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PY4A01 Solar System Science
Minimum mass for jovian planets and plutoMinimum mass for jovian planets and pluto
o Jupiter: Almost nebula composition due to gas capture ~20%.
=> Mnebula =100 / 20 Mjupiter ~ 5 Mjupiter is minimum mass required.
o Saturn: Cooler than Jupiter, with slightly different composition ~12.5%.
=> Mnebula = 100/12.5 Msaturn ~ 8 Msaturn
o Uranus: Less gas capture ~6.7% condensed to form planet.
=> Mnebula = 100/6.7 Muranus = 15 Muranus
o Neptune: ~5% of solar nebula material condensed to form planet.
=> Mnebula = 100/5 Mneptune = 20 Mneptune
o Pluto: Main fraction due to ices ~1.4 % => Mnebula = 100/0.14 Mpluto = 70 Mpluto
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PY4A01 Solar System Science
Minimum mass for jovian planetsMinimum mass for jovian planets
o What is the minimum mass required to make the Jovian planets?
o Total mass is therefore = 184450 x 1026 g = 3085 Mearth.
o This is minimum solar nebula mass required to make the Jovian planets.
Planet Mass
(x1026 g)
Factor Min Mass
(x1026 g)
Jupiter 19040 5 95200
Saturn 5695 8 55560
Uranus 890 15 13050
Neptune 1032 20 20640
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PY4A01 Solar System Science
Minimum nebula mass Minimum nebula mass
o The minimum mass required to condense the nine planets is therefore:
Planet M (x Mearth)
Terrestrial 500
Jovian 3085
Pluto 0.119
3585 Mearth
o This is the minimum mass required to produce the planets.
o As Msun ~ 2 x 1033 g, the mass required to make the planets is therefore ~0.01 Msun.
o Disk contained 1/100 of the solar mass.
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PY4A01 Solar System Science
Nebular surface density profileNebular surface density profile
o To make a more precise estimate, distribute min mass requirements over series of annuli, centred on each planet.
o Choose boundaries of annuli to be halfway between the orbits of each planet. i.e., Mercury @ 0.38 AU and Venus @ 0.72 AU = > (0.72-0.38)/2 = 0.17 AU.
o We therefore estimate that Mercury was formed from material within an annulus of 0.38±0.17 AU => 0.33 - 0.83 x 1013 cm.
o The surface density of an annulus, = mass / area,
where area = router2
- rinner2
= [(0.83 x 1013)2 - (0.33 x 1013)2] = 1.82 x 1026 cm2
o Surface density of disk near Mercury is therefore:1160x1026 / 1.82 x 1026 = 637 g cm-2
0.33x1013 cm
0.83x1013 cm
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PY4A01 Solar System Science
Nebular surface density profile Nebular surface density profile
o For Venus at 0.72 AU, Mercury is at 0.38 AU and Earth is at 1 AU => Venus’ annulus extends from
(0.72 - 0.38)/2 = 0.17 to (1 - 0.72)/2 = 0.14
o The material that formed Venus was located between0.72 - 0.17 AU and 0.72 + 0.14 or 0.55-0.86 AU. This is 0.83-1.29 x 1013 cm.
o Area is then = router2 - inner
2 = 3.06 x 1026 cm2.
=> = 13150 x 1026 / 3.06 x1 1026 = 4300 g cm-2.
o This is the approximate surface density of the disk where Venus formed.
o For Jupiter at 5.2 AU, the Asteroids are at 3 AU and Saturn is at 9.6 AU. The annulus therefore ranges from 4 - 7.2 or 6 - 11 x 1013 cm.
o As the area = 267 x 1013 cm2 => = 95200 x 1026 / 267 x 1026 = 356 g cm-2
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PY4A01 Solar System Science
Minimum mass and densityMinimum mass and density
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PY4A01 Solar System Science
Surface density of solar nebulaSurface density of solar nebula
o Surface density of the drops off as:
(r) = 0 r -
o 1 < < 2, 0 ~ 3,300 g cm-3.
o Local deficit of mass in asteroid belt. Mars is also somewhat deficient in mass.
o Inside Mercury’s orbit, nebula material probably cleared out by falling in on Sun or blown out.
o Outer edge may be due to a finite scale size of the original nebular condensation.
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PY4A01 Solar System Science
Surface density of solar nebulaSurface density of solar nebula
o Hayashi et al. (1981) widely used:
(r) = 1700 (r / 1AU)-3/2 g cm-2
o Weidenschilling (1977) produced figure at right which shows similar trend.
o Mars and asteroids appears to be under-dense.
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PY4A01 Solar System Science
Surface density of solar nebulaSurface density of solar nebula
o Desch (2007) - disk much denser.
o Disk much more massive:
o 0.092 M in 1-30AU vs 0.011 M
o Density falls steeply (as r-2.2) but very smoothly and monotonically. Matches to < 10%.
o Nepture and Uranus must be switched in position => planetary migration?
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PY4A01 Solar System Science
Minimum mass estimateMinimum mass estimate
o Can also estimate minimum mass from :
where RS is the radius of the Sun and RF is the max distance of Pluto.
o Assume that ( r ) = 3300 ( r / RE )-2, where RE = 1 AU. Therefore,
o Setting RE = 1.49 x 1013 cm, and RS = 6.96 x 1010 cm, and RF = 39 AU = >
M 0.02 MSun
o Ie approximately a factor of two of previous estimate.
€
M = σ (r)dA = σ (r)rdrdθRS
RF∫0
2π
∫∫
€
M = (3300(r /RE )−2)rdrdθRS
RF∫0
2π
∫
= 3300RE2 dθ r−1drdθ
RS
RF∫0
2π
∫= 6600πRE
2 ln(RF /RS )
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PY4A01 Solar System Science