o Topics to be covered: o Composition and …...PY4A01 Solar System Science Lecture 3 - Minimum mass...

PY4A01 Solar System Science

Lecture 3 - 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

Minimum Mass Solar Nebula (MMSN)

o  MMSN is not a nebula, but a protoplanetary disc.

Protoplanetary disk Nebula

o  Gives minimum mass of solid material to build the 8 planets.

Minimum 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

•  Figure from Page 115 of “Physics & Chemistry of the Solar System” by Lewis

Nebula 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

Si, Mg, Fe Refractories (metals, silicates)

Minimum mass for terrestrial planets o  Mercury:~5.43 g cm-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

Minimum 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

Minimum 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

Minimum 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

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.

Nebular 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

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 between

0.72 - 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 = � router

2 - � 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

Minimum mass and density

From “Physics and Chemistry of the Solar System” by Lewis

Surface density of solar nebula

o  Surface density of the drops off as:

� (r) = �0 r -�

o  � ~ 1.5 and �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.

Surface density of solar nebula

o  Hayashi (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.

Minimum Mass Extrasolar Nebula

o  Chuan & Laughlin (2013) used 1,925 exoplanets observed by Kepler (http://arxiv.org/abs/1211.1673)

o  Surface density near exoplanet is:

is best fit power-law relationship for the Solar System. o  Semimajor axis from Kepler’s laws is:

σ exoplanet =Mexoplanet

2πaexoplanet2

Mexoplanet = Rexoplanet /REarth( )2.06MEarth

aexoplanet = P / yr( )2 /3 AU

Minimum Mass Extrasolar Nebula

o  MMEN:

σ = 620 a0.2AU# $ %

& ' ( −1.6

Minimum 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)rdrdθ

RF∫0

2π∫

= 3300RE2 dθ r−1drdθ

RF∫0

2π∫

= 6600πRE2 ln(RF /RS )

o Topics to be covered: o Composition and …...PY4A01 Solar System Science Lecture 3 - Minimum mass...

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