Week 10: WS17 #2

2. The Minimum Mass Solar Nebula (MMSN): The MMSN is the minimum mass protoplanetary disk that could have produced the Solar System. In this problem we will calculate it with increasing detail. Table 1 provides the current locations and compositions of the Solar System planets, as well as the original composition (estimated from the present day composition of the Sun). 

a) Single planet-based estimate: How many Jupiter masses in original Solar Nebula material would it have taken to build up the present day Jupiter? 

First we have to figure out the "limiting reactant" in Jupiter's formation. We can do this by taking the ratio of Jupiter's composition to the Sun's composition for each individual component. 

H, He: \(\frac{93}{98.4}=0.95\)

C, N, O: \(\frac{6}{1.2}=5\)

Si, Mg, Fe...: \(\frac{1}{0.3}=3.33\)

Since the ratio is greatest for C, N, and O, these are the limiting reactants, and 5 times Jupiter's mass would be required in order to achieve high enough concentrations of those atoms to form the planet. Knowing that the Sun has a mass of \(3.33\times10^5M_\oplus\), we can calculate that mass in solar masses: 

\(\frac{5M_J}{1}\frac{320M_\oplus}{1M_J}\frac{1M_\odot}{3.33\times10^5M_\oplus}=4.8\times10^{-3}M_\odot\)

b) Planet-mass-based estimate: Make a table of the 8 planets. For each planet, calculate how many planet-masses of nebular material would have been needed to make that planet. Add another column with the mass of nebular material needed for each planet (in Solar masses). Which planet is the most important to consider for estimating the minimum mass of the Solar Nebula? What is the total (order of magnitude) minimum mass of the Solar Nebula taking into account all planets? 

The steps for the other planets follow the same order as what we just calculated for Jupiter: use ratios to figure out the limiting reactant and the number of planetary masses, and then convert to solar masses. The solutions for each planet are summarized in the table below. 

Planet	Distance (AU)	Mass (\(M_\oplus\))	Limiting reactant	Material needed (\(M_{planet}\))	Material needed (\(M_\odot\))
Mercury	0.4	0.06	Si, Mg, Fe…	333	6.00E-05
Venus	0.7	0.8	Si, Mg, Fe…	233	5.60E-04
Earth	1	1	Si, Mg, Fe…	233	7.00E-04
Mars	1.5	0.11	Si, Mg, Fe…	233	7.70E-05
Jupiter	5	320	C, N, O	5	4.80E-03
Saturn	9.6	95	C, N, O	6.67	1.90E-03
Uranus	19	15	C, N, O	58.3	2.63E-03
Neptune	30	17	C, N, O	70	3.57E-03

The total material needed is on the order of 10^-2 solar masses, about a third of which is contributed by Jupiter. 

c) Nebular surface density: The surface density describes the amount of material available to form a planet at a specific location assuming that a planet can accrete material regardless of how high up in the disk it is. In this problem we will calculate what the required surface density profile of the MMSN was to form the present-day Solar System. To achieve this, we will take the masses from (b) and distribute these minimum mass requirements over a series of annuli, centered on each planet. Choose boundaries of annuli to be halfway between the orbits of each planet. In the case of Mercury we would estimate that it formed from material within an annulus of 0.4±0.15AU. The surface density of an annulus is mass/area. Calculate the surface density needed for each planet in g/cm². Make a plot of surface density vs. radius. Notice a trend? Any deviations from that trend? How much does the surface density requirement decrease between 1 and 5 AU? Between 5 and 30 AU? 

In order to calculate the surface density, you need the inner and outer radii of each planet's annuli in order to calculate the area. You can then divide the mass, calculated above, by the area. For Mercury, we know that its reach is from 0.25-0.55AU. 

\(A_{Mercury}=\pi(0.55^2-0.25^2)=0.754AU^2\)

\(\frac{0.754AU^2}{1}(\frac{1.496\times10^{13}cm}{1AU})^2=1.69\times10^{26}cm^2\)

\(SD_{Mercury}=\frac{(6\times10^{-5}M_\odot)(2\times10^{33}g/M_\odot)}{1.69\times10^{26}cm^2}=711g/cm^2\)

The calculations for the other planets are summarized in the table below. 

Planet	Distance (AU)	Material needed (\(M_\odot\))	Annuli (AU)	Area (cm2)	Surface density (g/cm2)
Mercury	0.4	6.00E-05	0.25	1.69E+26	712
Venus	0.7	5.60E-04	0.55	2.95E+26	3793
Earth	1	7.00E-04	0.85	5.90E+26	2371
Mars	1.5	7.70E-05	1.25	6.32E+27	24
Jupiter	5	4.80E-03	3.25	3.00E+28	320
Saturn	9.6	1.90E-03	7.3	1.06E+29	36
Uranus	19	2.63E-03	14.3	2.78E+29	19
Neptune	30	3.57E-03	24.5	4.64E+29	15
			35.5

Plotting surface density vs. radius gives us this plot: 

Aside from Mercury and Mars which have very low surface densities, the trend looks like an exponential decay curve. There is a steep decline between 0 and 5 AU, with the surface density remaining relatively constant between 5 and 30 AU. 

d) A widely used expression for MMSN is \(1700(\frac{r}{1AU})^{-1.5}g/cm^2\). How does this expression compare with your answer in (c)? 

Plotting the planets based on this expression (orange) compared to what we calculated (blue) gives us this graph: 

They look pretty close, but I think it's easier to see the fit with a graph of the log of the surface density: 

Mercury and Mars are still outliers, but the trendlines are pretty similar.