ASTR 120 Problem Set 01 Solution Key

Problem 1. A Scale Model with Evidence Tags

Expected approach: multiply AU by 10.0 m for orbital distance. For scaled diameter, compute real diameter, convert the model scale consistently, and express the tiny value in millimeters.

Scaled distances: Mercury \(0.39\times10.0=3.9\) m; Earth \(1.00\times10.0=10.0\) m; Jupiter \(5.20\times10.0=52.0\) m; Neptune \(30.1\times10.0=301\) m.

For Earth, the real diameter is \(2R=12,742\) km. In this model, \(1\) AU = \(1.496\times10^8\) km becomes 10.0 m, so the scale is \(10.0\,\mathrm{m}/1.496\times10^8\,\mathrm{km}=6.68\times10^{-8}\,\mathrm{m\ km^{-1}}\). Earth's scaled diameter is \(12,742\,\mathrm{km}\times6.68\times10^{-8}=8.52\times10^{-4}\) m = 0.852 mm. Equivalent calculations for other worlds are acceptable if units are explicit.

Evidence tags should distinguish apparent positions and brightnesses as observations, spacecraft/radar-derived distances and radii as calibrated measurements, and composition categories as interpretations from density, spectra, samples, and formation models. Good limitations include finite measurement precision, incomplete sampling, model degeneracy, and the fact that bulk categories hide internal layering.

Common errors: using 1 AU = 10 km instead of 10 m, dropping the diameter factor of two, or treating composition as directly observed.

Problem 2. Density as a Comparative Planetology Tool

Expected approach: convert radius to meters, compute spherical volume, divide mass by volume, then compare to Earth by both absolute and ratio methods.

For the body, \(R=3390\) km = \(3.390\times10^6\) m. The volume is

\[V=\frac{4\pi R^3}{3}=\frac{4\pi(3.390\times10^6\,\mathrm{m})^3}{3}=1.63\times10^{20}\,\mathrm{m^3}.\]

Then \(\rho=6.4\times10^{23}\,\mathrm{kg}/1.63\times10^{20}\,\mathrm{m^3}=3.92\times10^3\,\mathrm{kg\ m^{-3}}=3.92\,\mathrm{g\ cm^{-3}}\).

For Earth, \(V_E=4\pi(6.371\times10^6\,\mathrm{m})^3/3=1.08\times10^{21}\,\mathrm{m^3}\), so \(\rho_E=5.97\times10^{24}/1.08\times10^{21}=5.51\times10^3\,\mathrm{kg\ m^{-3}}\). The density ratio is \(3920/5510=0.711\). The ratio method gives \((6.4\times10^{23}/5.97\times10^{24})/(3390/6371)^3=0.107/(0.532)^3=0.713\), agreeing to rounding precision.

Interpretation: the body is less dense than Earth, consistent with less metal fraction, more silicate/volatile material, or different compression. Density alone cannot uniquely determine core size, temperature, porosity, hydration, or detailed chemistry. Magnetic, seismic, gravity-field, moment-of-inertia, and composition measurements would reduce those degeneracies.

Common errors: forgetting to cube radius, mixing km and m, or overclaiming core composition from a single mean density.

Problem 3. Frost-Line Condensation Model

Expected approach: rearrange \(T=280(r/1\,\mathrm{AU})^{-1/2}\) to \(r=(280/T)^2\) AU.

For water ice, \(r=(280/170)^2\,\mathrm{AU}=2.71\) AU. For ammonia-rich ice, \(r=(280/90)^2\,\mathrm{AU}=9.68\) AU.

Mercury and Earth are inside the water frost line in this model, so abundant water ice would not condense locally. Jupiter at 5.20 AU is beyond the water line but inside the 90 K ammonia-rich line; water ice is plausible as solid building material, while ammonia-rich ice is less favored by this simple temperature law. Neptune at 30.1 AU is beyond both boundaries and is consistent with volatile-rich solids.

The limitation paragraph should note that present-day orbit is not necessarily formation location. Planet migration, scattering of icy planetesimals, radial drift of solids, time-varying luminosity, disk opacity, and local pressure all alter condensation boundaries.

Common errors: reversing inside/outside interpretation or using a linear rather than squared temperature-radius relation.

Problem 4. From Nebular Prediction to Testable Evidence

Expected approach: construct a falsifiable claim and evaluate three evidence streams without pretending any single measurement is decisive.

Example prediction: if condensation across the frost line shaped planetesimal composition, then bodies formed beyond roughly 2.7 AU should, on average, show lower bulk densities or direct volatile-rich compositions than bodies formed well inside 2.7 AU, after accounting for giant-planet gas accretion and later migration.

Evidence table should include: bulk density, with expected lower densities for ice-rich bodies but ambiguity from compression and H/He envelopes; meteorites/samples, with primitive or hydrated materials preserving nebular solids but ambiguity from parent-body alteration and collection bias; young-star disks, with rings/gaps/dust temperatures showing where solids exist but ambiguity in translating observed dust to final planets.

The Problems 1-3 data broadly support the prediction at a coarse level: inner Mercury/Earth are rocky/metal-rich, Jupiter is beyond the water frost line, and Neptune is beyond the colder volatile boundary. The support is limited because only four present-day objects are used, Jupiter's low-density envelope is not the same as a simple ice-rock mixture, and migration/scattering can move bodies after formation.

Common errors: presenting the model as proven, ignoring migration, or using disk images as direct photographs of finished planets.