Lecture 11 Solutions: Solar System Architecture & Formation

Practice Problem Solutions

Solutions to the Lecture 11 practice problems.
Author

Dr. Anna Rosen

Published

February 23, 2026

Note

Student note: These are model solutions written to show every step, units, and checks. I do not expect your work to be this detailed; shorter solutions are fine if your setup, units, and reasoning are correct.

HW note: Homework uses only a subset of these problems; the full set is included for future study and self-checking.

Core Problems

Problem 1: Kepler Application (Mars Period)

Restatement: Calculate Mars’s orbital period given its semi-major axis.

Given:

  • Semi-major axis: \(a = 1.52\) AU

Find: Period \(P\) in years.

Equation (Sun units): When \(P\) is in years, \(a\) is in AU, and the central mass is the Sun, Kepler’s Third Law simplifies to: \[P^2 = a^3\]

Solution:

Step 1 — Cube the semi-major axis: \[a^3 = (1.52 \text{ AU})^3 = 1.52 \times 1.52 \times 1.52 = 3.512 \text{ AU}^3\]

Step 2 — Since \(P^2 = a^3\), we have \(P^2 = 3.512 \text{ yr}^2\). Take the square root: \[P = \sqrt{3.512 \text{ yr}^2} = 1.87 \text{ yr}\]

Unit check: In Sun units (years, AU), \(P^2 = a^3\) returns years. ✓

Sanity check: Mars is farther than Earth but not hugely so (1.52 AU vs 1 AU), so a period somewhat longer than 1 year is reasonable. 1.87 years ≈ 1 year and 10.5 months. ✓

Answer: \(P \approx 1.87\) years.

What you should learn from this: In Sun units, Kepler III is a one-step calculation. The simplified form \(P^2 = a^3\) works only when \(P\) is in years and \(a\) is in AU — these “natural” units absorb the constants \(4\pi^2/(GM_\odot)\) that would otherwise appear. Mars oppositions (when Earth “laps” Mars) occur approximately every 2.1 years, consistent with this period.

Problem 2: Newton Application (Saturn’s Mass from Titan)

Restatement: Use Titan’s orbit to estimate Saturn’s mass and compare to Jupiter.

Given:

  • Titan’s orbital radius: \(a = 1.22 \times 10^6\) km
  • Titan’s orbital period: \(P = 16\) days
  • \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\)

Find: Saturn’s mass \(M\) in kg.

Equation: \[M = \frac{4\pi^2 a^3}{GP^2}\]

Solution:

Step 1 — Convert to SI units:

  • \(a = 1.22 \times 10^6 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 1.22 \times 10^9 \text{ m}\)
  • \(P = 16 \text{ days} \times \frac{86{,}400 \text{ s}}{1 \text{ day}} = 1.382 \times 10^6 \text{ s}\)

Step 2 — Compute the numerator (\(4\pi^2 a^3\)):

\[a^3 = (1.22 \times 10^9 \text{ m})^3 = 1.22^3 \times 10^{27} \text{ m}^3 = 1.816 \times 10^{27} \text{ m}^3\]

\[4\pi^2 \times 1.816 \times 10^{27} \text{ m}^3 = 39.48 \times 1.816 \times 10^{27} \text{ m}^3 = 7.17 \times 10^{28} \text{ m}^3\]

Step 3 — Compute the denominator (\(GP^2\)):

\[P^2 = (1.382 \times 10^6 \text{ s})^2 = 1.910 \times 10^{12} \text{ s}^2\]

\[GP^2 = (6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}) \times (1.910 \times 10^{12} \text{ s}^2) = 1.275 \times 10^{2} \text{ m}^3 \text{ kg}^{-1}\]

Step 4 — Divide:

\[M = \frac{7.17 \times 10^{28} \text{ m}^3}{1.275 \times 10^{2} \text{ m}^3 \text{ kg}^{-1}} = 5.6 \times 10^{26} \text{ kg}\]

Unit check:

  • Numerator: \(\text{m}^3\) (from \(a^3\); the \(4\pi^2\) is dimensionless)
  • Denominator: \(\text{m}^3 \text{ kg}^{-1} \text{ s}^{-2} \times \text{s}^2 = \text{m}^3 \text{ kg}^{-1}\)
  • Result: \(\frac{\text{m}^3}{\text{m}^3 \text{ kg}^{-1}} = \text{kg}\) ✓

Sanity check: Jupiter’s mass (from the worked example in the reading) is \(1.9 \times 10^{27}\) kg. Saturn should be less massive: \[\frac{M_{\text{Saturn}}}{M_{\text{Jupiter}}} = \frac{5.6 \times 10^{26}}{1.9 \times 10^{27}} \approx 0.30\]

Saturn is about 30% of Jupiter’s mass — roughly one-third. This matches the known value. ✓

Answer: \(M_{\text{Saturn}} \approx 5.6 \times 10^{26}\) kg, about 30% of Jupiter’s mass.

What you should learn from this: Newton’s form of Kepler III lets you weigh a planet by watching its moons. The calculation follows the same template as the Jupiter worked example: convert to SI, cube \(a\), square \(P\), divide. Unit tracking at every step catches errors before they propagate.

Problem 3: Wien Application (Neptune’s Temperature)

Restatement: Estimate Neptune’s temperature from its thermal emission peak.

Given:

  • Peak wavelength: \(\lambda_{\text{peak}} = 50{,}000\) nm (= 50 μm)
  • Wien’s Law constant: \(b = 2.9 \times 10^6\) nm·K

Find: Temperature \(T\) in Kelvin.

Equation: \[T = \frac{b}{\lambda_{\text{peak}}} = \frac{2.9 \times 10^6 \text{ nm·K}}{\lambda_{\text{peak}}}\]

Solution:

\[T = \frac{2.9 \times 10^6 \text{ nm·K}}{50{,}000 \text{ nm}} = \frac{2.9 \times 10^6}{5.0 \times 10^4} \text{ K} = 58 \text{ K}\]

Unit check: \(\frac{\text{nm·K}}{\text{nm}} = \text{K}\) ✓

Sanity check: Neptune is at 30 AU — very far from the Sun. A temperature of 58 K (\(-215°\)C) is extremely cold, as expected. For comparison: Earth’s equilibrium temperature is ~255 K, Jupiter’s is ~110 K, and Neptune is roughly half of Jupiter’s — consistent with being ~6× farther from the Sun. ✓

Answer: \(T \approx 58\) K.

What you should learn from this: Wien’s Law is a powerful thermometer. By measuring where thermal emission peaks (here, in the far-infrared at 50 μm), you can determine temperature without touching the object. The key is getting the units right: both \(b\) and \(\lambda_{\text{peak}}\) must use the same wavelength unit (here, nm).

Problem 4: Formation Concept (Orbital Alignment)

Restatement: Explain why all planets orbit the Sun in the same direction and in nearly the same plane.

Key insight: The solar system’s orderly pattern is inherited from the rotating disk of gas and dust that formed it.

Answer:

All planets formed from the same rotating disk of gas and dust (the solar nebula). Three key physics principles explain the pattern:

  1. Angular momentum conservation — The original molecular cloud had a slight rotation. As it collapsed under gravity, it spun faster (like a figure skater pulling in their arms, L5–L6 connection). This gave the entire system a single, well-defined rotation direction.

  2. Collisional flattening — The cloud started roughly spherical but flattened into a disk. Material moving vertically (toward the midplane) collided with material coming from the opposite direction. These inelastic collisions converted vertical kinetic energy into heat, which radiated away. But material moving horizontally (orbiting) was supported by angular momentum and couldn’t collapse inward. Result: vertical motion was damped, horizontal motion was preserved → a thin disk.

  3. Inheritance — Planets formed within this disk by accumulating material from it. They inherited the disk’s rotation direction and orbital plane.

This is why we see: all planets orbiting counterclockwise (viewed from above Earth’s north pole), all orbits nearly in the same plane (the ecliptic), and the Sun rotating in the same direction.

Sanity check: If planets formed independently (not from a shared disk), we’d expect random orbital orientations — some clockwise, some inclined. The fact that they’re aligned is strong evidence for a common formation disk. ✓

What you should learn from this: The orderly pattern of the solar system is not a coincidence — it’s a fossil of the disk from which everything formed. This is the Observable → Model → Inference pattern: we observe alignment, the model is a collapsing rotating cloud, and we infer a shared formation history.

Problem 5: Tool Choice (Rocky vs. Gaseous)

Restatement: Identify what measurements and Module 1 tools you need to determine whether an exoplanet is rocky or gaseous.

Key insight: Rocky vs. gaseous is a density question, and density requires mass and radius.

Answer:

To determine rocky vs. gaseous, you need the planet’s density: \(\rho = M / V\).

This requires two measurements:

Measurement 1 — Mass: For exoplanets, the most practical Module 1 tool is the Doppler effect (L10): measure the wobble the planet induces in its host star’s radial velocity. A larger wobble means a more massive planet. In principle, Newton’s version of Kepler’s Third Law (L5–L6) could also weigh a planet if you observed a moon orbiting it, but for most exoplanets the Doppler method is the realistic route.

Measurement 2 — Radius: Use transit observations. When the planet passes in front of its star, it blocks a fraction of the starlight. The fractional brightness dip tells you the planet’s cross-sectional area, and hence its radius: \(\Delta F / F \approx (R_{\text{planet}} / R_{\text{star}})^2\).

Decision rule: Calculate the volume (\(V = \frac{4}{3}\pi R^3\)), then the density (\(\rho = M/V\)). Compare:

  • Rocky planets: \(\rho \approx 3.9\)–\(5.5\) g/cm³ (like Earth at 5.5 g/cm³)
  • Gas giants: \(\rho \approx 0.7\)–\(1.6\) g/cm³ (like Jupiter at 1.33 g/cm³)

What you should learn from this: Classifying a planet requires combining multiple tools — no single measurement is enough. Transit gives radius, Doppler gives mass, and together they give density. This is why astronomers build multi-tool chains of inference.

Challenge Problems

Problem 6: Frost Line Reasoning (Luminous Sun)

Restatement: Determine how the frost line changes if the Sun were twice as luminous, and what consequences this has for planet types.

Given:

  • Current frost line: \(\sim 3\) AU
  • Hypothetical luminosity: \(L_{\text{new}} = 2 L_\odot\)
  • Equilibrium temperature scaling: \(T_{\text{eq}} \propto L^{1/4} / \sqrt{a}\)
  • Ice condensation temperature: \(T_{\text{frost}} \approx 150\) K

Find: New frost line distance and consequences for planetary architecture.

Equation: At the frost line, \(T_{\text{eq}} = T_{\text{frost}}\). Setting the two expressions equal:

\[\frac{T_{\text{eq,new}}}{T_{\text{eq,old}}} = \frac{L_{\text{new}}^{1/4}}{L_{\text{old}}^{1/4}} \times \sqrt{\frac{a_{\text{old}}}{a_{\text{new}}}} = 1\]

(We set this equal to 1 because both are at the frost line temperature.)

Solution:

Solve for the ratio of frost line distances:

\[\sqrt{\frac{a_{\text{old}}}{a_{\text{new}}}} = \frac{L_{\text{old}}^{1/4}}{L_{\text{new}}^{1/4}} = \frac{1}{2^{1/4}}\]

\[\frac{a_{\text{old}}}{a_{\text{new}}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}\]

\[a_{\text{new}} = a_{\text{old}} \times \sqrt{2} = 3 \text{ AU} \times 1.41 = 4.2 \text{ AU}\]

Unit check: AU × (dimensionless) = AU ✓

Sanity check: A more luminous Sun heats the disk more, so the frost line should move outward — it does. The change is modest (~40% farther) because temperature depends on \(L^{1/4}\), a weak dependence. ✓

Answer: The frost line moves from \(\sim 3\) AU to \(\sim 4.2\) AU.

Consequences:

  • The rocky planet zone extends farther out — more room for terrestrial worlds to form
  • Gas giants form farther from the star — there may be less material at greater distances, so gas giants might be smaller or take longer to form
  • The entire planetary architecture shifts outward — a more “spread out” system

What you should learn from this: The frost line is not a fixed property of planetary systems — it depends on the host star’s luminosity. This is why different stars produce different planetary architectures, a key theme in L12.

Problem 7: Integration Problem (Rocky Planets at 5 AU)

Restatement: Infer properties of a host star from the extent of its rocky planet zone.

Given:

  • Rocky planets extend to 5 AU in the exoplanet system
  • Our frost line: \(\sim 3\) AU at \(L = L_\odot\)

Find: Luminosity of the host star relative to the Sun.

Equation: The frost line distance scales as: \[a_{\text{frost}} \propto L^{1/2}\]

This comes from setting \(T_{\text{eq}} = 150\) K in the relation \(T_{\text{eq}} \propto L^{1/4}/\sqrt{a}\) and solving for \(a\).

Solution:

\[\frac{a_{\text{frost,star}}}{a_{\text{frost,Sun}}} = \left(\frac{L_{\text{star}}}{L_\odot}\right)^{1/2}\]

If we assume that the outer edge of the rocky-planet region is set mainly by the frost line (rather than migration or later reshuffling), then the frost line must be at or beyond 5 AU:

\[\frac{5 \text{ AU}}{3 \text{ AU}} = \left(\frac{L_{\text{star}}}{L_\odot}\right)^{1/2}\]

\[\frac{L_{\text{star}}}{L_\odot} = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \approx 2.8\]

Unit check: Ratio of distances (dimensionless) squared = ratio of luminosities (dimensionless) ✓

Sanity check: Rocky planets farther out means a hotter star — the luminosity is greater than \(L_\odot\), which is consistent. A factor of 2.8 is not extreme — it corresponds to a star modestly hotter and more massive than the Sun. ✓

Answer: Under that assumption, the host star is roughly 2.8 times more luminous than the Sun (\(L_{\text{star}} \approx 2.8 L_\odot\)).

Interpretation: The main physical takeaway is that the star would need to be more luminous than the Sun, so its disk would stay warm farther out. That pushes the frost line outward and gives rocky material more room before ices can condense.

What you should learn from this: The frost line is a powerful diagnostic for the host star. By observing where rocky planets end and giants begin, you can estimate the star’s luminosity — as long as you remember the key assumption that the system has not been strongly rearranged by migration.

Problem 8: Frost Line from First Principles

Restatement: Derive the frost line distance from the equilibrium temperature scaling and known values.

Given:

  • \(T_{\text{eq}} \propto 1/\sqrt{a}\) (for a fixed star)
  • \(T_{\text{eq,Earth}} = 255\) K at \(a_{\oplus} = 1\) AU
  • Ice condensation temperature: \(T_{\text{frost}} = 150\) K

Find: Frost line distance \(a_{\text{frost}}\) in AU.

Equation: \[T_{\text{eq}}(a) = T_{\text{eq,Earth}} \times \sqrt{\frac{a_{\oplus}}{a}}\]

Set \(T_{\text{eq}} = 150\) K and solve for \(a\).

Solution:

Step 1 — Substitute known values:

\[150 \text{ K} = 255 \text{ K} \times \sqrt{\frac{1 \text{ AU}}{a}}\]

Step 2 — Isolate the square root:

\[\frac{150 \text{ K}}{255 \text{ K}} = \sqrt{\frac{1 \text{ AU}}{a}}\]

\[0.588 = \sqrt{\frac{1 \text{ AU}}{a}}\]

Step 3 — Square both sides:

\[0.588^2 = \frac{1 \text{ AU}}{a}\]

\[0.346 = \frac{1 \text{ AU}}{a}\]

Step 4 — Solve for \(a\):

\[a = \frac{1 \text{ AU}}{0.346} = 2.89 \text{ AU} \approx 3 \text{ AU}\]

Unit check: K/K = dimensionless ✓ ; AU / (dimensionless) = AU ✓

Sanity check: The frost line should be between Mars (1.5 AU) and Jupiter (5.2 AU) — and 2.9 AU is exactly in that range. It sits within the asteroid-belt region and well outside Mars, which is the usual rough picture for our solar system’s early frost line. ✓

Answer: \(a_{\text{frost}} \approx 2.9\) AU \(\approx 3\) AU.

What you should learn from this: The frost line explains the rocky-inside, giants-outside pattern. The calculation uses only the proportionality \(T_{\text{eq}} \propto 1/\sqrt{a}\) and one reference point (Earth) — a beautiful example of how a scaling relation and a single measurement can predict a structural feature of the entire solar system.

Problem 9: Rocky or Gaseous? (Density Calculation)

Restatement: Calculate an exoplanet’s density from its mass and radius, and classify it.

Given:

  • Mass: \(M = 6.0 \times 10^{26}\) kg
  • Radius: \(R = 6.0 \times 10^{7}\) m (about 9.4 \(R_\oplus\))

Find: Average density \(\rho\) in kg/m³ and g/cm³; classify as rocky or gaseous.

Equations:

  • Volume of a sphere: \(V = \frac{4}{3}\pi R^3\)
  • Density: \(\rho = M / V\)
  • Unit conversion: \(1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3\)

Solution:

Step 1 — Calculate the volume:

\[R^3 = (6.0 \times 10^{7} \text{ m})^3 = 6.0^3 \times 10^{21} \text{ m}^3 = 216 \times 10^{21} \text{ m}^3 = 2.16 \times 10^{23} \text{ m}^3\]

\[V = \frac{4}{3}\pi \times 2.16 \times 10^{23} \text{ m}^3 = \frac{4}{3} \times 3.1416 \times 2.16 \times 10^{23} \text{ m}^3\]

\[V = 4.189 \times 2.16 \times 10^{23} \text{ m}^3 = 9.05 \times 10^{23} \text{ m}^3\]

Step 2 — Calculate the density:

\[\rho = \frac{M}{V} = \frac{6.0 \times 10^{26} \text{ kg}}{9.05 \times 10^{23} \text{ m}^3} = 663 \text{ kg/m}^3\]

Step 3 — Convert to g/cm³ for comparison with the table in Part 1:

\[663 \text{ kg/m}^3 \times \frac{1 \text{ g/cm}^3}{1000 \text{ kg/m}^3} = 0.66 \text{ g/cm}^3\]

Unit check:

  • \(R^3\): \(\text{m}^3\) ✓
  • \(V\): \(\text{m}^3\) (dimensionless × m³) ✓
  • \(\rho\): \(\text{kg} / \text{m}^3 = \text{kg/m}^3\) ✓
  • Conversion: \(\text{kg/m}^3 \times \frac{\text{g/cm}^3}{\text{kg/m}^3} = \text{g/cm}^3\) ✓

Sanity check: Compare to the table from Part 1 of the reading:

  • Rocky planets: 3.9–5.5 g/cm³
  • Gas/ice giants: 0.69–1.6 g/cm³

Our planet’s density (0.66 g/cm³) is below water (1.0 g/cm³) and almost identical to Saturn (0.69 g/cm³). ✓

Answer: \(\rho \approx 0.66\) g/cm³. This planet is gaseous — its density matches Saturn’s, not Earth’s. The large radius (9.4 \(R_\oplus\)) independently confirms this: rocky planets max out around 1–2 \(R_\oplus\).

What you should learn from this: Density is the key diagnostic for planet composition. The calculation is straightforward (\(V = \frac{4}{3}\pi R^3\), then \(\rho = M/V\)), but unit tracking is essential — especially the kg/m³ → g/cm³ conversion (divide by 1000) needed to compare with the familiar values in the reading.

Problem 10: Diagnosing a Mystery Planet (Multi-Tool Transfer)

Restatement: Use Wien’s Law and spectroscopy to determine whether a mystery planet is rocky or an ice giant.

Given:

  • Thermal emission peak: \(\lambda_{\text{peak}} = 20{,}000\) nm (20 μm)
  • Spectroscopy: strong CH₄ (methane) absorption, no CO₂
  • Wien’s Law constant: \(b = 2.9 \times 10^6\) nm·K

Find: Temperature; classify using temperature + composition evidence.

Solution:

Tool 1 — Wien’s Law (L8): Estimate the planet’s temperature.

\[T = \frac{2.9 \times 10^6 \text{ nm·K}}{20{,}000 \text{ nm}} = 145 \text{ K}\]

Unit check: \(\frac{\text{nm·K}}{\text{nm}} = \text{K}\) ✓

This is very cold — much more typical of an outer planet than of a warm inner rocky world.

Tool 2 — Spectroscopy (L9): Analyze the atmospheric composition.

  • Strong CH₄ (methane): Methane is abundant in the atmospheres of Uranus and Neptune. It absorbs red light, which is why those planets appear blue-green.
  • No CO₂: This means the spectrum does not look Venus-like or Mars-like. By itself, the absence of CO₂ is not a proof, but paired with the low temperature and strong methane it pushes us away from a warm rocky-world interpretation.

Comparison to known planets:

Clue Mystery Planet Rocky Inner Planet? Ice Giant?
Temperature 145 K Too cold for a typical inner rocky world Consistent with a cold outer planet
CH₄ Strong Usually weak or absent Common
CO₂ Absent Not the usual Venus/Mars pattern Often weak or absent

Unit check: All temperatures in K; all comparisons use the same units ✓

Sanity check: Three independent clues all point the same direction — cold temperature, methane present, no CO₂. If the clues disagreed, we’d need more data. That they agree gives us high confidence. ✓

Answer: Between the two choices in the prompt, this is much more likely to be an ice giant than a rocky inner planet. Three clues point the same way:

  1. Temperature (145 K) — too cold for the inner solar system, consistent with a distant planet
  2. Methane absorption — characteristic of ice giant atmospheres
  3. No CO₂ — not decisive by itself, but consistent with a methane-rich outer planet rather than a Venus-like or Mars-like rocky world

What you should learn from this: This problem demonstrates the power of combining multiple Module 1 tools. Wien’s Law gives temperature, spectroscopy gives composition, and together they constrain the planet’s identity far better than either tool alone. Real astronomers use exactly this approach — stacking independent lines of evidence to build a case.