Wrap blackboard derivations in admonition boxes for Lectures 1–7

Match the Lecture 8 format: full derivation content inside the
tip-styled admonition, with bold inline headings instead of
### subsections. Converts {note} directives to bold paragraphs
to avoid nested directive issues.

Co-Authored-By: Claude Opus 4.6 <noreply@anthropic.com>

Author: timlichtenberg

book/01_introduction/introduction.md

@@ -160,13 +160,12 @@ Planets are broadly classified by composition and structure:
 
 ## Blackboard derivation: Solar mass from planetary orbits
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: Solar mass from Kepler's third law
 :class: tip
 
 **Goal:** Use Kepler's third law to estimate the mass of the Sun from Earth's orbital parameters, then evaluate the planet-to-star mass ratio for the solar system.
-```
 
-### Setup
+**Setup.**
 
 Consider a planet of mass $M_p$ in a circular orbit of radius $r$ around a star of mass $M_*$. The gravitational force provides the centripetal acceleration:
 
@@ -176,7 +175,7 @@ $$
 
 where $v = 2\pi r / P$ is the orbital velocity and $P$ is the orbital period.
 
-### Derivation
+**Derivation.**
 
 Substituting $v = 2\pi r / P$ and cancelling $M_p$:
 
@@ -192,11 +191,9 @@ $$ (eq:kepler-mass)
 
 This is Newton's form of Kepler's third law (for $M_p \ll M_*$). The planet's mass cancels — the orbital period depends only on the central mass and the orbital radius.
 
-```{note}
-For elliptical orbits, the same relation holds with $r$ replaced by the semi-major axis $a$. The derivation of the general case requires the vis-viva equation, which we will cover in {ref}`lecture02`.
-```
+**Note:** For elliptical orbits, the same relation holds with $r$ replaced by the semi-major axis $a$. The derivation of the general case requires the vis-viva equation, which we will cover in {ref}`lecture02`.
 
-### Application
+**Application.**
 
 Using Earth's orbital parameters:
 
@@ -212,14 +209,16 @@ $$
 
 This agrees with the accepted value $\Msun = 1.989 \times 10^{30}$ kg — a remarkably accurate estimate from just two measurable quantities.
 
-### The planet-to-star mass ratio
+**The planet-to-star mass ratio.**
 
 The more precise form of Kepler's third law is $P^2 = 4\pi^2 a^3 / [G(M_* + M_p)]$, which gives $M_* + M_p$ rather than $M_*$ alone. The approximation $M_* + M_p \approx M_*$ is justified because the planet-to-star mass ratio is tiny:
 
 - Jupiter, the most massive planet: $\Mjup \approx 318 \, \Mearth \approx 1.90 \times 10^{27}$ kg, giving $\Mjup / \Msun \approx 9.5 \times 10^{-4}$.
 - Total mass of all eight planets: $\approx 446 \, \Mearth \approx 2.7 \times 10^{27}$ kg, giving $M_\mathrm{planets}/\Msun \approx 1.3 \times 10^{-3}$.
 
 The Sun contains **99.87%** of the solar system's total mass. Jupiter alone accounts for 71% of the planetary mass. This extreme concentration of mass in the central star is a fundamental property of planetary systems — and one that planet formation theory must explain ({ref}`lecture02`).
+```
+
 
 ## Comparative planetology as a methodology
 

book/02_formation_orbits/formation_orbits.md

@@ -258,13 +258,12 @@ The escape velocity is exactly $\sqrt{2}$ times the circular orbital velocity at
 
 ## Blackboard derivation: The vis-viva equation
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: The vis-viva equation
 :class: tip
 
 **Goal:** Derive the vis-viva equation from energy conservation in a Keplerian orbit, then apply it to calculate orbital velocities for Earth and Halley's comet.
-```
 
-### Setup
+**Setup.**
 
 Consider a planet of mass $m$ in an elliptical orbit around a star of mass $M$ (with $m \ll M$). The orbit has semi-major axis $a$ and eccentricity $e$. At any point, the planet is at distance $r$ from the star and moves with speed $v$.
 
@@ -276,7 +275,7 @@ $$
 
 We want to express $E$ in terms of $a$ alone, then derive a formula for $v(r)$.
 
-### Derivation
+**Derivation.**
 
 **Step 1: Evaluate energy at perihelion and aphelion.**
 
@@ -320,7 +319,7 @@ $$
 \boxed{v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)}
 $$
 
-### Application: Earth's orbital velocity
+**Application: Earth's orbital velocity.**
 
 For Earth ($a = 1$ AU $= 1.496 \times 10^{11}$ m, $e = 0.017$):
 
@@ -331,7 +330,7 @@ For Earth ($a = 1$ AU $= 1.496 \times 10^{11}$ m, $e = 0.017$):
 
 The variation is only about $\pm 0.5$ km s$^{-1}$ around the mean — because Earth's orbit is nearly circular. The mean orbital velocity is approximately $v \approx 2\pi a / P \approx 29.8$ km s$^{-1}$.
 
-### Application: Halley's comet
+**Application: Halley's comet.**
 
 For a dramatic contrast, consider Halley's comet ($a = 17.8$ AU, $e = 0.967$):
 
@@ -342,8 +341,7 @@ For a dramatic contrast, consider Halley's comet ($a = 17.8$ AU, $e = 0.967$):
 
 Halley's comet hurtles through the inner solar system at nearly twice Earth's speed, yet crawls beyond Neptune's orbit at less than 1 km s$^{-1}$. This enormous range is a direct consequence of the vis-viva equation applied to a highly eccentric orbit.
 
-```{note}
-The vis-viva equation is the generalisation of the Kepler mass formula derived in {ref}`lecture01`. There, we used a circular orbit ($r = a$) to get $v^2 = GM/a$, which is just the vis-viva equation with $r = a$.
+**Note:** The vis-viva equation is the generalisation of the Kepler mass formula derived in {ref}`lecture01`. There, we used a circular orbit ($r = a$) to get $v^2 = GM/a$, which is just the vis-viva equation with $r = a$.
 ```
 
 

book/03_heat_energy/heat_energy.md

@@ -147,13 +147,12 @@ The key distinction is one of **timescale**. Conduction is a diffusive process t
 
 ## Blackboard derivation: The conductive cooling timescale
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: The conductive cooling timescale
 :class: tip
 
 **Goal:** Derive the characteristic timescale for heat to conduct through a body of size $L$, and show that conduction alone cannot cool planet-sized bodies — motivating the need for convection.
-```
 
-### Setup
+**Setup.**
 
 The starting point is the **heat diffusion equation** (also called the heat equation), which describes how temperature evolves in a conducting medium:
 
@@ -169,7 +168,7 @@ $$
 
 This equation says that the rate of temperature change at any point is proportional to the curvature of the temperature profile — regions where the temperature profile is concave up ($\pdv{^2 T}{x^2} > 0$) heat up, and regions where it is concave down cool down.
 
-### Derivation
+**Derivation.**
 
 We use **dimensional analysis** to extract the characteristic timescale. The heat equation relates:
 - A time derivative: $\pdv{T}{t} \sim T/\tau$ (where $\tau$ is the timescale we seek)
@@ -189,7 +188,7 @@ $$ (eq:cooling-timescale)
 
 This is the **conductive cooling timescale**. It tells us how long it takes for a thermal disturbance to propagate a distance $L$ by conduction alone. The key feature is the **$L^2$ dependence**: doubling the size of a body increases its conductive cooling time by a factor of four.
 
-### Application
+**Application.**
 
 Using $\kappa \approx 10^{-6}$ m$^2$ s$^{-1}$ (typical for silicate rock), we can estimate the conductive cooling timescale for bodies of different sizes:
 
@@ -202,8 +201,7 @@ Using $\kappa \approx 10^{-6}$ m$^2$ s$^{-1}$ (typical for silicate rock), we ca
 
 A small boulder cools in days — consistent with everyday experience. A 100 km asteroid cools in hundreds of millions of years, which is long but shorter than the age of the solar system — so small asteroids should have cooled and solidified, consistent with the meteorite record. But for the Moon and Earth, the conductive cooling time far exceeds the age of the universe ($\sim 14$ Gyr). **Earth cannot cool by conduction alone.**
 
-```{note}
-This result tells us something profound: since Earth *is* losing heat at a rate of ~47 TW ({cite}`DaviesDavies2010`), there must be a more efficient transport mechanism operating in its interior. That mechanism is **convection**. The question of *when* convection occurs — and how vigorous it is — leads directly to the Rayleigh number.
+**Note:** This result tells us something profound: since Earth *is* losing heat at a rate of ~47 TW ({cite}`DaviesDavies2010`), there must be a more efficient transport mechanism operating in its interior. That mechanism is **convection**. The question of *when* convection occurs — and how vigorous it is — leads directly to the Rayleigh number.
 ```
 
 (rayleigh-nusselt)=

book/04_differentiation_magnetospheres/differentiation_magnetospheres.md

@@ -213,13 +213,12 @@ Schematic of the geodynamo mechanism. Convective motions in the electrically con
 
 ## Blackboard derivation: The magnetic Reynolds number
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: The magnetic Reynolds number $\mathrm{Rm}$
 :class: tip
 
 **Goal:** Derive the magnetic Reynolds number $\mathrm{Rm} = UL/\eta$ from the induction equation by dimensional analysis, and estimate $\mathrm{Rm}$ for Earth's outer core to demonstrate that advection dominates over diffusion — the fundamental criterion for dynamo action.
-```
 
-### Setup
+**Setup.**
 
 We start from the magnetic induction equation derived in the previous section (Eq. {eq}`eq:induction-equation`):
 
@@ -229,7 +228,7 @@ $$
 
 We want to determine which term dominates for a given flow. If advection wins, the flow can amplify magnetic field; if diffusion wins, any field decays away.
 
-### Derivation
+**Derivation.**
 
 We use **dimensional analysis** to estimate the magnitude of each term. Let $U$ be a characteristic flow velocity, $L$ a characteristic length scale, and $B$ the field strength.
 
@@ -265,7 +264,7 @@ When $\mathrm{Rm} \gg 1$, advection dominates: the magnetic field is "frozen in"
 
 For a self-sustaining dynamo, numerical simulations and theory indicate that the critical value is $\mathrm{Rm}_c \sim 10$–$100$, depending on the flow geometry.
 
-### Application: Earth's outer core
+**Application: Earth's outer core.**
 
 For Earth's outer core, the relevant parameters are {cite}`Schubert2001`:
 
@@ -291,8 +290,7 @@ $$
 
 This is much less than Earth's age (4.5 Gyr), confirming that the field cannot be a relic — it must be **continuously regenerated** by dynamo action.
 
-```{note}
-Paleomagnetic measurements of ancient rocks show that Earth has had a magnetic field for at least **3.4–3.5 billion years** {cite}`Tarduno2010`. This places a strong constraint on the thermal and compositional evolution of the core: the convective driving mechanism must have been sustained over most of Earth's history.
+**Note:** Paleomagnetic measurements of ancient rocks show that Earth has had a magnetic field for at least **3.4–3.5 billion years** {cite}`Tarduno2010`. This places a strong constraint on the thermal and compositional evolution of the core: the convective driving mechanism must have been sustained over most of Earth's history.
 ```
 
 

book/05_atmospheres_1/atmospheres_1.md

@@ -120,13 +120,12 @@ The barometric formula tells us that atmospheric pressure decreases **exponentia
 
 ## Blackboard derivation: The atmospheric scale height
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: Atmospheric scale height from hydrostatic equilibrium
 :class: tip
 
 **Goal:** Derive the atmospheric scale height $H = \kB T / (\mu \, m_u \, g)$ from hydrostatic equilibrium combined with the ideal gas law, and compute $H$ for Earth, Mars, Venus, Jupiter, and Titan.
-```
 
-### Setup
+**Setup.**
 
 We start from the equation of hydrostatic equilibrium (Eq. {eq}`eq:hydrostatic-equilibrium`) and the ideal gas law (Eq. {eq}`eq:ideal-gas-atm`):
 
@@ -136,7 +135,7 @@ $$
 
 Our goal is to find the characteristic length scale over which pressure varies — the **scale height**.
 
-### Derivation
+**Derivation.**
 
 Rearranging the ideal gas law to express $\rho$ in terms of $P$:
 
@@ -162,7 +161,7 @@ The scale height has a clear physical interpretation:
 - **Heavier molecules** (larger $\mu$) → smaller $H$: heavier molecules are harder to loft.
 - **Stronger gravity** $g$ → smaller $H$: stronger gravity compresses the atmosphere more.
 
-### Application: scale heights across the solar system
+**Application: scale heights across the solar system.**
 
 We can now compute scale heights for several bodies using their characteristic atmospheric temperatures, compositions, and surface gravities:
 
@@ -182,8 +181,7 @@ $$
 
 This matches our everyday experience: commercial aircraft cruise at $\sim$10–12 km altitude, where the pressure is roughly $0.2$–$0.3$ atm (about 1.2–1.4 scale heights up).
 
-```{note}
-Jupiter's scale height is large despite its strong gravity because $\mathrm{H_2}$ has a very low molecular weight ($\mu = 2.2$). Titan's scale height is large because its gravity is weak ($g = 1.35$ m s$^{-2}$) — Titan's atmosphere extends to a proportionally much greater height than Earth's, despite being colder.
+**Note:** Jupiter's scale height is large despite its strong gravity because $\mathrm{H_2}$ has a very low molecular weight ($\mu = 2.2$). Titan's scale height is large because its gravity is weak ($g = 1.35$ m s$^{-2}$) — Titan's atmosphere extends to a proportionally much greater height than Earth's, despite being colder.
 ```
 
 

book/06_atmospheres_2/atmospheres_2.md

@@ -53,13 +53,12 @@ The physics of cloud formation is the same in every case — the Clausius-Clapey
 
 ## Blackboard derivation: The Clausius-Clapeyron equation
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: The Clausius-Clapeyron equation
 :class: tip
 
 **Goal:** Derive the exponential dependence of saturation vapour pressure on temperature from thermodynamic phase equilibrium, and apply the result to predict cloud condensation conditions across the solar system.
-```
 
-### Setup: phase equilibrium
+**Setup: phase equilibrium.**
 
 Consider a substance (e.g., water) that exists in two phases: liquid and vapour. Along the **coexistence curve** in the $P$–$T$ diagram — the line separating the liquid and vapour phases — the two phases are in thermodynamic equilibrium. This means the **Gibbs free energy per unit mass** is equal in both phases:
 
@@ -69,7 +68,7 @@ $$
 
 where $g_\ell$ and $g_v$ are the specific Gibbs free energies of the liquid and vapour, respectively.
 
-### Derivation
+**Derivation.**
 
 If we move along the coexistence curve by an infinitesimal amount ($\dd T$, $\dd P$), the Gibbs free energies must remain equal:
 
@@ -118,7 +117,7 @@ $$ (eq:clausius-clapeyron)
 
 This is the **Clausius-Clapeyron equation**. The saturation vapour pressure depends **exponentially** on temperature through the ratio $L_v / R_v$, which has units of temperature and characterises the sensitivity of the phase transition.
 
-### Worked example: water on Earth
+**Worked example: water on Earth.**
 
 For water vapour:
 - $L_v = 2.50 \times 10^6$ J kg$^{-1}$ (latent heat of vaporisation at 0°C)
@@ -135,7 +134,7 @@ $$
 
 The measured value is 2.3 kPa — excellent agreement. The small discrepancy arises because $L_v$ decreases slightly with temperature (from $2.50 \times 10^6$ J kg$^{-1}$ at 0°C to $2.26 \times 10^6$ J kg$^{-1}$ at 100°C), which we neglected by treating $L_v$ as constant.
 
-### Application: condensing species across the solar system
+**Application: condensing species across the solar system.**
 
 The Clausius-Clapeyron equation applies to *any* vapour-to-liquid (or vapour-to-solid) transition. The following table lists the key condensing species in solar system atmospheres and their thermodynamic properties:
 
@@ -153,6 +152,7 @@ $^\dagger$Latent heat of sublimation (solid $\leftrightarrow$ vapour).
 Data from {cite}`Catling2017` and {cite}`dePaterLissauer2010`.
 
 The large $L_v/R_v$ ratio for $\mathrm{H_2SO_4}$ explains why Venus's sulfuric acid clouds occupy a relatively narrow altitude range — the exponential sensitivity confines condensation to a thin temperature band. Conversely, $\mathrm{CH_4}$ has a low $L_v/R_v$, meaning its saturation curve is flatter and methane clouds on Titan can extend over a wider altitude range.
+```
 
 
 ## Clouds across the solar system

book/07_surfaces/surfaces.md

@@ -65,13 +65,12 @@ This is roughly 100 times the energy of the largest nuclear weapon ever detonate
 
 ### Blackboard derivation: crater scaling law
 
-```{admonition} Blackboard derivation (~10 min)
+```{admonition} Blackboard derivation: Crater scaling law from dimensional analysis
 :class: tip
 
 **Goal:** Use dimensional analysis to derive how crater diameter $D$ depends on impact energy $E$, target density $\rho$, and surface gravity $g$, then apply the result to estimate the crater produced by a 1 km asteroid impact on the Moon.
-```
 
-#### Setup
+**Setup.**
 
 We want to find the diameter $D$ of the crater (a length) produced by an impact with kinetic energy $E$ into a target with density $\rho$ under surface gravity $g$. In the **gravity regime** (where crater size is limited by gravity rather than material strength — valid for craters larger than $\sim$100 m), dimensional analysis requires:
 
@@ -81,7 +80,7 @@ $$
 
 where $C$ is a dimensionless constant of order unity and $a$, $b$, $c$ are exponents to be determined.
 
-#### Dimensional analysis
+**Dimensional analysis.**
 
 Writing the dimensions of each quantity in terms of mass $M$, length $L$, and time $T$:
 
@@ -114,7 +113,7 @@ $$ (eq:crater-scaling)
 
 This is the **crater scaling law** in the gravity regime {cite}`Holsapple1993`. It tells us that crater diameter scales as the fourth root of impact energy — doubling the energy increases the crater diameter by only a factor of $2^{1/4} \approx 1.19$ (about 19%). This weak dependence on energy explains why craters have a relatively narrow size range even though impactor energies span many orders of magnitude.
 
-#### Worked example: 1 km asteroid on the Moon
+**Worked example: 1 km asteroid on the Moon.**
 
 For our 1 km asteroid ($E \approx 3 \times 10^{20}$ J) impacting the Moon ($\rho \approx 2500$ kg m$^{-3}$ for the regolith, $g = 1.62$ m s$^{-2}$):
 
@@ -125,6 +124,7 @@ $$
 This is consistent with the observed sizes of lunar craters formed by $\sim$1 km impactors. For comparison, the 85 km crater Tycho was formed by a much larger impactor ($\sim$8 km). The dimensionless prefactor $C$ in the full scaling law depends on target material properties and is of order unity for rocky surfaces.
 
 The more complete **pi-scaling framework** of {cite}`Holsapple1993` parameterises the transition between the gravity regime and the strength regime (where material cohesion, not gravity, limits crater growth) and accounts for target porosity and impactor properties.
+```
 
 
 ### Crater morphology