Worked examples — Space environment — LEO radiation (SAA, Van Allen), atomic oxygen, MMOD debris

This page is the practice arena for the parent topic. The parent gave you the physics; here we grind through every kind of question the four hazards (radiation, SAA, atomic oxygen, MMOD) can throw at you — including the sneaky degenerate cases and one exam-style twist.

Before any formula appears, remember the plain-language meaning of each symbol we reuse:

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The scenario matrix

Every problem in this topic is one of these cells. The examples below are tagged with the cell they cover.

Cell	What makes it distinct	Example
A. Normal flux → fluence → damage	ordinary positive numbers, one hazard	Ex 1 (AO), Ex 2 (radiation)
B. Two-region mixing	flux differs inside/outside SAA, must weight by time	Ex 3
C. Zero / degenerate input	flux = 0, or perfectly shielded, or ram angle = 90°	Ex 4
D. Small-$\lambda$ limit	rare-event probability, Poisson approximation	Ex 5
E. Large-$\lambda$ / near-certain limit	many small particles, opposite extreme	Ex 6
F. Real-world word problem	pick the right model from a messy story	Ex 7
G. Exam twist	a hidden trap (unit change, or a "which grows faster" question)	Ex 8

We deliberately hit both limits of the Poisson curve (cells D and E) and the zero case (cell C) so you never meet a scenario we did not show.

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Example 1 — Atomic-oxygen erosion (Cell A)

Forecast: the parent got 189 μm for 2 years. Guess the 1-year answer before reading on.

1. Convert 1 year to seconds. Why this step? Flux is per second, so time must be in seconds to cancel. $t=365\times 86400=3.1536\times 10^7\text{s}$
2. Fluence = flux × time. Why? Fluence is the running total; multiply the rate by the duration (see the ramp figure below). $F=10^{14}\times 3.1536\times 10^7=3.1536\times 10^{21}{\text{atoms/cm}}^2$
3. Depth = fluence × yield. Why? Each atom removes $Y$ of volume; volume-per-area is depth. $d=FY=3.1536\times 10^{21}\times 3\times 10^{-24}=9.46\times 10^{-3}\text{cm}=94.6\mu \text{m}$

Verify: exactly half the parent's 2-year value (189 μm) — because depth is linear in time. Units: $({\text{atoms/cm}}^2)({\text{cm}}^3/\text{atom})=\text{cm}$. ✓

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Example 2 — Total Ionizing Dose from a constant rate (Cell A)

Forecast: krad or Mrad? Guess the order of magnitude first.

1. Integral collapses when $\dot{D}$ is constant. Why? $\text{TID}={\int }_0^T\dot{D}dt$; a constant pulls out, leaving $\dot{D}\cdot T$. $T=3\times 365\times 86400=9.4608\times 10^7\text{s}$
2. Multiply. Why? Rate × time = accumulated dose. $\text{TID}=0.002\times 9.4608\times 10^7=1.892\times 10^5\text{rad}\approx 189\text{krad}$

Verify: 189 krad ÷ 10 krad ≈ 19× over the commercial limit → this part must be rad-hard. Sanity: 0.002 rad/s ≈ 63 krad/year, ×3 ≈ 189 krad. ✓ This ties to single-event effects as the dose companion to instantaneous upsets.

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Example 3 — Two-region SAA weighting (Cell B)

Forecast: Will the SAA or the "outside" time dominate the daily dose? Guess.

1. Orbits per day. Why? Everything scales with how many times we lap the Earth. $N=\frac{1440}{90}=16\text{orbits/day}$
2. Seconds in each region per day. Why? We must weight each rate by its own exposure time — the whole point of a two-region problem. $t_{\text{in}}=16\times 20\times 60=19200\text{s},t_{\text{out}}=86400-19200=67200\text{s}$
3. Weighted sum. Why? Total dose = (rate × time) added over regions. ${\text{TID}}_{\text{day}}=0.04(19200)+0.001(67200)=768+67.2=835.2\text{rad}$

Verify: the SAA term (768) is ~11× the outside term (67.2) despite being ~3.5× shorter in time — confirms the SAA dominates, matching the parent's SAA example. ✓ See ISS collision avoidance for why crewed vehicles time activities around these passes.

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Example 4 — Degenerate cases (Cell C)

Forecast: Can a formula ever give exactly zero? Guess before checking.

1. AO depth with $\Phi =0$. Why this step? Erosion depth $d=\Phi tY$. Any factor being zero zeroes the product. $d=0\times t\times Y=0\text{for all }t.$ Geometric reason: the flux a surface feels scales with $\cos \theta$ where $\theta$ is the angle between the surface normal and the velocity. At $\theta =90°$, $\cos 90°=0$ — no atoms strike the face. (This is exactly the "orient surfaces away from ram" mitigation.)
2. TID with $\dot{D}=0$. Why? $\text{TID}={\int }_0^{T}0dt=0$. A perfectly shielded part accumulates nothing.

Verify: both give exactly 0, and importantly independent of mission length — a limiting sanity check. In practice neither is truly zero (secondary particles, off-ram scattering), but the degenerate formula behaves correctly. ✓ This informs materials selection.

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Example 5 — Rare-event probability, small-$\lambda$ limit (Cell D)

Forecast: will this be closer to 0.1% or 10%? Guess.

1. Expected count $\lambda$. Why? Impacts are a Poisson process — random, independent hits. The expected number is flux × area × time. $\lambda =2\times 10^{-6}\times 30\times 12=7.2\times 10^{-4}$
2. Zero-impact probability. Why? Poisson $P(0)=e^{-\lambda }$ — the chance the storm misses us entirely. $P(0)=e^{-7.2\times 10^{-4}}\approx 0.99928$
3. Complement. Why? "At least one" = 1 minus "none". $P(\ge 1)=1-0.99928=7.197\times 10^{-4}\approx 0.072\%$

Verify: because $\lambda$ is tiny, $P(\ge 1)\approx \lambda$ (the small-$x$ approximation $1-e^{-x}\approx x$): $\lambda =7.2\times 10^{-4}$ vs computed $7.197\times 10^{-4}$ — they match to 3 decimals. ✓ This is the low-flux, large-particle corner of the map — see Kessler Syndrome for why this number is rising over time.

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Example 6 — Near-certain-impact, large-$\lambda$ limit (Cell E)

Forecast: near-zero or near-one? Guess before the algebra.

1. Expected count. Why? Same Poisson multiplication, new flux. $\lambda =2\times 10^{-2}\times 30\times 12=7.2$
2. Probability of at least one. Why? Same complement rule. $P(\ge 1)=1-e^{-7.2}=1-7.47\times 10^{-4}\approx 0.99925$

Verify: here the small-$x$ shortcut fails ($\lambda =7.2$ is not small), which is exactly why we needed the full $e^{-\lambda }$ formula. $P(\ge 1)\approx 99.9\%$ — near-certain, the opposite corner from Ex 5. Sanity: $e^{-7.2}$ is a tiny number, so $P$ hugs 1. ✓

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Example 7 — Word problem: choosing the model (Cell F)

Forecast: survives, or fails — and if it fails, roughly how soon? Guess.

1. Pick the model. Why? This is an AO erosion story (chemical, ram-facing), not radiation or debris — so use $d=\Phi tY$.
2. Solve for the failure time by setting depth = film thickness. Why? The film is "gone" when erosion depth equals its 25 μm thickness. Convert: $25\mu \text{m}=2.5\times 10^{-3}$ cm. $t_{\text{fail}}=\frac{d_{\text{film}}}{\Phi Y}=\frac{2.5\times 10^{-3}}{10^{14}\times 3\times 10^{-24}}=\frac{2.5\times 10^{-3}}{3\times 10^{-10}}=8.33\times 10^6\text{s}$
3. Convert to days. Why? Human-scale check. $t_{\text{fail}}=\frac{8.33\times 10^6}{86400}\approx 96.5\text{days}\approx 3.2\text{months}$

Verify: the film dies in ~3 months, not 2 years — the team is wrong by a factor of ~8. Consistent with the parent's "25 μm eroded in ~3 months" claim. A protective SiO₂ coating (a materials decision) is mandatory. ✓

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Example 8 — Exam twist: which mission killer wins? (Cell G)

Forecast: does atomic oxygen or radiation break the part first? This is the trap — most students only check one hazard.

1. AO time-to-limit. Why? Set erosion = 100 μm using the Ex 1 rate. $t_{\text{AO}}=\frac{100\mu \text{m}}{94.6\mu \text{m/yr}}=1.057\text{years}$
2. TID time-to-limit. Why? Convert 100 krad = $10^5$ rad, then divide by rate. $t_{\text{TID}}=\frac{10^5}{0.0008}=1.25\times 10^8\text{s}=\frac{1.25\times 10^8}{3.1536\times 10^7}=3.96\text{years}$
3. Compare. Why? The smaller time is the first failure. $t_{\text{AO}}=1.06\text{yr}<t_{\text{TID}}=3.96\text{yr}$

Verify: AO wins — the uncoated part is destroyed at ~1.1 years, long before radiation (~4 years) or the 5-year mission end. The twist: whichever hazard has the shortest time-to-limit governs, and here it is the "chemical sandpaper", not the radiation. Units check: rad ÷ (rad/s) = s ✓; μm ÷ (μm/yr) = yr ✓.

Recall Self-test the matrix

Which cell is a flux of exactly 0? ::: Cell C (degenerate / zero input) Why did $P(\ge 1)\approx \lambda$ work in Ex 5 but not Ex 6? ::: In Ex 5 $\lambda \ll 1$ so $1-e^{-\lambda }\approx \lambda$; in Ex 6 $\lambda =7.2$ is too large for the approximation. In Ex 8, which hazard governs and why? ::: Atomic oxygen — it has the shorter time-to-limit (1.06 yr vs 3.96 yr).