Spacecraft bus — structure, power, thermal, ADCS, C&DH, comms, propulsion: Worked examples

We reuse only the two ideas from the parent bus note:

Recall The two pillars we build on

Steady-state balance ::: $Q_{in} = Q_{out}$ — heat coming in equals heat leaving, so temperature stops changing. Radiated heat ::: $Q_{out} = ε σ A T^{4}$ — a warm surface glows away energy; the hotter it is, the far more it glows (fourth power).

Before any numbers, let us re-earn every symbol so a newcomer is never lost.

Definition Every symbol, in plain words

$S$ = solar constant, the power of raw sunlight per square metre at Earth's distance, $S = 1361 W/m^{2}$ . Picture a $1 m \times 1 m$ square held up to the Sun: 1361 watts pour through it.
$α$ = absorptivity, the fraction of arriving sunlight the surface swallows (0 to 1). Black paint swallows almost all; a mirror swallows almost none.
$ε$ = emissivity, the fraction of the ideal glow the surface actually radiates away (0 to 1). A good radiator has high $ε$ .
$A_{proj}$ = projected area, the shadow the spacecraft casts on a wall behind it — the flat silhouette the Sun actually sees. Only this catches sunlight.
$A$ = total radiating area, the whole skin that can glow heat outward.
$A_{Earth}$ = Earth-facing area, the part of the skin pointed at Earth; only this catches bounced sunlight and Earth's own glow.
$Q_{solar}$ = direct solar heat, the sunlight the surface actually swallows. Written $Q_{solar} = α A_{proj} S$ .
$Q_{dissip}$ = internal dissipation, waste heat the electronics make just by running (like a phone getting warm).
$Q_{albedo}$ = albedo heat, sunlight that Earth reflects back up onto the spacecraft. Written $Q_{albedo} = α A_{Earth} a S$ .
$a$ = Earth's albedo, the fraction of sunlight Earth bounces back, $a \approx 0.3$ (about a third).
$Q_{Earth}$ = Earth-IR heat, the infrared glow warm Earth itself sends up. Written $Q_{Earth} = ε A_{Earth} E_{Earth}$ .
$E_{Earth}$ = Earth-IR flux, the power per square metre of that Earth glow, $\approx 237 W/m^{2}$ .
$σ$ = Stefan–Boltzmann constant, $5.67 \times 1 0^{- 8} W/(m^{2} K^{4})$ — the fixed number nature uses to convert temperature into glow.
$T$ = surface temperature in kelvin (K). Kelvin is just Celsius $+ 273.15$ ; we use it because glow depends on temperature measured from absolute zero, not from the freezing point of water.

The scenario matrix

Every thermal problem this topic can throw is one of these cells. The examples below each carry a [Cell n] tag so you can see the whole space is covered.

Cell	Case class	What is special	Example
1	Sunlit, broadside	full solar input, hot case	Ex. 1
2	Eclipse (zero Sun)	$S = 0$ , cold limiting case	Ex. 2
3	Edge-on ( $A_{proj} \to 0$ )	degenerate projected area	Ex. 3
4	Coating sign-flip ( $α / ε$ )	ratio drives hot vs cold	Ex. 4
5	All four inputs on	albedo + Earth-IR included	Ex. 5
6	Limit $ε \to 0$	near-zero radiator, runaway heat	Ex. 6
7	Real-world word problem	pick a coating to hit a target	Ex. 7
8	Exam twist: solve for area	unknown is $A$ , not $T$	Ex. 8

Example 1 — Sunlit, broadside (hot case) · [Cell 1]

Worked example The baseline

A CubeSat panel of total radiating area $A = 0.03 m^{2}$ faces the Sun broadside with projected area $A_{proj} = 0.01 m^{2}$ . Black paint: $α = 0.95$ , $ε = 0.85$ . Internal dissipation $Q_{dissip} = 5 W$ . Ignore albedo and Earth-IR for now. Find the equilibrium temperature.

Forecast: Black paint drinks sunlight — guess: uncomfortably hot, well above 20 °C?

Step 1 — Solar input. $Q_{solar} = α A_{proj} S = 0.95 \times 0.01 \times 1361 = 12.93 W$ Why this step? Only the projected silhouette catches sunlight, and only the fraction $α$ is swallowed.

Step 2 — Total heat in. $Q_{in} = 12.93 + 5 = 17.93 W$ Why this step? At steady state everything that comes in must leave; sum every source first.

Step 3 — Fourth-root for temperature. $T = (\frac{17.93}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 330 K = 57 °C$ Why this step? Undo the $T^{4}$ glow law to ask "which temperature radiates 17.93 W away?"

Verify: Plug $T = 330$ K back: $0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 33 0^{4} = 17.9 W$ ✓ — matches $Q_{in}$ . Units: $W/m^{2} K^{4} \cdot m^{2} \cdot K^{4} = W$ ✓. Too hot for electronics — matching our forecast.

The picture below shows this one balance visually: two arrows of heat pouring in (orange solar, magenta dissipation) equal the violet arrows glowing out.

Figure 1 — Example 1's steady state: $12.93 W$ solar plus $5 W$ dissipation in equals $17.9 W$ radiated out at $T = 330$ K.

Example 2 — Eclipse: the Sun switches off · [Cell 2]

Worked example Zero-input limiting case

Same CubeSat, now in Earth's shadow: $S = 0$ , no albedo, no Earth-IR. Only $Q_{dissip} = 5 W$ keeps it warm. Coating $ε = 0.85$ , $A = 0.03 m^{2}$ . Find the temperature.

Forecast: With sunlight gone, it should be cold — below freezing?

Step 1 — Set the solar term to zero. $Q_{solar} = α A_{proj} \cdot 0 = 0$ Why this step? Eclipse is the degenerate input $S = 0$ ; the equation still holds, one term vanishes.

Step 2 — Only dissipation remains. $Q_{in} = 5 W$ Why this step? With every solar and Earth term switched off, the electronics' waste heat is the only thing left to radiate away — so the whole $Q_{in}$ collapses to $Q_{dissip}$ .

Step 3 — Fourth-root. $T = (\frac{5}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 243 K = - 30 °C$ Why this step? The glow must fall until it only radiates the 5 W the electronics make.

Verify: $0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 24 3^{4} = 5.0 W$ ✓. Cold, as forecast — and this is exactly why heaters exist: the hot case (57 °C) and cold case (−30 °C) bracket the survival range.

Example 3 — Edge-on to the Sun ( $A_{proj} \to 0$ ) · [Cell 3]

Worked example Degenerate projected area

Same panel, but turned edge-on so the Sun sees a razor-thin sliver: $A_{proj} = 0.0005 m^{2}$ . Black paint, $Q_{dissip} = 5 W$ . Find $T$ .

Forecast: Almost no sunlight caught — temperature should sit near the eclipse value, a touch warmer.

Step 1 — Tiny solar input. $Q_{solar} = 0.95 \times 0.0005 \times 1361 = 0.65 W$ Why this step? Projected area is the shadow the object casts; edge-on, that shadow shrinks toward zero.

Step 2 — Total in. $Q_{in} = 0.65 + 5 = 5.65 W$ Why this step? Add the tiny surviving solar input to the ever-present dissipation to get the full $Q_{in}$ that must be radiated away.

Step 3 — Temperature. $T = (\frac{5.65}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 250 K = - 23 °C$ Why this step? Take the fourth root to find the temperature whose glow exactly carries off $5.65$ W.

Verify: $25 0^{4} \times 0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 = 5.65 W$ ✓. As $A_{proj} \to 0$ we smoothly recover the eclipse case (243 K) — the limiting behaviour is continuous, no surprises.

Example 4 — Coating sign-flip: white vs black · [Cell 4]

Worked example The

$α / ε$ ratio decides everything Broadside sunlit case ( $A_{proj} = 0.01$ , $A = 0.03$ , $Q_{dissip} = 5$ W). Compare black ( $α = 0.95, ε = 0.85$ ) with white ( $α = 0.25, ε = 0.90$ ). Find both temperatures.

Forecast: White reflects sunlight yet radiates well — guess it lands far cooler than black.

Step 1 — Compute the driving ratio. $\frac{α}{ε}_{black} = \frac{0.95}{0.85} = 1.12, \frac{α}{ε}_{white} = \frac{0.25}{0.90} = 0.28$ Why this step? For a purely sunlit plate, $T^{4} \propto α / ε$ : high ratio → hot, low ratio → cold. This single number predicts the outcome.

Step 2 — White paint temperature. $Q_{solar} = 0.25 \times 0.01 \times 1361 = 3.40 W, Q_{in} = 3.40 + 5 = 8.40 W$ $T = (\frac{8.40}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 272 K = - 1 °C$ Why this step? Sum the reduced solar swallow with the dissipation to get $Q_{in}$ , then fourth-root it — the same master move, now with white paint's low $α$ letting far less sunlight in.

Step 3 — Compare with black (57 °C from Ex. 1). A swing of 58 °C from a paint choice alone.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 27 2^{4} = 8.4 W$ ✓. Low $α / ε$ ⇒ cold, exactly as the ratio warned. This is the sign-flip: crossing $α / ε = 1$ separates "runs hot" from "runs cold".

The curve below plots equilibrium temperature against the ratio $α / ε$ , with the black and white points marked — you can read off the 58 °C gap directly.

Figure 2 — Example 4's sign-flip: a sunlit plate's temperature rises with $α / ε$ ; black paint (ratio 1.12) sits hot, white paint (ratio 0.28) sits cold.

Example 5 — All four heat inputs on · [Cell 5]

Worked example The full LEO environment

A flat radiator in low Earth orbit, one face to the Sun and Earth. $A = 0.03 m^{2}$ (radiating), $A_{proj} = 0.01 m^{2}$ (to Sun), face area to Earth $A_{Earth} = 0.01 m^{2}$ . White paint $α = 0.25, ε = 0.90$ . Albedo $a = 0.3$ ; Earth-IR flux $E_{Earth} = 237 W/m^{2}$ . $Q_{dissip} = 5 W$ . Find $T$ .

Forecast: Adding Earth's warmth on top of the Sun — a bit hotter than the pure-sunlit white case (−1 °C)?

Step 1 — Solar. $Q_{solar} = 0.25 \times 0.01 \times 1361 = 3.40 W$ . Why: direct sunlight, as before.

Step 2 — Albedo (sunlight bounced off Earth). $Q_{albedo} = α A_{Earth} a S = 0.25 \times 0.01 \times 0.3 \times 1361 = 1.02 W$ Why: Earth reflects a fraction $a$ of sunlight; the surface absorbs fraction $α$ of what it sees.

Step 3 — Earth's own infrared glow. $Q_{Earth} = ε A_{Earth} E_{Earth} = 0.90 \times 0.01 \times 237 = 2.13 W$ Why: warm Earth radiates IR; the surface absorbs it with its IR efficiency $ε$ .

Step 4 — Sum and solve. $Q_{in} = 3.40 + 1.02 + 2.13 + 5 = 11.55 W$ $T = (\frac{11.55}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 293 K = 20 °C$ Why this step? Now that all four inputs are on, we add every one into a single $Q_{in}$ and take the fourth root — the same master move, just with no term switched off.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 29 3^{4} = 11.55 W$ ✓. Warmer than the sunlit-only white case (−1 °C), exactly as forecast — Earth's albedo and IR add ~21 °C.

Example 6 — Limit $ε \to 0$ : the runaway radiator · [Cell 6]

Example 7 — Word problem: choose a coating to hit 25 °C · [Cell 7]

Worked example Design, not just analysis

A sunlit instrument box ( $A_{proj} = 0.01 m^{2}$ , $A = 0.03 m^{2}$ , $Q_{dissip} = 5 W$ ) must sit at exactly $T = 298 K$ (25 °C). We can tune emissivity by mixing paints; fix $α = 0.30$ . What $ε$ is required?

Forecast: We want it neither hot nor cold — moderate $ε$ , maybe near 0.9?

Step 1 — Write balance with $ε$ unknown. $Q_{solar} + Q_{dissip} = ε σ A T^{4}$ Why this step? Now the coating is unknown and the temperature is fixed — the design inverse.

Step 2 — Left side is a known number. $Q_{in} = 0.30 \times 0.01 \times 1361 + 5 = 4.08 + 5 = 9.08 W$ Why this step? Everything on the heat-in side is now numeric, so collapse it to one value before isolating the unknown $ε$ .

Step 3 — Solve for $ε$ . $ε = \frac{Q _{in}}{σ A T ^{4}} = \frac{9.08}{5.67 \times 1 0 ^{- 8} \times 0.03 \times 29 8 ^{4}} = 0.677$ Why this step? Divide out everything but $ε$ ; the target temperature enters through $T^{4}$ .

Verify: $0.677 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 29 8^{4} = 9.08 W$ ✓ $= Q_{in}$ . A physically realisable $ε \in (0, 1)$ — a real grey paint would do it.

Example 8 — Exam twist: solve for radiator area · [Cell 8]

Worked example Unknown is

$A$ , not $T$ An electronics box dumps $Q_{dissip} = 40 W$ of waste heat in eclipse (no Sun). Its radiator ( $ε = 0.90$ ) must not exceed $T = 313 K$ (40 °C). What minimum radiating area $A$ is needed?

Forecast: A lot of heat, a modest temperature cap — the area can't be tiny.

Step 1 — Balance with $A$ unknown. $Q_{dissip} = ε σ A T^{4} \Rightarrow A = \frac{Q _{dissip}}{ε σ T ^{4}}$ Why this step? At the temperature limit, the radiator must be just big enough to glow all 40 W away.

Step 2 — Plug numbers. $A = \frac{40}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 31 3 ^{4}} = 0.0824 m^{2}$ Why this step? Larger $Q$ or lower allowed $T$ forces a bigger radiator — area scales as $1/ T^{4}$ , so a cool cap is expensive in area.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.0824 \times 31 3^{4} = 40.0 W$ ✓. Units: $W / (W/m^{2} K^{4} \cdot K^{4}) = m^{2}$ ✓. Bigger than the whole CubeSat panel — that is why high-power satellites carry deployable radiators.