Exercises — Spacecraft bus — structure, power, thermal, ADCS, C&DH, comms, propulsion

The camera is the payload — the mission-specific instrument. The bus is everything that keeps it alive and pointed. The seven bus subsystems are:

1. Structure
2. Power (EPS)
3. Thermal (TCS)
4. ADCS (Attitude Determination & Control)
5. C&DH (Command & Data Handling)
6. Communications
7. Propulsion

What we did: we sorted "does the real science" (payload) from "supports the science" (bus). Why: every mass/power/cost trade starts by knowing which side of that line a component sits on.

• ${\sigma }_y$ ::: yield strength — the stress at which a material starts to permanently deform (units: Pa). (Note the subscript $y$ — bare $\sigma$ is the Stefan–Boltzmann constant instead.)
• $\rho$ ::: density — mass per unit volume (kg/m³).
• $\alpha$ ::: absorptivity — fraction of incoming radiation a surface soaks up (0–1, no units).
• $ϵ$ ::: emissivity — how efficiently a surface radiates heat away (0–1).
• $S$ ::: solar constant — sunlight power per square metre at Earth's distance, $1361$ W/m².

Why this matters: $\alpha$ and $ϵ$ look like twins but answer opposite questions — one is "how much do I drink in?", the other "how much do I glow out?". Mixing them up is the #1 thermal error (see below).

Step 1 — the force. The spine of the rocket must carry the whole spacecraft's weight times the g-load: $F=m\cdot a=4\times 98=392\text{N}$ Why: launch is the harshest mechanical moment of the whole mission — nothing in orbit pushes this hard.

Step 2 — the area. Stress is force spread over area, ${\sigma }_y=F/A$ at the limit. We demand $F/A\le {\sigma }_y$, so: $A\ge \frac{F}{{\sigma }_y}=\frac{392}{240\times 10^6}=1.63\times 10^{-6}{\text{m}}^2=1.63{\text{mm}}^2$ Why: any smaller area concentrates the same force onto less material, pushing the stress past the yield limit — the wall would permanently deform. What it looks like: the left panel of the shared figure — the force arrow spread across the wall cross-section.

Answer: $A\ge 1.63{\text{mm}}^2$ (this is the strength minimum; real walls are far thicker for stiffness).

Step 1 — the day/night ratio. How much longer (relatively) is the deficit period? $\frac{T_{\text{eclipse}}}{T_{\text{sun}}}=\frac{33}{62}=0.532$ Why: this ratio is the heart of the sizing problem — it tells us how much extra energy the array must bank during sunlight to cover the dark stretch, relative to the running load. A bigger eclipse (or shorter daylight) forces a bigger surplus.

Step 2 — the sizing formula. From the parent note's energy see-saw: $P_{\text{SA}}=P_{\text{load}}\left(1+\frac{T_{\text{eclipse}}}{T_{\text{sun}}\cdot {\eta }_{\text{charge}}{\eta }_{\text{discharge}}}\right)=50\left(1+\frac{0.532}{0.85}\right)$ $P_{\text{SA}}=50\times 1.626=81.3\text{W}$ Why the extra term: the array must not only run the load in sunlight but also bank enough (through lossy batteries) to survive eclipse.

Answer: $P_{\text{SA}}\approx 81.3$ W.

Step 1 — the deciding quantity. For a strength-limited part, required area is $A=F/{\sigma }_y$, and mass is $m=AL\rho =\frac{FL\rho }{{\sigma }_y}$. So mass scales with $\rho /{\sigma }_y$ — equivalently, lightness scales with the structural efficiency: ${\eta }_{\text{struct}}=\frac{{\sigma }_y}{\rho }$ Why this and not just ${\sigma }_y$: a stronger material lets you use less area, but if it is also dense you gain nothing. Only the ratio tells you the true payoff.

Step 2 — compute for each. ${\eta }_{\text{Al}}=\frac{240\times 10^6}{2700}=8.89\times 10^4\text{N⋅m/kg}$ ${\eta }_{\text{CFRP}}=\frac{600\times 10^6}{1600}=3.75\times 10^5\text{N⋅m/kg}$

Step 3 — the ratio. $\frac{{\eta }_{\text{CFRP}}}{{\eta }_{\text{Al}}}=\frac{3.75\times 10^5}{8.89\times 10^4}=4.22$

Answer: CFRP is the lighter choice; for the same strength requirement it needs about 4.2× less structural mass. (Caveats: real parts are often stiffness-limited, and CFRP costs more and outgasses — but on pure strength-to-weight it wins.)

Step 1 — write the balance for a flat plate in sun (ignore Earth terms for the argument). Heat in equals heat out at steady state: $\alpha A_{\text{proj}}S=ϵ\sigma A{T}^4$ where $\sigma =5.67\times 10^{-8}{\text{W/(m}}^2{\text{⋅K}}^4)$ is the Stefan–Boltzmann constant. Step 2 — solve for $T$. Take the plate's sunlit and radiating areas equal, $A_{\text{proj}}=A$, so they cancel: $T={\left(\frac{\alpha }{ϵ}\cdot \frac{S}{\sigma }\right)}^{1/4}$ Why set $A_{\text{proj}}=A$: we are isolating the coating's effect, not the geometry's. By forcing the two areas equal they cancel out of the equation, leaving temperature to depend only on the material ratio $\alpha /ϵ$ — exactly the comparison the question asks for. (In a real spacecraft the two areas differ, but that changes only a geometric prefactor, not the $\alpha /ϵ$ dependence.)

The insight: $\alpha$ and $ϵ$ do not appear separately — only their ratio $\alpha /ϵ$ does. That ratio is the single knob a coating gives you.

Step 3 — compare.

• Black: $\alpha /ϵ=0.95/0.85=1.12$
• White: $\alpha /ϵ=0.25/0.90=0.28$

Since $T\propto (\alpha /ϵ{)}^{1/4}$, black paint sits at $(1.12/0.28{)}^{1/4}=1.41$× the absolute temperature of white. Black runs far hotter — which is exactly why radiators and sun-facing panels are usually white or silvered.

Step 1 — heat in. Two sources: absorbed sunlight and the electronics' own waste heat. $Q_{\text{solar}}=\alpha A_{\text{proj}}S=0.25\times 0.01\times 1361=3.40\text{W}$ $Q_{\text{in}}=Q_{\text{solar}}+Q_{\text{dissip}}=3.40+5=8.40\text{W}$ Why include dissipation: in space there is no air to carry electronics heat away — it all has to leave as radiation, so it counts on the "in" side of the balance.

Step 2 — heat out and solve. Set $Q_{\text{out}}=ϵ\sigma A{T}^4=Q_{\text{in}}$: $T={\left(\frac{Q_{\text{in}}}{ϵ\sigma A}\right)}^{1/4}={\left(\frac{8.40}{0.90\times 5.67\times 10^{-8}\times 0.03}\right)}^{1/4}$ $T={\left(5.487\times 10^9\right)}^{1/4}=272.2\text{K}$

Step 3 — convert and judge. $T=272.2-273.15=-0.95^{\circ }\text{C}$ This sits comfortably inside $-20^{\circ }\text{C}$ to $+50^{\circ }\text{C}$. Answer: $\approx 272$ K $\approx -1^{\circ }\text{C}$ — in limits. ✅

Step 1 — end-of-life array need (using the power-budget formula from 2.2): $P_{\text{SA,EOL}}=12\left(1+\frac{33}{62\times 0.85}\right)=12\left(1+0.626\right)=12\times 1.626=19.5\text{W}$ Why EOL first: the array must still meet the load in year 5, after it has degraded — so we design the worst (latest) case first.

Step 2 — back out beginning-of-life. We model degradation as linear: losing $2.5\%$ of the original output each year, so over 5 years the total loss is $2.5\%\times 5=12.5\%$ and the array retains $1-0.125=0.875$ of its start value. To have $19.5$ W left at the end: $P_{\text{SA,BOL}}=\frac{P_{\text{SA,EOL}}}{0.875}=\frac{19.5}{0.875}=22.3\text{W}$

Answer: design the array for $\approx 22.3$ W at beginning of life so it still delivers $\approx 19.5$ W after 5 years.

Assumption flag — linear vs. compound: we used a linear model, $\text{retained}=1-pN$ with $p=0.025$, $N=5$. Many space-solar analyses instead use a compound (exponential) model, $\text{retained}=(1-p{)}^N=0.975^5=0.881$, because each year's degradation acts on the already-degraded output, not the original. The two give $0.875$ vs $0.881$ here — nearly identical over 5 years, but the gap widens for longer missions. Always state which you use; we use linear throughout this page for arithmetic simplicity.

Part (a) — structure. Launch force: $F=ma=6\times (12\times 9.8)=6\times 117.6=705.6\text{N}$. Strength-limited area: $A=F/{\sigma }_y=705.6/(240\times 10^6)=2.94\times 10^{-6}{\text{m}}^2$. Structural mass: $m_{\text{struct}}=AL\rho =2.94\times 10^{-6}\times 0.3\times 2700=2.38\times 10^{-3}\text{kg}\approx 2.4$ g. Reality note: just as in the parent CubeSat example, this pure-strength number is tiny — real stiffness/vibration/safety-factor requirements multiply it by 10–20×. It sets the floor, not the design.

Part (b) — power. EOL array: $P_{\text{SA,EOL}}=40\left(1+\frac{35}{60\times 0.85}\right)=40\left(1+0.686\right)=40\times 1.686=67.5\text{W}$. Linear degradation over 3 yr: retains $1-0.025\times 3=0.925$. BOL: $P_{\text{SA,BOL}}=67.5/0.925=72.9\text{W}$.

Part (c) — thermal. The array generates its BOL power (worst case for waste heat), and it produces $1$ W of waste heat for every $8$ W generated: $Q_{\text{array}}=\frac{P_{\text{SA,BOL}}}{8}=\frac{72.9}{8}=9.1\text{W}$ Add the payload's internal dissipation, which (as in the L4 trap) must also be radiated away: $Q_{\text{reject}}=Q_{\text{array}}+Q_{\text{payload}}=9.1+6=15.1\text{W}$ Answer: structure floor $\approx 2.4$ g; BOL array $\approx 72.9$ W; the thermal system must reject $\approx 15.1$ W.

What we just did — the whole point of L5: one 6 kg number rippled through three subsystems. The launch load sized the walls; the same spacecraft's electronics sized the array; the array's own inefficiency plus the payload fed the thermal load. That coupling — where every subsystem's answer becomes another's input — is exactly what systems engineering is.

• The quantity that decides which material is lighter for a strength-limited part is ==${\sigma }_y/\rho$ (structural efficiency)==.
• Equilibrium temperature in sunlight depends on the ratio ==$\alpha /ϵ$==, not on $\alpha$ alone.
• Internal dissipation appears in both the power budget and the thermal budget.
• You size the array for end-of-life conditions first, then divide by the retained fraction to get beginning-of-life.
• Bare $\sigma$ on this page means the Stefan–Boltzmann constant, while ${\sigma }_y$ means yield strength.