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

This deep dive belongs to Spacecraft Bus — Structure, Power, Thermal, ADCS, C&DH, Comms, Propulsion. We go deeper than the parent's "Energy Balance in Orbit" formula — we earn every symbol.

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Step 1 — What is "power," and why does an orbit split into two halves?

WHAT. A satellite in Low Earth Orbit (LEO) circles the planet in about an hour and a half. For part of that loop it is bathed in sunlight; for the rest it passes through Earth's shadow.

WHY split it. Solar panels only make power when they see the Sun. So the orbit has a fundamentally different character in each half, and energy must be tracked separately in each.

PICTURE. The circle below is one orbit. The yellow arc is the sunlit phase (duration $T_{\text{sun}}$); the gray arc is the eclipse phase (duration $T_{\text{eclipse}}$). Together they add to one full orbit, $T_{\text{orbit}}$.

$T_{\text{orbit}}=\underset{\text{panels see Sun}}{\underbrace{T_{\text{sun}}}}+\underset{\text{in shadow}}{\underbrace{T_{\text{eclipse}}}}$

• $T_{\text{orbit}}$ — time for one full loop (seconds or minutes).
• $T_{\text{sun}}$ — how long the panels are lit.
• $T_{\text{eclipse}}$ — how long the satellite is in the dark.

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Step 2 — The load never sleeps

WHAT. We draw a flat line: the load demands the same $P_{\text{load}}$ watts every second of the orbit.

WHY it matters. During eclipse there is no sunlight, yet the load still demands its watts. That demand has only one possible source: a battery. So the panel isn't just feeding the load in daylight — it must have pre-charged the battery to cover the whole eclipse.

PICTURE. The orange line is the constant load. The yellow bar (daylight) is the panel's output; notice it towers above the load — the surplus is what charges the battery. In the gray region (eclipse) the panel gives nothing, and the battery alone carries the orange line.

• $P_{\text{load}}$ — constant demand, present in both halves.
• $P_{\text{SA}}$ — solar-array output, present only in daylight (this is what we're solving for).

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Step 3 — The eclipse deficit: how much energy the darkness costs

WHAT. Compute the energy the battery must supply during eclipse. Energy = power × time, so:

$E_{\text{need}}=P_{\text{load}}\times T_{\text{eclipse}}$

• $E_{\text{need}}$ — energy (joules) the battery must deliver in the dark.
• $P_{\text{load}}$ — the constant watts drawn.
• $T_{\text{eclipse}}$ — how long the darkness lasts.

WHY. This is the debt the daylight must have paid off in advance. It is a plain rectangle: height = load, width = eclipse duration. Area = energy.

PICTURE. The shaded gray rectangle below is exactly this energy — the "hole" the battery must fill.

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Step 4 — Batteries are leaky buckets: efficiency enters

WHAT. We correct the raw energies for these leaks.

WHY these two, and why divide vs. multiply?

• Charging: we put in $E_{\text{charge}}$, but only $E_{\text{charge}}\cdot {\eta }_{\text{charge}}$ is stored. We multiply by a number below 1 → less survives.
• Discharging: the load needs $E_{\text{need}}$, but the battery must give more to cover its own losses: $E_{\text{need}}/{\eta }_{\text{discharge}}$. We divide by a number below 1 → the requirement grows.

PICTURE. A funnel diagram: energy poured in shrinks (charge loss); energy demanded on the far side forces an even bigger draw (discharge loss). Both leaks make the panel's job harder.

$\text{stored}=E_{\text{charge}}{\eta }_{\text{charge}},\text{must draw}=\frac{E_{\text{need}}}{{\eta }_{\text{discharge}}}$

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Step 5 — The daylight surplus: what actually charges the battery

WHAT. In daylight the panel makes $P_{\text{SA}}$, the load eats $P_{\text{load}}$, and the leftover charges the battery:

$P_{\text{charge}}=P_{\text{SA}}-P_{\text{load}}$

• $P_{\text{charge}}$ — surplus watts flowing into the battery.
• $P_{\text{SA}}$ — solar-array output (the unknown).
• $P_{\text{load}}$ — the same constant demand as always.

Over the sunlit duration this surplus delivers energy $P_{\text{charge}}\cdot T_{\text{sun}}$, of which only ${\eta }_{\text{charge}}$ survives into the battery.

WHY. This is the supply side of the ledger — the energy we manage to bank while the Sun is up.

PICTURE. The tall yellow bar minus the orange load line leaves a green surplus band. Its area (times $T_{\text{sun}}$) is the energy we're saving up.

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Step 6 — Balance the ledger: stored = withdrawn

WHAT. For the battery to never run empty, the energy successfully stored in daylight must equal the energy the eclipse withdraws:

$\underset{\text{banked in daylight}}{\underbrace{P_{\text{charge}}{T}_{\text{sun}}{\eta }_{\text{charge}}}}=\underset{\text{withdrawn in eclipse}}{\underbrace{\frac{P_{\text{load}}{T}_{\text{eclipse}}}{{\eta }_{\text{discharge}}}}}$

WHY this equality. If the left side were larger, the battery would end each orbit fuller than it started — wasted panel. If smaller, it drains a little each orbit and eventually dies. Steady state = exact balance. This is the heart of the derivation.

PICTURE. A two-pan balance scale: the green "banked" area on one pan, the gray "withdrawn" area on the other, level.

Now substitute $P_{\text{charge}}=P_{\text{SA}}-P_{\text{load}}$ and solve for $P_{\text{SA}}$:

$(P_{\text{SA}}-P_{\text{load}})T_{\text{sun}}{\eta }_{\text{charge}}=\frac{P_{\text{load}}{T}_{\text{eclipse}}}{{\eta }_{\text{discharge}}}$

Divide through, add $P_{\text{load}}$ back, and factor it out:

$\boxed{P_{\text{SA}}=P_{\text{load}}\left(1+\frac{T_{\text{eclipse}}}{T_{\text{sun}}{\eta }_{\text{charge}}{\eta }_{\text{discharge}}}\right)}$

Read it term by term:

• The leading $P_{\text{load}}$: the panel must at least cover the running load.
• The $1$: that daytime load itself.
• The fraction: the extra fraction of panel needed to also refill the battery.
• $T_{\text{eclipse}}/T_{\text{sun}}$: longer dark or shorter light → bigger extra.
• ${\eta }_{\text{charge}}{\eta }_{\text{discharge}}$ in the denominator: leaky batteries → bigger extra.

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Step 7 — Edge and degenerate cases (never let the reader hit an unshown scenario)

Case A — No eclipse ($T_{\text{eclipse}}=0$). A satellite in permanent sunlight (e.g. a dawn-dusk Sun-synchronous orbit). The fraction vanishes: $P_{\text{SA}}=P_{\text{load}}(1+0)=P_{\text{load}}.$ Sensible: no darkness to prepare for, so the panel just matches the load.

Case B — Perfect battery (${\eta }_{\text{charge}}={\eta }_{\text{discharge}}=1$). $P_{\text{SA}}=P_{\text{load}}\left(1+\frac{T_{\text{eclipse}}}{T_{\text{sun}}}\right).$ The panel enlargement is purely the day/night time ratio — the cleanest possible case.

Case C — Vanishing sunlight ($T_{\text{sun}}\to 0$). The denominator $\to 0$, so $P_{\text{SA}}\to \infty$. Physically correct: with almost no light, no finite panel could bank enough energy. This warns designers that very short sunlit arcs are a hard limit.

PICTURE. A single plot of the enlargement factor $(1+\frac{T_{\text{eclipse}}}{T_{\text{sun}}{\eta }_c{\eta }_d})$ versus $T_{\text{sun}}$, showing it flatten toward 1 for long daylight and blow up as daylight shrinks — with the three cases marked.

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Step 8 — The worked number (the parent's LEO example, retraced)

Reveal-yourself check:

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The one-picture summary

The whole story on one graph: a tall yellow daylight bar overtops the flat orange load; the green surplus (banked, shrunk by ${\eta }_{\text{charge}}$) exactly fills the gray eclipse deficit (grown by $1/{\eta }_{\text{discharge}}$). Balance those two areas and out drops the boxed formula.

Recall Feynman retelling — tell it back in plain words

A satellite spends part of every orbit in sunlight and part in Earth's shadow. Its gadgets need a steady trickle of power the whole time. In sunlight the panels do double duty: they run the gadgets and stuff extra energy into a battery. In shadow the panels are useless, so the battery alone runs everything. Energy is just power multiplied by time, which on a power-vs-time graph is area. So the rule is simple: the energy area banked in daylight must equal the energy area drained in darkness. Batteries leak, though — some energy is lost going in, and even more coming out — so you have to bank a bit extra to cover the leaks. Solve that "areas must match" equation for the panel size and you get: the panel equals the load, plus a top-up that grows with how long the dark lasts, how short the light is, and how leaky the battery is. Push it to the extremes and it stays honest: no shadow → panel just equals the load; almost no sunlight → panel size explodes. For a real LEO bird pulling 50 W, that's about 81 W, bumped to ~93 W so it still works after five years of Sun damage.