Chapter 25: Power Systems

"We must conclude that the stars are drawing on some vast reservoir of energy by means unknown to us. This reservoir can scarcely be other than the sub-atomic energy which, it is known, exists abundantly in all matter." — Arthur Eddington, 1920

Overview

A spacecraft is, before it is anything else, a machine that runs on electricity. The radio that carries a picture of Pluto back across five billion kilometers, the reaction wheels that hold a telescope steady on a distant galaxy, the heaters that keep the propellant of Chapter 24 from freezing solid, the computer that flies the whole vehicle — none of them does anything without a steady supply of watts. And in space there is no grid, no outlet, no extension cord. You bring your own power plant, you bolt it to the vehicle, you pay the launch cost of every kilogram of it, and then it has to run — often for a decade or more — with no one to fix it.

That is the problem of this chapter, and it has the same shape as every other problem in this book. Sunlight is free once you are in space, but the panels that catch it have mass, and mass is the enemy (Chapter 3). Batteries let you survive the dark side of the orbit, but they are heavy, and every kilogram of battery is a kilogram of science you did not fly. Go far enough from the Sun and the sunlight thins to uselessness, so you carry a lump of plutonium instead — safe, reliable, decades-long, and staggeringly expensive. Want kilowatts in the dark, on the surface of Mars, or to feed an electric thruster? Then you are talking about a nuclear reactor, and a new set of trades. Every one of these choices is a negotiation between the power you need, the mass you can afford, and the environment that is trying to starve or freeze your spacecraft to death.

Eddington was writing about how stars shine, but his sentence names the two power sources this chapter is about: the Sun (fusion, caught by solar cells) and the atom (decay and fission, caught by thermocouples and turbines). Between them they have powered every spacecraft ever flown.

In this chapter, you will learn to:

• Turn solar flux into watts, and size a solar array for a real spacecraft load — including the sun-angle "cosine loss," the cell efficiency, and the fraction of every orbit spent in shadow.
• Age a solar array: account for radiation degradation so the array still meets its load at end of life, not just at launch.
• Size a battery from the eclipse it must survive, and see why a low-Earth-orbit battery and a geostationary battery are sized by completely different rules.
• Understand radioisotope thermoelectric generators (RTGs) and fission reactors, and know which missions cannot be flown any other way.
• Assemble a power budget — the ledger that makes sure generation beats demand in every mission phase — and choose a bus voltage that does not cost you a harness made of copper bricks.

Learning Paths

🚀 Space Enthusiast: Read 25.1 for how sunlight becomes watts, then 25.4 (RTGs — the plutonium that keeps Voyager talking) and the story of the lunar night in Case Study 2. You can skim the algebra of the array-sizing example and still get every idea.

📐 Engineering Student: Work every section and every worked example by hand. The array-sizing and battery-sizing calculations (25.1–25.3) are the core deliverables; the ⭐⭐/⭐⭐⭐ exercises turn them into a repeatable method you will reuse in the capstone.

🎮 KSP Player: Stock KSP models electric charge, solar panels that fade with distance, and batteries that run dry on the night side — this chapter is the physics behind that gauge. Focus on 25.1, 25.2, and the power budget in 25.6; then go watch your own craft brown out in Kerbin's shadow and size it properly.

🛰️ Industry Prep: Sections 25.5 (fission power is suddenly real again) and 25.6 (power budgets and bus voltage) are the language of a real power-subsystem design review. Build the power budget for your mission at the checkpoint and carry it to the capstone.

25.1 Solar power and array sizing

Start with the physical fact that makes solar power possible at all: sunlight carries energy, and a lot of it. At Earth's distance from the Sun — one astronomical unit — a surface held square to the Sun receives about

$$ S = 1{,}361\ \text{W/m}^2, $$

the solar constant (or solar flux), which you met as a thermal input in Chapter 24 and which is tabulated in Appendix B. That is over a kilowatt on every square meter, for free, forever — a hair dryer's worth of power falling on a dinner plate. The engineering question is how much of it you can turn into electricity, and how big a collector you need to run your spacecraft.

The device that does the converting is the photovoltaic cell.

Definition (photovoltaic cell). A photovoltaic (PV) cell, or solar cell, is a semiconductor device that converts sunlight directly into electricity: photons striking the cell knock electrons across a junction, producing a voltage and driving a current. A solar array is a large assembly of many such cells wired in series and parallel to reach a useful voltage and power.

Modern spacecraft do not use the humble silicon cells on a suburban roof (~15–18% efficient). They use multi-junction cells — typically three stacked semiconductor layers, each tuned to a different slice of the solar spectrum — which reach about 30% efficiency in space. That number, $\eta \approx 0.30$, is the single most important property of the cell: it is the fraction of incident sunlight that leaves as electricity.

Three things reduce the power below the naive "flux times area":

1. Efficiency $\eta$. Only ~30% of the light becomes electricity; the rest becomes heat (which the thermal system of Chapter 24 must then reject).
2. Cosine loss. A cell produces power in proportion to the sunlight it intercepts, which depends on the angle between the sunlight and the panel. If the array's normal points at angle $\theta$ away from the Sun, a panel of area $A$ intercepts only $A\cos\theta$ of sunlight. Point the array edge-on to the Sun ($\theta = 90^\circ$) and it makes nothing at all.
3. Everything else, folded in later: packing gaps between round cells, cover-glass reflection, high-temperature losses, and the slow radiation damage we treat in §25.2.

Putting the first three ideas together, the electrical power from an array of area $A$ at the beginning of its life is

$$ P = S \, \eta \, A \cos\theta . $$

💡 Intuition: Read this as a chain of "keep only a fraction" steps. Start with $1{,}361\ \text{W/m}^2$ of raw sunlight. Keep ~30% (the cell efficiency). Keep $\cos\theta$ (how squarely you face the Sun). Multiply by your area. What survives all three cuts is your electricity. A well-pointed 30% array delivers roughly $1361 \times 0.30 \approx 400\ \text{W}$ per square meter square-on — so a kilowatt spacecraft needs several square meters of panel, which is exactly why big satellites unfold those long wings.

The inverse-square fade

The solar constant is only constant at 1 AU. Sunlight spreads out over an expanding sphere, so its intensity falls as the inverse square of distance from the Sun. At a distance of $d$ astronomical units,

$$ S(d) = \frac{1{,}361}{d^{2}}\ \text{W/m}^2 . $$

This one relationship dictates which missions can use solar power at all. Run the numbers for a few destinations:

At Mars a solar array must be more than twice the size it would be at Earth to make the same power. At Jupiter you need twenty-seven times the area — the reason Jupiter missions were flown on plutonium for decades, until Juno finally proved (Case Study 1) that enormous arrays could do it. By Saturn, solar power is essentially hopeless; past there it is out of the question. The inverse-square law is why deep space belongs to the RTG (§25.4).

Worked example: size an array for a spacecraft

Strategy first. Sizing an array is a two-step calculation, and the order matters. Step 1: figure out how much electrical power the array must produce — which is more than the spacecraft's load, because the array only works in sunlight yet must power the vehicle around the whole orbit, storing energy for the eclipse and paying storage losses on the way. Step 2: convert that required power into an area, using the flux, efficiency, and cosine loss. In this section we do both at the beginning of life; §25.2 then ages the array so it still works years later.

Take a small communications satellite in a $500\ \text{km}$ low Earth orbit. From its power budget (which we build in §25.6) it needs an orbit-average electrical load of

$$ P_{\text{load}} = 1{,}000\ \text{W} = 1\ \text{kW}, $$

and we will assume for now that it draws this in sunlight and in shadow alike. Its orbit lasts about $95\ \text{min}$, of which roughly $35\ \text{min}$ is spent in Earth's shadow and $60\ \text{min}$ in sunlight (we derive these numbers in §25.2; for now take them as given).

Step 1 — how much must the array produce? During the $60\ \text{min}$ of sunlight the array has to do two jobs at once: run the spacecraft and charge the battery for the coming $35\ \text{min}$ of darkness. Energy is conserved over an orbit, but the paths are lossy. Following the standard method (Wertz, SMAD), let $X_d$ be the efficiency of the sunlit path (array → regulator → loads) and $X_e$ the efficiency of the eclipse path (array → battery → regulator → loads). The eclipse path is worse because it also pays the battery's charge and discharge losses; typical values are $X_d \approx 0.85$ and $X_e \approx 0.65$. The array output power $P_{sa}$ must satisfy

$$ P_{sa} = \frac{\dfrac{P_d\,T_d}{X_d} + \dfrac{P_e\,T_e}{X_e}}{T_d}, $$

where $P_d, P_e$ are the sunlit and eclipse loads and $T_d, T_e$ the sunlit and eclipse times. With $P_d = P_e = 1{,}000\ \text{W}$, $T_d = 60\ \text{min}$, $T_e = 35\ \text{min}$:

$$ P_{sa} = \frac{\dfrac{1000 \times 60}{0.85} + \dfrac{1000 \times 35}{0.65}}{60} = \frac{70{,}588 + 53{,}846}{60} = \frac{124{,}434}{60} \approx 2{,}074\ \text{W}. $$

Stop and feel that result: to support a $1{,}000\ \text{W}$ spacecraft, the array must generate about $2{,}074\ \text{W}$ — more than twice the average load. The factor of two is the price of night. The array works only 63% of the time ($60/95$), and the energy it stashes for the dark side leaks away at both ends of the battery.

Step 2 — how big is that array? Now turn watts into square meters. Per unit area, at the beginning of life, square-on, an array produces $S\,\eta = 1361 \times 0.30 = 408.3\ \text{W/m}^2$. But this array cannot always face the Sun perfectly; take a representative worst-case pointing/seasonal offset of $\theta = 25.8^\circ$, so $\cos\theta = 0.90$. Then it produces

$$ p_{\text{area}} = S\,\eta\cos\theta = 1361 \times 0.30 \times 0.90 = 367\ \text{W/m}^2 . $$

The required area is therefore

$$ A_{\text{BOL}} = \frac{P_{sa}}{p_{\text{area}}} = \frac{2{,}074}{367} \approx 5.6\ \text{m}^2 . $$

Sanity check. Is $5.6\ \text{m}^2$ believable for a 2 kW-class array? That is a panel about $2.4 \times 2.4\ \text{m}$, or two wings roughly $1.4 \times 2\ \text{m}$ each — entirely ordinary for a satellite of this size. Real arrays deliver on the order of $300\ \text{W/m}^2$ at the panel after all losses, and $2074/300 \approx 6.9\ \text{m}^2$, in the same ballpark. The number is sound. But we are not done: this is the array a brand-new spacecraft needs. Radiation has not yet touched it. In §25.2 we age it, and it grows.

🔄 Check Your Understanding 1. Your spacecraft's average load doubles from 1 kW to 2 kW, everything else unchanged. Does the array area double? 2. A designer proposes mounting the array flat on the spacecraft body, where its normal sits $60^\circ$ from the Sun on average. By what factor does the required area grow versus a Sun-pointing array?

Answers

1. Yes — the required array output $P_{sa}$ scales linearly with the load, and area scales linearly with $P_{sa}$, so the area doubles too (to ~11 m²). This linearity is exactly why power hunger translates so directly into array mass, and array mass into launch cost. 2. $\cos 60^\circ = 0.5$ versus $\cos 25.8^\circ = 0.90$, so the flat array makes only $0.5/0.90 = 0.56$ as much per square meter and must be $1/0.56 \approx 1.8\times$ larger. This is why high-power spacecraft pay for gimballed, Sun-tracking wings rather than body-mounted panels.

25.2 Degradation and eclipse periods

Two facts turn the clean array of §25.1 into a bigger, real one: the array spends part of every orbit in the dark, and it slowly dies in the radiation of space. This section quantifies both, because array sizing — choosing the panel area a mission actually needs — is not honest until it accounts for them.

Definition (array sizing). Array sizing is the process of determining the solar-array area (and hence mass) required so that the array meets the spacecraft's power demand under worst-case conditions and at end of life — after accounting for eclipse, sun angle, temperature, degradation, and system losses — with margin. Sizing to beginning-of-life, sunlit, square-on performance is a classic and dangerous mistake.

Eclipse: the fraction of the orbit spent in shadow

Every low orbit passes behind its planet once per revolution, into a shadow where the array makes nothing and the battery carries the whole load. How long is the dark? Model Earth's shadow as a cylinder of the planet's radius $R_\oplus$ (the umbra, ignoring the small penumbra). A satellite in a circular orbit of radius $r$ is in shadow while it lies within that cylinder; a little geometry gives the eclipse fraction — the share of the orbit spent in darkness, in the worst case when the Sun lies in the orbit plane:

$$ f_e = \frac{1}{\pi}\arcsin\!\left(\frac{R_\oplus}{r}\right). $$

For our $500\ \text{km}$ orbit, $r = 6378 + 500 = 6878\ \text{km}$, so

$$ f_e = \frac{1}{\pi}\arcsin\!\left(\frac{6378}{6878}\right) = \frac{1}{\pi}\arcsin(0.9273) = \frac{1.187}{\pi} = 0.378 . $$

About 38% of every orbit is night. The orbital period at this altitude is $94.6\ \text{min}$ (from Kepler's third law, Chapter 8), so the eclipse lasts $0.378 \times 94.6 \approx 35.8\ \text{min}$ and the sunlit arc $58.8\ \text{min}$ — which is where the "35 and 60 minutes" of §25.1 came from (rounded for legibility). A low-orbiting spacecraft therefore endures roughly 5,500 eclipses per year, each one a full discharge–recharge of its battery. Hold that number; it decides the battery design in §25.3.

🔗 Connection: Orbit choice and power are deeply coupled. A dawn–dusk sun-synchronous orbit (Chapter 9) rides the day–night terminator so the spacecraft is in near-continuous sunlight — almost no eclipse, so a smaller battery and a simpler power system. Geostationary satellites (Chapter 9) are eclipsed only near the equinoxes, about 90 times a year, never in between. The same orbit that a communications engineer picks for coverage, the power engineer inherits as an eclipse schedule. Choosing an orbit is choosing a power problem.

Degradation: the array ages

Space is an unforgiving environment, and it is unforgiving to solar cells in particular. Energetic protons and electrons — from the Van Allen belts, from solar storms, from the Jovian magnetosphere — knock atoms out of the semiconductor lattice, creating defects that quietly steal efficiency. Ultraviolet light darkens the cover glass; thermal cycling cracks solder joints. The array you launch is the best array you will ever have; from then on it fades.

Engineers model the fade as a steady annual loss. If the array loses a fraction $D$ of its output each year, then after $y$ years the surviving fraction — the life degradation factor — is

$$ L_d = (1 - D)^{y}. $$

Radiation-tolerant triple-junction cells lose on the order of $D \approx 0.5\%$ to $3.75\%$ per year, depending brutally on the orbit (a satellite that spends its life in the heart of the Van Allen belts ages far faster than one in low orbit). Take $D = 2.75\%$/yr over a $7$-year mission:

$$ L_d = (1 - 0.0275)^{7} = 0.9725^{7} = 0.823 . $$

The array will deliver only 82% at end of life of what it made new. So we must size it bigger, so that even after seven years of decay it still meets the load. Re-sizing our §25.1 array to end-of-life:

$$ A_{\text{EOL-sized}} = \frac{P_{sa}}{p_{\text{area}}\, L_d} = \frac{2{,}074}{367 \times 0.823} = \frac{2{,}074}{302} \approx 6.9\ \text{m}^2 . $$

The array grew from $5.6\ \text{m}^2$ to $6.9\ \text{m}^2$ — about 22% more panel, and 22% more mass — bought entirely to pay for damage that has not happened yet. That is the tax the space environment levies on every solar-powered mission.

🚪 Threshold Concept. You do not design a power system for the day it launches; you design it for its worst day, years later, in the dark. Once you see this, the whole subsystem reorganizes around it. The array is sized for end-of-life, degraded, at worst-case sun angle. The battery is sized for the longest eclipse of the mission at end of life, when it too has faded. The whole power system is a promise that the spacecraft's darkest, oldest, coldest moment will still have enough watts — because in space, one brownout can be fatal, and there is no one to swap the panel. Power engineering is the discipline of guaranteeing the worst case, not the average.

🔧 Engineering Reality: Cells also run hot, and hot cells are worse cells. A solar cell's voltage drops as its temperature rises, so an array baking at $+60\,^\circ\text{C}$ in full Sun produces measurably less than the same array at room temperature — an effect folded into the "inherent degradation" (~0.77) that real sizing (SMAD) applies on top of everything above. This is a direct handshake with Chapter 24: the thermal design sets the cell temperature, the cell temperature sets the power, and the waste heat the cells reject (70% of the incident sunlight) is a load the thermal system must plan for. Power and thermal are two views of the same energy.

🔄 Check Your Understanding 1. Why does a geostationary comsat need a much smaller battery relative to its load than a low-Earth-orbit satellite with the same load? 2. Two identical arrays fly, one in low Earth orbit and one through the Van Allen belts to a higher orbit. After five years, which makes less power, and why?

Answers

1. GEO is eclipsed only near the equinoxes and for at most ~72 minutes, and only ~90 times a year; a LEO satellite is eclipsed ~35 minutes every orbit, ~16 times a day. Battery energy is set by the eclipse it must cover, so GEO's rare, though longer, eclipses still demand less relative to a battery cycled 5,500 times a year at shallow depth (§25.3). 2. The one that flew through the belts. Degradation depends on accumulated radiation dose, and the Van Allen belts are the most intense trapped-radiation region near Earth; a belt-crossing orbit can age cells several times faster than low orbit, so its $L_d$ is lower and its output smaller.

25.3 Batteries

The array is useless in the dark, and the spacecraft still needs to run. That is what the battery is for: it stores energy in sunlight and releases it in shadow, and on many spacecraft it also covers peak loads (a burst of transmitter power, a thruster firing) that briefly exceed what the array can supply. Batteries are the second-heaviest part of most power systems, so sizing them well is squarely a fight against mass (theme 4).

Almost every modern spacecraft uses lithium-ion cells — the same chemistry as a laptop or an electric car, space-qualified for vacuum, radiation, and thousands of cycles. Lithium-ion won because of energy density: it stores roughly $100$–$160\ \text{W·h/kg}$, three to four times the nickel-cadmium and nickel-hydrogen batteries it replaced (Hubble and the early ISS flew NiH₂). More watt-hours per kilogram is, directly, less mass for the same night survived.

The number that governs battery life is the depth of discharge.

Definition (depth of discharge). Depth of discharge (DoD) is the fraction of a battery's total energy capacity that is withdrawn on a given discharge, expressed as a percentage: a battery of $100\ \text{W·h}$ capacity from which $30\ \text{W·h}$ is drawn each cycle runs at $30\%$ DoD. The complementary quantity is state of charge. The crucial fact: the more deeply you cycle a battery, the fewer cycles it survives. A lithium-ion cell that lasts many tens of thousands of cycles at 20% DoD might manage only a few thousand at 80%.

This single trade-off splits battery design along the same line the eclipse counts did. Because you only ever use a fraction $\text{DoD}$ of the battery's capacity, and the discharge path has efficiency $\eta$, the capacity you must install to deliver an eclipse's worth of energy is

$$ C = \frac{P_e\, T_e}{\text{DoD}\times \eta}. $$

Worked example: the LEO battery

Our $500\ \text{km}$ satellite draws $P_e = 1{,}000\ \text{W}$ through an eclipse of $T_e = 35\ \text{min} = 0.583\ \text{h}$. The energy it must deliver each night is

$$ E_e = P_e\, T_e = 1000 \times 0.583 = 583\ \text{W·h}. $$

But this satellite is eclipsed ~5,500 times a year — nearly 39,000 cycles over a 7-year life. To survive that many cycles, lithium-ion must be cycled shallowly; choose $\text{DoD} = 0.25$ (a typical LEO value) and a discharge-path efficiency $\eta = 0.90$:

$$ C = \frac{583}{0.25 \times 0.90} = \frac{583}{0.225} \approx 2{,}593\ \text{W·h} \approx 2.6\ \text{kW·h}. $$

Convert to mass — the number that actually matters. At $150\ \text{W·h/kg}$ (a good space-qualified lithium-ion value):

$$ m_{\text{batt}} = \frac{2{,}593}{150} \approx 17\ \text{kg}. $$

Sanity check and the trade. Seventeen kilograms of battery to carry a 1 kW satellite through the night is reasonable — comparable to the array's mass. But notice where it came from: we installed $2{,}593\ \text{W·h}$ to deliver only $583\ \text{W·h}$, a $4.4\times$ oversize, almost all of it to keep the DoD low enough to survive 39,000 cycles. Had this been a GEO satellite, cycled only ~1,350 times in 15 years, we could safely run $\text{DoD} = 0.7$ and install roughly a third as much capacity for the same nightly energy. Same job, one-third the battery mass — purely because of how often it is cycled. The cycle-life curve, not the energy, sizes a LEO battery.

🐛 Find the Error. An engineer sizing the LEO battery writes: "We need to deliver $583\ \text{W·h}$ each eclipse, and our cells hold $150\ \text{W·h/kg}$, so the battery masses $583/150 \approx 3.9\ \text{kg}$." The flight battery came in at $17\ \text{kg}$, over four times heavier. What did the estimate ignore?

Answer

Two things, both essential. First, depth of discharge: you cannot use 100% of a battery's capacity every cycle without destroying it in months, so for a LEO satellite cycled tens of thousands of times you install ~4× the deliverable energy and only skim 25% off the top. Second, the discharge-path efficiency (~90%): some of the stored energy is lost to internal resistance and regulation on the way to the load. Dividing $583\ \text{W·h}$ by $(0.25 \times 0.90)$ gives the real $2{,}593\ \text{W·h}$ and the honest $17\ \text{kg}$. Sizing a battery to its energy alone, ignoring DoD, is one of the most common and most punishing mistakes in the subsystem.

🔧 Engineering Reality: Batteries are also the power system's most temperamental component. Lithium-ion hates the cold (capacity plummets below $0\,^\circ\text{C}$) and hates the heat (it ages fast and, abused, can go into thermal runaway), so it sits inside the thermal control of Chapter 24, often with its own heaters — heaters that are themselves a power load in the eclipse budget. It must be protected from overcharge and over-discharge by the power electronics; a single cell driven to zero volts can be permanently killed. And because "everything must work," flight batteries are built from many cells with redundancy, so that losing one cell degrades rather than ends the mission.

25.4 RTGs: power from plutonium

Send a spacecraft to Jupiter, or into the permanent shadow of a lunar crater, or onto the surface of Mars where dust storms can blot out the Sun for weeks, and solar power fails. The flux is too weak, or the Sun too often absent. For six decades the answer has been to carry your own private, tiny, reliable star: a lump of radioactive plutonium whose heat you turn into electricity.

Definition (radioisotope thermoelectric generator, RTG). An RTG is a power source that converts the heat of natural radioactive decay directly into electricity. A block of plutonium-238 decays (emitting alpha particles) and grows hot; an array of thermocouples surrounding it exploits the thermoelectric (Seebeck) effect — a temperature difference across a junction of two materials produces a voltage — to generate a steady current. There are no moving parts, nothing to wear out, and it works in sunlight, shadow, or interstellar dark, day and night, for decades.

Why plutonium-238? It is chosen for a rare combination: it releases a lot of heat per gram (about $0.54\ \text{W}$ of heat per gram of the pure isotope), it emits mostly easily-shielded alpha radiation, and it has a half-life of 87.7 years — long enough to power a mission for decades, short enough to be usefully hot. Its heat is relentless and needs no ignition, no sunlight, no maintenance. The catch is the conversion: thermocouples are inefficient, turning only about 6–8% of the heat into electricity and dumping the rest as waste heat (which, conveniently, can warm the spacecraft — Chapter 24). An RTG is mostly a very reliable, very expensive space heater that leaks a little electricity.

Two RTGs have flown most of the famous deep-space missions:

Worked example: will the RTG last the mission?

An RTG fades for two reasons: the plutonium decays, and the thermocouples themselves degrade. Take a GPHS-RTG with $P_0 = 300\ \text{W}$ at the beginning of life (Tier 2), and ask what it delivers after 30 years of flight.

Decay of the heat source. Radioactive decay is exponential; after $t$ years the surviving thermal power is the initial times $\left(\tfrac{1}{2}\right)^{t/t_{1/2}}$ with $t_{1/2} = 87.7\ \text{yr}$:

$$ \frac{P_{\text{thermal}}(30)}{P_{\text{thermal}}(0)} = \left(\tfrac{1}{2}\right)^{30/87.7} = \left(\tfrac{1}{2}\right)^{0.342} = 0.789 . $$

So after 30 years, 79% of the heat remains — the plutonium is barely winded.

Decay of the thermocouples. The junctions sinter and crack over time, losing perhaps $0.8\%$ of their conversion efficiency per year (illustrative; the real rate is temperature- and design-dependent). Over 30 years that compounds to $(1 - 0.008)^{30} = 0.786$.

Combined electrical output:

$$ P_{\text{elec}}(30) = 300 \times 0.789 \times 0.786 \approx 186\ \text{W}. $$

After three decades the generator still delivers 62% of its original electricity — about $186\ \text{W}$. That is the almost unbelievable durability that lets Voyager 1 and 2, launched in 1977, still whisper to Earth from interstellar space nearly half a century later, powering down one instrument at a time as their generators slowly cool. No battery and no solar array could do this.

⚠️ Common Misconception: "Why not just fly a bigger RTG and get kilowatts?" Because plutonium-238 is one of the scarcest engineered materials on Earth. It does not occur naturally; it must be bred in a reactor and chemically separated, at enormous cost. After the Cold War the United States stopped making it and ran down its stockpile; production only restarted at Oak Ridge around 2015, at a target rate of roughly $1.5\ \text{kg}$ per year (Tier 2). To make $1\ \text{kW}$ of electricity from an RTG at 6% efficiency would need about $30\ \text{kg}$ of Pu-238 — twenty years of national production for a single spacecraft. RTGs are, and will remain, a few-hundred-watt technology. When a mission needs kilowatts in the dark, the answer is not more plutonium; it is fission (§25.5).

📜 From History: The first solar-powered satellite, Vanguard 1, went up in 1958 with six tiny silicon cells running a $5\ \text{mW}$ transmitter; the cells outlived the chemical battery beside them by years and proved solar power in space. The first RTG flew a year later. Since then the two technologies have divided the solar system between them along the line the inverse-square law draws: solar cells everywhere the Sun is strong, plutonium everywhere it is not. Every mission to Jupiter and beyond — Pioneer, Voyager, Galileo, Cassini, New Horizons — has carried a nuclear heart, because past the asteroid belt there is simply not enough sunlight to fly on.

25.5 Fission reactors: Kilopower and KRUSTY

An RTG passively harvests the trickle of heat from radioactive decay. A fission reactor actively splits uranium nuclei in a controlled chain reaction, releasing heat on a completely different scale — kilowatts to megawatts — and, unlike an RTG, it can be turned on when you arrive and does not fade on mission timescales. Fission is the only power source that can give a spacecraft or a surface base sustained kilowatts in the dark.

The physics is the reverse of the RTG's scarcity problem. A reactor runs on uranium-235, which — while still special — is vastly more available than plutonium-238, and a few kilograms of it hold enormous energy. The reactor's heat is converted not by feeble thermocouples but by heat engines: Stirling convertors or turbines that reach $\sim25$–$35\%$ efficiency, four to five times an RTG's. The price is complexity: a reactor has moving parts, needs control (neutron-absorbing rods to hold the chain reaction at criticality), and needs radiation shielding to protect the spacecraft and any crew.

The landmark recent demonstration is KRUSTY — the Kilopower Reactor Using Stirling Technology — which NASA and the Department of Energy tested in the Nevada desert in 2018. KRUSTY is a small, elegant machine: a solid cast core of uranium-235 the size of a paper-towel roll, a passive sodium heat-pipe network carrying heat to a ring of Stirling engines, and control by a single reflector. It ran for hours, throttled up and down, survived deliberately induced faults, and produced electricity without a hiccup. The Kilopower design family it validated targets $1\ \text{kW}$ to $10\ \text{kW}$ of electricity from a unit of a few hundred kilograms — the power a crewed surface base or a nuclear-electric tug would need.

Comparing the sources: specific power

The honest way to compare power sources is not watts, but watts per kilogram — specific power — because in spaceflight the mass is what you pay for (theme 4). Roughly, at Earth's distance:

The table shows why no single source wins everywhere. Near the Sun, solar is unbeatable on mass — nothing else comes close per kilogram — which is why every satellite in Earth orbit is solar-powered. In deep space, where solar collapses, the RTG's modest few W/kg is the only option below a kilowatt. And when a mission needs kilowatts in the dark — a Mars surface habitat running through dust storms and nights, or a nuclear-electric propulsion stage feeding power-hungry ion engines (Chapter 20) — only fission scales to meet it. Fission's specific power improves as you build bigger, which is exactly the regime, in Chapter 21, where nuclear-electric propulsion becomes attractive.

🧩 Productive Struggle. A crewed base near the lunar south pole needs $10\ \text{kW}$ of continuous electrical power, through a night that lasts fourteen Earth days. Before reading on, reason about which power source it should use, and roughly estimate why the "obvious" answer — solar panels and batteries, like every Earth satellite — fails badly here. (We work this exact problem in Case Study 2; a two-week night is where battery mass explodes.)

25.6 The power budget

Everything so far — array, battery, RTG, reactor — exists to feed a crowd of hungry subsystems, and the document that keeps generation ahead of demand is the power budget. It is the electrical twin of the mass budget of Chapter 23 and the delta-v budget of Chapter 3: a ledger that must balance, or the mission fails.

Definition (power budget). A power budget is an itemized accounting of the electrical power every subsystem consumes, tabulated for each mission mode and phase, compared against the power the source can supply. A design is power-positive only if, in every mode — including the worst-case combination of loads and the most degraded generation — supply meets or exceeds demand, with margin. The power budget is the master constraint of power-system design.

The essential subtlety is that a spacecraft does not draw one fixed power; it has modes, and the budget must close in all of them. A comsat transmitting at full rate, a telescope slewing to a new target on its reaction wheels, a lander firing its descent engine, a probe hibernating through cruise — each is a different column of loads. The worst case is often not the most active mode but the coldest: in eclipse the payload may switch off, yet the heaters roar on to keep the battery and propellant warm (Chapter 24), and heaters can dominate the eclipse budget.

Here is a representative power budget for our 1 kW-class communications satellite, in watts, for two modes:

  Subsystem              Sunlit (payload on)   Eclipse (cold case)
  -----------------------------------------------------------------
  Payload (transponders)         500                  500
  Attitude control (wheels)        90                   90
  Command & data (computer)        50                   50
  Communications (TT&C)            40                   40
  Thermal (heaters)                60                  180   <- driver
  Propulsion (valves, idle)        15                   15
  -----------------------------------------------------------------
  Subtotal                        755                  875
  Harness/distribution (+5%)       38                   44
  -----------------------------------------------------------------
  Delivered load                  793                  919
  Design margin (+25%)            198                  230
  =================================================================
  BUDGETED LOAD                   991                 1149

Read the budget as a story. The sunlit load is about $1{,}000\ \text{W}$ — the number we fed the array sizer in §25.1. But the eclipse column climbs to $\sim1{,}150\ \text{W}$, because the heaters triple to fight the cold of shadow even as the payload keeps transmitting. The eclipse, not the busy sunlit pass, is the sizing case — so a careful designer loops back and re-sizes the array and battery to the $1{,}150\ \text{W}$ worst case, growing both by ~15%. This feedback — budget sizes the source, source mass changes the spacecraft, changed spacecraft changes the budget — is mission design in miniature, and we formalize the margins in Chapter 29.

Bus voltage: why the harness has a voltage

The power leaves the source and reaches the loads over a bus — the spacecraft's electrical backbone — and the voltage of that bus is a real design choice with a mass consequence.

Definition (bus voltage). The bus voltage is the voltage at which the power system distributes electricity to the spacecraft's loads. The traditional standard is 28 V (inherited from aircraft); high-power spacecraft use higher buses — 50, 70, 100, or 120 V — and the International Space Station distributes primary power at 160 V. Downstream converters step the bus down to the 12 V, 5 V, and 3.3 V that electronics actually want.

Why raise the voltage? Because power is $P = V \times I$, so at a higher voltage the same power flows as a lower current — and current is what makes wires heavy. Resistive loss in the harness is $I^{2}R$, and to keep that loss small at high current you need thick, heavy copper. Consider delivering $2{,}000\ \text{W}$:

• At a $28\ \text{V}$ bus: $I = 2000/28 = 71\ \text{A}$.
• At a $100\ \text{V}$ bus: $I = 2000/100 = 20\ \text{A}$.

The current drops by $3.6\times$, and the resistive loss (going as $I^2$) by nearly $13\times$ — so for the same acceptable loss the wires can be far thinner and lighter. This is precisely why the power-hungry ISS distributes at 160 V and why every high-power satellite has crept upward from the old 28 V standard: at the kilowatts modern spacecraft demand, a low-voltage bus would need a harness of copper bars weighing more than the payload. Bus voltage is one more front in the war on mass (theme 4).

🔄 Check Your Understanding 1. In the budget above, which mode sizes the power system, and why isn't it the "busiest" mode? 2. A cubesat designer keeps a 5 V bus "for simplicity" as the spacecraft grows to draw 400 W. What goes wrong, and what is the fix?

Answers

1. The eclipse mode ($\sim1{,}150\ \text{W}$) sizes it, even though the payload is doing less, because the heaters spike to fight the cold — the worst-case load is a worst-case combination, not the most active subsystem. This is why budgets are built mode by mode. 2. At 5 V, 400 W means $I = 80\ \text{A}$ — enormous current demanding thick, heavy, lossy wiring and huge connectors, with large $I^2R$ losses heating the harness. The fix is to raise the bus voltage (e.g. to 28 V, cutting the current to ~14 A) and use local converters for the low-voltage electronics.

Mission Design Checkpoint: your power budget and power.py

This is a major increment: your Mission Design Review (MDR) gains its power budget, and astrotools gains its power module.

The design. Open your MDR and build a power budget for your chosen mission (Track A comsat, B lunar lander, C Mars orbiter, D asteroid rendezvous). Do it in three steps:

1. List the loads. Tabulate every subsystem's power draw — payload, attitude control, computer, comms, thermal/heaters, propulsion — for each mission mode (at minimum: an active mode and an eclipse/quiet mode). Add ~5% for harness losses and a ~25% design margin (this early, margins are generous). The largest mode total is your budgeted load.
2. Choose a source. If your mission stays where the Sun is strong (Tracks A, B, and C near Mars), size a solar array to your budgeted load using §25.1–§25.2: required array power, then area at end of life. If your mission goes to deep space or must survive long darkness (a Track D far-asteroid, a permanently-shadowed lander), pick an RTG or, for kilowatts, a fission reactor, and note the plutonium-scarcity limit.
3. Size the storage. From your worst eclipse (or the lunar/Mars night), size the battery with §25.3: eclipse energy, a DoD appropriate to your cycle count, and a mass at $150\ \text{W·h/kg}$. Record array area, battery capacity and mass, and chosen bus voltage in the MDR.

The code. Create astrotools/power.py with the sizing tools of this chapter (canonical signatures, kept stable for the capstone):

import math

S_1AU = 1361.0  # solar constant at 1 AU, W/m^2

def solar_flux(au):
    """Solar flux (W/m^2) at a distance of `au` astronomical units."""
    return S_1AU / au**2

def required_array_power(p_day, t_day, p_ecl, t_ecl, x_day=0.85, x_ecl=0.65):
    """Array output power (W) needed to run the loads and recharge for eclipse.
    Times in consistent units; x_day/x_ecl are path efficiencies (SMAD)."""
    return (p_day * t_day / x_day + p_ecl * t_ecl / x_ecl) / t_day

def array_area(p_required, flux=S_1AU, eff=0.30, cos_theta=0.90, life_factor=1.0):
    """Solar-array area (m^2) to produce p_required watts, given cell efficiency,
    sun-angle cosine loss, and an end-of-life degradation factor (<=1)."""
    return p_required / (flux * eff * cos_theta * life_factor)

def battery_capacity_wh(p_ecl, t_ecl_h, dod=0.25, eff=0.90):
    """Battery capacity (W*h) to deliver an eclipse, given depth of discharge
    and discharge-path efficiency."""
    return p_ecl * t_ecl_h / (dod * eff)

if __name__ == "__main__":
    # Track A comsat, 500 km LEO: 1 kW load, 60 min sun, 35 min eclipse, 7-yr life.
    p_sa = required_array_power(1000, 60, 1000, 35)
    life = (1 - 0.0275) ** 7                       # 82.3% at end of life
    area = array_area(p_sa, life_factor=life)
    batt = battery_capacity_wh(1000, 35 / 60)
    print(f"array output needed: {p_sa:.0f} W")
    print(f"array area (EOL-sized): {area:.2f} m^2")
    print(f"battery capacity: {batt:.0f} Wh ({batt/150:.0f} kg)")
    # Expected output:
    # array output needed: 2074 W
    # array area (EOL-sized): 6.86 m^2
    # battery capacity: 2593 Wh (17 kg)

Point these functions at your mission's numbers. The array area and battery mass they return feed straight into the mass budget of Chapter 23 — power is not free real estate, it is kilograms — and into the capstone MDR of Chapter 40.

Summary

Power is generated, stored, distributed, and every watt costs mass. Carry these forward:

Numbers worth memorizing: $S = 1361\ \text{W/m}^2$; space cells ~30%; array ~300 W/m² at the panel; Li-ion ~150 W·h/kg; Pu-238 half-life ~88 yr; RTG ~6–7% efficient.

Spaced Review

Retrieval strengthens memory. Answer from memory before checking, then look back at the cited chapter.

1. (Ch. 24) A spacecraft's equilibrium temperature comes from balancing absorbed sunlight against radiated heat. Write the radiative balance for a blackbody sphere at 1 AU and estimate its temperature. Why is this the same solar flux we used to size the array?
2. (Ch. 24) What does multi-layer insulation (MLI) do, and why is the eclipse the thermal cold case — the same eclipse that drives the battery and heater loads in this chapter?
3. (Ch. 23) The array and battery we sized (~7 m² and ~17 kg) go into the spacecraft's mass budget. What is a mass budget, and why does every subsystem track margin against it?
4. (Ch. 23) Solar arrays must survive the launch loads of Chapter 23 while folded, then deploy. Name one launch load a stowed array must be designed against.

Answers

1. Balance absorbed power $\alpha S \pi R^2$ against radiated power $\varepsilon \sigma T^4 (4\pi R^2)$; for a blackbody ($\alpha = \varepsilon = 1$), $T = \left(\tfrac{S}{4\sigma}\right)^{1/4} = \left(\tfrac{1361}{4\times5.67\times10^{-8}}\right)^{1/4} \approx 278\ \text{K}$ (about $5\,^\circ\text{C}$). It is the same $1361\ \text{W/m}^2$ because the sunlight that warms the spacecraft is the same sunlight the array converts — power and thermal are two accountings of one energy flow. 2. MLI is many thin, reflective layers ("space blankets") that nearly stop radiative heat transfer, drastically cutting heat loss (or gain) so the spacecraft holds its temperature with little heater power. The eclipse is the cold case because the Sun — the dominant heat input — vanishes while the spacecraft keeps radiating, so temperatures fall and heaters must switch on, spiking the eclipse power load. 3. A mass budget is the itemized accounting of every component's mass against the vehicle's mass allocation; margin is tracked because mass only ever grows during design, and the rocket equation (Ch. 3) punishes every overrun exponentially. 4. Any of: quasi-static acceleration (several g at max thrust), violent random vibration and acoustic loading, or shock at stage separation — a stowed array is a large, floppy structure that must not resonate or tear loose during ascent.

What's Next

We have given the spacecraft a beating heart of watts — sunlight caught, energy stored, plutonium glowing in the dark, every load accounted for in a budget that balances. But a spacecraft that cannot talk is a spacecraft that might as well not exist: the data it gathers is worthless until it reaches Earth, and the commands that steer it must somehow cross the void. And the radio that does this talking is one of the hungriest loads on the very power budget we just built — a deep-space transmitter can draw hundreds of watts to push a whisper across billions of kilometers. In Chapter 26 we follow that whisper: the link budget that decides whether a signal is heard, the giant dishes of the Deep Space Network, and how a probe at Pluto sends home a picture on less power than a refrigerator bulb.