Satellite internet · Session 5 of 8
Closing the energy budget
Where does the energy come from, where does it go, and how big must the battery be?
A satellite at 550 km laps the planet every 95.5 minutes, and for part of almost every lap it passes through Earth's shadow. Count what that means: about fifteen sunrises a day, and fifteen nights, each up to 35.6 minutes long. During each of those nights the spacecraft collects nothing at all and still has to keep its radios, computers and heaters alive. Every watt it spends in the dark was collected earlier and stored — and the battery has to be sized before launch, because nobody is coming to change it.
By the end
- Calculate available solar power from the solar constant, panel area, efficiency and the cosine law.
- Size a battery from eclipse duration, load and depth of discharge, keeping units explicit throughout.
- Distinguish power generation, energy storage, communication, attitude control, thermal control and propulsion as separate functions that cannot substitute for one another.
The idea
Above the atmosphere, sunlight arrives at 1,361 W/m² — the solar constant. A high-grade triple-junction space solar cell converts roughly 30% of that to electricity, so about 408 W of electrical power per square metre of panel, facing the Sun squarely. A spacecraft drawing 2 kW therefore needs around 5 m² of panel, which is why these satellites unfold into something far larger than the body that launched.
“Facing squarely” is doing real work in that sentence. Power collected falls with the cosine law, P = P₀ cos θ, where θ is the angle between the panel's normal and the incoming sunlight. At 30° off you keep 86.6%; at 60° you keep half. This is why attitude control — which way the spacecraft points — is a power system concern as much as a pointing concern, and why the panels are on a mechanism that tracks the Sun independently of the antennas.
Then the Sun goes out. The eclipse fraction follows from geometry: the half-angle of Earth's shadow as seen from orbit is arcsin(R / r) = 67.0° at 550 km, so in the worst case the satellite spends 2 × 67.0 ÷ 360 = 37.2% of each orbit in darkness — 2,133 s, or 35.6 minutes, of a 95.5-minute lap.
Now the budget closes. A 2 kW load for 2,133 s needs 2,000 W × 2,133 s = 4.27 MJ = 1,185 Wh of stored energy. But a battery that is fully drained every lap dies fast, so designers limit the depth of discharge — often to about 30% in low orbit — which means installing 1.185 ÷ 0.30 = 3.95 kWh. The battery is sized by cycle life, not by capacity: fifteen laps a day for five years is roughly 27,500 charge–discharge cycles, perhaps thirty times what a phone battery ever sees.
Every watt that does not leave as radio has to leave as heat, and in vacuum there is no air to carry it. The only exit is thermal radiation, governed by the Stefan–Boltzmann law P = εσAT⁴, with σ = 5.67 × 10⁻⁸ W/(m²·K⁴). A 2 m² radiator at 300 K with emissivity 0.85 sheds about 780 W. Because of the fourth power, a radiator that runs slightly hotter sheds dramatically more — and a spacecraft that cannot reject its heat fails just as surely as one that runs out of power.
Propulsion is a separate system again, and it does a job no other part can. At 550 km, drag steadily lowers the orbit, so the satellite must push back; it must also dodge debris and, at end of life, lower itself deliberately. Modern satellites of this class use Hall-effect thrusters, which accelerate ionised krypton or argon electrically. Their thrust is tiny — tens of millinewtons — but their specific impulse of 1,000 to 2,500 s is several times a chemical rocket's, so the same change in velocity costs far less propellant mass.
Where the energy goes
Gather
Reused cardboard, paper panels, tape, labels, protractor and calculator. Optional: a rated educational solar-cell kit with a matching load, a multimeter and a lamp, used strictly to its instructions.
Build and investigate
- Build a cardboard spacecraft and label six subsystems: power generation, energy storage, communications, attitude control, thermal control and propulsion. Draw energy paths in one colour and information paths in another; they are not the same arrows.
- Give each subsystem a sentence of the form “without this, the spacecraft fails when …”. Identify which subsystem carries the load during eclipse, and which one is useless then no matter how large you make it.
- Calculate the panel area your design needs. Choose a load in watts, then use area = load ÷ (1361 × efficiency). Draw the panels to scale on your model and note how much larger they are than the body.
- Size the battery. Compute the eclipse duration as 37.2% of the 5,730 s orbit — work in seconds, not minutes, so the rounding does not bite — then multiply by your load for the energy in watt-hours, then divide by your chosen depth of discharge. Record every unit conversion as a separate line.
- Test the cosine law. If you have a kit, fix lamp distance, load and ambient light, then measure at 0°, 30° and 60°, three readings each, and plot measured power against cos θ. Without a kit, predict all three values from P₀ cos θ and have someone check your arithmetic.
- Revise one subsystem using your numbers — more panel area, a deeper discharge, a different orbit — and state what the change costs elsewhere. Label every value in your table as calculated, measured or assumed.
Run the numbers
Energy balance is bookkeeping: what arrives, what is stored, what is spent, what must be radiated away. Keep watts and watt-hours apart — one is a rate, the other an amount.
| Relation | Expression | What it means |
|---|---|---|
| Solar power | P = S · A · η · cos θ | S = 1361 W/m²; η ≈ 0.30; θ from the panel normal |
| Electrical power | P = V · I | voltage alone is not power |
| Energy | E = P · t | watts × seconds = joules; ÷ 3600 for watt-hours |
| Eclipse fraction | f = arcsin(R / r) ÷ 180° | worst case; 37.2% at 550 km |
| Battery size | E_batt = P · t_eclipse ÷ DoD | DoD = depth of discharge, often ≈ 0.30 |
| Heat rejection | P = εσAT⁴ | σ = 5.67 × 10⁻⁸ W/(m²·K⁴) |
| Rocket equation | Δv = I_sp · g₀ · ln(m₀ / m₁) | g₀ = 9.81 m/s² |
Core · Grade 7–8
- A satellite needs 2,000 W. Using S = 1361 W/m² and η = 0.30, calculate the panel area required with the Sun square on. Then recalculate at θ = 45° and state the extra area needed.
- Eclipse lasts 37.2% of a 5,730 s orbit. Find the eclipse duration in seconds and in minutes, then the energy in watt-hours a 2,000 W load consumes across it.
- Limit depth of discharge to 30%. What capacity must the battery have? Express the answer in kWh, and compare it with a typical laptop battery of about 0.06 kWh.
Deeper · Grade 9–12
- Derive the 37.2% eclipse fraction. The shadow's angular half-width seen from orbital radius r is arcsin(R / r). Evaluate it at r = 6921 km, double it, divide by 360°, and explain why this is a worst case rather than an average — what geometry makes some orbits eclipse-free for weeks?
- A 2 m² radiator with ε = 0.85 must reject 780 W. Using P = εσAT⁴, solve for the operating temperature in kelvin and convert to Celsius. Then find the temperature needed to reject 1,560 W from the same radiator, and explain why it is not twice as hot.
- A Hall thruster has I_sp = 1,500 s against a chemical thruster's 300 s. Using Δv = I_sp·g₀·ln(m₀/m₁), calculate the propellant mass fraction each needs for Δv = 100 m/s. State the mass saved on a 300 kg satellite, and what is given up in exchange.
Measure and record
Record every quantity with its unit and its status. For the cosine test, record panel angle, cos θ, the predicted power, and — if measuring — voltage, current and the product V × I for three trials per angle. Voltage on its own is not a measurement of power.
Worked example
Size the battery for a 2,000 W satellite at 550 km. Eclipse: 0.372 × 5,730 s = 2,133 s, which is 35.6 min. Energy: E = 2,000 W × 2,133 s = 4.27 × 10⁶ J, and dividing by 3,600 gives 1,185 Wh = 1.185 kWh. Now apply the 30% discharge limit: 1.185 ÷ 0.30 = 3.95 kWh installed, more than three times the energy actually used each night. The margin is not timidity — at fifteen laps a day for five years the cell sees about 27,500 cycles, and shallow cycling is what makes that survivable. Check the generation side too: at 1361 W/m² and 30% efficiency a panel yields 408 W/m², so 2,000 W needs 4.9 m² square-on, rising to 6.9 m² if the Sun sits 45° off the normal, since cos 45° = 0.707.
Pause and explain
Your design misses its energy budget by 20% during eclipse. Name three distinct fixes, give the formula each one acts on, and state the cost of each.
Check your reasoning after your attempt
Several are defensible; the reasoning matters more than the choice. Increase battery capacity — acts on E_batt = P·t ÷ DoD, costs mass and launch price. Deepen the allowed discharge from 30% to 40% — acts on the same formula with no extra mass, but shortens cycle life, which is what sets the mission duration. Cut the eclipse load by shutting non-essential subsystems — acts on P, costs capability for part of every lap. Raise the orbit to reduce the eclipse fraction — acts on f = arcsin(R/r) ÷ 180°, but increases slant range, path loss and delay, and lengthens post-failure decay time. What is not defensible is adding panel area, which changes nothing while the spacecraft is in shadow.
Save in your mission notebook
The labelled subsystem diagram with energy and information paths separated; panel area and battery sizing with every unit conversion shown; the cosine-law table with predicted and, if available, measured values; one revision with its cost named.
Parent / teacher notes
Hold the line between watts and watt-hours; nearly every failed budget in this lesson is a unit error rather than a physics error, and insisting on a conversion line each time catches it. Step 2's question — which subsystem is useless in eclipse — is the one that produces the misconception worth having, so ask it before the arithmetic. The kit is genuinely optional: the cosine-law prediction can be checked on paper, and a wrong prediction is as useful as a reading. Grade 7–8 can do the panel-area and battery sizing. Grade 9–12 should do the eclipse-fraction derivation and the Stefan–Boltzmann inversion, which needs a fourth root.
Suggested timing: 10 minutes predict, 15 read and explain, 30 build and investigate, 25 measure and calculate, 10 independent explanation. Adapt the activities rather than rush unfinished reasoning.
Take it further
Model the full lap minute by minute. Collect S·A·η·cos θ in sunlight with θ varying as the satellite rotates, spend a constant load throughout, and track the battery's state of charge. Find the condition under which the battery returns to full before the next eclipse begins — and show that a design which merely survives one eclipse can still fail on the tenth.