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

1. Structure Subsystem

Purpose

The primary structure provides:

Mechanical support for all other subsystems
Load path during launch (high g-forces, vibration)
Attachment to launch vehicle via payload adapter

Key Design Drivers

Formula Structural Mass Fraction

The structure typically consumes 15–25% of total spacecraft mass. We want to minimize this while meeting strength requirements.

Derivation from first principles: Launch acceleration imposes axial load: $F_{\text{axial}} = m_{\text{total}} \cdot a_{\text{launch}}$

For a cylindrical structure with cross-sectional area $A$ and yield strength $\sigma_y$ : $\sigma = \frac{F}{A} \leq \sigma_y$

Solving for minimum area: $A \geq \frac{m_{\text{total}} \cdot a_{\text{launch}}}{\sigma_y}$

Mass of structure: $m_{\text{struct}} = A \cdot L \cdot \rho$

where $L$ is height, $\rho$ is material density.

Structural efficiency (higher is better): $\eta_{\text{struct}} = \frac{\sigma_y}{\rho}$

This is why we use aluminum (6061-T6) or composites (CFRP): high strength-to-weight ratio.

Worked example Example: CubeSat Structure Sizing

Given:

Total mass: $m = 4$ kg (3U CubeSat)
Launch acceleration: $a = 10g = 98$ m/s²
Aluminum6061-T6: $\sigma_y = 240$ MPa, $\rho = 2700$ kg/m³ Find: Minimum wall thickness for a 10 cm × 10 cm square cross-section frame.

Step 1: Calculate axial load. $F = 4 \times 98 = 392 \text{ N}$ Why this step? Launch is the harshest mechanical environment; structure must survive without buckling.

Step 2: Find required cross-sectional area of load-bearing walls. $A_{\text{req}} = \frac{392}{240 \times 10^6} = 1.63 \times 10^{-6} \text{ m}^2 = 1.63 \text{ mm}^2$ Why this step? This is the minimum material area to stay below yield stress.

Step 3: Distribute over 4 walls (each 10 cm long). $A_{\text{per wall}} = \frac{1.63}{4} = 0.41 \text{ mm}^2$ Wall thickness: $t = \frac{0.41}{100} = 0.004$ mm Reality check: This is absurdly thin! Real CubeSats use 1–2 mm walls because:

Vibration modes require stiffness (not just strength)
Handling, thermal cycling, and safety factors multiply the requirement by 10–20×

2. Electrical Power System (EPS)

Function

Generate, store, regulate, and distribute electrical power to all subsystems.

Components

Power Budget

Formula Energy Balance in Orbit

Derivation: A spacecraft in LEO experiences day/night cycles. Must balance energy over one orbit period $T_{\text{orbit}}$ .

Daytime phase (duration $T_{\text{sun}}$ ):

Solar array generates power $P_{\text{SA}}$
Load consumes $P_{\text{load}}$
Excess charges battery: $P_{\text{charge}} = P_{\text{SA}} - P_{\text{load}}$

Eclipse phase (duration $T_{\text{eclipse}} = T_{\text{orbit}} - T_{\text{sun}}$ ):

Battery discharges at $P_{\text{load}}$

Energy balance over full orbit — the energy stored in the battery during daylight (after charging losses) must equal the energy drawn during eclipse (with discharge losses): $E_{\text{charge}} = E_{\text{discharge}}$

where:

$E_{\text{charge}} = P_{\text{charge}} \cdot T_{\text{sun}} \cdot \eta_{\text{charge}}$ = usable energy put into the battery
$E_{\text{discharge}} = \dfrac{P_{\text{load}} \cdot T_{\text{eclipse}}}{\eta_{\text{discharge}}}$ = energy that must be pulled from the battery (including discharge inefficiency)

Setting them equal: $P_{\text{charge}} \cdot T_{\text{sun}} \cdot \eta_{\text{charge}} = \frac{P_{\text{load}} \cdot T_{\text{eclipse}}}{\eta_{\text{discharge}}}$

A common simplification assumes the array is sized so daytime generation only just covers both the load and the recharge, giving the standard sizing form: $P_{\text{SA}} = P_{\text{load}} \left(1 + \frac{T_{\text{eclipse}}}{T_{\text{sun}} \cdot \eta_{\text{charge}} \cdot \eta_{\text{discharge}}}\right)$

Typical Li-ion efficiencies: $\eta_{\text{charge}} = 0.9$ , $\eta_{\text{discharge}} = 0.95$

Worked example Example: Power Sizing for LEO Satellite

Given:

Orbit: 500 km altitude, $T_{\text{orbit}} = 95$ min
Eclipse fraction: 35% → $T_{\text{eclipse}} = 33$ min, $T_{\text{sun}} = 62$ min
Average load: $P_{\text{load}} = 50$ W
Battery efficiency: round-trip 85% ( $\eta_{\text{charge}} \times \eta_{\text{discharge}} = 0.85$ ) Find: Required solar array power.

Step 1: Calculate charge/discharge ratio. $\frac{T_{\text{eclipse}}}{T_{\text{sun}}} = \frac{33}{62} = 0.532$ Why this step? The array must generate enough surplus during sunlight to cover the eclipse deficit.

Step 2: Apply energy balance. $P_{\text{SA}} = 50 \left(1 + \frac{0.532}{0.85}\right) = 50 \times 1.626 = 81.3 \text{ W}$ Why this step? This accounts for battery losses and the unequal day/night durations.

Step 3: Add margin for degradation. Solar cells degrade ~2.5% per year in LEO (radiation damage). For 5-year mission: $P_{\text{SA,BOL}} = \frac{81.3}{(1 - 0.025 \times 5)} = \frac{81.3}{0.875} = 92.9 \text{ W}$ BOL = Beginning of Life

3. Thermal Control System (TCS)

Purpose

Maintain all components within operational temperature limits (typically −20°C to +50°C for electronics).

Heat Sources & Sinks

Formula Thermal Balance Equation

Derivation: Spacecraft in space has no convection. Only radiation and internal dissipation matter.

Energy balance at steady state: $Q_{\text{in}} = Q_{\text{out}}$

Heat inputs:

Solar flux: $Q_{\text{solar}} = \alpha \cdot A_{\text{proj}} \cdot S$ $Q_{solar} = α \cdot A_{proj} \cdot S$
- $\alpha$ = absorptivity (0–1)
- $A_{\text{proj}}$ = projected area facing Sun
- $S$ = solar constant (1361 W/m² at Earth)
Albedo (reflected sunlight from Earth): $Q_{\text{albedo}} = \alpha \cdot A \cdot a \cdot S$ $Q_{albedo} = α \cdot A \cdot a \cdot S$
- $a$ = Earth's albedo ≈ 0.3
Earth IR emission: $Q_{\text{Earth}} = \epsilon \cdot A \cdot E_{\text{Earth}}$ $Q_{Earth} = ϵ \cdot A \cdot E_{Earth}$
- $E_{\text{Earth}}$ ≈ 237 W/m²
Internal dissipation: $Q_{\text{dissip}}$ (waste heat from electronics)

Heat outputs: $Q_{\text{out}} = \epsilon \cdot \sigma \cdot A \cdot T^4$ where:

$\epsilon$ emissivity (0–1)
$\sigma$ = Stefan-Boltzmann constant = $5.67 \times 10^{-8}$ W/(m²·K⁴)
$T$ = spacecraft temperature (Kelvin)

Key insight: By choosing surface coatings, we control $\alpha$ (how much heat we absorb) and $\epsilon$ (how much we radiate). Want low $\alpha/\epsilon$ ratio for hot environments, high ratio for cold.

Worked example Example: Thermal Coating Selection

Scenario: CubeSat in LEO must keep electronics at 20°C. Internal dissipation = 5 W. Exposed area = 0.03 m².

Option A: Black paint ( $\alpha = 0.95$ , $\epsilon = 0.85$ ) Option B: White paint ( $\alpha = 0.25$ , $\epsilon = 0.90$ )

Step 1: Calculate solar input (sun-facing side, 0.01 m² projected). Option A: $Q_{\text{solar}} = 0.95 \times 0.01 \times 1361 = 12.9$ W Option B: $Q_{\text{solar}} = 0.25 \times 0.01 \times 1361 = 3.4$ W Why this step? High absorptivity means more solar heating.

Step 2: Total heat to reject. Option A: $Q_{\text{total}} = 12.9 + 5 = 17.9$ W Option B: $Q_{\text{total}} = 3.4 + 5 = 8.4$ W

Step 3: Required equilibrium temperature. $Q_{\text{out}} = \epsilon \sigma A T^4 = Q_{\text{total}}$ Option A: $T = \left(\frac{17.9}{0.85 \times 5.67 \times 10^{-8} \times 0.03}\right)^{1/4} = 330$ K = 57°C ❌ Too hot! Option B: $T = \left(\frac{8.4}{0.90 \times 5.67 \times 10^{-8} \times 0.03}\right)^{1/4} = 281$ K = 8°C ✓ Conclusion: White paint keeps the spacecraft cooler because low $\alpha/\epsilon$ ratio favors heat rejection.

Passive vs Active Control

Passive: Coatings, multi-layer insulation (MLI), heat pipes
Active: Heaters, louvers, fluid loops (for large spacecraft)

4. Attitude Determination & Control System (ADCS)

Purpose

Know where the spacecraft is pointing (determination) and change/maintain that orientation (control).

Components

Physics of Reaction Wheels

Formula Angular Momentum Exchange

Derivation: Conservation of angular momentum in an isolated system.

Total angular momentum: $\vec{L}_{\text{total}} = \vec{L}_{\text{spacecraft}} + \vec{L}_{\text{wheel}}$

In space (no external torques): $\frac{d\vec{L}_{\text{total}}}{dt} = 0$

Therefore: $\frac{d\vec{L}_{\text{spacecraft}}}{dt} = -\frac{d\vec{L}_{\text{wheel}}}{dt}$

For spacecraft: $I_{\text{sc}} \frac{d\vec{\omega}_{\text{sc}}}{dt} = -I_{\text{wheel}} \frac{d\vec{\omega}_{\text{wheel}}}{dt}$

To rotate spacecraft by angle $\theta$ about an axis:

Spin up wheel in opposite direction
Spacecraft rotates to conserve momentum
Wheel stores angular momentum $L = I_{\text{wheel}} \omega_{\text{wheel}}$

Wheel saturation: Eventually $\omega_{\text{wheel}}$ reaches max speed. Must desaturate using magnetorquers or thrusters to dump momentum.

Worked example Example: Slew Maneuver Calculation

Given:

Spacecraft moment of inertia: $I_{\text{sc}} = 10$ kg·m²
Reaction wheel: $I_{\text{wheel}} = 0.01$ kg·m², max speed = 6000 RPM
Desired rotation: 90° in 60 seconds Find: Required wheel acceleration and check if it saturates.

Step 1: Calculate required spacecraft angular acceleration. Using $\theta = \frac{1}{2} \alpha t^2$ for constant acceleration over first half, then deceleration: $\alpha_{\text{sc}} = \frac{4\theta}{t^2} = \frac{4 \times (\pi/2)}{60^2} = 1.75 \times 10^{-3} \text{ rad/s}^2$ Why this step? We need to accelerate to mid-point, then decelerate symmetrically.

Step 2: Find required wheel acceleration (opposite direction). $I_{\text{sc}} \alpha_{\text{sc}} = I_{\text{wheel}} \alpha$ $\alpha_{\text{wheel}} = \frac{10 \times 1.75 \times 10^{-3}}{0.01} = 1.75 \text{ rad/s}^2$ Why this step? Conservation of angular momentum dictates this ratio.

Step 3: Check saturation. Maximum wheel speed during maneuver (at $t = 30$ s): $\omega_{\text{wheel,max}} = \alpha_{\text{wheel}} \times 30 = 52.5 \text{ rad/s} = 501 \text{ RPM}$ This is only 8.4% of max speed (6000 RPM) → ✓ No saturation

5. Command & Data Handling (C&DH)

Function

The onboard computer that:

Receives commands from ground (via comms subsystem)
Executes command sequences (autonomy)
Collects telemetry from all subsystems
Stores science/housekeeping data
Performs onboard processing (compression, fault detection)

Architecture

Why radiation-hardening? Space radiation (cosmic rays, solar protons) causes single-event upsets (bit flips). Rad-hard chips use larger transistors, error-correcting memory, and triple-modular redundancy.

6. Communications Subsystem

Purpose

Radio link to ground stations for:

Uplink: Commands (low data rate, high reliability)
Downlink: Telemetry and science data (high data rate)

Link Budget

Formula Friis Transmission Equation

Derivation: How much power arrives at the receiver?

Transmitter: Radiates power $P_t$ (watts) through antenna with gain $G_t$ (dimensionless).

Free-space spreading: Signal spreads over sphere of area $4\pi d^2$ , where $d$ is distance.

Power density at receiver: $\text{Power density} = \frac{P_t G_t}{4\pi d^2}$

Receiver antenna: Collects power proportional to its effective area $A_{\text{eff}}$ : $A_{\text{eff}} = \frac{\lambda^2 G_r}{4\pi}$ where $G_r$ is receiver antenna gain, $\lambda$ is wavelength.

Received power: $P_r = \frac{P_t G_t}{4\pi d^2} \cdot \frac{\lambda^2 G_r}{4\pi}$

Simplifying: $P_r = P_t G_t G_r \left(\frac{\lambda}{4\pi d}\right)^2$

In decibels (dBW = $10\log_{10}(P/1\text{W})$ ): $P_r[\text{dBW}] = P_t[\text{dBW}] + G_t[\text{dBi}] + G_r[\text{dBi}] - L_{\text{path}}[\text{dB}]$

where path loss: $L_{\text{path}} = 20\log_{10}\left(\frac{4\pi d}{\lambda}\right) = 20\log_{10}(d) + 20\log_{10}(f) - 147.55$ ( $d$ in meters, $f$ in Hz)

Worked example Example: Downlink Budget for CubeSat

Given:

Range to ground station $d \approx 2000$ km (max slant range from 500 km orbit)
Frequency: 437 MHz (UHF) → $\lambda = 0.686$ m
Transmit power: $P_t = 2$ W = 3 dBW
Spacecraft antenna gain: $G_t = 2$ dBi (dipole)
Ground station antenna gain: $G_r = 15$ dBi (Yagi array)
System losses: $L_{\text{sys}} = 3$ dB
Receiver noise temperature: $T_{\text{sys}} = 200$ K

Find: Data rate achievable with bit-error-rate (BER) = 10⁻⁶.

Step 1: Calculate path loss. $L_{\text{path}} = 20\log_{10}(2 \times 10^6) + 20\log_{10}(437 \times 10^6) - 147.55$ $= 126.0 + 172.8 - 147.55 = 151.3 \text{ dB}$ Why this step? This is the dominant loss term—radio waves spread out in 3D space.

Step 2: Received power. $P_r = 3 + 2 + 15 - 151.3 - 3 = -134.3 \text{ dBW}$ Why this step? This is the signal power arriving at the ground station.

Step 3: Receiver noise spectral density. $N_0 = k T_{\text{sys}} = 1.38 \times 10^{-23} \times 200 = 2.76 \times 10^{-21} \text{ W/Hz} = -205.6 \text{ dBW/Hz}$ where $k = 1.38 \times 10^{-23}$ J

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, is note ka core idea bahut simple hai — ek spacecraft ko tum apne body ki tarah samjho. Jo actual kaam karta hai (camera, telescope, ya science instrument) woh hai payload, matlab dimaag. Baaki sab kuch jo us dimaag ko zinda rakhta hai, sahi direction mein point karwata hai, aur Earth se baat karwata hai — woh hai "bus". Bus ke andar 7 subsystems hote hain: structure (haddi ka dhaancha), power (bijli), thermal (temperature control), ADCS (attitude yaani orientation control), C&DH (command aur data handling), communications, aur propulsion. Bina bus ke tumhara mehenga payload sirf ek useless space junk ban jaata hai.

Ab why-it-matters samjho. Jab rocket launch hota hai, tab bahut zyada g-forces aur vibration lagti hai, isliye structure ko strong hona zaroori hai. Note mein jo formula diya hai — $A \geq \frac{m \cdot a}{\sigma_y}$ — woh basically keh raha hai ki structure ka cross-section area itna hona chahiye ki launch ke load mein material yield (toot) na jaaye. Aur isiliye hum aluminium ya carbon-fibre composites use karte hain, kyunki inka strength-to-weight ratio ( $\sigma_y / \rho$ ) high hota hai — matlab kam weight mein zyada strength. Space mission mein har extra kilogram ka matlab lakhon rupaye ka launch cost, isliye engineers minimum mass mein maximum strength dhoondte hain.

Ek important lesson jo example se milta hai — jab CubeSat ka wall thickness nikaala toh answer aaya 0.004 mm, jo practically bilkul absurd hai. Real life mein 1-2 mm walls lagti hain. Kyun? Kyunki sirf strength hi kaafi nahi — vibration modes ke liye stiffness chahiye, handling aur thermal cycling jhelni padti hai, aur safety factors se requirement 10-20 guna badh jaati hai. Toh yaad rakho, textbook ka pure calculation aur real engineering mein farak hota hai. Aakhir mein power system (EPS) bhi important hai — orbit mein day-night cycle hota hai, toh solar panels se din mein bijli banti hai aur battery charge hoti hai, taaki eclipse (raat) mein bhi spacecraft chalta rahe. Ye poora energy balance ka game hai.

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

What Is the Spacecraft Bus?

1. Structure Subsystem

Purpose

Key Design Drivers

2. Electrical Power System (EPS)

Function

Components

Power Budget

3. Thermal Control System (TCS)

Purpose

Heat Sources & Sinks

Passive vs Active Control

4. Attitude Determination & Control System (ADCS)

Purpose

Components

Physics of Reaction Wheels

5. Command & Data Handling (C&DH)

Function

Architecture

6. Communications Subsystem

Purpose

Link Budget

Concept Map

Hinglish (regional understanding)

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