Discussions of space data centers often produce two opposing claims: "space is cold, so cooling is easy" and "heat has nowhere to go in a vacuum, so a data center is impossible." Neither is correct. Although a vacuum rules out convection to the surrounding air, heat can still escape as infrared radiation. A facility that already needs thousands of square meters of solar panels may also be able to accommodate radiators (heat-rejection panels) covering roughly a thousand square meters—at least in terms of area. This article considers a 1 MW-scale solar-powered data center operating at roughly Earth's distance from the Sun and estimates the radiator area it would require.
1. Radiator area and mass on the ISS
The International Space Station (ISS) provides a useful comparison: it operates large solar arrays alongside radiators. Its original solar arrays cover about 2,247 m²[1].
By June 2023, six roll-out iROSA arrays had been installed to supplement the eight original main arrays[9]. The 2,247 m² figure is the reference for the original arrays; it excludes these additions.
To compare the size of these structures, I use projected area: the area of a panel seen face-on. Heat rejection depends on the surface area available to emit radiation. If both faces are fully available, this radiating surface area is twice the projected area.
The US segment has six main radiators and four radiators for its power equipment. NASA's dimensions and ratings give a combined projected area of about 645 m² and a nominal maximum heat-rejection capacity of about 126 kW[2]. These large structures occupy less area than the original solar arrays.
The main Heat Rejection Subsystem (HRS) has six radiator assemblies, each measuring 23.3 m × 3.4 m. Together, they provide about 475 m² of projected area and a nominal maximum capacity of 70 kW. Each assembly contains eight panels and can be replaced in orbit as an Orbital Replacement Unit (ORU). The four photovoltaic radiators (PVRs) cool power-system equipment; each measures 3.12 m × 13.6 m and is rated for up to 14 kW[2].
Adding both groups gives about 645 m² and 126 kW. NASA's 2025 MARVL presentation lists roughly 1,300 m² of radiating surface, approximately twice this projected area[10].
Mass matters as well as area. NASA's nominal figures give about 9.7 t for the radiator assemblies[2] and at least 8.7 t for the eight original Solar Array Wings[3]. Both sets of structures weigh several metric tons before adding the rest of the cooling and power systems.
The 9.7 t total covers the six HRS units and four PVRs. It excludes pumps, tanks, rotary joints, heat exchangers, and external plumbing[2].
NASA's STS-119 press kit gives the mass of each original Solar Array Wing as more than 2,400 lb (about 1.09 t). Eight wings therefore weigh at least \(2{,}400\times8\ \mathrm{lb}\simeq8.7\ \mathrm{t}\), or at least about 3.9 kg/m² over the 2,247 m² array area[3]. This excludes the supporting truss, rotary joints, power-conversion equipment, and added iROSA arrays.
Figure 1: Solar panels and main radiators of the ISS
Orange arrows mark the solar panels; blue outlines mark the six main radiators. The 645 m² total also includes four power-system radiators outside the outlines. Annotations have been added to a NASA image by Thomas Pesquet / ESA.
The original photograph is NASA ID "jsc2021e064215_alt," taken on November 8, 2021[11]. It is used for informational purposes in accordance with NASA's media usage guidelines[12].
In a vacuum, no surrounding air carries heat away by convection, but coolant can still transport heat from the computers to a radiator. Heat then flows through the radiator material by conduction and leaves the surface as infrared radiation[4].
For a uniform surface temperature, the Stefan–Boltzmann law gives the emitted power:
\[
P_\mathrm{emit}=\varepsilon\sigma A_\mathrm{emit}T^4,
\]
where \(A_\mathrm{emit}\) is the radiating surface area, and infrared emissivity \(\varepsilon\) measures how readily the surface emits. Temperature \(T\) is in kelvin (Celsius plus 273.15), and \(\sigma\simeq5.67\times10^{-8}\ \mathrm{W/(m^2\,K^4)}\). With area and emissivity fixed, emitted power rises with the fourth power of absolute temperature, so a warmer surface emits more power.
Net cooling capacity is emitted power minus absorbed power. A surface facing only deep space absorbs very little. Sunlight and radiation from Earth or nearby structures add heat that the radiator must also shed, leaving less capacity to cool the equipment.
If the surroundings can be represented by a uniform radiation temperature \(T_\mathrm{sink}\), net heat rejection is
\[
P=\varepsilon\sigma A_\mathrm{emit}\left(T^4-T_\mathrm{sink}^4\right).
\]
For a radiator around 0–100 °C facing only deep space, the incoming background radiation is negligible. In low Earth orbit, radiation from the Sun, Earth, and surrounding structures must also be included in the heat balance.
Even an object with no internal heat source can become warm in sunlight. Consider a flat plate at Earth's distance from the Sun, facing it directly and absorbing all incident sunlight. With uniform temperature across both faces, infrared emissivity of 1, and no other heat input, emission balances absorption at an equilibrium temperature of about 58 °C.
NASA gives a solar irradiance of \(S=1361.6\ \mathrm{W/m^2}\) at Earth's distance, based on measurements during the 2019 solar minimum[6]. For an object heated only by sunlight,
\[
\alpha S A_\mathrm{sun}=\varepsilon\sigma A_\mathrm{emit}T^4,
\]
where \(\alpha\) is the fraction of sunlight absorbed, \(A_\mathrm{sun}\) is the projected area as seen from the Sun, and \(A_\mathrm{emit}\) is the radiating surface area. Assuming uniform temperature and infrared emissivity of 1, changing the shape or reflectivity gives the following temperatures:
Object
Solar absorptivity
Sun-facing projected area : radiating surface area
Equilibrium temperature
A flat plate facing the Sun, radiating from both faces
1
1 : 2
about 58 °C
A sphere
1
1 : 4
about 5 °C
A sphere reflecting 30% of sunlight
0.70
1 : 4
about −18.5 °C
A sphere intercepts sunlight over \(\pi r^2\) but radiates over \(4\pi r^2\). Geometry and reflectivity alone produce a temperature difference of more than 70 °C between these cases. The last case also approximates Earth's radiation balance; the resulting effective radiating temperature is distinct from Earth's surface temperature.
Solar panels convert part of the energy they absorb into electricity. Computers turn nearly all that electricity into heat, so a 1 MW load needs about 1 MW of cooling. The panels radiate the remaining absorbed energy from their own surfaces.
At steady state, the panel's energy balance is
\[
P_\mathrm{solar,absorbed}=P_\mathrm{electric,out}+P_\mathrm{panel,thermal}.
\]
When no electrical power is drawn, the panel must radiate all the absorbed energy, so it settles at a higher temperature. NASA thermal analyses describe this difference between supplying electrical power and operating with no electrical output[13]. Excessive panel temperatures reduce performance and service life[5].
For the same electrical power draw, a computer doing useful work produces essentially as much heat as a resistor.
If computation stops and no power is drawn from the panels, the computers produce less heat and the panels radiate more. Drawing electrical power from the panels changes where the absorbed solar energy becomes heat; the energy ultimately leaves the facility as radiation in either case.
For the estimates below, I assume a continuous 1 MW electrical load covering computers, power converters, communications, and pumps, with 1 MW of heat rejected by the data-center radiators. If the computers alone use 1 MW, supporting equipment adds to both electricity demand and heat output. Section 4 accounts for energy-storage losses separately.
4. Solar-panel area for a continuous 1 MW load
Space-grade multijunction solar cells use several semiconductor layers to capture a wider range of wavelengths. NASA's 2026 review gives a nominal efficiency of about 30%, with some cells reaching roughly 34%[5]. For this estimate, I assume the entire panel operates at 30% efficiency and faces the Sun directly. The incident sunlight supplies 1,361.6 W/m²[6], giving an electrical output of about 408 W/m².
Supplying 1 MW in continuous sunlight requires
\[
A_\mathrm{PV}=\frac{1{,}000{,}000\ \mathrm{W}}{0.30\times1361.6\ \mathrm{W/m^2}}\simeq2{,}450\ \mathrm{m^2}.
\]
This estimate omits gaps between cells, wiring losses, temperature effects, and aging.
The panels must also charge storage if the orbit crosses Earth's shadow. Consider a 90-minute orbit with 55 minutes of sunlight and 35 minutes of eclipse. If storage returns 90% of its charging energy, powering the load and charging storage during sunlight requires
\[
P_\mathrm{sun}=1\ \mathrm{MW}\times\left(1+\frac{35}{0.90\times55}\right)\simeq1.71\ \mathrm{MW}.
\]
Panel area increases by the same factor, to about 4,180 m².
Storage losses produce about 43 kW of heat averaged over the orbit. Adding this to the 1 MW estimate in Section 6 increases radiator area by about 4.3% at the same temperature and environmental heat input.
The load consumes \(1\ \mathrm{MW}\times35/60\ \mathrm{h}\simeq0.583\ \mathrm{MWh}\) during the eclipse. Storage requires \(0.583/0.90\simeq0.648\ \mathrm{MWh}\) to supply that energy, losing about 0.065 MWh per orbit. Averaged over 90 minutes, the heat output is
\[
P_\mathrm{loss}=1\ \mathrm{MW}\times\frac{35}{90}\left(\frac{1}{0.90}-1\right)\simeq43\ \mathrm{kW}.
\]
A thermal design must also accommodate changes in heat output during charging and discharging.
The two examples require about 2,450 m² and 4,180 m² of solar panels. The area for a particular orbit depends on its eclipse duration[4], and practical arrays need additional capacity to cover losses and degradation.
5. Coolant and radiator surface temperatures
The radiation calculation uses the radiator's surface temperature. The coolant must be warmer to transfer heat to that surface. In NASA's ISS reference case, the average coolant temperature is about 2.4 °C, while the average panel temperature is about 0.8 °C[7].
In the reference case, ammonia enters the radiator at 283.3 K (about 10.2 °C) and leaves at 267.7 K (about −5.5 °C). The arithmetic mean of these temperatures is 275.5 K (about 2.4 °C); the study reports an average panel surface temperature of 273.9 K (about 0.8 °C). Each assembly rejects about 11.68 kW, giving six assemblies a total of about 70 kW[7].
The −40 °C value in NASA's ATCS overview is a target outlet temperature used to control radiator orientation. The roughly 2.8 °C supply temperature comes from mixing cold radiator outlet flow with warm bypass flow before sending it to the heat exchangers. Neither describes the radiator surface as a whole[2].
Some data-center racks accept coolant at an inlet temperature of 45 °C[8]. The radiator temperature depends on how much the coolant warms in the rack and the temperature drop between coolant and surface.
NVIDIA's March 2026 description of Vera Rubin-generation MGX racks specifies a coolant inlet temperature of 45 °C[8]. This temperature applies before the coolant absorbs heat from the rack's components.
In Section 6, I assume radiating temperatures of 40 °C and 60 °C to compare the required areas. Both require a cooling system that supplies heat at those temperatures; a 60 °C surface needs coolant hotter than 60 °C.
6. Radiator area for 1 MW of heat
6.1. Temperature and environmental heating
With infrared emissivity 0.90, the Stefan–Boltzmann law gives about 491 W per square meter of radiating surface at 40 °C and 629 W at 60 °C. Raising the temperature by 20 °C increases emitted power by about 28%.
The power absorbed from the surroundings reduces the net cooling capacity. A surface that reflects sunlight and readily emits infrared helps keep this heat input low. Even so, a surface absorbing 15% of sunlight takes in about 204 W/m² when facing the Sun, compared with 491 W/m² emitted at 40 °C. Radiator orientation must also limit exposure to sunlight and radiation from Earth and nearby structures.
AZ-93, a white thermal-control paint used on spacecraft, has a nominal solar absorptivity of 0.15 and infrared emissivity of 0.91[14]. Sunlight and thermal infrared occupy different wavelength bands, allowing these properties to coexist. Facing the Sun directly, the surface absorbs \(0.15\times1361.6\simeq204\ \mathrm{W/m^2}\).
NASA Ames' design standard gives a direct solar flux of 1,322–1,414 W/m² for Earth-orbit designs and a global annual-average terrestrial infrared flux of 234 ± 7 W/m²[15]. The power absorbed by a radiator also depends on its orientation, how much Earth fills its view, and its surface properties. Sunlight reflected by Earth and clouds adds another contribution.
I assume both faces radiate under the same conditions, with 80% of each face active and the rest occupied by supports and other inactive parts. This active area fraction and an environmental heat input of about 148 W/m² per face are based on the ISS reference design[7]. The heat input is per square meter of the full face, including inactive parts, and stays fixed across temperatures.
NASA's 1996 design study lists 129.8 m² of radiating surface per HRS assembly, about 82% of the 158.4 m² two-sided area calculated from its external dimensions. I round that fraction to 80%. The study represents the surroundings with an equivalent sink temperature of 245.4 K (about −27.8 °C)[7]. With emissivity 0.90, this gives an environmental heat input per square meter of the full face of
\[
q_\mathrm{env}=0.80\times0.90\times\sigma\times(245.4\ \mathrm{K})^4\simeq148\ \mathrm{W/m^2}.
\]
I use this value derived from the ISS study as a common environmental condition for comparing temperatures. The environmental heat input for a specific data center would depend on its orbit and layout.
For a surface whose temperature varies, the effective radiating temperature is the uniform temperature that would produce the same total emission. At 40 °C, multiply the emitted power by the active area fraction, subtract the absorbed power, and add the net heat rejection from both faces:
\[
2\times(0.80\times491-148)\simeq489\ \mathrm{W/m^2}.
\]
The result is the net cooling power per square meter of projected area. Dividing 1 MW by this value gives about 2,040 m². At 60 °C, the same calculation gives about 1,410 m².
For uniform emissivity, the effective radiating temperature is defined by
\[
T_\mathrm{eff}^4=\frac{1}{A_\mathrm{active}}\int_{A_\mathrm{active}}T(\mathbf{x})^4\,\mathrm{d}A,
\]
where \(A_\mathrm{active}\) is the active radiating area and \(T(\mathbf{x})\) is local surface temperature in kelvin. Averaging the fourth power makes this different from a simple mean surface temperature.
For two faces with the same temperature, active area fraction \(f_A\), and environmental heat input \(q_\mathrm{env}\), net rejection per square meter of projected area is
\[
q_\mathrm{net}=2\left(f_A\varepsilon\sigma T_\mathrm{eff}^4-q_\mathrm{env}\right).
\]
The active area fraction multiplies only the emission term because \(q_\mathrm{env}\) already includes that fraction and is expressed per square meter of the full face. Dividing the heat load by \(q_\mathrm{net}\) gives the required projected area.
The table includes a 40 °C case without environmental heating for comparison. All three cases use two radiating faces, emissivity 0.90, and 80% active area.
Effective radiating temperature
Heat input per m² of each face
Net cooling per m² of projected area
Projected area for 1 MW
40 °C
0 W/m²
about 785 W/m²
about 1,270 m²
40 °C
about 148 W/m²
about 489 W/m²
about 2,040 m²
60 °C
about 148 W/m²
about 710 W/m²
about 1,410 m²
At 40 °C, including environmental heating increases the required area by about 60%. Raising the temperature to 60 °C with that heat input held fixed reduces the area by about 30%.
Using only one face doubles the projected area if its conditions stay the same, giving about 4,090 m² for the 40 °C case with environmental heating. If the faces experience different conditions, calculate their net heat rejection separately and add the results.
The ISS main radiators have a nominal maximum capacity of 70 kW over about 475 m², or 147 W/m² of projected area. Scaling the area in proportion to the heat load gives about 6,790 m² for 1 MW. Their lower operating temperatures require more area than the 40–60 °C cases.
NASA's 1996 study (Section 3.1.4, printed page 57) gives an average panel temperature of \(T_\mathrm{panel}=273.9\ \mathrm{K}\), fin efficiency \(\eta_\mathrm{fin}=0.88\), and equivalent sink temperature \(T_\mathrm{sink}=245.4\ \mathrm{K}\)[7]. Fin efficiency accounts for the temperature drop as heat travels through the fins. The following calculation closely reproduces the tabulated 11.68 kW per assembly:
\[
P_\mathrm{ORU}=129.8\ \mathrm{m^2}\times0.90\sigma\eta_\mathrm{fin}\left(T_\mathrm{panel}^4-T_\mathrm{sink}^4\right)\simeq11.67\ \mathrm{kW}.
\]
To compare this model with the area estimate above, I incorporate fin efficiency into an equivalent temperature:
\[
T_\mathrm{eq}=\left[T_\mathrm{sink}^4+\eta_\mathrm{fin}\left(T_\mathrm{panel}^4-T_\mathrm{sink}^4\right)\right]^{1/4}\simeq270.9\ \mathrm{K}\quad(-2.2\ ^\circ\mathrm{C}).
\]
This equivalent temperature incorporates fin losses into the model; it differs from both the reported mean panel temperature and the effective radiating temperature defined from a surface temperature distribution. Using 80% active area gives about 144 W/m² across both faces, or about 6,950 m² for 1 MW. The roughly 2% difference from scaling nominal capacity mainly comes from rounding the active area fraction from about 82% to 80%.
Using the mean coolant temperature of 2.4 °C as the surface temperature and omitting inactive area, temperature drops, fin losses, and environmental heating gives about 588 W/m² across both faces. That is four times the nominal 147 W/m². Using coolant temperature in the ideal radiation equation therefore substantially underestimates the area required by this hardware.
Powering a continuous 1 MW load with sunlight requires thousands of square meters of solar panels. In this estimate, assuming effective radiating temperatures of 40–60 ℃, an active area fraction of 80%, and environmental heat input of about 148 W/m² per face derived from the ISS benchmark, two-sided radiators require a projected area of about 1,400–2,050 m². Both systems are large structures with areas on the order of a thousand square meters; heat rejection does not require equipment vastly larger than the power-generation system.
Keeping radiators compact requires warm surfaces and low environmental heating. A working design must also account for orbit, orientation, surface degradation, and failures. Cooling-system mass, deployment, maintenance, and resistance to micrometeoroids and orbital debris remain challenges for long-term operation beyond this area estimate.
