With the recent success of Starship’s 5th and 6th test flights, mining Bitcoin in space and data centers in space have gotten renewed attention. But every time this is discussed, people inevitably raise the objections that “space isn’t cold”, “keeping things cool in space is almost impossible”, and “latency is too high”.

These people are wrong. They’ve read too much over-simplified “pop science”, and haven’t done the physics for themselves.

So we’re going to do that physics now. We’re going to understand the thermal behavior of a solar power Bitcoin miner, what heats it up, and what cools it down, as well as the impact of latency on stale rate. I’m going to assume you’ve done enough high-school-level physics to understand what units are and some basic math. You don’t have to remember it all — I’ll link to Wikipedia where appropriate so you can refresh your memory.

What we’ll find out is that there’s nothing fundamentally stopping space mining from being possible. That’s not to say it’s going to actually happen — launch costs need to be reduced. But if launch costs can in fact be reduced enough, the economics of it might make sense.

How Cold is Space Anyway?

On average, very cold!

What do we mean by “cold”? Well, there’s a bunch of definitions of temperature that get used in various contexts. But for our purposes, we’re going to use the simple definition most relevant to engineering: if we put a non-heat-producing object in space and leave it there, what temperature will it eventually reach?

The equilibrium temperature is reached when the non-heat-producing object is losing heat as fast as it is absorbing heat. Since space is a vacuum, we know that the only non-trivial mechanism by which our object must be losing heat is thermal radiation. Equally, the only way our object can gain heat is also radiation, such as the Cosmic Microwave Background, and light from stars and other astronomical objects.

The Cosmic Microwave Background is equivalent to a black body radiator at \(2.7K\), just slightly above absolute zero. As for starlight, just look at a night sky: it’s black! Space is incredibly empty, so while individual stars are very hot, the total energy you receive from them must be tiny.

For our purposes — engineering a Bitcoin miner — exactly how tiny doesn’t matter much. We’re quite confident that the number is essentially zero when compared to the energy from sunlight. For example, Pluto’s surface gets as cold as \(-240^{\circ}C\), or \(33K\), even though Pluto is heated by the sun, as well as internal radioactive decay, and the remaining heat of formation¹.

Suppose that Pluto is a perfect black-body radiator with emissivity, \(\epsilon = 1\), in thermal equilibrium with the radiation from interstellar space. That would mean that the radiant exitance \(M\) of the surface would be equal to the flux density \(E\) of the night sky. That is, rate at which energy is leaving the surface, \(M\), is equal to the rate at which energy is being absorbed, \(E\).

While we know that this is not true — Pluto is heated by other sources, like the sun, internal heat energy from formation, radioactive decay, etc. — we can use the known temperature at the surface with the Stefan–Boltzmann law to calculate an upper bound on what the flux density of the night sky could be:

\[\begin{align} E = M &= \epsilon \sigma T^4 \\ M &= 5.67 \times 10^{-8} \frac{W}{m^2 K^4} (33K)^4 \\ M &= 0.067 \frac{W}{m^2} \end{align}\]

In comparison to the flux density of sunlight at Earth’s orbit, \(1380 \frac{W}{m^2}\), the night sky is pretty much zero.

Space is very cold!

Why Do People Say Space is Hot?

Mostly because space is empty. Here on Earth you can easily get rid of large amounts of waste heat by transferring it to the atmosphere. But in space, the only way to expel waste heat with a closed system is to expel it via thermal radiation.

Secondly, if you are in space at a similar distance to the sun as Earth is, sunlight is significant: \(1380\frac{W}{m^2}\). Without an atmosphere to dump heat to, this makes spacesuit design tricky.

A typical adult male has a resting metabolic rate of about \(P = 100W\), and a body surface area of about \(A = 2m^2\). In daylight, about half that surface area would be exposed to the sun, \(A_E = 1m^2\). Suppose our spaceman wears a sci-fi looking jet black spacesuit, a perfect black body radiator with with \(\epsilon = 1\). For our spaceman to be in thermal equilibrium, the energy he is absorbing from the sun, and the energy released by his metabolism, must be equal to the energy released via thermal radiation:

\[\begin{align} P + A_E E = A M \end{align}\]

Applying Stefan-Boltzman’s law again:

\[\begin{align} P + A_E E &= A M = A \epsilon \sigma T^4 \\ T &= \left( \frac{P + A_E E}{A \epsilon \sigma} \right)^{\frac{1}{4}} \\ &= \left( \frac{100W + 1m^2 1380\frac{W}{m^2}}{2m^2 \times 5.67 \times 10^{-8} \frac{W}{m^2 K^4}} \right)^{\frac{1}{4}} \\ &= 338K = 64^{\circ}C \end{align}\]

Oops! Our spaceman is going to be cooked alive! To solve this problem real-life spacesuits are usually white, to absorb less heat, and typically reject heat to space via the evaporation of water²; spacesuits are not closed systems.

However, if he were an computer chip, he’d survive just fine at those temperatures…

Flat-Pack Solar Powered Space Miner

Unlike most other computing tasks, Bitcoin mining is simple, highly parallelizable, and requires essentially no communication between compute elements. This means we can reduce launch mass to a minimum while still expelling heat effectively with a flat-pack design: mount the computation elements to the backside of the solar panels powering them.

Since this design is solar powered, our total heat budget per square-meter is always exactly equal to the flux density, \(M = 1380\frac{W}{m^2}\) at Earth’s orbital distance. The electrical usage and efficiency of the solar panels doesn’t matter: either solar energy is converted to electricity, and turned into heat via the computation, or solar energy is absorbed by the panel and converted directly to heat. Some solar energy is reflected by the panels. But the absorption ratio and emissivity of currently available solar panels is pretty close to 1. So for our rough analysis, we can round up to 1.

How hot will this design get? Since a flat panel radiates on both sides, our energy budget is:

\[\begin{align} E = 2M \end{align}\]

Applying Stefan-Boltzman’s law again:

\[\begin{align} E &= 2 \epsilon \sigma T^4 \\ T &= \left( \frac{E}{2 \epsilon \sigma} \right)^{\frac{1}{4}} \\ &= \left( \frac{1380\frac{W}{m^2}}{2 \times 5.67 \times 10^{-8} \frac{W}{m^2 K^4}} \right)^{\frac{1}{4}} \\ &= 332K = 59^{\circ}C \end{align}\]

Not a problem at all! Silicon runs just fine at those temperatures.

But for sake of argument, what if that number wasn’t good enough? There’s a very easy solution: tilt the panels. Rather than make them perpendicular to the sun, simply put them at some angle \(\theta\). Now we can reduce the solar flux absorbed, \(E_\theta\), to whatever fraction of \(E\) that we want by changing the angle, giving us:

\[\begin{align} E_\theta = \cos(\theta) E &= 2M \\ \cos(\theta) E &= 2 \epsilon \sigma T^4 \\ T &= \left( \frac{\cos(\theta) E}{2 \epsilon \sigma} \right)^{\frac{1}{4}} \end{align}\]

The Parker Solar Probe uses this technique to avoid overheating its solar panels while it’s close to the sun.

Heat Spreading

Our previous calculation has an assumption: that heat is generated evenly across the entire surface of the panel. While this is true for the solar flux that is absorbed directly as heat, it might not be true for the percentage of sunlight that is turned into electricity and used to actually power the computation.

A Bitmain Antminer S21 has 324 ASIC chips in total consuming 3500W in total, 11W per chip. Each chip is about \(1cm^2\) in size. That works out to a power density of \(110\frac{kW}{m^2}\), about two orders of magnitude more than the sun. Without any heat spreading, and assuming a black body, that would lead to a surface temperature of:

\[\begin{align} P &= 2 \sigma T^4 \\ T &= \left( \frac{E}{2 \epsilon \sigma} \right)^{\frac{1}{4}} \\ &= \left( \frac{110\frac{kW}{m^2}}{2 \times 5.67 \times 10^{-8} \frac{W}{m^2 K^4}} \right)^{\frac{1}{4}} \\ &= 992K \end{align}\]

We don’t even need to convert that to Celsius to know that’s unacceptable! If we’re going to use chips in that package, we need to spread that heat. Roughly speaking, we need to spread the heat from a \(1cm^2\) area to a \(80cm^2\), or a circle with a radius of \(5cm^2\).

It’s obviously possible to do this with a sufficiently thick heat spreader. Unfortunately unlike our previous calculation, there’s no easy way to derive a closed form equation. It’s also a question of economics: since the cost of the heat spreader can be reduced spreading the heat generation across a higher number of ASIC chips, the optimal design will likely be a combination of heat spreading and reducing the power intensity of the ASICs. What these trade-offs are depends on the mass of the heat spreader, launch costs, ASIC costs, etc.

Regardless, the fact that off-the-shelf planar heat pipes exist that can handle \(1000W\) in an area that large — significantly more than the 11W we need — is sufficient to prove that the idea is plausible.

Latency

The final objection to space mining and putting data centers in space is communication latency due to the speed of light. For the needs of data centers in general, latency may or may not matter depending on the application — bulk tasks like AI training do exist that can tolerate hours of latency.

For Bitcoin mining, if the space miner in question has a small percentage of total hash power, latency translates into a higher stale rate and thus less profit; if the space miner has a large percentage (roughly more than 30%) of total hash power, higher latency results in their competitors having a higher state rate. This is why space mining is potentially very dangerous for Bitcoin as a whole³.

In summary, latency is not a valid objection. But let’s analyze it in depth anyway.

Let’s assume we have negligible hash power — low enough that the probability of finding two blocks in a row is negligible. Secondly, we’ll treat all other hash power as a uniform entity, with negligible communication latency between other miners. Let \(t\) be our communication latency, which we will assume is symmetric in both directions. For a block found by us to not be stale, two conditions must be true:

1. No blocks were found prior to our block. If we assume we switch instantly upon learning of a new block, the time window for this to happen has duration \(t\).
2. No blocks are found after our block, before our block gets to the rest of the hash power. Again, the time window for this to happen has duration \(t\).

If \(t\) is small relative to the block rate \(r\) — the number of blocks per time interval; the inverse of the block interval — we can approximate this probability linearly:

\[P(\mathrm{stale}) \approx 2tr\]

This approximation is commonly used, and, for example, AI tools like Grok tend to give it as an answer. However, this approximation overestimates the stale rate as \(t\) increases: as \(t\) increases the probability reaches 1, even though there is always a chance that other miners fail to find blocks regardless of how large the communication latency is.

We can do better by applying the Poisson distribution:

\[P(N = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]

Thus the probability of no blocks being found, \(N=0\), in a time period \(t\) is:

\[\begin{align} \lambda &= tr \\ P(N = 0) &= \frac{\lambda^0 e^{-rt}}{0!} = e^{-rt} \end{align}\]

In our case the win condition is no blocks being found in two different \(t\) time periods. Thus our probability of a stale is:

\[P(\mathrm{stale}) = 1 - P(N = 0)^2 = 1 - e^{-2rt}\]

As an aside, by taking the derivative of \(P(\mathrm{stale})\) with respect to \(t\) we can see why the linear approximation is very good for small \(t\):

\[\frac{d}{dt}P(\mathrm{stale}) = 2r e^{-2rt}\]

If \(rt\) is small, \(e^{-2rt} \approx 1\). For example, assume \(r = \frac{1}{600s}\) and \(t = 133s\), the average speed of light delay from Earth to Venus. The approximation gives us:

\[P(\mathrm{stale}) \approx 2tr = 44.3\%\]

…while the poisson equation gives us:

\[P(\mathrm{stale}) = 2r e^{-2rt} = 35.8\%\]

Our approximation overestimates the stale rate by only 24%.

Now, for a more realistic example, let’s suppose that our solar powered space miners are at the L1 or L2 Lagrange points, 1.5 million km from Earth. Even that far out, the communication delay is only \(5s\), corresponding to a stale rate of \(1.7\%\). Assuming your space mining operation is sufficiently profitable to be worth doing at all, it’ll probably be profitable enough that the latency hit doesn’t matter.

Furthermore, you probably don’t have to go that far out anyway. A sun-synchronous orbit is sufficient to ensure that the panels are always in sunlight, and those are typically no more than 1000km in altitude. If anything, that would be sufficiently low that atmospheric drag would be an issue. Though it may be feasible to use atmospheric-breathing electric propulsion to counteract drag.

Footnotes