Giant Cyclers for Radiation Safe Earth to Mars Journeys

Nextbigfuture commenter, Goatguy, did interplanetary travel calculations to check the numbers that I had cited. I had estimated that 1 meter of water surrounding astronauts would be enough to protect them from space radiation. Definitely, 5 meters of water would be enough. Mass production of SpaceX Starships is likely by 2030. There is a Raptor…
Giant Cyclers for Radiation Safe Earth to Mars Journeys

Nextbigfuture commenter, Goatguy, did interplanetary travel calculations to check the numbers that I had cited. I had estimated that 1 meter of water surrounding astronauts would be enough to protect them from space radiation. Definitely, 5 meters of water would be enough.

Mass production of SpaceX Starships is likely by 2030. There is a Raptor engine factory that will make 4000 engines per year. Each Super Heavy Starship stack would use about 40 engines. Full production is 100 Super Heavy Starships per year.

Each SpaceX Starship has 1000 cubic meters of volume. One hundred of them would have 100,000 cubic meters of volume. Ocean going cruise ships are measured in terms of gross tonnage (GT), which is a measurement of the internal volume of a ship. The average cruise ship has a gross tonnage of 120,000 GT.

300 meters by 30 meters by 11 meters would be 99000 cubic meters. The picture above is of Space Shuttle External fuel tanks arranged into a rotating space station configuration. There are about 24 tanks. Space Shuttle fuel tanks were similar in size to the SpaceX starship. There are about 20 tanks going around the outside ring and about 4 bundled in the center. If there were 100 Starships, then there could be a triple ring going around the outside and bundles of 25 in the interior. Three stack of eight bundled together. There could be Super Heavy Boosters or other engines in the central part.

If we wanted to reduce the surface area, then a 50-meter-tall cylinder with 50 meters of diameter would be almost 97000 cubic meters. The outer one meter would be about 12000 cubic meters. If we wanted to reduce weight, then we could move the water shield from the outer wall to an inner set of walls or around “safe-room” living areas.

Each SpaceX Starship is 50 meters tall and 9 meters in diameter. The least surface area for 100 would be to pack them in a single-layer bundle.

There could be more of the Starships stack end to end in the central area. It depends how one would want the weight distributed around propulsion or for a rotating system for centrifugal force in place of gravity.

There are some other configurations where one could imagine using Starship modules to form a larger structure.

If mass production is further scaled then 1000 Super Heavy Starships per year could be made. There could be a stream of say forty-two Mars Cyclers in constant cycling orbits of Mars and Earth. Mars cyclers would travel from Earth to Mars in 146 days (4.8 months) and spend the next 16 months beyond the orbit of Mars. This would be almost 21-month cycle. 42 cyclers would mean they would be two weeks apart. There would be nine or ten between Earth and Mars and 32 beyond the orbit of Mars.

Detailed Calculations of Space Radiation and Shielding

Earth’s atmosphere, with a ground-level pressure of about 100,000 pascals (N/m²) corresponds to (÷ 9.81) = 10,200 kg or 10.2 tons of atmosphere ‘above our heads’ per m².

Placing 1 ton of water between the inside and outside of the proposed radiation-proofed capsule is ¹⁄₁₀th that radiation protection. Space radiation consists largely of 4 components: solar protons, neutrons, gamma/x-rays and so-called cosmic rays.

The Cosmic Ray Radiation Dose in Interplanetary Space – Present Day and Worst-Case Evaluations (2005).

Is a particularly good estimator of actual space radiation in inter-planetary space. Apologies for being highly technical, but ‘the answers and concerns’ are well laid out. In a nutshell, 50 cSv (or 50 RAD, which are dose-by-absorption-energy adjusted) per year is probable during Solar Minimum.

And what does 1.0 m of H₂O shielding do? It definitely quenches the lowest energy GCRs (galactic cosmic rays) as well as ALL protons. Many neutrons succumb too. Around 90% of them. From another source

Beating 1 Sievert: Optimal Radiation Shielding of Astronauts on a Mission to Mars

Key Points on Optimal Shielding

* Space missions to Mars should be scheduled to be launched during solar max

* Optimal spacecraft shielding is ~30 g/cm2, which allows long-duration flights of ~4 years

* Increase of shielding thickness beyond ~30 g/cm2 results in dose increase

* The best time for launching a human space flight to Mars is during the solar maximum, as it is possible to shield from SEP particles. Our simulations show that an increase in shielding creates an increase in secondary radiation produced by the most energetic GCR, which results in a higher dose, introducing a limit to a mission duration. We estimate that a potential mission to Mars should not exceed approximately 4 years

Continuing Goatguy Calculations

We get a pretty good summary of the effectiveness expected of shielding. Essentially, it cites that too much shielding is counterproductive, since the incoming GCRs spall off all sorts of secondary particles, radiation in their own right. In any case, compared to Earth’s atmosphere, shielding walls (however you want to look at them) are far inferior to air itself.

50 cSv/year. That’s the shielding take-away, with perhaps 50% less having water shielding over aluminum. Hydrogen is a good thing. 1 m/m² thickness corresponds to 1,000,000 g ÷ 100² cm = 100 g/cm², which is arguably adequate.


Léts see if the cylindrical figure is right. Volume of cylinder is ¼π𝒅²𝒍 where 𝒅 = 3.5 meters, and 𝒍 = 20 meters. So, the area of a pair of cylinders subtracted (smaller from larger) gives the total water. 1 meter thickness, uniformly spread out.

Vol₁ = ¼π⋅3.5²⋅20

Vol₁ = 192 m³

Vol₂ = ¼π⋅(3.5 ⊕ 2)²⋅(20 ⊕ 2)

Vol₂ = 522 m³

ΔVol = 330 m³ = 330 tons.

Excellent! The math works.


I think underestimating the transit cost of all that shielding is the real problem to solve. If we recall Tsiolkovsky’s Rocket equation

ΔV = Isp • G₀ • ln( M₀ / M₁ )

Where M₁ is the end-of-acceleration total ship+fuel+etc mass, and the M₀ is the beginning … and if one considers that some 70% of the mass needs to drop on the outbound acceleration leg, leaving 22% left for deceleration, and until we come up with more energetic propulsion means, that Isp’s greater than 2,000 (ions) with nuclear power are optimal, then:

ΔV = 2,500 × 9.81 • ln( 100% / ( 100% – 70% ) )

ΔV ≈ 30 km/s

Sounds good, right? It is. But getting to Mars is a long way to go … something on the order of 100,000,000 km. Just a linear fit to that is 100×10⁶ ÷ 30 = 3.3×10⁶ seconds (40 days) But … there’s Sol’s gravity well to overcome. Sling-shots around Luna might help some, but sling-shots around Venus wouldn’t really save a whole lot of time, and exacerbate the approach problem (deceleration). Turns out more like a 200 day trip.

And there’s the problem of creating enough nuclear power, safely enough, to accelerate 70% of the rest-mass of the whole shebang, quickly enough, to not ruin the time-savings in so doing. 200 to 300 days. A year.


In that year the cosmonauts would almost irreducibly absorb some 50 cSv, which is about ⅓ of the NASA and ESA recommended lifetime exposure limit. Depending as the article said on ‘age and physical condition’.

Goatguy agrees with Brian Wang: the possibility of making the trip in a reasonable amount of time, with a reasonable exposure to GCR and ESR (solar) has been well worked out. The real gotcha is coming up with the energy-production plan and engineering the system.


And what would be the energy production needed? Let’s say that 5,000 ton interplanetary cruiser. It needs to accelerate 70% of its rest mass (70% × 5,000,000 kg = 3,500,000 kg) to accelerate. At an Isp of 2,500 (equiv to 25,000 m/s) at maybe 70% electrical-to-ion efficiency (VASIMR), then you have

2 × ½ • mv² = mv² = 3,500,000 kg × 25,000² → 2×10¹⁵ joules or 600,000 megawatt-hours

of electrical input energy. Nuclear power. If its to be produced over 4 months (⅓ the trip), continuously, then thats 600,000 MWh / (4 × 30 × 24) = 200 MW of specific output energy. That’s a pretty potent nuclear reactor, but not ridiculous.

Feels do-able, to this old physics goat. The actual acceleration would be 30,000 m/s / (4 × 30 × 24 × 60 × 60) = 0.0029 m/s² or about 0.30 milli-G

Imperceptable. Stately. But effective over months. Never stops, basically.

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