If using water or ice as a radiation shield how thick/deep would it need to be? Would it make a difference if the water was frozen (water ice being less dense then liquid)?


I would like to know how much shielding would be required for an indefinite stay. Preferably at Terrestrial limits per Code of Federal Regulations (29 CFR 1910.96), but at least at levels per National Council on Radiation Protection and Measurements (NCRP) presented in its Report 98 A beginning place for research is the Spaceflight Radiation Health Program at JSC A listing of all different types of radiation NASA "What is space radiation?"


2 Answers 2


We know from the nuclear power industry that spent fuel storage pools are pretty safe places to be around, radiation-wise. They're actually safe to swim in, to a point, because they're serviced routinely by human divers. They just can't get too close to the spent fuel.

We use these pools for short-term storage because water is a really good radiation shield. How good? Well, according to a report on the topic prepared for the DoE back in 1977, a layer of water 7 centimeters thick reduces the ionizing radiation (rays and particles) transmitted through it by half (the remainder is captured or moderated to non-ionizing energy levels, mainly heat). Freshly discharged nuclear fuel puts out about 100,000 R/hour as measured from one foot away in air (at that rate, certain death is about 5 minutes' exposure and you'd fall into a coma in about 10). Background ionizing radiation levels on Earth's surface are about .000001 R/hour (1 mSv/hr), while a "safe dose" to live with long-term is about .0004 R/hr. A halving represents about .3 of a power of 10, so in rough terms, to reduce a fresh fuel rod's radioactivity to safe levels, you would need about 2 meters (8/0.3 * 7 / 100), and through more than 2.5m the radioactivity of the fuel rods is indistinguishable from background radiation. In fact, according to the link from the comment, diving about 6 feet down would expose you to less radiation than at the surface of the pool.

According to Wikipedia, the upper estimate for a dose equivalent received by unshielded astronauts operating outside Earth's magnetic field (such as a mission to Mars) is about 90,000 R/yr or about 10 R/hour. If we assume the energy levels are comparable, reducing that to lower than Earth background radiation would only require a layer of water around 1 meter thick.

However, let's do some more math. Let's say that the Mars vehicle that will get them there and back is a cylinder roughly 3.5m by 20m (same as was used for the MARS-500 experiments; that is a very small tin can in which to spend 3 1/2 years with 4 or 5 other people). With 1m of shielding water around all surfaces of that cylinder, the outer hull would be about 5.5m by 22m.

The volume of shielding water needed is the difference between those two cylinders, or $22\cdot\pi\cdot2.75^2 - 20\cdot\pi\cdot1.75^2 \approx 522.68 - 192.42 = 330.26 m^3$. As one cubic meter of water weighs 1 metric ton (1,000kg), that's 330,260kg to get into space.

Putting that in perspective, the current record holder for payload-to-LEO is the Saturn V rocket, which had a maximum LEO payload of 120,000kg (said payload being the S-IVb, including CSM, LEM and Earth departure stage, for most of its missions). To put the volume of water we'd need into orbit would require 3 Saturn V rockets. The planned-but-never-built Ares V was spec'ed to have 188 tonnes P2LEO capacity, which would have cut down the number of launches to only 2. Doing it with Space Shuttles (25 tonnes cargo to LEO) would take 14 missions. The SLS Block II (130 tonnes payload to LEO) would take just about 3 launches. Doing it with any orbital rocket currently in service, manned or unmanned (Soyuz II, Soyuz FG, Delta IV, Atlas V, Falcon 9) would require between 50 and 100 launches.

Given that we could achieve getting this much mass into LEO, getting it to break orbit and out into interplanetary space is that much harder; going to Mars using a Hohmann transfer orbit takes about as much delta-V as getting to LEO in the first place, so all the fuel expended to get the craft and its water shield into space has to be brought up into orbit, requiring many more launches. Using a gravity assist, say from Venus, would be a logistical nightmare requiring all three planets to be in alignment as of departure from LEO, and while it would save fuel it would require us to cover much more distance and take much more time, possibly placing the mission further beyond our current capabilities.

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    $\begingroup$ There are several types of harmful radiation in space, does the 7 centimeter halving apply to all of them? $\endgroup$ Sep 7, 2013 at 0:28
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    $\begingroup$ Probably not; the data I have is most concerned with gamma rays and neutron radiation, the two forms most commonly seen from nuclear waste (and the two hardest to shield against). Ionizing cosmic radiation is mostly gamma rays and protons, which are overall very similar to the radiation from nuclear waste. Protons are slightly lighter and so would have less energy at a given velocity (more easily stopped); but, because they're protons, if captured by the water they'd basically become H+ ions which could acidify the shielding water in time (not seen as much with nuclear waste). $\endgroup$
    – KeithS
    Sep 7, 2013 at 0:38
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    $\begingroup$ "going to Mars using a Hohmann transfer orbit takes about as much delta-V as getting to LEO in the first place" -- Getting to LEO takes about 9 km/s. From LEO trans mars insertion takes about 3.6 km/s. On arrival at Mars, a 1 km/s burn suffices to park the spacecraft in an elliptical orbit whose periapsis passes through Mars' upper atmosphere. The delta V is about half that of getting to orbit. $\endgroup$
    – HopDavid
    May 17, 2014 at 14:30
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    $\begingroup$ If astronauts spend half their time (sleeping and a few more hours per "day") in a 1 meter diameter hole inside a 3 meter diameter cylinder with the length 2 meters filled with one meter water or fuel or waste around, that would weigh about 13 tons (not thousands of tons). And that alone would cut the radiation exposure in half. $\endgroup$
    – LocalFluff
    Jul 24, 2014 at 12:16
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    $\begingroup$ This What If? article by Randall Munroe goes into detail about the "effects" of swimming in a nuclear spent fuel pool. It's a great read for people asking these kinds of questions. $\endgroup$
    – Ellesedil
    Sep 1, 2015 at 16:43

A 1 km cube with room for 1,000,000 people each with 16 rooms of 20 Square meters, would require 18 tons per person for a 3 meter shield, which would reduce radiation levels to less than the .000001 R/Hr of the Earths surface.

Allocating each person 1000 cubic meters. If a floor height is 3 meters the 1000 cubic meters provides 333 square meters of floor space. This space can be divided 16 rooms of 20 square meters (200 square feet). The assumption is that people will share part of that space for common areas.

A 1 km cube has 6 sides each 1000 x 1000 meters. The total surface area is thus 6 * 1000 * 1000 or 6 million square meters . Divided among the 1 million occupants means 6 meters of shielding each. Three meters of shielding is required for each square meter of surface. This is a Stack of 3, 1 meter cubes. A cubic meter of water weighs/has a mass of 1 metric tonne (very close to 1 ton). The stack of 3 * surface area of 6 totals 18 tons.

For a 100 meter cube (1,000 people) the figure is 180 tons per person. The surface area is 6 * 100 * 100 = 60,000 square meters , which is 60 square meters for each of the 1000 people. Thus 3 * 60 or 180 tons.

The 3 meter shield size is derived from the information in keithS's post for the same question.

It is required to reduce 100,000 R/hr to .000001 (1 millionth ) R/hr which is a reduction of 100 thousand million times - which is under $2^{38}$.

Each 7cm (http://www.nist.gov/pml/data/xraycoef/index.cfm) of water divides the radiation by a factor of 2. Thus the total required is 38 * 7cm or 266 cm. Use 300 cm / 3 meters to make calculation easier and it provides an extra safety margin which makes the calculation easier.

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    $\begingroup$ Hi, welcome to Space.SE. Interesting figures, would you mind showing us how you arrived at them? $\endgroup$
    – Hobbes
    Oct 28, 2015 at 15:54
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    $\begingroup$ Re your edit, somehow I don't think it was requested that you prove your geometry as much as how you came to the conclusion that a 3 m thick shield of water would reduce radiation levels to less than the .000001 R/Hr of the Earths surface. You also don't mention for which orbital altitude are your figures. Also note that the question doesn't ask about the mass but directly about shield thickness. So most of your answer, which is about mass of water required to shield some arbitrary number of occupants per some arbitrary living area volume per person, is missing the point. $\endgroup$
    – TildalWave
    Oct 28, 2015 at 21:48
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    $\begingroup$ The reference you give is for photons with energies less than 20 MeV. Cosmic rays are atomic nuclei with energys mostly between 100 MeV and 1 GeV, and the worst damage is done by heavy particles like iron nuclei. This paper has a good overview - int.washington.edu/PHYS554/2011/chapter9_11.pdf $\endgroup$
    – kim holder
    Nov 1, 2015 at 22:12

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