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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)?

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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?"

James Jenkins
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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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KeithS
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    There are several types of harmful radiation in space, does the 7 centimeter halving apply to all of them? – James Jenkins Sep 07 '13 at 00:28
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    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). – KeithS Sep 07 '13 at 00:38
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    "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. – HopDavid May 17 '14 at 14:30
  • Don't radiation thickness requirements (for water shields) change according to distance from the Sun.? I'm assuming a spacecraft at Mercury would need thicker shielding than one at Mars. . . – MercuryPlus May 17 '14 at 15:48
  • Very probably. For an Earth-Mars transition using a Hohman orbit (which would result in the spacecraft staying between the orbits of the two planets rather than slingshotting around a closer orbit to the Sun), the numbers we already have for lunar excursions represent the highest long-term dose as they represent the dose at the closest point of the excursion to the Sun (outside Earth's protective magnetic field). As you get closer to the Sun, you are exposed to a greater arc of its radiation, and thus a greater amount of power that you must moderate with additional shielding. – KeithS Jun 17 '14 at 00:05
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    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. – LocalFluff Jul 24 '14 at 12:16
  • @HopDavid - the crew may want to return to Earth at some point. – Russell Borogove Nov 21 '14 at 17:29
  • @RussellBorogove If "going to Mars" were the same as "going to Mars and coming back", you'd have a point. But KeithS' last paragraph isn't describing a round trip. He seems to be saying LEO to Mars is 9 km/s -- which is wrong. – HopDavid Nov 21 '14 at 17:47
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    If I say "I'm going to the store" or "I'm going to work" or "I'm going to New York", the default assumption is that I'm coming back. – Russell Borogove Nov 21 '14 at 18:07
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    Many Mars architectures call for refueling at Mars. And some call for one way trips. So, no, that's not the default assumption here. If I say I have enough gas to get to work, the default assumption is that I have enough to get to work, not get to work and come back. – HopDavid Nov 21 '14 at 23:43
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    @LocalFluff and if we could make the shielding directional we might be able to reduce that down to ~2 tons. – NPSF3000 Nov 23 '14 at 10:27
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    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. – Ellesedil Sep 01 '15 at 16:43
  • @LocalFluff It depends on how much damage is acceptable. Spending 1/2 your time shielded would still mean 40 days total exposure for an 80 day trip. – Jason Goemaat Oct 11 '16 at 09:50
  • How long would that water last? I assume some of it is changed to heavy or tritiated water over time. – Martin Schröder Feb 27 '18 at 12:24
  • I'm wondering more about cosmic rays and water's effects on those, because those are the more damaging to humans; correct? – Magic Octopus Urn Jun 29 '18 at 20:56
  • "1 mSv/hr" - background radiation number seems not correct, it more like annual dose 1.26mSv, hourly doses on level of 0.1-0.18 uSv (10^-6 Sv), not sure if it affects any other number statement in the post, maybe need revaluating for that – MolbOrg May 04 '19 at 16:01
  • Just recently oct 22 2021 spinlaunch test flung a suborbital 'rocket' tens of thousands of feet up with a centrifuge. Pure conjecture but I see potential to relativly cheaply fling ice up for constructing spacecraft shielding and a refueling station in low earth orbit to process some of the ice to hydrogen and oxygen. – Keith Reynolds Nov 13 '21 at 23:11
  • @Keith Reynolds spinlaunch only acts as the first stage of propulsion. To get to orbit the spin launched payload has to be attached to a "second" stage rocket – Slarty Sep 28 '22 at 21:07
  • @Slarty Yup. Spinlaunch platform throws up a rocket with apayload to near orbit. Then the rocket fires in the second phase, but is itself a single stage rocket. My main point was that water ice payload inside the rocket (figuratively buckets) could be spinlaunched to space comparitivly cheaply and used for shielding, or Rather than use massive rockets to get a massive deep space rocket and rocket fuel to orbit for a deep space mission. Spinlaunch watet to orbit, process into fuel, and in orbit transfer the the rocket fuel to smaller rockets launched from the ground intended for deep space, – Keith Reynolds Feb 17 '23 at 03:05
  • An interesting idea, but I suspect that if Starship comes up to expectations it will be capable of taking its own water - cheap to reuse and capable of lifting 100-150 tonnes to low earth orbit (spin launch can manage 200kg). – Slarty Feb 17 '23 at 23:08
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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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    Hi, welcome to Space.SE. Interesting figures, would you mind showing us how you arrived at them? – Hobbes Oct 28 '15 at 15:54
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    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. – TildalWave Oct 28 '15 at 21:48
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    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 - http://www.int.washington.edu/PHYS554/2011/chapter9_11.pdf – kim holder Nov 01 '15 at 22:12