What thickness/depth of water would be required to provide radiation shielding in Earth orbit?

Question

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)?

Edit

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?"

Answer 1

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.

Answer 2

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.