NASA Developing Small Nuclear Reactor For the Moon
marshotel writes "NASA astronauts will need power sources when they return to the moon and establish a lunar outpost. NASA engineers are exploring the possibility of nuclear fission to provide the necessary power, and they are taking initial steps toward a non-nuclear technology demonstration of this type of system."
I think it depends on the reactor type. Some can use liquid sodium, etc. Think "micro-reactor" similar to the proposals by the Japanese space program or Toshiba for small output, "4S":
http://en.wikipedia.org/wiki/Toshiba_4S
A 40kw reactor like they discuss in the article would use a small amount of uranium, probably less volume of radioactive material than used for the RTGs in the cassini probe. Whereas we have tons and tons of nuclear waste to dispose of, not just spent fuel rods, but reactor internals, coolant, and so on.
Except for the fact that it would be dark at your moonbase for nearly two straight weeks at a time, solar power would be great.
Sig this!
Night time on the moon is kinda long (weeks). What do you do then? Batteries that can store weeks worth and PV arrays that run at over 2x capacity are not really going to work all that well. Well not as well as a 24/7 nuke plant.
If information wants to be free, why does my internet connection cost so much?
Nuclear waste is not really waste. It simply needs to be used in a different reactor. Storing this waste and doing nothing with it is really a waste.
UNIX/Linux Consulting
How about some perspective on that reality?
http://en.wikipedia.org/wiki/Image:Fy2008spendingbycategory.png
Here's a hint: The NASA slice is the 0.6% one. Double NASA's budget and you're still not up to the level of "Other Off-Budget Discretionary Spending."
IWARS.
People, in general, disappoint me. Politicians even more so.
Well the sun is a hellish inferno of radiation as it stands, dumping a million tonnes of the nastiest crap we can find into it would be like spitting into niagara falls.
Um let's see...
m = 1 kg
v = 12 km/sec = 12000 m/sec
KE = 1/2*m*v^2 = 1/2 * 1 * 12000^2 = 72 MJ for a 1kg object
I was always happy when my lab partner and I came within an order of magnitude of the correct answer in my EE lab.