Curious , why not send spent nuclear rods to the sun ?

Discussion in 'General Science & Technology' started by river, Jan 10, 2014.

  1. Kaiduorkhon Registered Senior Member

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    552
    Well, it's a bright idea, but the glitch here is: until we get organized enough to at least build a launch pad on the moon, the Earth's gravity is prohibitively expensive to be shuttling anything, let alone the density of spent nuclear power plant carbon rods to any location, let alone a place that takes seven and one half minutes to get here. Of course, everyone is entitled to their own levititous or gravitationally weighty opinion. Certainly that liberty shines extensively, throughout much of this thread.

    If you've a rich uncle, or are independently wealthy (with the gold standard), it might fly?

    Post Script:
    Have you ever heard of the 'K.I.S.S. principle'?
     
    Last edited: Jan 28, 2014
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  3. Kaiduorkhon Registered Senior Member

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    Well, yours is a bright idea, but: the glitch is, we're first required to build a launch pad on the moon. The Earth's gravity makes it prohibitively expensive to transfer inherently weighty stuff like spent nuclear power plant carbon rods, to a place that takes seven and one half minutes to get here. And that's really light-weight photon jetsam; smoking along at light-speed.

    If you've a rich uncle, or are independently wealthy (by the gold standard), it might fly?

    Post Script:
    Have you ever heard of the 'K.I.S.S. principle'?
     
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  5. billvon Valued Senior Member

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    How about we just sit on it forever? It is a lot cheaper to maintain a waste repository forever than to do any of the "launch waste into the sun" nonsense.
     
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  7. Arne Saknussemm trying to figure it all out Valued Senior Member

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    How about reading a post before you reply to it? I never said anything about the sun,
     
  8. ElectricFetus Sanity going, going, gone Valued Senior Member

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    There are several more viable solutions then hurling nuclear waste at greater than 30 km/s, that the speed needed to cancel earth orbital velocity and drop into the sun, plus more to make it out of earth gravity field and atmosphere, we could cut that down in half by hurling it around Jupiter or at Jupiter, heck just throwing nuclear waste in outer space in a stable solar or even high earth orbit would be just as good as into the sun and require far less delta V. A stable earth orbit beyond 40 Mm would last for hundreds of thousands of years at least, long enough for nuclear waste to decay and become inert, and if the orbit does decay it would decay outwards and crash into the moon not the earth. Anyways all that is ridiculously expensive compared to:

    1) Reprocessing: recycling nuclear fuel after use can consume 90% of nuclear waste
    2) Fast neutron reactors, breeder reactors and assisted sub-critical reactors: reactor can be built that consume 99% nuclear waste, they would need to be generation 4 reactors with non-water coolants.
    3) Bury it under the sea floor
     
  9. Arne Saknussemm trying to figure it all out Valued Senior Member

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    Not on my sea floor!
     
  10. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    Earth has an orbital velocity around the sun of c.30km/s.
    To get something to impact the sun would need to reach Earth's escape velocity, which from the surface is about 11 km/s, and then lose orbital speed with respect to the sun until the orbital perigee is within the sun's radius.
    So let's say 41km/s delta-V required.

    To get something to land on moon requires roughly around half of this, I think: 11 to escape earth, a bit to get into transfer orbit, about 4 to then get in to Mars orbit etc.

    So purely from energy point of view, probably only 4 times as much is needed to shoot for the sun (although my calls are very simplistic).
    However, there are more efficient ways for the solar shot, such as shooting to Jupiter and doing a slingshot.

    That said, even just sending something into a highly elliptic orbit might do the job... Or at least put it out of our concern for 100+ years.
    But it's ultimately about cost, as such transport is not cheap.


    Heh! Teach me for not reading all the posts: seems ElectricFetus has already said almost exactly the same.

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  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I'm still too lazy but Think a better way to estimate how much energy is required to hit Mars is to know that gravitational potential energy (with far from the source defined as zero) goes as -1/r the separation from the source (for r > greater than the source radius, of course). And also know that the sun's gravity on the moon is stronger than that of the Earth, so neglect the sun's gravity until waste get nearly out to the moon's distance away from earth and compute how much (most) of the 11km/sec would be used up climbing to that point where rest of the climb is mainly against the pull of the sun. Then ignore the earth and compute how much higher in the sun's gravitational well, Mars is than Earth as the final "hill" to climb to cross Mars' orbit (well timed to hit it as it passes).

    I may be wrong, but think if you just want to hit Mars, I think you can ignore Earths orbital speed and just worry about climbing two gravitational hills first against Earth's pull and then against the sun's pull using the fact gravitational potentials goes as 1/r to make that relatively easy / quick and dirty / estimate.
     
  12. billvon Valued Senior Member

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    12,804
    Sorry, or "launch it into deep space." I'd much rather have nuclear waste sitting safely on a pad in the desert than flying over my head.
     
  13. Arne Saknussemm trying to figure it all out Valued Senior Member

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    In very tiny, well shielded quantities, very gradually?
     
  14. billvon Valued Senior Member

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    12,804
    They would be far safer stored terrestrially in very tiny, well shielded quantities. Besides, what would you do with it when you got there?
     
  15. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    23,198
    See: http://www.sciforums.com/showthread...s-to-the-sun&p=3152232&viewfull=1#post3152232 - A re-posting of the idea I suggested more than five years ago, is safer than current practices* (and its life cycle cost is cheaper, by more than a factor of 10!) It is somewhat like ElcetricFetus suggested in (3) you are objecting to, but no need to bury the waste on the sea floor, if that spot is well chosen. Boat just steams over deep ocean trench and automatic disk thrower, much like used for shot-gun practice with clay pigeons, hurls the approximately foot in diameter, inch thick glassified waste disks off the stern. Read link or my 5-year old post for more details.

    * Mainly as no need to protect waste against terrorists wanting material for a "dirty bomb" for hundreds of years.
     
  16. Arne Saknussemm trying to figure it all out Valued Senior Member

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    I already said.
     
  17. billvon Valued Senior Member

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    Well, you said "LaGrange points between stars." They don't really exist since stars don't orbit each other. You could just dump it in deep space, but again, having it float around randomly in deep space isn't notably safer than storing it terrestrially.
     
  18. Arne Saknussemm trying to figure it all out Valued Senior Member

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    I asked if there were such things as LaGrange points between stars. There's not? Okay, thank you for clearing that up. I can't do the math, but surely having tiny containers, or, better yet, vitrified units of nuclear waste in deep space light years away from any planetary system would be safer than storing it in a similar way on Earth. The chances would be remote of anyone coming across it accidentally in such vastness (and beacons could warn them off) and, in such vastness, the hundreds of thousands years of their half-lives could easily pass with no interference from anyone, and the waste not interfering with any thing. And remember there would be little risk in launching it as it is only a handful and 'wrapped' for export. I am imagining the age of interstellar flight would be eons long, and if we are patient, we can do away with ALL the waste eventually.
     
  19. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    No it would be more than 10,000 times more expensive and at least dozens of times more risk to people as launch rockets do explode compared to suggestion I made 5-years ago and reposted idea in this thread here: http://www.sciforums.com/showthread...s-to-the-sun&p=3152232&viewfull=1#post3152232

    From a life cycle cost POV my suggestion is at least 10 times cheaper (and also safer) than current practice.

    Here is link to the 5-year old post for more details: http://www.sciforums.com/showthread.php?50995-Nuclear-waste&p=934547&viewfull=1#post934547
     
  20. Arne Saknussemm trying to figure it all out Valued Senior Member

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    I mean that the flights would be being made anyway: interstellar passengers or freight would be going from one star system to another for reasons other than waste disposal, and the vitrified bit could be ejected mid-flight at some predetermined point. And it would not necessarily be a rocket. Who knows what interstellar ships will be like, if we ever develop them, or how they will set off? Remember also the amount of radioactive material on any given flight would be minimal, and contained as safely as possible.
     
  21. billvon Valued Senior Member

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    12,804
    Only if there are tens of thousands of spacecraft flying by to deposit a significant amount of it. And if there are tens of thousands of spacecraft flying by floating nuclear waste dumps, yes, that's a problem. Much better to contain it in one place where it won't get in the way of spacecraft 10,001.

    If it is so benign that it could fall out of the sky, land in a town and not harm anyone, then surely it would be thousands of times safer in a secure location in an uninhabited desert.

    If we are patient, we can do away with all the waste eventually by putting it in the desert. After a few thousand years its radioactivity decays to a relatively safe level. (i.e. similar to the original material.)
     
  22. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    The Earth's gravitational pull on the moon is stronger than the sun's. The moon is c.380,000 km (on average) from earth, and the Earth's sphere of influence is c.950k km before the influence of the sun takes over on a celestial body.
    The earth is the important factor initially - you have to break free of the sphere of influence, and you need 11km/s delta-V from the surface to do that. You can't hope to go anywhere but orbit the earth if you don't - whether you want to go to the sun or the moon.

    To fall in to the sun you then need to negate most of the 30km/s orbital speed you have with respect to the Sun.
    To get to Mars you need to increase your orbital speed and by doing so achieve a higher orbit with respect to the Sun, until you intersect with Mars.
    That takes about 4 km/s just to get into Mars sphere of influence I think. So absolute minimum of c.15km/s from the surface. (I previously double-counted the bit to get into transfer orbit - which is actually included in the 11km/s).

    If you only want to hit Mars then sure, it's possibly not as much as landing robots on the surface, as you don't need to control your descent, which was the example given.

    The difference between the two is the "gravitational hills" as you describe them, but to start you must get out of Earth's sphere of influence, and that takes 11km/s from the surface. Therefore it does need to be included in the comparison if you're comparing energies required from the surface. Otherwise you could end up comparing X to Y when it should be (11+X) to (11+Y).
     
  23. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Not true. The force of gravity goes as M/(R^2). So with lower case letters m and r for the earth and capital letters for the sun, the Sun to Earth force ratio is (M/m)(r/R)^2. Put the numbers in and you will see that the M/m is so large it rules over the smaller (r/R)^2 to make (M/m)(r/R)^2 > 1. {when you do this calculation, please post the ratio - I have done it several times, but forget the result, but think it is something like 1.4 from memory.}

    The tidal force (moon on earth, or conversely) is larger than that of the sun on either as tidal forces go as inverse r^3.

    That greater force from the sun makes the moon's acceleration ALWAYS TOWARDS THE SUN. I.e. the moon's trajectory is ALWAYS curving towards the sun. Only towards the Earth when the Earth happens to be closer to the sun than the moon is.
     

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