This article confuses me a great deal, and IAANP (grad student). They say "one hundred thousandth of a becquerel per cubic centimeter of xenon-133 and xenon-135 was detected in gas samples.", that means one decay per second in every 1/10 of a cubic meter. This is a very low rate. U-238 undergoes spontaneous fission in about 1 in 10^5 radioactive decays whether it is in a reactor or not,and about 1% of those fissions produces a Xe-135 (either directly, or after decay of one of its parents like I-135). If I do a back of the envelope calculation, I find that for 10 tons (a guess) of U-238 sitting there being nice, about 100,000 Xe-135 will be produced every second. Thus, unless the air volume they are sampling from is much larger than 10,000 cubic meters, this sounds like what I would expect WITHOUT criticality.

Am I missing something here?

Kepler's laws only apply for the 1 planet + 1 star case. They're really just some consequences of the more general Newton's laws, applied to that simple scenario. If the two stars are much closer together than their distance from the planet, then Kepler's laws will probably be a fine approximation for the short term, just using the center of mass of the stars and their average mass. Eventually (or soon, if the stars aren't so close to each other), deviations from the simple elliptical orbit from Kepler's laws will show up. You could get the planet changing its orbital radius, eccentricity, and precession of the major axis.

I'm betting it's the precession that will cause the predicted loss of signal for Kepler. Maybe the planet's orbital plane is not the same as the stars, and we're just lucky that the point where they all line up is pointed at Earth, but as the orbit precesses that direction will shift elsewhere.

I did the calculation, after finding the details of the planet on the Kepler website. They don't have a mass value for 19b, just an upper limit at 14 earth masses. I just plugged in a value of 10 earth masses for my calculation, and I get 10^30 J, or about 200 zettatons of TNT equivalent, or enough energy to accelerate 3.6 billion pounds of bacon to the speed of the LHC beam.

I know about Lagrangian points perfectly well, and they are not stable. At best they are pseudostable. I'm talking about long-term stability, having it not wander around and smack us millennia after it leaves. Please point out in the article where it tells us that one of these points is suitable over very long timescales.

The plan of bringing it in so it is barely bound and letting it wander off some years later sounds much more dangerous than bringing it into a proper orbit.

What will that asteroid do over the following centuries/millenia? We would have to monitor it forever and might need to nudge it again later. I'm also not sure if there are any truly stable orbits around the Earth, given the size of our moon driving it. Maybe there is some resonance with the moon's orbit that is safe. If so, that seems the best place to put it, and leave it there forever.

That's the "Seigniorage" aspect to this. The coin doesn't have to have metal worth $5T in it, it just has to have "$5T" printed on it. Many coins are worth much less in their metal than in the face value. The presidential dollars, for example. Also, all paper bills are essentially worthless for their material, but that doesn't stop them from being worth $100 if that is printed on them. This plan must use a coin though, because there are laws in place governing paper bills that keep the President from doing it with paper.

Now I've been out of the space exploration loop for a few years, but it strikes me that the U.S.A. does not have civilian or manned launch capabilities at the moment. That leaves the civilian program contracting out launches to the Russians, E.S.A., and their military. And quite frankly I don't see that changing in the near future since I don't think that they have the political will to change it.

I think what you are trying to say is that the US has no capacity to put anything at all into orbit, which is just false. For example, Boeing and Lockheed make a variety of rockets which are frequently launched from within the US. Space X is also entering the market, and will soon be launching manned spacecraft. The US only needs to contract out manned launches, and only for a short while until Dragon is ready.

We often see them coming thanks to satellites like the one that made these movies. It takes hours-days for the flare to get from the sun to Earth, so there is time to prepare. I think it's hard to be sure exactly how hard any given flare will hit the Earth, though.

I'm not sure if your foil-on-computer question is an analogy or not. On the personal scale I expect that your regular surge protector is sufficient. The disaster planning needs to be centered on the large-scale power grid, because it's the long power lines that build up the overvoltage, not your living room. We're not worried as much about your computer as we are half the power substations on earth exploding within an hour of each other.

Even more importantly, will we be able to keep playing Wings of Liberty multiplayer without buying Heart of the Swarm? I couldn't find an answer to this in any of the links. If we can't and they charge $60 for the expansion, you can bet I'll never buy anything from Blizzard again.

Well how are you going to make the electricity? Solar power is worthless outside the inner solar system, so you'll need a generator of some kind and that will need fuel. The generator fuel is going to have mass, and once you use it it is dead weight, so you might as well use it as propellant.

You just took 10 and divided it by pi and wrote that down, all in base 10. Also, pi in base pi would be 10, not 1 (like 2 is 10 in binary), and pi^2 is 100.

I'm not certain how other numbers in a non-integer base would be written down, but I think 10 in base pi would be 100.010221222... (pi^2 + pi^-2 + 2*pi^-4 + 2*pi^-5...) There may be multiple representations of the same number. For example, I think 10 could also be 30.121...

At random screening, for every person correctly identified, 794 were misidentified.

So 1 out of 794 people is a threat to aircraft security?

So to your first post, the claim is not that you cannot collect all the light nor that you are in fact able to focus it to exactly the intensity of the source. In your system the translucent sphere serves to prevent us from focusing the light anywhere near the intensity of the original source, so there is nothing wrong with that. Further, we would now find ourselves unable to focus the light to a spot any more intense than the surface intensity of the translucent sphere. If one were to envelop the whole system with collectors one could in principle collect all the light, just not at one spot.

To your 2nd post, "the system" must include the energy radiated away. The energy is not decreasing, but the entropy is. For a thought experiment, assume you are able to focus the light as tightly as you wish, and heat an object to greater than the source's temperature. You could then connect the two objects together and heat would flow from the hotter one to the cooler because that is what maximizes the entropy. Therefore we had violated the 2nd law when we heated the object above the source's temperature.

Your setup with the elliptical reflector would not work. Because the sun has finite size, you would find that at the other focus the light isn't directed to a single point, but at a sphere the same size as the sun. In this most extreme possible scenario, you could heat an object there to exactly the sun's temperature, but no higher. Within that sphere the light is not traveling radially inward to the focus, rather most of the light is missing the focus because it was emitted from the sun's surface, not its center.

There's a theorem in imaging that says you cannot focus a light source to create a beam any more intense then at the surface of what is emitting the light. A consequence of this is that you cannot heat something to hotter than the surface temperature of the sun by concentrating sunlight in any way, even if you had a lens the size of the solar system. The spot size that you get will just keep getting bigger.

Incidentally if you were able to do this it would violate the 2nd law of thermodynamics, because you would be moving energy from a cooler object to a warmer one without doing any work, thus decreasing the total entropy of the universe.

Depends on how you weigh it. If you truly measure weight, then yes. If you are really measuring mass, then no. For example: a spring scale will show a difference because the gravitational force is different. If you use a pan balance you will not see a difference, because both the subject and reference masses change their weights by the same fraction. Same goes for any true measurement of mass, such as penning traps or RFQ's.

You would need a good scale, but not extraordinarily good. A 1 kg weight would weigh ~100 mg weight different between the max and min.