Space Habitat

How much does it cost to build a space habitat? --Cosmic Cat


99¢ $9.95 $129.99 A hell of a lot of money. There are two approaches to figuring out the exact figure. One way is to take the price of a larger space-based object and divide that. Luckily, there is one rather large object that is (not) in space that has an estimated price: the Death Star. The steel for the Death Star would cost $852,000,000,000,000,000 in 2012 dollars.[1] It's probably safe to say that the cost of everything else doesn't add more than maybe one digit to that figure. The Death Star is big^{[original research?]}; much bigger than a typical space habitat, though. Let's assume that a space habitat will provide room for 10,000 people, since that seems to be minimum of two designs in a book I once saw.(fill in source) If we assume 1,000 square feet of living area per person plus an extra 5,000 square feet per person of non-living space, then a space habitat has a floor area of 60,000,000 square feet. Assuming that there are ten feet between floors, the total volume of the space habitat is 600,000,000 cubic feet.

Now, what is the volume of the Death Star? It's a sphere 160 kilometers in diameter[2], so the radius is 80 kilometers, which is 262,464 feet. Thus, the volume is 75 quadrillion cubic feet. This is about 100 million times the volume of our space habitat, so I'll assume it's 100 million times as expensive. This comes out to $8.52 billion, but building materials aren't everything, so by completely unsourced speculation, I'd estimate the habitat to cost $40-80 billion. Note that this doesn't count the cost of actually getting it anywhere and keeping it running. But hey, you asked how much it costs to build a space habitat, not how much it costs to maintain one.

There are bigger space habitats existing planned though. The O'Neill Cylinder is twenty miles long and could hold two or three million people. Since the cylinder would be twenty miles long and possibly about as wide (it doesn't really matter if it's a few miles wider or narrower). Its volume is 924 trillion feet, 81 times smaller
than the Death Star. Using the calculations above, this cylinder would cost $10.5 quadrillion. Like the antimatter car (link to first blog), it isn't something Bill Gates could afford. Interesting digression: the EPA has an estimate on the value of a human life: $7.4 million in 2006 dollars.[3] From this calculation, I can conclude that the entire population of earth is worth less than a fleet of six or so O'Neill Cylinders.

Yes, that is a very badly drawn alien. This is a science blog, not an art blog.

Now for the second method: take the cost of a smaller object and multiply it. The International Space Station cost somewhere in the neighborhood of $100 billion.[4] The ISS has roughly the living space of a six-bedroom house, which (presumably) is habitable for six people.[5] Multiplying upwards, the first habitat
(with 10000 people) would cost $167 trillion and the second habitat (with 2-3 million people) would cost $250 quadrillion. Those numbers to seem more reasonable than the first, though they (like pretty much every other number I come up with on this blog) are probably off by a few orders of magnitude.

Shoot Down an Asteroid

Would it be possible to stop an asteroid from hitting the earth by firing bullets at it? --Myself

I'll be analyzing one of my own thoughts this time. I'm sure literally everyone^[who?] has wondered about this. For the purposes of this post, I'll assume that we're talking about the asteroid 4942 Munroe. It's big enough to be interesting (6 to 10 kilometers across[1], so I'll go with 8 kilometers), but not big enough to actually destroy the earth. Asteroids typically travel at 25 kilometers per second.[2] Asteroid densities vary widely[3], but let's assume that 4942 Munroe has a density of 4 grams per cubic centimeter. From this, I can calculate that the momentum of 4942 Munroe is 1.28*10^19 kilogram-meters per second.

Now it's time to think about the guns. An AK-47 fires 600 rounds per minute at a velocity of 715 meters per second.[4] Each round weighs 7.9 grams.[5] If one AK-47 is fired straight at the asteroid, the asteroid will lose 715*0.0079*600=3,389.1 kilogram-meters per second every minute. At this rate it would take slightly over 7 billion years to bring 4942 Munroe to a halt. Chances are that there will be no earth to hit by then anyway.

Luckily for humanity, it's possible to do better, but it'll take some thinking. The ideal gun for this sort of task would have massive bullets with high velocity and a high rate of fire. The heavy artillery of World War I and World War II had massive ammunition and often that ammunition had a high velocity. The railway gun Krupp K5 had a shell mass of 255 kilograms and a muzzle velocity of 1120 meters per second. The low firing rate (0.25 rounds per minute) might be a problem, but let's see. 1120*255*0.25=71,400 kilogram-meters per second lost per minute. This would bring the time down to 341 million years, which is actually a lot better than the AK47 (but still impractical). Some cannons are better. The immense V-3 cannon has a far higher rate of fire and muzzle velocity, but only a slightly lower shell size. A V-3 cannon being fired at the asteroid would cause it to lose momentum at a rate of 1,050,000 kilogram-meters per minute, which brings the time down to under 24 million years. The anti-aircraft gun GAU-8 Avenger (link to WP) is slightly superior (13 million years). None of these are going to cut it though.

More modern weapons can do still better. Consider railguns, which are moving from experimental devices into ones with practical uses. The US Navy plans in 2016 to test a railgun with a firing rate of 10 rounds per minute, a muzzle velocity of 2500 meters per second, and a bullet mass of 10.5 kilograms[6] Even that is not enough: the stopping time would be nearly 100 million years. The last thing that I'll try is this Assuming a muzzle velocity of 1000 meters per second, it would take 2.4 million years for one of those to stop 4942 Munroe. Giving one to every person on earth would stop the asteroid in just under three hours, but that wouldn't be practical.

I guess we're doomed if a huge asteroid comes our way. Or not?

Freefall

If you jumped into a hole in the center of the earth, how long would it take to come to a stop? --Sodium

I feel compelled to explain why this is not entirely practical. For one thing, it's entirely likely that such a hole would collapse, but let's assume it's lined with a magic extra-strong material that also keeps the tunnel from reaching temperatures of thousands of degrees. The air pressure would also be a problem. We currently have only 62 miles (100 kilometers) of air above our heads, so having several thousand miles of air about our heads would likely crush us.

With that out of the way, let's turn to the actual scenario. I'll assume that this tunnel is a vacuum (so that the air pressure doesn't crush the jumper [and because physicists like working in a vacuum]) and that it's sealed at both ends (so the tunnel isn't filled with wind). I'll also assume that your aim is good enough that you don't hit any walls, which would (obviously) kill you.

The exact formula for figuring this out is more complicated than traditional gravitational acceleration since the mass that's actually pulling on you will decrease as you approach the center of the earth.[1] On average, the jumper's acceleration will be about 5 meters per second squared, but it does change some.[2] I'll assume that the change in acceleration (jerk) is constant, even though it isn't, to simplify the equations. Because I'm lazy!

I can't resist going off-topic to describe some rather interesting facts. As everyone (who's read this far at least) knows that velocity, acceleration, and jerk are the 1st, 2nd, and 3rd derivatives of motion. What you may not know is that jounce (or snap), crackle, pop, lock, drop, shot, and put are the 4th, 5th, 6th, 7th, 8th, 9th, and 10th derivatives of motion, respectively. Just an awesome fact that was worth sharing. I don't know why anyone would use these though.

Anyway, back to business. I wrote a bit of code which tells me the average jerk and also tells me that it would take about 22 minutes to fall to the center of the earth. This implies that it would take 21 minutes, so my math can't be that far off. It doesn't really matter though, since your momentum would carry you all the way back up to the surface, at which point you'd just repeat it all over again.[3]

I wonder how long it would take for a person to get bored of falling assuming they didn't know that they'd never hit the ground...

Apologies

I just want to apologize for the long delay with no posts. I have been busy with other things lately, but I will try to make a new post soon. Thanks,

--Jakob

Saving Electricity

Assuming you could get 100% citizen compliance in this country, and assuming you could come up with a number for "average household" use, then average household energy savings... What amount of energy could be saved, and what might that savings amount to in dollars? -- Sonya

Well, to start idle devices use as much as 10 percent of a typical household's energy use.[1] A typical household uses 10,837 kilowatt-hours per year and there are 116,716,292 households in the United States as of the 2010 US census.[2][3] (There are three random facts in the last two sentences. I love the internet.)

If each one unplugged their idle devices, the total energy saved would be...

...116,716,292*10,837*0.1 = 126,485,445,640 kilowatt-hours, or 3.5162954 exajoules.

That's the raw math, but how much energy is that really? It's more than twice the annual electricity consumption of South Korea and between two and three percent of the annual electricity consumption of the United States. It is also more than 15 times the explosive yield of the largest nuclear bomb ever detonated, the Tsar Bomba.[4]

The cost of all the electricity that would be saved? Well, in residential areas, electricity costs an average of 13.01 cents per kilowatt-hour, as of August 2014.[5] With 126,485,445,640 kilowatt-hours of electricity being saved, the total amount of money saved would be approximately 1.64 trillion cents (more pennies than are currently in circulation), or $16,455,756,477.80. That would look like 16 of these (those are 100 dollar bills) and would take 80 years to print out with a typical printer (again, assuming that you're printing 100 dollar bills).[6]

That sum of money would also be enough money to pay for electricity for a single residence for 11,671,635 years and 211.7 days. Which, by the way, is probably longer than anyone has owned a house for.^{[original research?]}

Penny Statue

How large of a sculpture could you make from melted-down pennies? -Myself

Today I'll be sharing a question that I considered myself: how large of a statue of a person could be made from melted-down US pennies?

While one might think of a penny as a disk, it is technically cylindrical. (OK, technically it's not cylindrical due to the relief images on both sides and whatnot. But that doesn't matter.) The diameter of a penny is 19.05 millimeters and the height of a penny is 1.52 millimeters.[1] From this, it is apparent that the volume of a penny is 285.02 cubic millimeters. So next I needed to find out how many pennies are in circulation. The answer (wait for it):  200,035,318,672.[2]

Given this, the total volume of all pennies currently in circulation is 57 trillion cubic millimeters, or 57,014 cubic meters. That is a lot of cubic meters.^{[citation needed]} It would be enough to fill up 57 Olympic swimming pools and still have enough to fill 14 refrigerators.[3] I love Wikipedia's "orders of magnitude" lists. The average height of an American male over 20 is 69.3 inches.[4] The average width of a human is 18 inches and the average...um...length? thickness? depth? of a human is 9.5 inches.[5] Thus, a person's volume is 11,850.3 cubic inches, or 0.194 cubic meters. Admittedly I fudged this a bit since people aren't rectangular prisms^{[dubious--discuss]}, but it's close enough.

The pennies have a volume 293886.5979 times that of a human. The cuberoot of that is about 66.48, so the statue would be 66.48 times taller, 66.48 times wider, and 66.48 times deeper than life-size. Plugging in the numbers shows that this statue would be 384 feet (117 meters) tall, 100 feet (30 meters) wide, and 53 feet (16 meters) deep. In other words, this statue would be large enough to rank among the largest statues ever built, but not large enough to blatantly defy reality.

Derived from this diagram, created by Jdcollins13. Image is licensed under the CC-BY-SA.

(The statue made of pennies is not anatomically accurate or precise)

With the math out of the way, let's turn to the physics of this statue.

These days, pennies are mostly zinc and the density of zinc is 7.140 grams per cubic centimeter and thus 7,140 kilograms per cubic meter according to Theodore Gray's The Elements. With a volume of 57,014 cubic meters, the statue would have a weight of 407,079.96 metric tons. That's a manageable weight, since many structures that size have been built. For instance, the Great Pyramid of Giza is nearly ten times heavier.[6]

This would cause quite a bit of pressure on the feet of the statue. Some source[7] says that the average 19 year old male foot is 26.15 centimeters long. I also estimate based on a chart here that the typical male foot is 10 centimeters wide. Thus, the area of a both human feet combined is 261.5 square centimeters, or 0.02615 square meters. The statue's feet would have a combined area of 1155.72 square meters. Thus, the pressure that the statue would exert on the ground it is standing on would be 352,230 kilograms per square meter, or about 3.453 megapascals. I suspect this may cause some problems, but they could probably be alleviated by putting the statue on a large concrete base.

In case you want to try this, just be warned that it would be pretty expensive (and apparently illegal[8]). The cost of the pennies would only be part of the cost of building the statue, since you'd have to melt the copper and shape it as well.

Atom Or Universe?

Is a person proportionally closer in size to an atom or the universe?

This is a good question for using Fermi Estimation.

A person is about one meter across. At atom is about a hundred picometers across and the universe is about 100 sextillion kilometers across. If we convert these values to scientific notation, an atom is
10^-10 meters across, a person is 10⁰ meters across, and the universe is 10²⁶ meters across. Thus, the universe is 36 orders of magnitude larger than an atom, so an object that is in between the two in size would be 18 orders of magnitude larger than an atom and 36 orders of magnitude smaller than the universe. 26-18 (and for that matter -10+18) is 8, so an in-between object would have a diameter of 100,000,000 meters 100,000 kilometers. This is smaller than Jupiter and Saturn, but larger than the other two gas giants (Uranus and Neptune) and much larger than Earth.[1] It's also probably larger than you are unless you happen to be a gas giant or star (as far as I'm aware, there aren't any reading my blog).

Oh. Right. Sorry. Anyway, I'll bring some real numbers into the equation and see if that changes anything. The atomic radius of hydrogen (the most common atom in the
universe) is 53 picometers[2] so the diameter would be 106 picometers. The average height of an American male over 20 is 69.3 inches[3], or 1.76022 meters. The obeservable universe is 92 billion light years across.[4]

There are 9.4605284 quadrillion meters in a light year, so the observable universe is 8.703686128*10²⁶ meters across. So far, it seems that my Fermi Estimation was
pretty close to the actual sizes (within an order of magnitude or so, which is pretty close considering the roughness of the estimates). Let's plug in these new values.

A hydrogen atom is 1.06*10^-12 meters across, a human is 1.76022*10⁰ meters acorss, and the observable universe is 8.703686128*10²⁶ meters across. Thus, the observable universe is 9.225907296*10³⁸ times the size of a hydrogen atom. Some plugging in of numbers on my calculator shows that an in-between object is 3.037417867*10^19 times the size of a hydrogen atom.
This comes out to 28,654,885.53 meters. It's a bit closer to the size of a human than the result I got with Fermi Estimation. It's twice the size of the largest terrestrial planets and half the size of the smallest gas giants.[5]

Does this make you feel small and insignificant? (You are small and insignificant) Well...

I was counting by diameter in the previous examples. But since humans are full of mass and the universe is rather diffuse, maybe counting by mass will bring a more satisfactory result. A typical human weighs 62 kilograms[6], but weights of individuals can vary wildly.^{[citation needed]}. A hydrogen atom weighs 1.673534*10^-27 kilograms.[7] The mass of a hydrogen atom tends not to vary so much. The typical obsevable universe has a mass of 10^53 kilograms (the average of all known observable universes).[8] This figure isn't known to differ from individual to individual.

So the mass of the universe is 5.98*10^79 times heavier than the mass of a hydrogen atom. Thus, an object with a mass in between that of a hydrogen atom and that of the universe would be very roughly 10^40 times heavier than a hydrogen atom, or 1.673534*10^13 kilograms. This result is the best of all. It's far, far smaller than any planet (though obviously nowhere near as small as a person). This Wikipedia article tells me that such a mass would be about as heavy as a fairly large mountain.

I guess that still makes you feel small and insignificant. Oh well.