When I got my copy of Randall Munroe's (of the amazing webcomic xkcd) new book What If?, I had been holding out hope that in it he would finally answer some of the (multiple) questions I've asked him. Unfortunately, he didn't (although he did answer a very similar one from someone else). So I guess the only thing left for me to do is to take a crack at answering them myself. Obviously, I am not a webcomic artist or an astrophysicist who used to work at NASA, so my answers won't be nearly as funny, well-illustrated, or (possibly) correct as Randall's would be. But hopefully they'll still be worth it. Let's go back through my old Emails...
If people packed shoulder-to-shoulder on every floor of a tall building and jumped, would they have any chance of bringing the building down?
Great question, me! As our example skyscraper, I'll use the shiny new One World Trade Center; since it was built to survive being hit by a hijacked airplane or a truck bomb, it should have at least as much of a chance as any other skyscraper.
According to Wikipedia, the "Freedom Tower" has 86 usable floors. The base is a 200-foot square footprint whose edges get chamfered as they rise up the building's side, ending in a smaller square rotated 45 degrees that fits in the first. The bottom floor (above the base) has an area of 40,000 ft² and the top has an area of half that, 20,000 ft². Let's assume that the average usable floor is 30,000 ft² ≈ 2,755 m². Let's further assume that 95% of that area can be stood on (i.e. is not a wall, elevator shaft, etc.), or 2617 m². If we assume three people can fit in a square meter, that's 7851 people per floor, or 675206 people total (nevermind the hellish logistics of getting them all in there, which would be almost as bad as Randall explains in his similar answer here).
If I then assume that an average adult weighs 75 kg and can jump 0.3 meters in the air (per this article), they would be traveling at v = √(2gh) = √(2 * 9.8 * 0.3) = 2.42 m/s when they came down. The force exerted by an impact is the change in momentum divided by time. How much time does it take for you to stop moving after you hit the ground while jumping? It's hard to find good figures, but from this solid-looking question from a physics textbook and some testing that would no doubt confuse my roommate if he saw it, I'm going to be conservative and say about 0.1 seconds if you're smart and bend your knees.
So, the force exerted by a 75kg person when landing from a jump is about 2.42 * 75 / 0.1 = 1,815 N, decelerating the person at about 2.5 g. The force from all 675,206 people would then be about 1.225 GN (giganewtons), the equivalent of about 125,000 metric tons. All of this force will be transmitted to the Freedom Tower's base. Will it survive? Well, considering how each of the old towers weighed about 500,000 tons (I can't find data on the new one), it's safe to say that this additional load should be well within the structure's margin of safety. That [architect of One World Trade Center] David Childs really thought of everything.
I hope it isn't in bad taste that I originally answered this question on September 11th.
What would happen if you somehow brought a cubic meter of neutron star matter to earth? (Both the actual case where it would probably explosively decompress, and the hypothetical case where it stayed together in a solid unit)
Randall almost answered this one in his new book. The actual question was, "If a bullet with the density of a neutron star were fired from a handgun (ignoring the how) at the Earth's surface, would the Earth be destroyed?"
My guess of what would actually happen was right. It would indeed explosively decompress into superhot normal matter, apparently releasing more energy than a nuclear bomb. So Randall assumes it somehow stays in its superdense state; the bullet would weigh as much as the Empire State Building.
It wouldn't matter much if the bullet were fired or dropped. It would immediately burrow its weight to the center of the earth, forming an underground shooting star, and would then sit there pretty uneventfully. He then explores what would happen if you could somehow keep it on the earth's surface. (Answer: if you tried to touch it, it would try to rip your arm off with gravity; surrounding it with water would allow buoyancy to cancel out the gravity and maybe, just maybe, allow you to touch it) It was a pretty cool question. But my question was about considerably more neutron star material. Let's see what happens...
According to Wikipedia, neutron stars have an average density of 3.7–5.9 × 1017 kg/m3. So our cubic meter of neutron star matter would weight about three to six hundred million billion kilograms. The Empire State Building weighs 365,000 imperial tons, or about 332,000 metric tons. So (if we take a middling estimate of the neutron star matter's density, 4.5 × 1017 kg/m3), our sample would have the mass of about 1.2 billion Empire State Buildings. Incidentally, it appears that Randall was (gasp!) wrong in his answer about the mass of the bullet. Assuming its volume is a teaspoon, it would weigh about as much as 7,410 Empire State Buildings, not just one. Sadly, this means you probably wouldn't be able to get within about twenty meters of it.
Anyway, our sphere (I'll assume it forms a sphere as expected rather than a cube) of neutron star matter is much bigger and much more massive. This means that its gravity becomes much more appreciable. The results are hard for me to imagine. The surface of the sphere will have a gravitation acceleration of nearly eight million g—four times that of the best ultracentrifuges. At ten meters, the sphere's gravitational force would still be equal to about thirty thousand g. This decreases to (only) 306 g at 100 meters and 3 g at a kilometer. An easier measure might be that due to its smaller mass and radius, our neutron sphere has an escape velocity of 9.8 km/s, nearly equal to that of the earth. This also means that at six kilometers out, the gravity of the sphere would make you feel like you were on a slope with a grade of about 1:11.75—steep enough to constitute a violation of the Americans with Disabilities Act.
If you somehow kept this sphere together, it would, of course, fall to the center of the earth, probably causing a good deal more damage on the surface as it did so. If you managed to keep it on the surface as well, things would get pretty weird. If we assume the ground around the sphere is made of reasonably firm dirt with an angle of repose of 45°, all the dirt within about 1.7 kilometers of the sphere (along with anything on top of it) would avalanche towards it. As it turns out, because of the inverse-square law, an object just five meters from the sphere (the equivalent of dropping something to earth from beyond geostationary orbit) will, ignoring air resistance, have picked up 95% of its escape velocity when it impacts. It's hard for me to specify exactly what would happen, but the energy created by the hypersonic impact would probably be contained by the continuing avalanche of dirt, eventually resulting in a large, hot hill developing around the sphere. It puts me in mind of a certain song by Megadeth: "High Speed Dirt".
What would the atmospheric pressure be at the bottom of the Mariana Trench if you took out all the water?
This is much easier than the last question. The relationship between atmospheric pressure and altitude is common knowledge. The side effects of draining the world's oceans (which Randall studies exhaustively in the book) would slightly change things since there would no longer be oceans to displace the atmosphere upward, but this won't significantly change our results (I think). Plugging in the depth of Mariana Trench gives us a pressure of 3.65 atmospheres at the bottom. Apparently, if you stayed in this pressure for several hours, you could develop pulmonary oxygen toxicity, whose main symptom is respiratory inflammation. We all know that humans need supplementary oxygen to survive at high altitudes, but apparently exceedingly low altitudes have problems of their own.
The more interesting part has to do with the atmospheric lapse rate, or how quickly it gets colder as you go higher. It is about 6.5 K per kilometer. As you might guess, it also works in reverse. With the Mariana Trench being 10.911 kilometers deep, the air at the bottom should be about 70(!) degrees Celsius hotter than at sea level—potentially hot enough to spontaneously boil water, if water didn't boil at 140° C due to the increased pressure. Also, due to the Mariana Trench being over five times the depth of the Grand Canyon, it's hard to predict what kind of climatic effects will moderate this temperature increase (I don't imagine the bottom of the trench would get much sunlight).
The Earth has come closer to this scenario than you might think. During the Messinian salinity crisis, which began about six million years ago, the Strait of Gibraltar closed off and the Mediterranean Sea dried up into a sort of super Dead Sea, a hypersaline lake surrounded by a desertified abyssal basin where temperatures may have reached 80° C. Rivers that fed into the basin, like the Nile, cut deep gorges as they ran down to several kilometers below their current mouths. African species like hippopotami migrated across the basin before it got too hot and dry, then were stranded on cooler highlands like Malta and underwent island dwarfism. The crisis finally ended with the Zanclean flood, in which the sea refilled through the Strait of Gibraltar at a rate of about a thousand times the discharge rate of today's Amazon River.
A common trope in anime, video games, or other media is to depict the moon as much larger in the sky than in real life, often seeming to fill half the sky. If the moon were actually this large (or, alternately, this close to the earth), how would it affect life on earth, gravity, the tides, etc.?
Moon has a maximum angular size of about 0.57°.
This can't end well.
If we increase the size of the moon to match this new angular size but keep its distance the same, we get a new lunar mean radius of about 33,300 km, as opposed to 1,700 km for the real Moon. Or 6,300 km for the Earth. This truly super-Moon would be by far the largest rocky body in the Solar System. It would not orbit around the Earth; the Earth would orbit around it in just under three days. If we naively scale up the real moon in its proportions, this truly super Moon would have the mass of 87 Earths, nearly as much as Saturn, and a surface gravity of over 3 g. This moon would also have a gravity differential over the earth more than seven thousand times that of the regular moon, which would probably cause mile-high tides or something. (In real life, the earth would be tidally locked to the super-Moon just as the regular moon is to the earth)
If, on the other hand, we move the moon closer to the Earth so it appears this large, we get a new semimajor axis of 20,000 km, about 5% of the old semimajor axis of 384,000 km. This is just over three Earth radii and perilously close to the Moon's Roche limit, the distance at which tidal forces from the Earth tear the Moon apart and turn it into a ring system. It will have a new orbital period of less than eight hours and produce even higher tides than the super-Moon.
Here is an extreme example. Considering the fisheye effect in use here, let's suppose this moon has an angular size of 30°. Now things get really ridiculous. Scaling the moon up to these proportions gives it a ludicrous radius of 103,000 km, over a quarter of the distance to the Earth, and about 150% of the radius and eight times the mass of Jupiter. Its gravitational pull would cause your weight to tangibly fluctuate with the tides, which would be tens or hundreds of thousands of their current proportions. Again, I don't know the science involved with packing this must dirt together with these kinds of forces, but this is probably astrophysically impossible.
Moving the moon this close to earth puts it at a distance of just 6,500 km, giving it a new orbital period of less than 90 minutes and off-the-scale tides, but that doesn't matter because before you can get the Moon this close it will collide with the Earth and kill us all.
|This is actually NASA's conception of the impact that created the Moon, but the actual result would be similar.|
What would happen if you could somehow connect two planets (say, Earth and Jupiter) with an unbreakable, unstretchable tether? Or an unbreakable, rigid girder?
Bad, bad things.
I wasn't sure how to answer this from a purely physical standpoint, so I wrote a quick Python simulation to model the situation. The results are interesting. If you connect the two planets at their point of conjunction (so they are about 4.2 AU apart), Earth basically acts a a pendulum hanging towards the Sun from Jupiter. Meanwhile, Jupiter's distance from the sun varies surprisingly regularly from about 5.3 AU to 4.64 AU over a 16-year year period; I think the Earth's swinging motion off the tether (which gets faster or slower as it gets closer to or farther from the Sun) acts somewhat like pumping your legs on a swing to go higher or lower; the force of the Sun's gravity on Earth, transmitted to Jupiter through the tether, either pulls it higher or lower in its orbit.
What this means for Earth is that instead of a normal year, it has a pendulum-like swing cycle that lasts about 10 months at its/Jupiter's furthest point from the Sun (where its distance varies from about 1 AU to 1.7 AU, beyond the orbit of Mars) and 4 months at their nearest approach (where the distance varies from about 0.45 to 1 AU, within the orbit of Mercury) with about 8 years elapsing between the high and low points of Earth/Jupiter's orbit. For reference, the habitable zone of the Solar System is (very roughly) around 0.75 AU to 1.4 AU. Earth's "orbit" will take it close enough to the Sun to boil the oceans and far enough away to freeze them. Presumably the atmosphere would exert some kind of moderating effect on these wild temperature swings, but things look pretty grim. There is also the risk of planetary collision with Mercury, Venus, and Mars to worry about.
If the Earth and Jupiter start out in opposition, then the tether obviously does nothing. (Even assuming it is indestructible and can survive passing through the Sun) What about if Earth and Jupiter start out 90° apart in their orbits?
|That's not good.|
What (roughly) would Mars look like with all the water we drained from the Earth's oceans on it? How would the water affect its climate?
Randall actually answered this one! (As asked by someone else)
What if every human being on earth used all their mechanical power (say, on exercise bicycles) to heat and boil the oceans? Would this have any noticeable effect on water levels or the weather? What if we also turned the power we generate from other sources (turbines, generators, cars, etc.) to this purpose?
The average power output of someone working hard is about 500 W. Assuming the fit people are able to balance out the infirm/children, the human race should be able to produce about 7 × 109 × 500 = 3.5 TW. Impressively, this is about a fifth of the total power consumption of the world, so the answers of the two parts of the question are more similar than I expected. Assuming we are boiling water from the surface of the ocean, which has an average temperature of 17 °C, we could boil about 1,334 tons of seawater per second. This becomes about 8,000 tons/second if we include our other means of generating power, which works out to roughly 250 km3 of water per year—just 0.05% of the global evapotranspiration caused by the sun. So it looks like the Sun wins this one. If we were smart enough to actually capture all that distilled water instead of letting it escape into the atmosphere and return as rain, though, we could solve California's water woes 34 times over. This sounds nice, but put another way it means that distilling enough water to satisfy California would require about 615 gigawatts, 3.8% of the energy generated worldwide. With that much power, we could power over 500 time machines to just go back in time and tell California to use less water.
|In homage to https://xkcd.com/656/.|
This question was inspired by this one, in which Randall describes the catastrophic consequences of speeding the Earth's rotation so that a day lasts one second. So I wondered, what about some more moderate day lengths? What is the shortest day the Earth could have and still have things remain "normal"? Jupiter has a ten-hour day; at this angular speed, the Earth's surface would be moving at around 1.1 km/s instead of the usual 0.46 km/s. The effects of the centrifugal force would not be enough to noticeably counteract Earth's gravity, so the most catastrophic effect would be having to adjust to a ten-hour day.
A two-hour day would cut the Earth's apparent gravity in half. This would likely be awesome and extremely fun. Swimming and sports would be more exciting, people would travel by bunny-hopping everywhere, and we might already have flying cars. The effects would probably be similar to those of increased gravity as described by Randall in this answer, only reversed. On the other hand, the Coriolis effect would be much stronger, potentially increasing the incidence of hurricanes, and the atmosphere would be less dense, which might make it harder to enjoy your newfound antigravity powers.
You can't go much further than a two-hour day. At about 84 minutes (just over the length of a day in Skyrim), the centrifugal force at the equator equals the force of gravity. Long before this, the Earth would deform into an oblate spheroid (more noticeably than it already has) and accordingly slow its rotation. With a one-hour day or less, the Earth's mass around the Equator would break off and fly out into space (moving at escape velocity), though without the awesome consequences of the one-second day.
What would be the effects on Earth if its axial tilt were 90°, like that of Uranus?
Increasing the Earth's axial tilt from 23.5° to 90° would have some pretty drastic effects on the climate. As commentors in this discussion say, this would basically mean that the Equator and the Arctic/Antarctic circles would become the same. Everywhere on earth would get experience the midnight sun and polar night for part of the year. I'll break down what the day-night-year cycle would look like at a few select latitudes:
- The Equator: Exactly 12 hours of daylight every day of the year (just like the real Equator), except on the solstices. The Sun's maximum elevation during these days ranges from 90° at the equinoxes (the Sun would travel directly across the middle of the sky) to very low around the solstices; it would just barely peek about the horizon, albeit for longer than it does at the real-life poles (still 12 hours). At the solstices, the Sun would circle the entire horizon without actually rising above it, creating 24-hour twilight. This Equator would probably have more extreme seasons than the real one; it would likely get quite cold at the solstices, and would be hot around the equinoxes much like in real life.
- 30°: Two months of midnight sun, from about May 21st to July 21st, polar night from about November 21st to February 21st. Lahaina Noon on about August 21st and April 21st. At the Summer solstice, the Sun would hang at 30° in the sky all day; consequently, Summers would be surprisingly cool. Spring and Fall would be very hot due to the high insolation and winter would be bitterly cold, but this might end up being one of the more habitable latitudes.
- 45°: Three months of midnight sun from about May 6th to August 6th, polar night from about November 6th to March 6th. The days of maximum insolation would also be May 6th and August 6th; on these days the Sun rises from the horizon at midnight to directly overhead at midday; these days would be sweltering. At the Summer solstice the Sun would stay at 45° all day. Late Spring and Summer would be brutal, but at least you get to look forward to three months of icy darkness to make up for it!
- 65°: Near the real-life Arctic/Antarctic circles. Four-plus months of midnight sun, from about April 16th to August 26th, and polar night from about October 16th to February 24th. At the Summer solstice the Sun is at a constant 65° of elevation, and during the entire month of June it is at over 45° 24 hours a day. On May 26th and about July 14th, the Sun rises from 45° to directly overhead. In case you haven't noticed, things are getting worse the closer we get to the poles.
- The North Pole: The Sun's elevation is constant throughout the day, every day. Six months of midnight sun, six months of polar night. We have no Earthly analogue for what the Summer solstice would be like: the Sun would stay over nearly the same point on the Earth's surface for weeks, causing unimaginable amounts of heating and evaporation. Meanwhile the Winters would be even colder than those of our poles.
In a nutshell, the seasons would become much more extreme than those of the real Earth the further you go away from the Equator. Closer to the Equator, the equinoxes would be hot (as hot as the real-life Equator) and the solstices would be cold, but not uninhabitably so. Most people would probably live close to the Equator. I don't feel qualified to speculate about the plant and animal life that would inhabit this alternate Earth. I really think Randall should answer this one; it would be a good counterpart to his fascinating article Cassini.