Would You Notice If An Atom Of Anti-Hydrogen Annihilated In Your Room?

If an atom of anti-hydrogen came into existence in the room you are in, would you notice?
One of the first lead-lead collisions at the Large Hadron Collider, recorded by the ALICE detector in November 2010. In this collision of lead nuclei at a small impact parameter (central collision), 1209 positively-charged (darker tracks) and 1197 negatively-charged (lighter tracks) particles are produced. Image credit:  CERN, for the benefit of the ALICE Collaboration (License:  CC-BY-4.0 )

One of the first lead-lead collisions at the Large Hadron Collider, recorded by the ALICE detector in November 2010. In this collision of lead nuclei at a small impact parameter (central collision), 1209 positively-charged (darker tracks) and 1197 negatively-charged (lighter tracks) particles are produced. Image credit:  CERN, for the benefit of the ALICE Collaboration (License: CC-BY-4.0)

Originally posted at Forbes!

Anti-hydrogen is the antimatter equivalent of the hydrogen atom. The simplest atom in our Universe, hydrogen is usually made of a single proton and a single electron. Hydrogen is also one of the most abundant elements in our Universe by a large margin, but its antimatter counterpart has been rather difficult to study. Anti-hydrogen was recently in the news, as the folks at CERN have only recently succeeded in measuring the spectrum of light that a positron (anti-electron) produces when it is bound to an antiproton. So far as we can tell, anti-hydrogen makes exactly the same spectrum as regular old hydrogen which is good news for the current model of particle physics.

Anti-matter’s main feature is that it will rather catastrophically annihilate if it comes into contact with regular matter. These matter/antimatter annihilations are cases for Einstein’s E=mc^2 equivalence, as the energy that is produced by the destruction of the antiparticle and the particle itself is equivalent to the mass of the two particles, multiplied by c, the speed of light, squared. If an anti-hydrogen came into being in the room that you’re standing in, the first thing we would expect it to do is annihilate as soon as it runs into an air molecule. (I must note that an atom of anti-hydrogen spontaneously coming into being in your living room is extremely unlikely, or the ATLAS team would have a much easier time trying to coax anti-hydrogen atoms together, instead of doing the painstaking work to collect atoms together.)

A positron, our antimatter equivalent of the electron, when it annihilates with an electron, creates gamma radiation. This is about as clean an annihilation as you’re going to get -- the two particles convert their mass directly into light, with no intermediary cascade of other particles. Gamma radiation is generally bad for humans in large doses, but a single positron/electron annihilation event only produces two gamma ray photons. That’s it. And the photons go in opposite directions, so you’re only ever going to get hit by one of them.

This track is an example of simulated data modelled for the CMS detector on the Large Hadron Collider (LHC) at CERN, which will begin taking data in 2008. The Higgs boson is produced in the collision of two protons at 14 TeV and quickly decays into four muons, a type of heavy electron which is not absorbed by the detector. The tracks of the other products of the collision are shown by lines and the energy deposited in the detector is shown in blue. Image credit: Lucas Taylor/CERN (License: CC-BY-SA-4.0)

This track is an example of simulated data modelled for the CMS detector on the Large Hadron Collider (LHC) at CERN, which will begin taking data in 2008. The Higgs boson is produced in the collision of two protons at 14 TeV and quickly decays into four muons, a type of heavy electron which is not absorbed by the detector. The tracks of the other products of the collision are shown by lines and the energy deposited in the detector is shown in blue. Image credit: Lucas Taylor/CERN (License: CC-BY-SA-4.0)

Proton/antiproton annihilation tends to be a bit messier, with more of a cascade of intermediary, unstable particles. In the end, these events also produce high energy light particles, though they’re not usually gamma rays.

To get a handle on how much of a radiation dose you could potentially have from a single atom of antihydrogen, I’m assuming that the entire atom annihilates into light, with no leftover particles like neutrinos (which are a common byproduct of proton/antiproton collisions). This means that we’d be left with the maximal amount of energy as gamma radiation. If you convert the entirety of a hydrogen atom into energy, you get 1.5 x 10^-10 Joules. That looks like a very small number, and it is; but gamma radiation is usually bad for you, right? Let’s calculate the dose.

UGC 1382 appeared to be a simple elliptical galaxy, based on optical data from the Sloan Digital Sky Survey (SDSS). But spiral arms emerged when astronomers incorporated ultraviolet data from the Galaxy Evolution Explorer (GALEX). Combining that with a view of low-density hydrogen gas (shown in green), detected at radio wavelengths by the Very Large Array, scientists discovered that UGC 1382 is a giant. Image credit: NASA/JPL/Caltech/SDSS/NRAO

UGC 1382 appeared to be a simple elliptical galaxy, based on optical data from the Sloan Digital Sky Survey (SDSS). But spiral arms emerged when astronomers incorporated ultraviolet data from the Galaxy Evolution Explorer (GALEX). Combining that with a view of low-density hydrogen gas (shown in green), detected at radio wavelengths by the Very Large Array, scientists discovered that UGC 1382 is a giant. Image credit: NASA/JPL/Caltech/SDSS/NRAO

Sieverts are used to calculate radiation doses, and they’re measured in Joules/kilogram. If you are a person, on average, you are made of 80 kg of material. If you divide our tiny amount of joules by 80 kilograms, you get a radiation dose of 1.87 x 10^-12 Sievert. A small number here is good, because a single Sievert of radiation exposure is usually considered to be not something you should do every day. It's actually the career limit for cumulative exposure for a female astronaut at age 25. A milliSievert (0.001 Sievert) per year is the recommended threshold to stay underneath if you’re a member of the general public.

However, our number is so low that it is below the Banana Equivalent Dose, which is the amount of extra radiation your body endures in the time immediately after eating a banana, and is a comically small amount of radiation. The Banana Equivalent Dose is 9.8 x 10^-8 Sieverts; our atom of antihydrogen is 1,800 times less of a dose than eating a banana.

To give you a sense of how little banana we’re working with, I weighed a banana I had: 137 grams. Dividing that by 1800 gives us 0.07 grams of banana. My kitchen scale doesn’t work on precision that small, so here’s my analogy; eating a piece of banana about the size of the last digit of your little finger is 26 times more of a radiation dose that you’d be exposed to from a single atom of antihydrogen annihilating in your room.

You wouldn’t notice.

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If Heat Is Energy, How Does My Window Let The Cold In?

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How Much Science Is Behind Star Wars' Starkiller Base?

Is there any real science behind the weapon, “Star Killer” from Star Wars: Force Awakens?
Artist’s impression of a black hole feasting on matter from its companion star in a binary system. Material flows from the star towards the black hole and gathers in a disc, where it is heated up, shining brightly at optical, ultraviolet and X-ray wavelengths before spiralling into the black hole. Part of the disc material does not end up onto the black hole but is ejected the form of two powerful jets of particles. On 15 June 2015, the black-hole binary system V404 Cygni started showing signs of extraordinary activity, something that had not happened since 1989. The system consists of a black hole about twelve times more massive than the Sun and a companion star about half as massive as the Sun. The renewed activity is likely caused by material slowly piling up in the disc, until eventually reaching a tipping point that dramatically changes the black hole’s feeding routine for a short period. Since the first signs of such unusual activity, astronomers worldwide have been observing this exceptional system with ground-based telescopes and space-based observatories, monitoring the source at many different wavelengths across the electromagnetic spectrum. Image credit: ESA/ATG medialab

Artist’s impression of a black hole feasting on matter from its companion star in a binary system. Material flows from the star towards the black hole and gathers in a disc, where it is heated up, shining brightly at optical, ultraviolet and X-ray wavelengths before spiralling into the black hole. Part of the disc material does not end up onto the black hole but is ejected the form of two powerful jets of particles. On 15 June 2015, the black-hole binary system V404 Cygni started showing signs of extraordinary activity, something that had not happened since 1989. The system consists of a black hole about twelve times more massive than the Sun and a companion star about half as massive as the Sun. The renewed activity is likely caused by material slowly piling up in the disc, until eventually reaching a tipping point that dramatically changes the black hole’s feeding routine for a short period. Since the first signs of such unusual activity, astronomers worldwide have been observing this exceptional system with ground-based telescopes and space-based observatories, monitoring the source at many different wavelengths across the electromagnetic spectrum. Image credit: ESA/ATG medialab

Originally posted at Forbes!

Here’s the thing with looking for science in the Star Wars universe: you’re already in a world where lightsabers work, The Force is a real thing, and faster-than-light travel is possible. This is not a universe where our laws of physics are consistently applied, so a healthy dose of suspension of belief is required. When I go into a Star Wars movie I’m not really looking for perfect physics. So what I’m going to do here is lay out a couple of issues with Starkiller Base, were it to be found in our Universe, which it isn’t. Star Wars fans, please feel free to invoke your favorite technobabble to get around any/all of these problems.

With that caveat out of the way, we can certainly look at the feasibility of Starkiller Base. Starkiller Base, in a nutshell, is a planet which has had some of its planetary innards removed and replaced with a fairly sizable weapon. This weapon is charged by finding a star, and vacuuming the star in its entirety into the planet. This star-energy is held inside the planet by plot devices, and then the star’s energy is fired out of the planet to vaporize unsuspecting planets.

The puzzling, fascinating surface of Jupiter’s icy moon Europa looms large in this newly-reprocessed color view, made from images taken by NASA’s Galileo spacecraft in the late 1990s. Image credit: NASA/JPL-Caltech/SETI Institute

The puzzling, fascinating surface of Jupiter’s icy moon Europa looms large in this newly-reprocessed color view, made from images taken by NASA’s Galileo spacecraft in the late 1990s. Image credit: NASA/JPL-Caltech/SETI Institute

Firstly; even though Starkiller Base is covered in ice, it’s still a rocky planet, which we know because we saw it starting to come apart at the seams towards the end of the movie. There’s no explicit size for the planet given, so I’m going to assume that it’s about Earth-size. I have two justifications for this assumption: the first is that everyone is walking around on it normally, which means its gravity has to be about normal, and second, it makes my life easy.

If Starkiller Base is about Earth sized, then that trench going around its equator is really deep. I did some rough math, based on the size of the planet in images and the depth of the trench in images, and scaling to the size of the earth. If Starkiller is Earth-sized, that trench is 581 miles deep. For scale, if you started at the southern border of Colorado, and drove north until you had hit the northern edge of Wyoming, you would have driven 560 miles. Google Maps informs methat’s over an 8 hour drive.

Fun fact: The Earth’s crust, at its deepest, is a little under 44 miles thick. This trench, if we were to install it on our own Earth, would be more than ten times deeper than our deepest rock, and would plant the base of it firmly in the mantle, which is made of a plastic-y form of melted rock. Installing this on the Earth would almost definitely cause some havoc with our plate tectonics, assuming of course that we can insulate such a structure from our planetary heat to keep it from melting immediately. But perhaps Starkiller Base doesn’t have tectonics (or an internal mantle), as it’s covered in ice and we didn’t see any volcanoes lurking on the horizon.

Secondly; Starkiller base can absorb an entire star in a pretty short period of time. Our universe does contain an object which can siphon material away from a star, but that object is a black hole. Starkiller base is definitely not a black hole, even at its core. A black hole would be gravitationally distressing to walk near, and it is a terrible place to put spacecraft, which would have to spend a huge amount of energy to escape its gravitational influence. Furthermore, even a black hole, as the most extreme gravitational object possible, is not very good at killing stars.

Artist’s rendering showing the black hole and its accretion disk with a disk wind fully established. Image credit: NASA/Goddard Space Flight Center/CI Lab

Artist’s rendering showing the black hole and its accretion disk with a disk wind fully established. Image credit: NASA/Goddard Space Flight Center/CI Lab

Black holes are frequently found in binary systems with other stars in our universe. This happens if two stars have spent their whole lives circling each other, having formed near each other, but one of them was slightly more massive than the other.  The more massive the star, the shorter its lifespan. The more massive star can go through its death throes and create itself again as a black hole, all without moving from its orbital dance with the other star. These binary black holes do drain the outer layers of their companion stars away, but it is a long, slow, tedious, & inefficient process, which drags on over millennia. It’s certainly not the quick deflation of the star we saw in the movie.

I’m going to skip over the whole “storing of an entire star’s worth of energy within a much smaller planet” bit, because that’s even less plausible than the rest of the premise of Starkiller Base. The movie seems to recognize this, so there’s a whole plot device dedicated to fixing this problem; the thermal oscillator. In the movie it’s a handy device that lets you not destroy the planet with all the matter/energy you’re storing, while also not compressing your matter/energy into a black hole. Thermal oscillators do not exist, so we’re well into Star Wars Physics and not our own Universe’s physics with this part of the story.

However, there was one piece of glorious physics to Starkiller base- right at the end. After the heroes manage to explode all possible mechanisms of keeping a whole star contained within the planet, physics as we know it returned to the scene. If you have a star’s worth of matter, no longer contained artificially, in a region of space which is dense enough to create a star, but not dense enough to collapse into a black hole, what you get is a star!  So as soon as all of the pieces keeping Star Wars physics operational were removed, it makes perfect sense for Starkiller Base to be consumed in the making of a brand new star, right next to where the old one had been.

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What would actually happen to the moon at the end of the movie Iron Sky?

In the movie Iron Sky, towards the end of the movie they put a massive crater in the moon. Would gravity pull the moon back into a sphere, and how long would that take? What would happen to the moon’s orbit?

I have to start by saying that, from a scientific perspective, there are a couple of really major problems with this scenario, before we even get to the physics of the energy required to make such a crater.

The first is that looking at the picture above, the moon is blocking out the sun, and yet, a quarter of the surface is illuminated. This is geometrically impossible; the sun illuminates the surface of the moon that faces the sun - there’s no way for sunlight to be reflected off the surface facing away from the sun.

Secondly, the way the crater appears, it looks like the moon is a flat disk that had a notch bitten out of it, like a cookie. The same is true if you watch the clip of the actual explosion, where they shoot a hole in the horizon so that they can see the Earth. The problem with this is that the moon is a sphere, not a disk, so any divot like that would have to be a cylindrical trench in the surface of the moon, not a spherical one like normal craters make. This cylindrical hole would only appear as a semicircular divot from a few positions in space, so that final shot would not see the chunk taken out of the moon as so clean a semicircle, since it’s looking from a different angle than the original firing.

To figure out what kind of damage the creation of a crater like this would do to the moon, I had to make a couple of simplifying assumptions. The major one was that the depth of the hole we see in the final shot is the depth of the crater, and that the width was equal to the width of the crater. I also assumed that the crater was round, and not a tunnel, since the explosion looked round in the clip. (It’s really hard to make cylindrical holes without a lot of prior planning.) With this information, I could calculate the volume of the crater, and using the average density of the moon, I can find the mass of the moon that had to be removed to make the crater. The above picture of the moon was very useful for this calculation, since I could measure the scale of the crater relative to the entire moon, which allowed me to work out that the crater is 1390 km wide, and 463 km deep. With the assumption of the crater being spherical, and the average density of the moon of 3.34 grams per cubic centimeter, this leads us to a total mass removed from the moon of 1.347 x 10^21 kg. This seems like a huge amount of matter (and it is), but since the entire moon weighs 7.348 x 10^22 kg, this is only a removal of about two percent of the total mass of the moon.

Since the final shot of the moon is some time after the explosion, and there’s not a lot of debris hanging around, I worked out the amount of energy required to both lift that much mass, and the amount required to fling that much mass permanently away from the moon. It’s a lot of energy. In fact, it’s more energy than could be generated by the entire world’s nuclear stockpile detonating at once, by a factor of a billion (10^9) - and that’s just to lift it. To permanently remove the mass, the current world nuclear stockpile is a factor of a quadrillion (10^15) too weak. The Götterdämmerung only fires two bullets - we can safely say that these bullets are physically impossible.

As far as whether the moon would eventually become a sphere again, the answer here is yes, it would, but it would do so in a way that’s very different from what’s depicted in the film. The film essentially depicts this massive detonation as resulting in what’s called a “simple” crater. A simple crater is just a bowl shaped cavity on the surface of a planet or moon, and the moon has plenty of them already. But when a large amount of energy is injected into the surface of a planet, a lot of the energy is lost into heating up the rock; that heat will temporarily liquify the surface. The surface will ripple, and then cool down, resulting in a series of rings and (often) a central peak - this is called a complex crater. This liquefaction and rippling means that the craters are not as deep as they would be if every crater was a simple one. The bottom of the crater, being a liquid, rebounds up towards what would have been the surface, and a lot of the material that could have been thrown out of the crater would stay on the surface as a liquid. These events are still quite destructive, and a large amount of material would be thrown out of the crater, but the moon would never have become quite as non-circular as it was depicted.

One of Saturn’s moons, Mimas, has a really massive complex crater on its surface that earned it the nickname the Death Star Moon. Mimas was thought to have been nearly shattered by this impact, and has a bunch of what is technically called “weird terrain on the exact opposite side of the moon. Weird terrain is where the seismic shock waves from the impact all meet up and collide on the opposite side of the object, releasing their energy. This fractures the surface in a particularly chaotic way, and can trigger volcanism if the planet or moon has magma hanging around under the surface. Mimas doesn’t, and our moon only has a small molten core, but an impact like this would certainly jumble the surface at the antipode of the moon.

The moon’s orbit would be remarkably not affected by this. The material thrown out (as shown in the film) from the moon is one fiftieth the mass of the remaining moon - and the laws of momentum tells us that an object fifty times larger is fifty times harder to move. Assuming that the material blown off the moon reaches escape velocity, the rest of the moon will receive a kick of 44.5 m/s. If the material doesn’t make it to escape velocity, the kick will be weaker. The moon’s average orbital speed is a touch over 1 km/s, which means that this additional velocity is only an alteration of 4% to its speed. However, we have to consider the geometry of the situation here - if the moon had been hit along the direction of travel, it would gain or lose speed in its orbit around the earth, and that might change the distance of its orbit by a small amount. But the hole is in the “top” of the moon, which means that the velocity boost to the moon is pointed “down”. This won’t change the distance of the moon’s orbit, but will make the orbit slightly more tilted. Only slightly, though, since the velocity kick still isn’t very strong.

Overall, this crater would not dramatically change the orbit of the moon, though it would make it more tilted, and if this had happened in the real universe, a lot more of the energy would have gone into melting the surface of the moon, and a lot less into making a crater.

Something here unclear, or have your own question? Feel free to ask!

Could you really have survived the end of Portal 2?

Could Chell really have survived her trip to the moon at the end of Portal 2?

Let’s review exactly what happens to our protagonist at the end of Portal 2. (If you have not played this game and are planning to, stop, go play it, and then come back. We’ll wait. Everyone has the above achievement? Good.)

So. Wheatley booby trapped the Stalemate Resolution Button with a bomb, blasting Chell through a metal grill, back into the main chamber and off her feet. Chell was meant to have been killed by this blast and wasn’t, but we can safely assume she was not feeling 100% after being bodily flung through a metal grate. Chell then puts a portal on the moon, as one does. This portal then sucks pretty much everything that isn’t bolted down out of the room and onto the moon, including Chell, who has the presence of mind to grab onto Wheatley’s handles before she gets pulled through the portal. Chell then spends 30 seconds clinging on to Wheatley until GLaDOS takes over control, bashes Wheatley off the system, and hauls Chell back through to Earth by her wrist with a metal clamp, and the portal is shut down. Can a person survive this?

There are a few concerns here; firstly, the moon has no air. Chell, fortunately, will not need to worry about the air to breathe - she’s only about two feet away from the portal, and there’s lots of air rushing past her. Nor would she need to worry about the average temperature on the moon being -53 C (-63 F), again because she’s surrounded by a vast rushing of room temperature air, and all the wind chill equations I could find indicate that room temperature air (starting at 70F) will only get down to 60F or so. So we’ll assume that she’s perfectly capable of breathing while on the moon. Even if there wasn’t, Chell could have survived the vacuum of space for about a minute - but we’ll assume that she won’t pass out due to a lack of air, since there’s so much of it nearby.

Chell’s main problem is going to be the wind speed. There’s a handy equation called the Ensewiler formula to convert a pressure difference into a wind speed, which is written out as P= 0.002496 v^2. P is the pressure difference in pounds per square foot, and the velocity comes out in miles per hour. The moon portal is a window between standard air pressure (1 atmosphere, or 2116.216 pounds per square feet) and the moon, which has no atmosphere, and thus no atmospheric pressure at all. So if you put in a difference of one atmosphere into the Ensewiler equation, we get a velocity of 920 miles per hour.

920 mph is kind of a problem. This is 1480 km/h, or 411 m/s, and faster than the speed of sound. For some context, the fastest recorded wind speeds on Earth are 253 mph (recorded in Tropical Cyclone Olivia, which hit Australia in 1996) and an F5 tornado that hit Oklahoma in 1999, which clocked in at 302 mph. Felix Baumgartner’s jump from the edge of space got him to clock in at 843 miles per hour before he pulled his parachute, but he was wearing a massive protective suit, specifically designed to keep him safe, and he crossed the sound barrier quite high in the atmosphere, where the air was not very dense. Chell has a tank top. However, the shock from entering a 920 mph wind wouldn’t have killed her immediately - a shock wave from a bomb only becomes lethal at about 1500 mph. It’s still not doing her any great favors, but she wouldn’t have died instantly.

Wheatley’s cabling would have to bear quite a lot of tension - if they were 3 cm x 3 cm cables, holding a weight of 135 pounds (my assumption for the weight of Chell + the weight of the core) should have been fine for steel cabling. However, those cables are also holding up 15,154 Newtons of force pulling Chell in the other direction. A Newton is defined as the amount of force required to push a kilogram one meter in one second. This number of 15,154 Newtons is assuming that the gravity of the moon would have partially counteracted the insane drag from the wind, but since the force of the wind is so huge, and the force of gravity on the moon is so weak, the moon’s gravity doesn’t really help out much. Even if Chell were on the Earth, the Earth’s gravity would only lessen the total force by 500 Newtons. Steel cabling of 3 cm x 3 cm should still be able to hold up to this wind, even through this is a lot of force. But it does mean that Chell genuinely would have been blown perpendicular to the surface of the moon by the wind - 920 mph is well above the terminal velocity for a human on Earth.

A bigger issue would have been Chell’s ability to hold on to Wheatley’s convenient handles. The average person has grip strength of approximately 500 Netwons of force. If the drag force from 920 mph is 15,154 Newtons in the other direction, 500 Newtons is not going to be enough to keep ahold of a bar against that pressure. 500 Newtons will keep you holding on to something in the face of 182 mph (293 km/h) winds, but nothing more than that. No matter how much Chell would have tried to cling to the handles, the sheer force of the wind would have blown her away.

This is assuming, of course, that Chell would be able to keep her hands gripping the bar. In actuality, being dragged out of that portal would be very much like being thrown out of a jet travelling at 900 mph, and told to hang on to a trapeze bar. The first thing that actually would have happened is that both of her shoulders would have dislocated. Shoulders dislocate with 325 Newtons of force, well below the force the wind is placing on her. If your shoulder is dislocated, there’s a very low probability that you’re going to be able to hold on to much of anything, since the major nerves travelling down the arm are being stretched a lot more than they usually are, and nerves don’t like being stretched. While we’re thinking about dislocations, if any part of Chell’s legs had been twisted at all by the wind, it’s very likely she could have also dislocated one or both knees or ankles.

Even worse, if any of those boxes flying out of GLaDOS’ chamber weighed about 10 pounds and happened to collide with Chell, they would inflict about 1860 Newtons of force on her, which is more than enough to break a finger bone. A five pound object would cause a wrist fracture similar to the kind of injuries that boxers can get.

There is a story about a fighter jet pilot who ejected from his aircraft at 800 mph. His shoulder was dislocated, his knee was dislocated, his other ankle was broken, and all the capillaries in his face were broken from the force of the wind - his head swelled up to the size of a basketball, and his lips swelled enormously to the point where he had a hard time moving them. He survived, but his clothes were torn to bits, and he was in the hospital and undergoing physical therapy for six months afterwards before he felt back to normal.

Chell’s wind is 120 mph stronger, and she wasn’t immediately slowing down after her trip to the moon - she was in 920 mph winds for 30 seconds, not the 3 seconds of Capt. Udell. It’s probably safe to assume that the capillary breakage suffered by Capt. Udell would also pose a problem for Chell, meaning that by the time she got back to Earth, her face would have been one enormous bruise, and she’d have been nigh unrecognizable from swelling in a matter of minutes.

In all likelihood, the most reasonable thing that happened was that after she was dragged back into GLaDOS’ chamber by her wrist (which almost certainly would have broken her wrist, if it hadn’t already been broken by passing debris) is that Chell promptly passed out. However, she would have needed extensive medical care and certainly morphine-strength painkillers to deal with her injuries, instead of being back up on her feet after a few minutes.

What of Wheatley in this scenario? Well, rather tragically for the Space Core, 920 mph is not enough to escape the gravitational pull of the moon. Escape velocity for the moon is 2.5 km/s - 920 mph corresponds to 0.411 km/s, six times too slow to escape. We can calculate how long Wheatley would spend flying in space before crashing down to the surface of the moon - since he was flung out perpendicular to the surface, he’d plop back down exactly where the portal was. If we assume that Wheatley was sped up to wind speed, he has 8 and a half minutes of flight (reaching a grand old height of 52 km above the surface of the moon) before crash-landing on the moon again at 920 miles per hour. Since the Space Core left the portal about 10 seconds before Wheatley was released, the space core would land about 10 seconds before Wheatley, pretty close to in the same place. Assuming neither of them smash upon impact, Space Core would then undoubtably drive Wheatley insane by repeating “I’m a moon base! Moon base! Spaaaaaaaace” ad infinitum.

Something here unclear? Have your own space question? Feel free to ask!