Space-walks: a few basics
Over on fanfiction.net, I have this story set after the ending of Prometheus (the 2012 film). And just because I can, I had one of the main characters, Elizabeth Shaw, do a spacewalk outside the Doughnut of Death 2.0 and using a Colt .45 semiautomatic to propel herself around. I was asked, is any of this even possible? Read on!
1. Cruisin' for a bruisin': going out won't separate you from the ship, but watch out for meteorites
The first thing I was asked was, can Elizabeth move around the ship, if the ship is moving at some significant fraction of light speed?
Well, while inside the ship, Elizabeth and the vessel are moving with the same velocity (same speed, same direction). Ok, maybe Elizabeth is pacing somewhere or other, but that velocity difference is simply walking pace. If Elizabeth is exiting the ship, since there's nothing in the vacuum of space to cause drag the way air or water would, she'd have basically the same velocity she had while inside the ship.
If no force acts on the ship while she exits, then the ship velocity stays constant. In particular, in my story, the engines are down and the ship has no thrust. It cruises at almost constant velocity. Almost, because there still is gravity acting on the ship- stars pull on it, but they also pull on Elizabeth as well. Since the distance between the ship and Elizabeth isn't too large (a few tens of meters, tops), while the distance from the ship (or Elizabeth) to any star is more than a million times larger than that, any star's gravity essentially acts on both ship and Elizabeth in pretty much the same way. It imparts the same acceleration on them, so it won't cause a difference in velocity to build up between ship and Elizabeth.
There's another subtle point here. Elizabeth steps out by pushing herself against the ship. That push not only knocks her forward- it also knocks the ship back a little, similar to the way ejecting gas from a rocket engine changes the velocity of the rocket. So by going out, Elizabeth must change the ship's velocity. However, since her mass is small compared to that of the Doughnut of Death 2.0, since the velocity difference imparted by a step is very small, and since this only happens once, there's only a very small difference in velocity between space-walker Elizabeth and the ship. That difference in velocity is simply the speed of the walking, running or jumping that Elizabeth used to exit- so a few meters per second, tops, is the difference between her speed and that of the ship's.
This is in marked difference from exiting a speeding vehicle while on Earth. Get out of a fast car and you're setting yourself up for a rude braking. That's because a host of things, like air and the road, are itching to stop you via friction. In space however, there's no need to fear that you will lag behind a ship that's not using its engines to provide thrust for itself.
In fact, if you're tethered to the ship, there's little concern that you'd lag behind it when it's thrusting either. If the ship is accelerating so strongly that it can break a tether like Elizabeth's spelunking rope (those things can support two metric tons of weight or more), then it's accelerating too strongly for anyone inside to be healthy. Imagine that the ship were accelerating such that it could break Elizabeth's spelunking rope, which, let's assume, only is able to hold two metric tons. This means that Elizabeth is pulled away from the ship with a force equal to the weight of two metric tons, but it also means that someone in the ship is pressed in the chair, or against a wall, or whatever obstacle, by a force equal to the weight of two metric tons. Think you can hold two metric tons on you for any significant amount of time?
(Magic stuff like faster than light travel physics not included in the above, of course.)
The real problem with performing a spacewalk is the stuff you might bump into, like micrometeorites and other pieces of debris that happen to be floating around. Relative to the ship and Elizabeth, these may be moving quite fast and as such would be, effectively, high speed bullets. If in deep interstellar space however, this is a minor concern.
2. Burst the bubble: space suits and what they absolutely must do
I made the leap in assuming that the suits in the Prometheus film were actual space suits, rather than simply a kind of hazmat suit. I'll try to explain the difference here and excuse that leap, somewhat.
So, what is a spacesuit supposed to do? First and foremost, it must provide air for the cosmonaut, and that air is likely to be pressurized at one atmospheric pressure, or some fairly large fraction of that. One atmospheric pressure means that the gas pushes on each square centimeter of its container with a force approximately equal to the weight of one kilogram (1.03kg actually).
That's the main difference between a hazmat and a space suit, because a hazmat suit is only intended to provide a breathable, safe atmosphere for the wearer while being surrounded, on the outside, by gases at equal pressure. While the air in Elizabeth's bubble helmet pushes outward on each square centimeter of the bubble with a force equal to the weight of 1kg, the gas outside also pushes inwards with the same force on each square centimeter. It all balances out and everything's fine, even if the bubble isn't very sturdy.
In the vacuum of space, the pressure provided by the outside gas is effectively zero, so the bubble must be strong enough to hold that inner pressurized gas without bursting. I simply assumed that it could, because it made sense to me for the Prometheus crew to be using space suits on a space expedition. You never know when you may need to do a space walk, after all.
3. Cold eternal darkness: space is cold, but not that chilling
Space is typically taken to be cold, but how hot can nothing be? Typically, the temperature of space is taken to be equal to that of the cosmic microwave background radiation, which is about three degrees above absolute zero. Brr.
Counter-intuitively, something taken out into space will not instantly chill to that temperature however. That's because of the three mechanisms for heat transfer, only one works in space.
Heat may be transferred by conduction. That means, some material stuff, which happens to be good at conducting heat, does so. Then, there's convection, where the stuff that does the transfer is not a good conductor, but is a fluid into which currents appear, and those currents take cold stuff to hot places and vice-versa. Finally, there's radiation, in which heat is transformed into electromagnetic radiation, usually infrared rays but also visible light if the temperature is high enough; that's where "red hot" and "white hot" come from.
In space, there's not enough stuff for either conduction or convection, so radiation alone remains. Assuming there's no infrared or light shining on someone who, for some reason, decided to go au naturel in the vacuum of space, then according to this handy radiation calculator, they'd lose heat at a rate of about 979 Watts. To put some meaning to that figure, the basal metabolic rate (the amount of heat produced by a human body at rest) is 90 Watts. So a bit more than ten times that is rather uncomfortable, but to see how much so, consider another scenario. The same person, now doing skinny dipping in salt water at -2 degrees Celsius, loses about 4400 Watts (using the heat transfer coefficient number found and formulas found here), more than four times as much. Victims of the Titanic sinking, while not skinny dipping, were swimming in wet clothes not suited for hardcore insulation, in -2 degree water, and died of hypothermia in 15 to 30 minutes. The fact that people do go ice swimming in nothing but trunks shows that, if the exposure to cold is short enough, it is survivable.
But not only is space four times less effective at cooling than cold water, it's also not the case that Elizabeth traipses naked through the vacuum. Though I'm sure there'll be kinkmemes on that, eventually. Anyway, keeping her, or any cosmonaut warm, is not a particularly demanding task for a space suit.
Again a couple of minor points here. I've ignored evaporative cooling on the nude cosmonaut, since I wanted to focus on the effect of radiation, and since evaporative cooling would not have been an issue with Elizabeth, who's decent enough to wear her suit. Also, the colder something gets in space, the less heat it will lose per unit of time. So if the nude cosmonaut initially loses 979 Watts when their skin is at 34 degrees Celsius, they will lose 929 when their skin temperature drops to 30 degrees. But that, I'd say, is a quibble. The point I'm making is that heat loss is not so dramatic in space, and that even going in the buff won't result in instant freezing. That's even if you factor in evaporative cooling, because as it turns out, some tests carried out on animals, and the experiences of a few unlucky humans, allowed us to see what would happen in case of a vacuum exposure. No instant freezing, no instant explosion, nothing dramatic or irreversible if the exposure to vacuum isn't so prolonged that the lack of oxygen takes its toll.
Incidentally-
4. Working better under pressure: in which the physiological need for outside pressure is elaborated on, and solutions to provide said pressure described
We've been living at the bottom of an ocean of air, so we are accustomed to being pressed by it. In vacuum, moisture in the mouth or nose would boil, as lack of pressure decreases boiling temperature. Boiling saliva doesn't sound that peachy, but at least blood doesn't boil inside the arteries and veins. Well, not at first, because of pressure provided by the skin. Eventually though, if exposure to zero pressure is long enough, blood will boil and that's definitely not nice at all. Further, if there's air inside a pair of lungs, then there's pressure in them, and there's a serious risk of injury if someone tries to hold their breath while being blown out into space. Lungs work best with a bit of pressure on them from the outside as well.
So one of the things a space suit should do is provide that outside pressure on the body. The typical space suits of today are essentially bubbles loosely shaped as a human being, that contain, beside a cosmonaut, air at atmospheric pressure. There's another design that is becoming feasible however- the space activity suit. The idea is that the suit needs to be elastic and form fitting, so that it can squeeze on the body. You'd have a bubble on the head for breathable air at one atmospheric pressure, and the rest of the suit pressing at some fraction of that on the body, much like in the film. The suit won't press with one atmospheric pressure- it would be awkward to put on if it did- but just enough to allow safe, healthy operation.
It turns out those skin tight suits in sci-fi pinups actually have a justification after all, besides glamour. Space activity suits can be made lighter, and the user is more agile in them than they would be in a gas filled suit. A particularly smart design method is given by Dava Newman of the MIT: she's traced 'lines of non-elongation' on the surface of the body. Elastic bands placed there will not be stretched by usual movements of the body, and as such they can be used to produce the necessary pressure. The resulting 'Biosuit' is oddly reminiscent of the ones in Prometheus, including one detail- the gloves. Current versions of the Biosuit use gas filled gloves because the joints in the hand are too minute to allow a line-of-non-elongation approach to be applied there. I would not be surprised if the similarities between the suits in the Prometheus film and the Bio suit are not entirely coincidental.
5. Radiation scares: what a suit can, and can't, do about radiation
Cosmic rays, oh my. Surely a space suit must keep the wearer protected from them. Well, that's only partly true.
A space suit should prevent ultraviolet rays from burning the cosmonaut's skin, and it should keep them cool while in sunlight, meaning that it should reflect away as much visible light and infrared as possible. That's why space suits of today are so white, and the Prometheus suits' bluish tint is not exactly a good sign. Still, keeping incoming light, UV and infrared away is fairly easy.
Then, there's higher frequency electromagnetic light (usually known as "X-rays" and "gamma rays"), which tends to be more penetrating and needs some kind of dense material, like metal, to be screened. Quite thick screens, in the case of gamma radiation. Which is why, for the most part, space suits are not required to protect against it. If it can pass through the suit, it can likely pass through the cosmonaut without incident, and the danger only comes after prolonged exposure. Since a cosmonaut will spend relatively little time doing space walks, it's considered unnecessary to encumber the suit with heavy screening.
A lot of the other types of radiation in space also tend to be in the highly penetrating category, so one can say that it's not the job of the space suit to protect against them. That's the role of the ship, where the cosmonaut will spend a lot of their time, especially on long journeys. The thing is that while penetrating radiation is likely to pass without incident through a ship and its passengers, every so often one particle will bump with something in the ship and its crew. These events accumulate over time and may result in a dangerous dose of radiation. So, while putting heavy screening on the suit may be avoided, not putting screening on a ship meant for long journeys is not at all wise.
As an aside, some types of nuclear radiation are very easy to shield against. Alpha radiation (fast helium nuclei) can be stopped simply by a lair of air, and beta radiation (fast electrons) can be screened by a layer of neoprene or spandex. Both kinds will be stopped by Elizabeth's suit quite efficiently, but she doesn't know that. (It's a minor plot point later.)
6. Have Colt, will travel: from guns to jet-packs
For me this was the most interesting bit when writing the spacewalk scene. See, once outside, without anything to push or pull against (except for the tether), Elizabeth discovers that moving around is very counter intuitive and very restricted. Weightlessness in empty space is not at all liberating, as it makes one a slave to inertia. Rotate an arm one way, the rest of the body tumbles the other, and just by flailing one's limbs around, one is unable to affect the movement of their center of mass.
The center of mass, by the way, is "the average position" of the mass of a body or system of bodies. It's natural to say that the movement of the center of mass is the (translational) movement of the body, and rotation of the body is naturally described as rotation around the center of mass. Because of momentum conservation laws, a system of bodies cannot change the motion of their center of mass, nor can they change their angular momentum around it, not without outside help. Which, if one floats in vacuum, isn't forthcoming.
However, Elizabeth, like any spacewalking cosmonaut, needs to move around the ship. Grabbing onto it and 'climbing' would be one way. Having some sort of jetpack is another, and in my story she uses a Colt .45 as a replacement for one.
First, yes, guns will fire in space. Why do I even need to link to Straight Dope for that?! Whatever. All the oxygen needed for rapid combustion is already present in the explosive cartridge, and the mechanism doesn't need gravity or outside air pressure to operate. So the basics are taken care of. The idea is then that the bullet being ejected by the gun is similar to the jet of a rocket engine. Pushing the bullet one way results in Elizabeth being pushed the other.
How big that push is can be calculated using the conservation of momentum. Let's start with a simple case. Let's assume a spherical, frictionless Elizabeth Shaw ... sorry, lil' in-joke there. Let's assume that Elizabeth is a point mass, and that she (and the bullet) are not moving initially, ie. their center of mass is stationary.
Momentum for the bullet and Elizabeth system is the sum of their masses multiplied by the speed of their center of mass which is zero. So momentum is zero.
Once Elizabeth fires the bullet however, it is speeding in one direction at 373 m/s (if it's a .45 ACP Bonded Defense). The bullet weighs 12g, and let's say Elizabeth and all her remaining kit weighs 60kg. Then, since linear momentum must be zero, we have that:
Elizabeth's mass x Elizabeth's speed + Bullet mass x bullet speed = 0.
It's straightforward then to see what Elizabeth's speed needs to be then. It will be a negative number, signifying that it's in the opposite direction to the way the bullet is going. Note that the center of mass of the Elizabeth and bullet system stays unmoving. The center of mass of one of the parts of that system (say, Elizabeth) will move however.
Ok, I should add the gases being ejected from the cartridge in that as well, but it's known that the recoil of a blank cartridge is much less than that of a live one, so one can say that the gases don't add much. So, since I only want an approximation, I can say that firing one bullet from her Colt will change Elizabeth's speed by about 10cm/s. Firing all seven bullets from the Colt will result in, roughly, 70cm/s difference to her original velocity, which is referred to as the delta-V of her 'jetpack'. (Elizabeth-and-kit's mass decreases with each bullet fired, but not by much, and after firing seven bullets, it's close enough to its starting value that I can ignore changes if I only need approximate values).
In real life though, Elizabeth is not a point mass and this complicates things a little. Suppose she aims as one usually does, close to eye level, and fires. Then recoil pushes through her her arms into her shoulders so one would expect to see her pushed back but also tumbling as a result. That's because the recoil doesn't push directly through her center of mass, but around it. Note, here I'm talking about -her- center of mass, not the center of mass of the system made of her and the fired bullet.
What kind of effect does firing with an offset from the center of mass have? The recoil is simply a force acting on some point of Elizabeth's body, let's say, the shoulder. We'd want to see what equivalent force and torque it produces at the center of mass, and once we see that, we can see how Elizabeth's movement is affected. For simplicity of exposition, I'll consider that only 2D movement is allowed: Elizabeth may move vertically, horizontally, and rotate around her center of mass.
To compute the equivalent forces and torques at the center of mass, one starts from what 'equivalent' is supposed to mean here: forces are equivalent if they have the same effect on the body they are applied to. In particular, this means they would inflict the same change to that object's movement, which further implies that they would cause the same change in that object's kinetic energy. Change in kinetic energy as result of a force is referred to in physics as '(mechanical) work', and is the product of the force with the displacement that the object moves through.
Imagine that the recoil from the gun pushes Elizabeth for a very small distance. Over that distance, that recoil does an 'elementary' (ie., very small) amount of work, dL. The equivalent force and torque acting on the center of mass must do the exact same amount of elementary work, so we have that
dL = {Force at shoulder} x {linear displacement at shoulder} = {Force at center of mass} x {linear displacement at center of mass} + {Torque at center of mass} x {angular displacement around center of mass}
Since we know where the shoulder is relative to the center of mass, we can express displacements at the shoulder in terms of displacements at and around the center of mass. Going full algebra on this problem:
That equality is valid for any arbitrary small displacements, so one could, say, set all but one of the small displacements to zero and there'd still be an equality. Therefore it makes sense to group force factors around the displacements and this gives the expressions for the equivalent forces and torque at the center of mass:
So the change in Elizabeth's center of mass velocity is the same wherever she fires the gun, however there's also some torque applied to her, and that amount of torque depends on the distance between the recoil line and the center of mass. That particular bit is probably not obvious from the formula for the torque at the center of mass in the form I've derived it in, so here's the explanation.
First, notice that the segment from shoulder to center of mass, and the point where the perpendicular to the recoil line that passes through the center of mass meets the recoil line, forms a right triangle. In that right triangle, one angle is equal to the angle alpha between the recoil line and the line between shoulder and center of mass. So far, obvious, and it means that the distance from the center of mass to the recoil line is simply the distance from shoulder to center of mass multiplied by the sine of alpha.
Next, notice that there are relationships between the magnitude of the recoil force and its components, one of which is neatly expressible in terms of the direction of the recoil line:
Substituting that into the expression for the torque, and using the formula for the sine of angle difference, gives an expression that's easier to understand, as it reveals the connection to the distance between the center of mass and the recoil line:
How fast exactly will Elizabeth tumble, though? This requires a calculation involving her moment of inertia. If I approximate that by a cylinder (let's say, all her mass, which I take to be 60kg, is in her head, torso, abdomen and legs; and the cylinder has 0.3m radius), using an approximate measure for her shoulder to center of mass distance (let's say, 0.7m), a value of height that's slightly taller than Ms. Noomi Rapace, say 1.7m, and assuming she fires straight ahead (alpha is ninety degrees, sine of alpha is 1), results in an angular velocity around the center of mass of about 0.27 radians per second. Meaning, one revolution every 23 seconds or so. If she fired straight above her however alpha would be zero, and so would the torque she would receive from the recoil. She wouldn't tumble at all in that case.
(Another paranthesis for nitpicking. In all the numerical calculations I've simplified things and said that the recoil is an 'impulsive force', which is a little trick engineers use to speed up calculations. Usually, forces act for a measurable, non-zero amount of time and accelerate the body over that period. Forces are measured in Newtons, or if you prefer, kilograms times meter per second squared, so when I calculate a force I should report the acceleration it produces. Likewise when presenting the effect of a torque I should report the angular acceleration it causes Elizabeth to undergo. Impulse forces/torques on the other hand are ficticious devices that act instantly to change the velocity of something, which is why I report velocity and angular velocity values. In many cases, like this one, it matters not for the result which of the methods I choose to use, but it's much easier to work with impulsive forces, as I don't need to worry what the time evolution of the recoil force looks like. And recoil operates for a short time and is almost an impulsive force anyway.)
There's yet another effect that firing a gun like the Colt will have. Since the Colt is rifled, it twists the bullet as it fires it, which means the bullet twists the gun and its holder in return. So firing a bullet will also impart some spin on the axis of firing. How much depends on how fast the bullet is spun (I haven't got data on this) and how Elizabeth fires it.
The point is, the bullet, originally, has no angular momentum. After it's fired and spinning, it has some angular momentum, so Elizabeth gains the opposite of that angular momentum in return. Angular momentum for spinning around an axis is
Angular momentum = {angular velocity around axis} x {moment of inertia around axis}
Moment of inertia around an axis increases proportionally with the mass of the object, but it's also affected by how tightly 'bunched' the object is around that axis. A bullet is a narrow object tightly surrounding its axis of spin, so it's angular momentum will be very small. Elizabeth is much more massive and much larger than a bullet so the change in her angular velocity because of the bullet's spinning will be very small. I'll skip doing computations on how small it is, because I'd need to justify why moment of inertia formulas look as they do and this post is getting long as it is. However, a simple experiment may give one the intuition needed. Fill a bottle with water and lay it horizontal on the floor. Try to spin it by grabbing in the middle, then by grabbing one of the ends. Make sure you have a comfortable grip in both cases and only look at the force (well, torque) you need to put in. Spinning from the center will be easier, because the mass of the bottle tends to be closer to the point where you apply the spin.
I'll end this with a comparison with reality. I said that the delta-V of the Colt is 70 cm/s. How does that compare to current day jetpacks for astronauts? Well, the SAFER (Simplified Aid for Extra Vehicular Rescue) has a delta-V of 3m/s, which means that if one kept it running constantly in one direction it would change speed by 3m/s, but it's meant for emergency use only. The Manned Maneuvering Unit, which is current generation technology or thereabouts, can do a bit over 24m/s delta-V, and is meant for general purpose extra vehicular activities.
The first maneuvering unit was the Hand-held Maneuvering Unit used in the Gemini 4 mission of 1965. It was a gun ejecting nitrogen gas, so functionally it wasn't too different from my Colt jet-pack. Of course, the HH-MU could do continuous jets, unlike the gun, and would not have the spin caused by rifling. But apart from those minor differences, it's the same, and here's what one of its users had to say about it:
(from the interview with James A. McDivitt, one of the astronauts on Gemini 4, conducted by Doug Ward on the 29th of June, 1999):
So yeah. The first Extra Vehicular Activity conducted by NASA sounds like an improvized job, but hey, it worked fine with some small measure of control on the astronaut's movement (it was Ed White, James A. McDivitt's colleague, who actually got to be the first American doing a spacewalk). The first E.V.A., which was done by Soviet cosmonaut Alexei Leonov in the Voskhod 2 mission, only had a tether.
Like Ed White, Elizabeth's E.V.A. wasn't about actually doing anything besides taking a look outside, so I'd say her gun is a plausible maneuvering unit. Even if a very, very awkward one to use.
1. Cruisin' for a bruisin': going out won't separate you from the ship, but watch out for meteorites
The first thing I was asked was, can Elizabeth move around the ship, if the ship is moving at some significant fraction of light speed?
Well, while inside the ship, Elizabeth and the vessel are moving with the same velocity (same speed, same direction). Ok, maybe Elizabeth is pacing somewhere or other, but that velocity difference is simply walking pace. If Elizabeth is exiting the ship, since there's nothing in the vacuum of space to cause drag the way air or water would, she'd have basically the same velocity she had while inside the ship.
If no force acts on the ship while she exits, then the ship velocity stays constant. In particular, in my story, the engines are down and the ship has no thrust. It cruises at almost constant velocity. Almost, because there still is gravity acting on the ship- stars pull on it, but they also pull on Elizabeth as well. Since the distance between the ship and Elizabeth isn't too large (a few tens of meters, tops), while the distance from the ship (or Elizabeth) to any star is more than a million times larger than that, any star's gravity essentially acts on both ship and Elizabeth in pretty much the same way. It imparts the same acceleration on them, so it won't cause a difference in velocity to build up between ship and Elizabeth.
There's another subtle point here. Elizabeth steps out by pushing herself against the ship. That push not only knocks her forward- it also knocks the ship back a little, similar to the way ejecting gas from a rocket engine changes the velocity of the rocket. So by going out, Elizabeth must change the ship's velocity. However, since her mass is small compared to that of the Doughnut of Death 2.0, since the velocity difference imparted by a step is very small, and since this only happens once, there's only a very small difference in velocity between space-walker Elizabeth and the ship. That difference in velocity is simply the speed of the walking, running or jumping that Elizabeth used to exit- so a few meters per second, tops, is the difference between her speed and that of the ship's.
This is in marked difference from exiting a speeding vehicle while on Earth. Get out of a fast car and you're setting yourself up for a rude braking. That's because a host of things, like air and the road, are itching to stop you via friction. In space however, there's no need to fear that you will lag behind a ship that's not using its engines to provide thrust for itself.
In fact, if you're tethered to the ship, there's little concern that you'd lag behind it when it's thrusting either. If the ship is accelerating so strongly that it can break a tether like Elizabeth's spelunking rope (those things can support two metric tons of weight or more), then it's accelerating too strongly for anyone inside to be healthy. Imagine that the ship were accelerating such that it could break Elizabeth's spelunking rope, which, let's assume, only is able to hold two metric tons. This means that Elizabeth is pulled away from the ship with a force equal to the weight of two metric tons, but it also means that someone in the ship is pressed in the chair, or against a wall, or whatever obstacle, by a force equal to the weight of two metric tons. Think you can hold two metric tons on you for any significant amount of time?
(Magic stuff like faster than light travel physics not included in the above, of course.)
The real problem with performing a spacewalk is the stuff you might bump into, like micrometeorites and other pieces of debris that happen to be floating around. Relative to the ship and Elizabeth, these may be moving quite fast and as such would be, effectively, high speed bullets. If in deep interstellar space however, this is a minor concern.
2. Burst the bubble: space suits and what they absolutely must do
I made the leap in assuming that the suits in the Prometheus film were actual space suits, rather than simply a kind of hazmat suit. I'll try to explain the difference here and excuse that leap, somewhat.
So, what is a spacesuit supposed to do? First and foremost, it must provide air for the cosmonaut, and that air is likely to be pressurized at one atmospheric pressure, or some fairly large fraction of that. One atmospheric pressure means that the gas pushes on each square centimeter of its container with a force approximately equal to the weight of one kilogram (1.03kg actually).
That's the main difference between a hazmat and a space suit, because a hazmat suit is only intended to provide a breathable, safe atmosphere for the wearer while being surrounded, on the outside, by gases at equal pressure. While the air in Elizabeth's bubble helmet pushes outward on each square centimeter of the bubble with a force equal to the weight of 1kg, the gas outside also pushes inwards with the same force on each square centimeter. It all balances out and everything's fine, even if the bubble isn't very sturdy.
In the vacuum of space, the pressure provided by the outside gas is effectively zero, so the bubble must be strong enough to hold that inner pressurized gas without bursting. I simply assumed that it could, because it made sense to me for the Prometheus crew to be using space suits on a space expedition. You never know when you may need to do a space walk, after all.
3. Cold eternal darkness: space is cold, but not that chilling
Space is typically taken to be cold, but how hot can nothing be? Typically, the temperature of space is taken to be equal to that of the cosmic microwave background radiation, which is about three degrees above absolute zero. Brr.
Counter-intuitively, something taken out into space will not instantly chill to that temperature however. That's because of the three mechanisms for heat transfer, only one works in space.
Heat may be transferred by conduction. That means, some material stuff, which happens to be good at conducting heat, does so. Then, there's convection, where the stuff that does the transfer is not a good conductor, but is a fluid into which currents appear, and those currents take cold stuff to hot places and vice-versa. Finally, there's radiation, in which heat is transformed into electromagnetic radiation, usually infrared rays but also visible light if the temperature is high enough; that's where "red hot" and "white hot" come from.
In space, there's not enough stuff for either conduction or convection, so radiation alone remains. Assuming there's no infrared or light shining on someone who, for some reason, decided to go au naturel in the vacuum of space, then according to this handy radiation calculator, they'd lose heat at a rate of about 979 Watts. To put some meaning to that figure, the basal metabolic rate (the amount of heat produced by a human body at rest) is 90 Watts. So a bit more than ten times that is rather uncomfortable, but to see how much so, consider another scenario. The same person, now doing skinny dipping in salt water at -2 degrees Celsius, loses about 4400 Watts (using the heat transfer coefficient number found and formulas found here), more than four times as much. Victims of the Titanic sinking, while not skinny dipping, were swimming in wet clothes not suited for hardcore insulation, in -2 degree water, and died of hypothermia in 15 to 30 minutes. The fact that people do go ice swimming in nothing but trunks shows that, if the exposure to cold is short enough, it is survivable.
But not only is space four times less effective at cooling than cold water, it's also not the case that Elizabeth traipses naked through the vacuum. Though I'm sure there'll be kinkmemes on that, eventually. Anyway, keeping her, or any cosmonaut warm, is not a particularly demanding task for a space suit.
Again a couple of minor points here. I've ignored evaporative cooling on the nude cosmonaut, since I wanted to focus on the effect of radiation, and since evaporative cooling would not have been an issue with Elizabeth, who's decent enough to wear her suit. Also, the colder something gets in space, the less heat it will lose per unit of time. So if the nude cosmonaut initially loses 979 Watts when their skin is at 34 degrees Celsius, they will lose 929 when their skin temperature drops to 30 degrees. But that, I'd say, is a quibble. The point I'm making is that heat loss is not so dramatic in space, and that even going in the buff won't result in instant freezing. That's even if you factor in evaporative cooling, because as it turns out, some tests carried out on animals, and the experiences of a few unlucky humans, allowed us to see what would happen in case of a vacuum exposure. No instant freezing, no instant explosion, nothing dramatic or irreversible if the exposure to vacuum isn't so prolonged that the lack of oxygen takes its toll.
Incidentally-
4. Working better under pressure: in which the physiological need for outside pressure is elaborated on, and solutions to provide said pressure described
We've been living at the bottom of an ocean of air, so we are accustomed to being pressed by it. In vacuum, moisture in the mouth or nose would boil, as lack of pressure decreases boiling temperature. Boiling saliva doesn't sound that peachy, but at least blood doesn't boil inside the arteries and veins. Well, not at first, because of pressure provided by the skin. Eventually though, if exposure to zero pressure is long enough, blood will boil and that's definitely not nice at all. Further, if there's air inside a pair of lungs, then there's pressure in them, and there's a serious risk of injury if someone tries to hold their breath while being blown out into space. Lungs work best with a bit of pressure on them from the outside as well.
So one of the things a space suit should do is provide that outside pressure on the body. The typical space suits of today are essentially bubbles loosely shaped as a human being, that contain, beside a cosmonaut, air at atmospheric pressure. There's another design that is becoming feasible however- the space activity suit. The idea is that the suit needs to be elastic and form fitting, so that it can squeeze on the body. You'd have a bubble on the head for breathable air at one atmospheric pressure, and the rest of the suit pressing at some fraction of that on the body, much like in the film. The suit won't press with one atmospheric pressure- it would be awkward to put on if it did- but just enough to allow safe, healthy operation.
 |
| MIT's Biosuit (showing the lines of nonelongation in orange) next to the Mars Mark III suit. The Biosuit is a 'space activity suit' providing pressure on the body by elastic constriction. The Mars Mark III is a suit containing pressurized gas. Image obtained from wikipedia, which says: this file is in the public domain because it was solely created by NASA. NASA copyright policy states that "NASA material is not protected by copyright unless noted". |
5. Radiation scares: what a suit can, and can't, do about radiation
Cosmic rays, oh my. Surely a space suit must keep the wearer protected from them. Well, that's only partly true.
A space suit should prevent ultraviolet rays from burning the cosmonaut's skin, and it should keep them cool while in sunlight, meaning that it should reflect away as much visible light and infrared as possible. That's why space suits of today are so white, and the Prometheus suits' bluish tint is not exactly a good sign. Still, keeping incoming light, UV and infrared away is fairly easy.
Then, there's higher frequency electromagnetic light (usually known as "X-rays" and "gamma rays"), which tends to be more penetrating and needs some kind of dense material, like metal, to be screened. Quite thick screens, in the case of gamma radiation. Which is why, for the most part, space suits are not required to protect against it. If it can pass through the suit, it can likely pass through the cosmonaut without incident, and the danger only comes after prolonged exposure. Since a cosmonaut will spend relatively little time doing space walks, it's considered unnecessary to encumber the suit with heavy screening.
A lot of the other types of radiation in space also tend to be in the highly penetrating category, so one can say that it's not the job of the space suit to protect against them. That's the role of the ship, where the cosmonaut will spend a lot of their time, especially on long journeys. The thing is that while penetrating radiation is likely to pass without incident through a ship and its passengers, every so often one particle will bump with something in the ship and its crew. These events accumulate over time and may result in a dangerous dose of radiation. So, while putting heavy screening on the suit may be avoided, not putting screening on a ship meant for long journeys is not at all wise.
As an aside, some types of nuclear radiation are very easy to shield against. Alpha radiation (fast helium nuclei) can be stopped simply by a lair of air, and beta radiation (fast electrons) can be screened by a layer of neoprene or spandex. Both kinds will be stopped by Elizabeth's suit quite efficiently, but she doesn't know that. (It's a minor plot point later.)
6. Have Colt, will travel: from guns to jet-packs
For me this was the most interesting bit when writing the spacewalk scene. See, once outside, without anything to push or pull against (except for the tether), Elizabeth discovers that moving around is very counter intuitive and very restricted. Weightlessness in empty space is not at all liberating, as it makes one a slave to inertia. Rotate an arm one way, the rest of the body tumbles the other, and just by flailing one's limbs around, one is unable to affect the movement of their center of mass.
The center of mass, by the way, is "the average position" of the mass of a body or system of bodies. It's natural to say that the movement of the center of mass is the (translational) movement of the body, and rotation of the body is naturally described as rotation around the center of mass. Because of momentum conservation laws, a system of bodies cannot change the motion of their center of mass, nor can they change their angular momentum around it, not without outside help. Which, if one floats in vacuum, isn't forthcoming.
However, Elizabeth, like any spacewalking cosmonaut, needs to move around the ship. Grabbing onto it and 'climbing' would be one way. Having some sort of jetpack is another, and in my story she uses a Colt .45 as a replacement for one.
First, yes, guns will fire in space. Why do I even need to link to Straight Dope for that?! Whatever. All the oxygen needed for rapid combustion is already present in the explosive cartridge, and the mechanism doesn't need gravity or outside air pressure to operate. So the basics are taken care of. The idea is then that the bullet being ejected by the gun is similar to the jet of a rocket engine. Pushing the bullet one way results in Elizabeth being pushed the other.
How big that push is can be calculated using the conservation of momentum. Let's start with a simple case. Let's assume a spherical, frictionless Elizabeth Shaw ... sorry, lil' in-joke there. Let's assume that Elizabeth is a point mass, and that she (and the bullet) are not moving initially, ie. their center of mass is stationary.
Momentum for the bullet and Elizabeth system is the sum of their masses multiplied by the speed of their center of mass which is zero. So momentum is zero.
Once Elizabeth fires the bullet however, it is speeding in one direction at 373 m/s (if it's a .45 ACP Bonded Defense). The bullet weighs 12g, and let's say Elizabeth and all her remaining kit weighs 60kg. Then, since linear momentum must be zero, we have that:
Elizabeth's mass x Elizabeth's speed + Bullet mass x bullet speed = 0.
It's straightforward then to see what Elizabeth's speed needs to be then. It will be a negative number, signifying that it's in the opposite direction to the way the bullet is going. Note that the center of mass of the Elizabeth and bullet system stays unmoving. The center of mass of one of the parts of that system (say, Elizabeth) will move however.
Ok, I should add the gases being ejected from the cartridge in that as well, but it's known that the recoil of a blank cartridge is much less than that of a live one, so one can say that the gases don't add much. So, since I only want an approximation, I can say that firing one bullet from her Colt will change Elizabeth's speed by about 10cm/s. Firing all seven bullets from the Colt will result in, roughly, 70cm/s difference to her original velocity, which is referred to as the delta-V of her 'jetpack'. (Elizabeth-and-kit's mass decreases with each bullet fired, but not by much, and after firing seven bullets, it's close enough to its starting value that I can ignore changes if I only need approximate values).
In real life though, Elizabeth is not a point mass and this complicates things a little. Suppose she aims as one usually does, close to eye level, and fires. Then recoil pushes through her her arms into her shoulders so one would expect to see her pushed back but also tumbling as a result. That's because the recoil doesn't push directly through her center of mass, but around it. Note, here I'm talking about -her- center of mass, not the center of mass of the system made of her and the fired bullet.
What kind of effect does firing with an offset from the center of mass have? The recoil is simply a force acting on some point of Elizabeth's body, let's say, the shoulder. We'd want to see what equivalent force and torque it produces at the center of mass, and once we see that, we can see how Elizabeth's movement is affected. For simplicity of exposition, I'll consider that only 2D movement is allowed: Elizabeth may move vertically, horizontally, and rotate around her center of mass.
To compute the equivalent forces and torques at the center of mass, one starts from what 'equivalent' is supposed to mean here: forces are equivalent if they have the same effect on the body they are applied to. In particular, this means they would inflict the same change to that object's movement, which further implies that they would cause the same change in that object's kinetic energy. Change in kinetic energy as result of a force is referred to in physics as '(mechanical) work', and is the product of the force with the displacement that the object moves through.
Imagine that the recoil from the gun pushes Elizabeth for a very small distance. Over that distance, that recoil does an 'elementary' (ie., very small) amount of work, dL. The equivalent force and torque acting on the center of mass must do the exact same amount of elementary work, so we have that
dL = {Force at shoulder} x {linear displacement at shoulder} = {Force at center of mass} x {linear displacement at center of mass} + {Torque at center of mass} x {angular displacement around center of mass}
Since we know where the shoulder is relative to the center of mass, we can express displacements at the shoulder in terms of displacements at and around the center of mass. Going full algebra on this problem:
That equality is valid for any arbitrary small displacements, so one could, say, set all but one of the small displacements to zero and there'd still be an equality. Therefore it makes sense to group force factors around the displacements and this gives the expressions for the equivalent forces and torque at the center of mass:
So the change in Elizabeth's center of mass velocity is the same wherever she fires the gun, however there's also some torque applied to her, and that amount of torque depends on the distance between the recoil line and the center of mass. That particular bit is probably not obvious from the formula for the torque at the center of mass in the form I've derived it in, so here's the explanation.
First, notice that the segment from shoulder to center of mass, and the point where the perpendicular to the recoil line that passes through the center of mass meets the recoil line, forms a right triangle. In that right triangle, one angle is equal to the angle alpha between the recoil line and the line between shoulder and center of mass. So far, obvious, and it means that the distance from the center of mass to the recoil line is simply the distance from shoulder to center of mass multiplied by the sine of alpha.
Next, notice that there are relationships between the magnitude of the recoil force and its components, one of which is neatly expressible in terms of the direction of the recoil line:
Substituting that into the expression for the torque, and using the formula for the sine of angle difference, gives an expression that's easier to understand, as it reveals the connection to the distance between the center of mass and the recoil line:
How fast exactly will Elizabeth tumble, though? This requires a calculation involving her moment of inertia. If I approximate that by a cylinder (let's say, all her mass, which I take to be 60kg, is in her head, torso, abdomen and legs; and the cylinder has 0.3m radius), using an approximate measure for her shoulder to center of mass distance (let's say, 0.7m), a value of height that's slightly taller than Ms. Noomi Rapace, say 1.7m, and assuming she fires straight ahead (alpha is ninety degrees, sine of alpha is 1), results in an angular velocity around the center of mass of about 0.27 radians per second. Meaning, one revolution every 23 seconds or so. If she fired straight above her however alpha would be zero, and so would the torque she would receive from the recoil. She wouldn't tumble at all in that case.
(Another paranthesis for nitpicking. In all the numerical calculations I've simplified things and said that the recoil is an 'impulsive force', which is a little trick engineers use to speed up calculations. Usually, forces act for a measurable, non-zero amount of time and accelerate the body over that period. Forces are measured in Newtons, or if you prefer, kilograms times meter per second squared, so when I calculate a force I should report the acceleration it produces. Likewise when presenting the effect of a torque I should report the angular acceleration it causes Elizabeth to undergo. Impulse forces/torques on the other hand are ficticious devices that act instantly to change the velocity of something, which is why I report velocity and angular velocity values. In many cases, like this one, it matters not for the result which of the methods I choose to use, but it's much easier to work with impulsive forces, as I don't need to worry what the time evolution of the recoil force looks like. And recoil operates for a short time and is almost an impulsive force anyway.)
There's yet another effect that firing a gun like the Colt will have. Since the Colt is rifled, it twists the bullet as it fires it, which means the bullet twists the gun and its holder in return. So firing a bullet will also impart some spin on the axis of firing. How much depends on how fast the bullet is spun (I haven't got data on this) and how Elizabeth fires it.
The point is, the bullet, originally, has no angular momentum. After it's fired and spinning, it has some angular momentum, so Elizabeth gains the opposite of that angular momentum in return. Angular momentum for spinning around an axis is
Angular momentum = {angular velocity around axis} x {moment of inertia around axis}
Moment of inertia around an axis increases proportionally with the mass of the object, but it's also affected by how tightly 'bunched' the object is around that axis. A bullet is a narrow object tightly surrounding its axis of spin, so it's angular momentum will be very small. Elizabeth is much more massive and much larger than a bullet so the change in her angular velocity because of the bullet's spinning will be very small. I'll skip doing computations on how small it is, because I'd need to justify why moment of inertia formulas look as they do and this post is getting long as it is. However, a simple experiment may give one the intuition needed. Fill a bottle with water and lay it horizontal on the floor. Try to spin it by grabbing in the middle, then by grabbing one of the ends. Make sure you have a comfortable grip in both cases and only look at the force (well, torque) you need to put in. Spinning from the center will be easier, because the mass of the bottle tends to be closer to the point where you apply the spin.
I'll end this with a comparison with reality. I said that the delta-V of the Colt is 70 cm/s. How does that compare to current day jetpacks for astronauts? Well, the SAFER (Simplified Aid for Extra Vehicular Rescue) has a delta-V of 3m/s, which means that if one kept it running constantly in one direction it would change speed by 3m/s, but it's meant for emergency use only. The Manned Maneuvering Unit, which is current generation technology or thereabouts, can do a bit over 24m/s delta-V, and is meant for general purpose extra vehicular activities.
 |
| Hand-held Maneuvering Unit used on the Gemini 4 mission. The first maneuvering unit used for extra vehicular activity. Image obtained from wikipedia, which says: this file is in the public domain because it was solely created by NASA. NASA copyright policy states that "NASA material is not protected by copyright unless noted". |
 |
| Ed White aboard the Gemini 4 mission, doing the first extravehicular activity that used a maneuvering unit. Image obtained from wikipedia, which says: this file is in the public domain because it was solely created by NASA. NASA copyright policy states that "NASA material is not protected by copyright unless noted". |
The first maneuvering unit was the Hand-held Maneuvering Unit used in the Gemini 4 mission of 1965. It was a gun ejecting nitrogen gas, so functionally it wasn't too different from my Colt jet-pack. Of course, the HH-MU could do continuous jets, unlike the gun, and would not have the spin caused by rifling. But apart from those minor differences, it's the same, and here's what one of its users had to say about it:
(from the interview with James A. McDivitt, one of the astronauts on Gemini 4, conducted by Doug Ward on the 29th of June, 1999):
"{pg. 34} McDivitt: Well, this is one of those ad hoc things. You sort of made it up as you went along. The gun—that little maneuvering gun that we used fortunately didn’t have much thrust, because it was a hopeless device! I mean, there was no way you could really control yourself. You could control yourself on the air table in two axes; but unfortunately when you’re in space, you’re in three axes in six degrees of freedom. So, it would have been hopeless to try to maneuver around much with just it. But we didn’t have much gas in it and the tether wasn’t too long, and we couldn’t get in a lot of trouble. But the things that were important were getting the hatch open, getting the hatch closed, getting out, getting back in, the equipment that you needed, what the thermal protection was going to be, what the micrometeorite protection was going to be. Just the fact that we could go out and do that stuff was very important.
{pg. 39} McDivitt: I think that we overreached a lot in the EVA thing to start with. We didn’t do anything. Like I told you, that gun {the HH-MU} was utterly useless. ... As soon as we got up in space, the only way you could make that gun work is if you fired it exactly through your center of gravity. And trying to find your center of gravity—we didn’t know where our center of gravity was; so there was no way we could’ve done it accurately. Yeah, that was a oversimplified thing. And fortunately, we didn’t have much gas in the gun and didn’t put out much thrust; so it couldn’t hurt us much."
So yeah. The first Extra Vehicular Activity conducted by NASA sounds like an improvized job, but hey, it worked fine with some small measure of control on the astronaut's movement (it was Ed White, James A. McDivitt's colleague, who actually got to be the first American doing a spacewalk). The first E.V.A., which was done by Soviet cosmonaut Alexei Leonov in the Voskhod 2 mission, only had a tether.
Like Ed White, Elizabeth's E.V.A. wasn't about actually doing anything besides taking a look outside, so I'd say her gun is a plausible maneuvering unit. Even if a very, very awkward one to use.
Comments
Post a Comment