This series looks to take a deep dive into some of the events and places of the Halo franchise and determine how sciencey they are. I make a point to give as much leeway to the fiction, so long as it does not directly contradict with science, since this is, you know, fiction. I always make a point to warn people if there might be spoilers ahead, even if they are incredibly minor, because there are new people getting into the Halo lore every day, and what might be basic and obvious to a lore veteran, may not for a new recruit. If nothing else, this article might spur some interest in the extended Halo universe, so if at the end you want to learn more, check out the related media for a good place to start.
RELATED MEDIA
While minor, the following article makes reference to and could potentially contain spoilers for Halo: The Fall of Reach, Halo: Warfleet, Halo 2, and Halo: Reach.
THE QUESTION
WHAT WOULD HAPPEN TO A MAGNETIC ACCELERATOR CANNON (MAC) ROUND IF IT MISSED ITS TARGET, AND ARE THERE HUNDREDS OF MAC ROUNDS FLYING AIMLESSLY IN SPACE, WAITING TO DESTROY A RANDOM SHIP IT COMES INTO CONTACT WITH?
SOME BACKGROUND
This is a question originally posed on the r/HaloStory subreddit by u/VaDiSt, and I thought it was an interesting thought experiment. The idea is that there were undoubtedly thousands of magnetic accelerator cannon rounds fired in space over the course of the history of the weapon, and of those, a small percentage must have missed their targets. We have been given at least a couple accounts of MAC rounds flying clear of their targets and disappearing into the blackness of space, so we know undoubtedly that it has happened. The real question is, what happens to those MAC rounds once they miss? Would they just vanish, never to be seen from again, or would they become a navigational hazard for any ship in the system, orbiting endlessly until they came crashing through some unsuspecting ship, or raining down upon a populated planet? From what I know of the Halo universe, that has never happened, so either this is something that doesn't get mentioned, or it is something that doesn't happen.
There are a couple things to consider before we get started. First, how does a MAC round destroy a target? A MAC slug is very simply a dense hunk of metal, accelerated to incredibly fast speeds and directed at whatever you want to go boom. The amount of energy imparted by the MAC is dependent on the total kinetic energy of the round, which is (1/2) * mass * velocity squared (1/2·m·v²). To add more energy to the system, we would have to either make the round more massive, or make the cannon fire the round faster. Depending on the mass and velocity of the round, this could be like a large bullet impacting a ship, or a massive nuke going off. For reference, the US Navy is testing a ship-based railgun in real life, and while not as large or powerful as the ones in the Halo universe, they definitely pack a punch.
The Electro-Magnetic Laboratory Rail Gun, a US Navy experimental weapon that can fire a projectile at 5,600 mph.
The question we want to know, however, is what will happen to the round itself if it doesn't hit anything. This will be dependent on something called escape velocity. Escape velocity is the speed a body (MAC round, asteroid, etc) has to be travelling away from the object in question (planet, star, galaxy), to escape the reach of the body's gravity and never return. This will vary for each object based on the distance from the surface and mass of the body in question, but we can use some simple assumptions to make the question a lot easier.
The last thing to consider is that we are, for the most part, talking about rounds fired in space, not in atmosphere. Since space is mostly empty, there is effectively nothing to stop or slow a fired round. Ever. Unless it comes into contact with a ship or planet of course. This is known as Newton's first law of motion, or colloquially as "an object in motion tends to stay in motion".
If you aren't quite sure how empty space is, we can quickly break it down into three categories. The first is the region between planets, known as the Interplanetary medium, which has about 5 particles per cubic centimeter. From there it only gets less dense, with the interstellar medium between solar systems at around 1 particle per cubic centimeter, and the intracluster medium at around 0.001 particles per cubic centimeter. To try and put that into perspective, a particle travelling through the interstellar medium would take on average 500 years before in collided with another particle. Even more unimaginable is that a particle in the intracluster medium has a mean free path, the average distance a particle travels before interacting with another particle, of almost a light-year. That is thousands or millions of years of travel before anything happened to it. So while there isn’t technically nothing in space, there is so much nothing that it is a very safe assumption that it is a perfect vacuum and won’t slow down any MAC round an appreciable amount.
ESTABLISHING CONDITIONS
To get to the heart of the answer, we will need to know a few things. Well, actually we need to know a lot of things, but for the sake of simplicity, we will make a number of assumptions. I will be as conservative as possible with my assumptions too, so whatever result we come to should hold up to closer scrutiny. Here are the things we need to know to figure out what happens to a MAC round that misses its target.
The mass of the MAC round
The velocity of the MAC round
The direction of the shot
The escape velocity of the nearby planet or solar system
I will take each of these in order, then at the end we can put it all together and figure out,
Where have all the MAC rounds goooone???
THE MASS AND VELOCITY OF A MAC ROUND
Coming up with a single mass and velocity of a MAC round is impossible since there are so many different sizes of magnetic accelerator cannons, so we will have to start by limiting this analysis to a few different MAC sizes. For the purposes of simplicity and because canonical information on MACs is somewhat limited, we will look at three different MACs, small, medium, and large.
SMALL MAC
First, the smallest I know of would be the round from a Mark 2488 MAC, or Onager, which is the MAC used at the end of Halo: Reach by Noble Team’s Emile-A239, and then SPARTAN-B312. This is a ground-mounted cannon, and we know from Halopedia and Halo 4: The Essential Visual Guide that the MAC round for the Onager is 15 cm long, and is made of some sort of ferric-tungsten composite. We aren't given anything else for mass, so I will have to make some assumptions to get the mass of the slug.
The Mark 2488 MAC, or Onager, defending the UNSC Pillar of Autumn during the Fall of Reach, August 30th, 2552.
I'll assume the slug is pure Tungsten for simplicity and to make it as massive as possible. Tungsten has a density of 19.3 grams per cubic centimeter. If the slug is 15 centimeters long, it won't be as wide. I think a fair assumption is to imagine the slug is a 5 cm diameter cylinder. Doing some quick math for the volume of a cylinder (pi * radius² * length), or 295 cubic centimeters. Using the density for Tungsten, the mass of a Mark 2488 MAC slug is 5,684 grams, or 5.7 kg, or 12.5 pounds. Not that heavy for a MAC round, but with magnetic accelerator cannons, the power of the weapon is in the kinetic energy of the projectile, not just its weight.
We aren’t given the velocity of the Onager projectile directly, but in this case we don’t need it. Also from Halo 4: The Essential Visual Guide, we know the Onager fires the projectile using 1.1 gigajoules of energy. Using the mass we calculated from above and the equation for kinetic energy (1/2·m·v²), we can determine the velocity of the round by doing the following math:
MASS (M) = 5.7 kg
KINETIC ENERGY (KE) = 1.1 gigajoules = 1,100,000,000 joules = 0.263 tons of TNT
KE = (1/2) · M · VELOCITY (V)²
1,100,000,000 = (1/2) · 5.7 · V²
V = √(1,100,000,000 / 2.85) = 19,646 m/s = 70,726 kph = 43,947 mph
This gives us all the data we need for the small MAC, so lets move on to the medium MAC. If you are interested in following along without doing the math yourself, check out an online kinetic energy calculator.
MEDIUM MAC
For the medium-sized MAC, I will use the most common MAC in the UNSC, the standard frigate-mounted MAC. Luckily, we know from Halo: The Fall of Reach novel that the standard UNSC frigate fires a 600 metric ton projectile at 30,000 meters per second, or 108,000 kilometers per hour, or 67,000 miles per hour. This is actually all we need to answer the question posed, but for the sake of comparison, lets do the math to calculate the kinetic energy of a Frigate-mounted MAC:
MASS (M) = 600 metric tons = 600,000 kg
VELOCITY (V) = 30,000 m/s = 108,000 kph = 67,000 mph
KINETIC ENERGY (KE) = (1/2) · M · V² = 270 terajoules = 64.5 kilotons of TNT
Three UNSC frigates including the UNSC Forward unto Dawn firing their MACs against the Portal at Voi during the Battle for Earth, November 17th, 2552 (Halo 3).
BIG MAC
The largest MAC platform in the Halo universe with good data are the orbital “Super” MACs that were in orbit around Reach, and still are (as far as we know at the time this article was published) in orbit around Earth. Again, from The Fall of Reach novel, the Super MAC fires a 3,000 metric ton projectile at 4% the speed of light, or 11,991,698 meters per second, or 43,170,114 kph, or 26,824,665 mph. The novel also gives us the energy of a Super MAC firing at 51.6 gigatons of TNT, which is good, because that allows us to compare our calculated numbers to those presented in the lore, providing a check that we are at least in the ballpark when it comes to lore-accurate. So doing the math again, we get the following:
MASS (M) = 3,000 metric tons = 3,000,000 kg
VELOCITY (V) = 11,991,698 m/s = 43,170,114 kph = 26,824,665 mph
KINETIC ENERGY (KE) = (1/2) · M · V² = 215.7 exajoules = 51.6 gigatons of TNT
Well what do you know, the numbers work out exactly. That’s a relief. I would not have been happy to find out that we did the math wrong to this point and had to start over.
Orbital Defense Platform Cairo Station, part of the defense force orbiting Earth, just prior to the Battle for Earth, October 20th, 2552. (Halo 2)
As a side note, at speeds approaching the speed of light, the kinetic energy equation we have been using starts to become less accurate, and relativity has to start being considered. At the speeds we are talking about the effect is negligible, but if you are interested in seeing the difference, you can use the relativistic kinetic energy calculator here.
MAC ROUND SUMMARY
For a quick summary of the three cases we will consider for this analysis, here is the data for the small, medium, and large MAC rounds:
CALCULATED REFERENCE MAC ROUND KINETIC ENERGY
| SIZE | MASS | VELOCITY | ENERGY |
| Small | 5.7 kg | 70,700 kph | 0.26 tons TNT |
| Medium | 600,000 kg | 108,000 kph | 64.5 kilotons TNT |
| Large | 3,000,000 kg | 43,170,114 kph | 51.6 megatons TNT |
It is pretty impressive the difference in energy between the Onager, the frigate-mounted MAC, and the orbital Super MAC. While the Onager round energy seems small (a Tomahawk cruise missile has a yield around double the Onager) it is applying that energy over a much smaller area, meaning the shot would still be able to penetrate any unshielded aircraft. It also fires many rounds, so it would be far more destructive than a single missile.
The frigate, despite its relatively small size, packs quite the punch, hitting its target with around four times the energy of the bomb that wiped out Hiroshima. As we know from the Halo lore, this still isn’t enough to take out Covenant shields, but it certainly does a lot of destruction to anything unshielded, including other human spacecraft. For better reference, here is what a fully charged frigate MAC impact would look like if it was directed at the ground.
US Nuclear weapons test Upshot-Knothole Climax, a 61 kiloton nuclear detonation. THIS is the lore-accurate size of a fully-charged frigate MAC impact.
Meanwhile, the Super MAC is somewhat hard to comprehend. It is hard to relate to numbers at this scale, so I am going to do my best. The frigate-mounted MAC is about 250,000 TIMES more powerful than the Onager. Its the same technology, but fires a much heavier slug faster. The Super MAC’s slugs are also heavier, but only 5 times more than the frigate MAC. The Super MAC gets all its energy from the sheer speed of the projectile, travelling 400 times faster than the frigate MAC. And since the velocity term is squared in the kinetic energy equation (KE = (1/2) · M · V²), that difference in speed translates into a MAC shot that is 800,000 TIMES more powerful. To put that into context, that is 1700 times larger than the Tunguska event in 1908, 1000 times larger than the Tsar Bomba, the largest ever nuclear blast in current human history, and 0.2% the size of the blast that created the Chicxulub crater and likely led to the extinction of the dinosaurs. I know these are still not great comparisons, but at scales like this, it is hard to find historical events that match the scale of what we are talking about.
The Tsar Bomba, a 50 megaton nuclear weapon, the largest ever tested. The mushroom cloud reached 40 miles into the atmosphere. The Super MACs are 1000 TIMES more powerful.
As a brief aside, if an insurrectionist group were able to commandeer an orbital defense platform, turn it 180 degrees, and fire it at the surface of the Earth, it would devastate a massive portion of the planet. Every shot would be like detonating thousands of nuclear bombs at once. Using an asteroid impact calculator and entering in our data we can determine just how much devastation would result. The crater would be over a mile DEEP, and nearly five miles wide. One shot would easily vaporize an entire city. Even if they only got one shot off, its effects would be incalculable.
THE DIRECTION OF THE SHOT
So getting back to the question at hand, what would happen to a MAC round that missed, we can make some simplifying assumptions as to the angle and direction of the shot. While you would think that the angle you fire a projectile away from the gravitational body in question would affect the analysis, it turns out it really doesn’t. So long as the MAC round doesn’t actually impact the planet or atmosphere, it will accelerate towards the planet and decelerate away from the planet. Because of this, we can assume everything happens from the surface of the planet or star since gravity is strongest there and that will bound our analysis without needed to make the math more complicated.
For the more visual learners, here is a quick demonstration of orbital versus escape velocities.
THE ESCAPE VELOCITY OF THE NEARBY PLANET OR STAR
As I mentioned at the start of all this, every object in the universe has an escape velocity, which is just a speed at which another object like a spaceship or MAC round will never slow down completely and return to the object in question. Put another way, at any speed slower than the escape velocity, a spaceship or other projectile would eventually fall back towards the planet or star it was fired away from. At any speed faster than the escape velocity, the projectile would fly away forever, never to return to the launch site. Knowing this, the assumption we made above that we start at the surface of the planet or star, and the speed of the MAC round at launch, it is fairly easy to look at each MAC round type and see what happens if it were fired from different locations.
The only numbers we are still in need of are the actual escape velocities of different planets and stars. Since these are already known, it is as easy as just Googling ‘escape velocity’ to generate a table of different planets and stars and their respective escape velocities.
VARIOUS CELESTIAL BODY ESCAPE VELOCITIES
| CELESTIAL BODY | ESCAPE VELOCITY |
| Moon | 2.4 km/s |
| Jupiter's moons | 2 - 3 km/s |
| Mars | 5.03 km/s |
| Earth | 11.2 km/s |
| Sun (from Earth, with orbit) | 16.6 km/s |
| Sun (from Jupiter, perpendicular) | 18.5 km/s |
| Sun (from Mars, perpendicular) | 34.1 km/s |
| Saturn | 36.1 km/s |
| Sun (from Earth, perpendicular) | 42.1 km/s |
| Jupiter | 60.2 km/s |
| Sun (from Earth, against orbit) | 73 km/s |
| Milky Way (from Sol system) | 500 - 600 km/s |
| Sun (from surface) | 618 km/s |
And just so you don’t have to scroll back up, here are the velocities of the different MAC rounds we looked at:
REFERENCE MAC ROUND MUZZLE VELOCITIES
| MAC TYPE | MUZZLE VELOCITY |
| Onager (small) | 19.6 km/s |
| Frigate-mounted (medium) | 30 km/s |
| Super-MAC (large) | 11,990 km/s |
ESCAPE VELOCITIES OF VARIOUS INTRASOLAR BODIES
THE RESULTS
Right off the bat, we can determine some things without doing any real math.
BIG MAC
Starting with the Super MAC round, its easy to see that those rounds are never coming back to Earth. Not only that, they are never coming back to the Sol system at all, and in fact they will leave the entire galaxy given enough time. It really doesn’t matter which direction you fire one of those things, if the round misses, it is gone forever. And while we don’t have a lot of data on rounds that actually hit their target, I’d imagine a good number of them would go right through their target and keep on going out of the galaxy. They are going 20 times faster than the required escape velocity of the milky way at Sol distances, so even if they lost 90% of their speed, they would still have plenty left over to keep on flying forever. So as a quick answer, no, there are not any Super MAC rounds left in orbit anywhere to crash into an unsuspecting spacecraft.
MEDIUM MAC
The next ones get a little more complicated, but they shouldn’t be too bad. I will continue to take them in reverse order, so next up is the frigate-mounted MAC.
EARTH
At 30 km/s, the frigate-mounted MAC is MUCH slower than the orbital defense platform MAC guns, but it is still incredibly fast at 30 km/s. Using the table above, you can see that the frigate MAC will easily escape the Earth’s gravity. Even if you factor in the worst case scenario of a ship in orbit and firing in the opposite direction of the spin (the ship would be traveling faster than the planet), that only adds in 5-6 km/s of required escape velocity at most. Unless the frigate fires at less than half power, the MAC round will be well above the Earth’s escape velocity.SUN
When we look at the escape velocity for the Sun, however, the numbers don’t make it quite so easy. From the surface of the Sun, the frigate-mounted MAC round isn’t going to get very far. With an escape velocity of over 600 km/s, the gravitational force of the Sun is far too strong for the MAC round to escape. From the Sun’s surface, the MAC round would just fall back and burn up rather quickly.Of course, how many UNSC frigates are hanging out near the Sun? Probably not any. Much more likely is that they are in the vicinity of a human colony. In the case of the Sol system, that means either Earth/Luna, Mars, or one of the Jovian moons. Of course, with orbital mechanics, nothing is as simple as you want it. We will start with Earth as an example.
SUN/EARTH SYSTEM
We already determined the frigate-based MAC will escape Earth’s gravity. When it comes to escaping the Sun’s gravity, however, that isn’t quite so simple. If the frigate fired a MAC round directly away from the Earth and Sun, it would not be going fast enough to escape the Sun’s gravity. At a little over 42 km/s, the Sun’s escape velocity from Earth is still too great. The round would end up in some sort of orbit around the Sun, but escape the Earth’s gravity.
Interestingly, though, the MAC round still could escape the Sun’s gravity. As it turns out, the Earth is rotating the Sun at about 30 km/s. So if the frigate were in orbit around the Earth and fired in the direction of the Earth’s rotation, the MAC round would be traveling at nearly 60 km/s (30 km/s Earth’s orbit + 30 km/s muzzle velocity). Put another way, the minimum escape velocity for the Earth/Sun system would only be 16.6 km/s, and at 30 km/s the MAC round would be going plenty fast enough to leave the Sol system entirely.
If the round were fired in the opposite direction of the Earth’s orbit, the round would have a net orbital velocity of just about 0 km/s (30 km/s Earth’s orbit - 30 km/s muzzle velocity). This would still escape the grasp of Earth, but it would basically drop straight into the Sun.
A frigate firing in the opposite direction of Earth’s orbit would basically be like this Mythbusters video (from the perspective of the Sun).
SUN/MARS SYSTEM
Being that Mars is much smaller, a frigate MAC would easily leave the confines of the planets gravity. Like orbiting from the Earth, however, the MAC round would still have to escape the Sun’s gravity as well. Just like the case of the Sun/Earth system, the results will depend on the direction the frigate fires. Fired directly away from Mars and the Sun, the round would not have enough velocity to completely escape. It would end up falling into some sort of elliptical orbit around the Sun just like in the Sun/Earth example. Fired with the orbit of Mars, the MAC round would have enough velocity to escape the Sun, as Mars orbits the Sun at around 24 km/s. Fired in the opposite direction, the slug would end up in a highly elliptical orbit around the Sun, and potentially crash into it.IO/JUPITER SYSTEM
For the sake of not making this even more complicated than it already is, looking at a MAC round fired around one of the human colonies in the Jovian moons, we get a similar answer to the above. The only difference is there would be a chance the round left the solar system, fell into a solar orbit, fell into a Jovian orbit, or crashed into Jupiter’s surface. With so many potential scenarios and so many variables, all we can really say with certainty is that all of these options are possible.
SMALL MAC
The ground-based Onager has the potential to make this even more complicated since it fires in-atmosphere, but I am going to make it fairly simple. While the Onager round has the velocity to get into a solar orbit or possibly escape the solar system depending on the trajectory, because it is fired in-atmosphere means it isn’t going to make it into orbit before the slug completely disintegrates. If one of these, or something similar, were fired in space, like the newer-model Onagers aboard the UNSC Infinity, then we would be looking at a scenario similar to the frigate-based MAC. While much smaller, the muzzle velocity of an Onager is 70% that of the frigate-mounted MAC, so all of the options above are still viable.
THE ANSWER
So that was a lot of typing and math to just get to the heart of the question: If a MAC round missed its target, what would happen to it, and would that pose a danger to space traffic? We have established that there probably are a number of MAC rounds somewhere in orbit around the sun and other stars, and that it is very unlikely that any are in orbit around Earth or any colony. In addition, there are also likely at least a handful of MAC rounds from the Orbital Defense Platforms around Earth and Reach that are flying through space, not bound by the system’s gravity or even the galaxy’s. But is this a hazard to space travel?
The short answer is no. It could be argued that MAC rounds in Earth orbit would pose a danger, but they would necessarily be travelling at orbital velocities, which are much slower than typical MAC muzzle velocities. Any debris in orbit would also be travelling at the same speeds, so it would pose as much of a hazard as any stray MAC round. Even if the math had shown that MAC rounds were in orbit around Earth, the sheer amount of debris from the Battle for Earth, the Invasion of Earth, and the New Phoenix Incident would be a much larger hazard by several orders of magnitude.
As for any rounds in orbit around the sun, consider this: When NASA plans a deep space mission taking a spacecraft through the asteroid belt, they don’t bother to plan a flight path around the asteroids because there is pretty much no chance of hitting one. In the entirety of the asteroid belt, there is an estimated 4% of the mass of Earth’s moon, spread over an region around four times the volume of a sphere contained within the orbit of the Earth (or the volume contained within the Sarcophagus). And when we are talking about maybe a few dozen MAC rounds orbiting somewhere in a similar volume, the chances of impact are even less likely. The probably isn’t technically zero, but for all intents and purposes it is.
To help with the scale of the solar system, there is a great website that really hammers home how empty the solar system is. If you just want the video, here it is.
A tediously accurate scale model of the solar system.
Then there is the rest of the galaxy, which is somehow even less dense than the solar system. As an example, our galaxy, the Milky Way, and our neighbor the Andromeda galaxy, are expected to collide in four billion years. Based on calculations, the chances of even a single stellar collision between the two galaxies is negligible. I really like how the Wikipedia page on this galactic collision is worded:
“For example, the nearest star to the Sun is Proxima Centauri, about 4.2 light-years... away. If the Sun were a ping-pong ball, Proxima Centauri would be a pea about 1,100 km (680 mi) away...
Although stars are more common near the centres of each galaxy, the average distance between stars is still 160 billion... km... That is analogous to one ping-pong ball every 3.2 km (2.0 mi). Thus, it is extremely unlikely that any two stars from the merging galaxies would collide.”
A simulation of the Milky Way and Andromeda galaxies colliding in a few billion years. There are not expected to be any collisions despite the hundreds of billions of stars whizzing by each other.
So basically, the chances of randomly colliding with a stray Super MAC round are zero. The chances of winning the lottery are likely much better, and those chances are effectively zero too. I suppose I could have started here and saved some time, but then we wouldn’t really know whether there are any MAC rounds floating out there somewhere at all. Plus, we got to explore the idea of an insurrectionist group commandeering an orbital defense platform and firing it at Earth, which is a story I would definitely be interested in.
CONCLUSION
This article has ended up much longer than I expected, so I will summarize our findings very briefly.
QUESTION ONE
What would happen to a MAC round if it missed its expected target?
ANSWER - DEPENDS
Depending on what launched it and in what direction, the round would either:Leave the entire galaxy (Super MAC, ~100% probability)
Leave the solar system (medium ship-based MACs, ~20% probability)
Fall into orbit around the system’s star (medium ship-based MACs, ~>75% probability)
Fall into the host star (medium ship-based MACs, ~<5% probability)
Burn up in the atmosphere (small ground-based platforms, ~>95% probability)
Fall back to the surface (small ground-based platforms, ~<5% probability)
QUESTION TWO
Are there hundreds of MAC rounds flying aimlessly through space?
ANSWER - YES
Though the number is unknown. It would highly depend on the percentage of misses, and the total number of rounds fired. Dozens seems like a more reasonable estimate, though that would be dozens over a significant region of the galaxy.
QUESTION THREE
Is it likely that a rogue MAC round would strike a spacecraft?
ANSWER - NO
The probability of such an event, while we didn’t calculate it specifically, is effectively zero. There is just so much empty space and so few MAC rounds that it isn’t reasonable to expect any of them to manage to hit a single ship, even one the size of the Infinity.
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