# Can you shoot someone with a bullet… from orbit?

I remember a character I once created who was a really good sniper. When describing him, I said he could pull off a headshot on a terrestrial target from orbit. Now, in this case it was with a laser, and it was a long time ago so I didn't know as much science as I do now, but I'd like to revisit the concept in case I can ever bring it up in a future story.

So let's say you're perched atop the ISS and you have a gun capable of shooting a 50-caliber-sized bullet into some unlucky guy's head, with lethal force. I realize that the chances of hitting the target are astronomically low, but I'd still like to know some things, such as:

1. What should the bullet be made of, how should it be shaped, and what does it look like before and after?
2. Say the target is on the equator. At what point is it best to shoot him, and where do you aim?
3. Does anything particularly interesting happen upon impact?

I'd like for answerers to assume that this can be done; that is, if there's something that makes it impossible for a regular sniper rifle, please consider a bigger or more specialized gun that can get the job done, if one can exist.

That said, the projectile should be a bullet; no guidance or stabilization after firing, and preferably just a single chunk of metal, no stages or layers. How you accelerate the bullet is up to you, as long as it's scientifically sound.

• @SJuan76 How do you figure a geostationary orbit has high speed relative to the surface of the Earth? The speed is, ideally, zero. – Samuel Jul 24 '15 at 18:27
• @Samuel the angular speed of a geostationary satellite and a point in the surface of Earth is the same, but since the surface of Earth is 6700 km from the center of rotation while the satellite is aprox 35000 km from that same center, the lineal speed is of the satellite is way higher. And that lineal speed is that the rifle has to compensate for. – SJuan76 Jul 24 '15 at 18:33
• Would a bullet actually be able to survive reentry? I would think that it would burn into a little bit of light... – bowlturner Jul 24 '15 at 19:09
• Is this a worldbuilding question? It sounds more like a physics question to me. – Monica Cellio Jul 24 '15 at 22:34
• I'm putting this on hold based on the reaction to my previous comment, despite the absence of any close votes from the community. I've started a discussion on meta: meta.worldbuilding.stackexchange.com/q/2353/28 – Monica Cellio Jul 27 '15 at 12:47

# Projectile Atmospheric Entry Simulations

I put together a bit of code to compute the trajectory of projectiles as they fall through the Earth's atmosphere. I made the following simplifying assumptions:

• All projectiles were fired from a circular orbit equal in altitude to the ISS (about $400~\text{km}$).

• The atmosphere is non-rotating with no winds, and is identical to the standard atmosphere.

• All projectiles fall perfectly straight: there are no lateral aerodynamic forces.

• None of the projectiles' properties are altered by reentry.

I computed two scenarios for each projectile. The red trajectory is for a bullet fired towards nadir (straight down), and the blue trajectory is for a bullet fired retrograde (opposite the direction of orbit).

## .50 BMG (Ball, M33)

The 50-caliber Browning Machine Gun cartridge was originally developed to fulfill an anti-aircraft role, but later became a popular round for snipers. Its high mass helps it keep its speed and accuracy, even at distances of over one mile. In fact, more than half of the 15 longest sniper kills were made with the .50 BMG.

• Projectile mass: $650~\text{gr}~(42~\text{g})$

• Projectile diameter: $0.510~\text{in}~(13.0~\text{mm})$

• Muzzle velocity: $3030~\text{fps}~(920~\text{m}/\text{s})$

• Drag computed based on data (pdf) from the Ballistic Research Laboratory at Aberdeen (see figure 19).

(The dashed line shows the orbit that the projectiles were fired from.) The impact points are:

• Retrograde: $2600~\text{mi}~(4180~\text{km})$ downrange in $12~\text{minutes,}~30~\text{seconds}$.

• Nadir: $2050~\text{mi}~(3300~\text{km})$ downrange in $9~\text{minutes,}~15~\text{seconds}$.

(The dashed line is the lower boundary of the stratosphere at $12~\text{km}$, slightly above the cruising altitude of most airliners.) Note that both shots impact at a steep angle. Although the .50 is heavy compared to most reentry vehicles (in terms of ballistic coefficient) it still loses almost all of its velocity to drag and ends up pretty much falling at terminal velocity.

(Above the dashed line the speed is measured in kilometers per second; below the line, the speed is measured by Mach number.) Again, we see that most of the velocity is lost in the tenuous upper atmosphere.

This plot shows the projectiles' total energy loss rate. Not all of this energy will heat the projectile, however: in fact a good portion is used ionizing and heating the air the projectile encounters.

Calculating the peak heating according to this Institute for Defense Analysis document (pdf), the projectile will encounter peak temperatures of over $4000-5000~\text{K}$. Modern ablative heatshields for spacecraft are built to withstand up to around $2600~\text{K}$ (although classified heatshield technologies for ballistic missile warheads may have moderately better performance). This means that the projectile would surely disintegrate upon entry.

## 30×173 mm (Armor-piercing incendiary w/ DU penetrator, PGU-14/B)

This massive shell is used by several autocannons and chain guns, most notably the GAU-8/A Avenger: the primary armament of the Fairchild Republic A-10 Thunderbolt II "Warthog", an anti-tank air support and attack aircraft (and a personal favorite of mine). The projectile consists of an aluminum "jacket" surrounding a $10\frac{1}{2}~\text{oz}~(300~\text{g})$ depleted uranium penetrator.

Although typical accuracy of the GAU-8 autocannon is only 40 yards at 4000 yards distance, the projectile is probably capable of accuracy similar to typical sniping rounds if fired from an appropriate gun (although such a weapon may be too large for a single operator, and would certainly not be man-portable).

• Projectile mass: $1~\text{lb}~8\frac{1}{2}~\text{oz}~(695~\text{g})$

• Projectile diameter: $1.18~\text{in}~(30.0~\text{mm})$

• Muzzle velocity: $3030~\text{fps}~(1010~\text{m}/\text{s})$

• Drag computed based on data (pdf) presented by the University of Sarajevo (see figure 9).

The impact points are:

• Retrograde: $2550~\text{mi}~(4100~\text{km})$ downrange in $10~\text{minutes,}~55~\text{seconds}$.

• Nadir: $2030~\text{mi}~(3260~\text{km})$ downrange in $7~\text{minutes,}~50~\text{seconds}$.

Although the shots now have distinguishably different trajectories, again they both impact at nearly the same angle (although both are more horizontal than the .50).

This time the projectiles maintain their velocity down into the stratosphere, but are still limited to terminal velocity at impact.

With the bulk of the deceleration occurring in a denser portion of the atmosphere, peak temperatures are now on the order of $9\,000~\text{K}$.

## 120 mm APFSDS-T (DM13)

The Armour-piercing fin-stabilized discarding-sabot round is a kinetic energy penetrator designed to defeat modern vehicle armor. For maximum penetration, the projectile is more dart-shaped than bullet-shaped. The projectile I found data on is the DM13, a non-DU round similar to the 120 mm M829 round fired by the United States' main battle tank, the M1 Abrams.

Since the fins stabilize the projectile in flight, APFSDS rounds are fired from smoothbore guns that allow increased muzzle velocity. This allows superior accuracy; modern tank crews can make a kill shot on another tank at several miles. But again it is unlikely that such a weapon could be operated by a single person.

• Projectile mass: $9~\text{lb}~12~\text{oz}~(4423~\text{g})$

• Projectile diameter: $1.50~\text{in}~(38~\text{mm})$

• Muzzle velocity: $5000~\text{fps}~(1500~\text{m}/\text{s})$

• Drag computed based on data (pdf) from the Ballistics Research Laboratory.

The impact points are:

• Retrograde: $1950~\text{mi}~(3140~\text{km})$ downrange in $8~\text{minutes,}~40~\text{seconds}$.

• Nadir: $1340~\text{mi}~(2150~\text{km})$ downrange in $4~\text{minutes,}~40~\text{seconds}$.

Both shots have extremely straight trajectories and impact at shallow angles, indicating that they have maintained speed down to the surface.

This time the projectiles impact at hypersonic speed, delivering a devastating blow beyond the capability of any modern non-explosive round.

However, peak temperatures are now on the order of $13\,000~\text{K}$. Even though heating only occurs for a few seconds (in-atmosphere the projectile loses around a kilometer of altitude every second) the intensity is so great that the projectile will be molten by the time it reaches the surface.

# Postmortem

We can see that there is a fundamental tradeoff between heating and impact speed. The projectile must be large and lightweight in order to decelerate slowly and avoid burning up; however, it must also be heavy and dense to retain its speed. There is no middle ground between the two.

This means that a successful projectile would have to change aspect during its trajectory, and since we're limited to "a single chunk of metal, no stages or layers," this is not possible.

Another issue is accuracy. Since the projectile takes several minutes to hit the target, there is no possibility of a second, more accurate shot after using the first to 'scope in.' This, combined with the fact that the impact point is thousands of miles away means that all the targeting will be computerized.

The position of the shooter can be pretty precisely determined (within meters) by GPS (even on-orbit). GPS also gives precise timing. Star trackers can give milliarcsecond angular resolution with reaction wheels for pointing. The main challenge is aerodynamic perturbations:

• The first problem is the orientation of the projectile. In the diagrams above, the projectile is shot right or down, but enters the atmosphere heading left. Thus the projectile would have to be shot backwards.

• The second issue is stability. In the very thin upper atmosphere spin-stability will not work if the projectile is long and thin. Explorer-1 (the United States' first satellite) was designed to spin about its long axis, without accounting for the (mathematically difficult) dynamics of free rotation in 3D, and it predictably transitioned quickly into a 'flat spin.' This pretty much guarantees that an inert bullet would tumble as it enters the atmosphere, no matter how we try to stabilize it.

• The third issue is wind: namely, the shooter would have to account for winds through the entire height of the atmosphere; and be able to predict the (highly chaotic) winds nearly ten minutes in advance.

# An Alternative

If you're really set on shooting someone from space, you're going to need an active projectile. The projectile should consist of a $1.5~\text{m}$ long tungsten penetrator surrounded by a large sabot made of a lightweight insulating ceramic with an ablative coating. The sabot will need a large, flat front. Finally, the projectile should include a large solid motor and some variant of solid-state attitude thrusters.

• First the projectile would be programmed with its target and released from the launch platform at a relatively low speed.

• Once well-separated from the launcher, the solid motor will execute a deorbit burn.

• As the projectile enters the atmosphere, the ablative coating protects the penetrator from heating.

• Just after peak heating, the sabot (along with the solid motor) is jettisoned with pyrotechnic fasteners.

• At this point, the penetrator begins to use aerodynamic maneuvering (like a missile) as it falls at its terminal velocity of Mach 2.

• The penetrator uses GPS for terminal guidance and impacts within meters of the target.

• As for accuracy of unguided reentry vehicles, I think it's rather telling that ballistic missile warheads are usually quoted as having a circular error probable in the hundreds of meters range. – 2012rcampion Jul 26 '15 at 21:39
• Fantastic answer. – HDE 226868 Jul 27 '15 at 15:01
• Seconded. This is a phenomenal answer. – Green Aug 24 '15 at 20:02
• +1 Awesome entry, but a note for the last section. At the point described, it pretty much stops being a bullet and instead becomes a missile. – Anoplexian - Reinstate Monica Sep 8 '16 at 14:37
• 1.5 m long tungsten penetrator a little rod from God. en.wikipedia.org/wiki/Kinetic_bombardment – RonJohn May 13 '17 at 8:22

It couldn't be done, no matter how improbably skilled the sniper, because he would lack information required to do the aiming.

Regular snipers have to consider more than the direction their firing, they need to consider the weather. Snipers will use... well effectively flags to determine wind speed and direction so they can adjust for it; because otherwise the bullet will be blown off course as it flies. The longer the trip, the more time the wind has to blow the bullet off course.

Now imagine a shot from orbit. It's going to be traveling a much longer distance, and thus have a much longer time to be blown off of course. It's also going to be traveling through high atmosphere where wind speeds are much greater, and far more chaotic. The wind and thermals are going to have a significant effect on the bullets trajectory.

The problem is that he is firing so far that the wind speeds and direction will change as the bullet flies. This is particularly true because he is firing down through atmosphere. The speeds and directions of winds are not only far greater at the upper atmosphere, but they change significantly as you travel through the atmosphere. There are many winds, thermals, and other weather patterns his bullet will be traveling through.

Even if we assume your sniper was a robot with an impossibly perfect AI that could instantly adjust for complex math such as orbital rotation, rate of the bullet's falling, and even change in mass of the bullet (it will be losing mass, at the speeds it is traveling, and that change would have a noticeable effect) your robot can not calculate the proper angle to fire at unless it knows all the weather patterns from here to your target. It's not enough to use a single flag, you're going to need to know information about weather speeds all over the atmosphere. He would need dozens of data points, at minimum, to have enough information about the wind speeds along the bullet's path to have enough raw data to make such information calculable.

Unless he has dozens of weather balloons flying in a rightly diagonal line leading to his target, he simply does not have enough raw data to calculate where to shoot with any remotely reliable accuracy, no matter how perfect he is.

In addition there is an even more boring problem. Even if he shoots perfectly he won't hit his target, because his target won't be there when it arrives. The ISS is quite a ways away from the ground, 400 km to be exact, and it takes time to travel that far.

Assuming your bullet is traveling at subsonic speeds (if it's going at supersonic speeds you have a whole different set of issues) it can travel no faster than 342 m/s. Even if we give it the benefit of the doubt and assume that you are firing straight down, and the bullet manages to stay at exactly the sonic barrier for the entire trip, it would still take over 19 minutes to reach the ground. You're not going to be able to anticipate your target's location 19 minutes in advance when you fire. You would need your bullet to reliably move at mach 10 speeds for the entire trip to get to your target in under 2 minutes, and even that is really too long to have any reliable chance of anticipating where he will be.

• I think the key to this answer is really in the travel time. Even if you had all the math worked out to be able to put the bullet exactly where you intended to, you have no way of making your target remain in position long enough for the bullet to get there, except maybe if they're asleep... and even then, they could turn over in their bed during those minutes. – Dan Henderson Jul 25 '15 at 7:08
• @DewiMorgan - OP says "you're perched atop the ISS". The ISS orbit is held at about 400 km. Furthermore, the bullet is launched at fairly shallow angle, so distance to target will be much more than 400 km. Standard muzzle velocity for a .50 BMG ball round is just under 1 km/sec. – WhatRoughBeast Jul 25 '15 at 22:53
• what's the terminal velocity of a bullet? – njzk2 Jul 25 '15 at 23:06
• @Mazura the difference is, at the distances Chris Kyle shot from, the total travel time of the bullet was insignificant at the scale of his targets' velocity, and thus, leading the target was all that was needed. But neither humans nor terrestrial vehicles have enough inertia to enable a multiple-minute lead. – Dan Henderson Jul 25 '15 at 23:53
• Some fancy Doppler radar systems can give you volumetric motion for the gasses in a specified space, so that's not really even that hard of a hurdle. The big challenge is predicting the wind at the point the bullet passed through a point, since the wind may change between the firing and the point the bullet reaches the point in question. – Fake Name Jul 26 '15 at 2:51

Only if the bullet is really big.

The problem with hitting someone with a bullet from space is that any tiny current of air that you did not perfectly predict is going to throw off your aim. If you're firing a bullet through the entire atmospheric column, there are going to be such air currents. Over a few seconds, through a few miles of air, it may be possible to accurately predict wind behavior, but at a flight speed of five times the speed of sound (at sea level), your sniper bullet will be travelling for around five minutes before it hits its target. You not only need to know exactly what the wind conditions below you are: you also need to know how they will act during that time.

Of course, with a big enough gun firing big enough bullets, you don't need to have perfect aim. So long as you release enough kinetic energy when you hit to vaporize everything withing a few meters or tens of meters of the impact point, your bullet can be buffeted around by the wind a bit without affecting anything.

With a big enough gun, it won't even matter if the target is buried deep in an underground silo, just find a gun that can shoot these:

Pictured: a very large bullet.

Addendum: the Chicxulub meteor is probably a bit large to count as a 'bullet', but smaller projectiles will work as long as you get them moving fast enough. Something like a 200-gigajoule railgun will probably do the trick.

• This is cheat. You just destroy the whole planet for the sake of simplicity. Why don't we just use a nuclear bomb bullet and explode the Earth? – Ooker Jul 26 '15 at 16:08
• @Ooker A bullet is specified as being 'a single chunk of metal' in the original question, which rules out nuclear bombs. – ckersch Jul 27 '15 at 14:28
• Upvote for the caption. – jdunlop May 23 '18 at 17:56

1- What should the bullet be made of, how should it be shaped, and what does it look like before and after?

The munition will need to be made of a material that will properly handle the heat of reentry. Tungsten or some kind of specialized ceramic should do the trick. Getting the hypersonic shape right is difficult but a pole shaped object with a small frontal cross section will bleed off energy less than the round flat frontal surfaces used in human rated reentry vehicles.

2- Say the target is on the equator. At what point is it best to shoot him, and where do you aim?

As anyone who plays Kerbal Space Program will tell you, deorbiting a projectile from orbit so that it hits a specific point is difficult. Here's a video of Scott Manley attempting to do precision hits with a telephone pole made of tungsten from orbit. It takes him multiple tries to hit a building sized target. Deorbiting a small projectile that can be blow off course even easier than a tungsten telephone pole. Hitting a moving target is a ba-zillion to one shot. Simple aiming from orbit will require ridiculous amounts of data to account for orbital rotation, thermals in the atmosphere, exact flight characteristics during reentry, exact orbital height, charge characteristics in the ionosphere and stratosphere, exact target position etc etc. Adding terminal guidance to the munition is your only bet of actually hitting the poor sap you want to kill.

I heard once that the accuracy requirements for the Hubble space telescope are equivalent to sticking a laser pointer on top of the Washington Monument in Washington DC and hitting a dime placed on top of the Empire State Building in NYC. Given the distances involved, the accuracy requirements for this gun in space are pretty similar.

3- Does anything particularly interesting happen upon impact?

It makes a big bloody mess. Anything with sufficient mass to maintain momentum deccelerating through the atmosphere from orbit while staying on target and going a significant portion of orbital velocity will absolutely wreck whatever it hits. Kinetic energy is calculated with:

$E_\text{k} =\tfrac{1}{2} mv^2$

where m is mass and v is velocity. So, an object in LEO has a speed of 7.8km/s. Let's assume the munition is 10kg and has lost 2.8km/s to reentry drag. So 0.5 x 10kg x 5.0 m/s^2 = 125 000 000 joules = 125 megajoules . Using this calculator to translate between joules and tons of TNT, we get 29.5kg of TNT. (The page doesn't note which kind of explosive is used, but TNT is a safe assumption.)

So you don't have to get pinpoint close but getting close is really hard.

• 10kg is a massive 'bullet' – depperm Jul 24 '15 at 20:29
• Wouldn't the bullet be slowed down by atmospheric breaking, reducing the total energy of the impact by converting a lot of that energy to heat and reducing mass? – Dan Smolinske Jul 24 '15 at 20:31
• using E=.5*m*v^2 with 10kg wouldn't the energy be more like 125000 Joules .5*10000*25 – depperm Jul 24 '15 at 20:32
• @depperm, thank you. I missed the .5 when doing the math. – Green Jul 24 '15 at 20:38
• Given the relatively shallow reentry angle, you should assume the bullet will have reached terminal velocity when it hits the target. For a .50 caliber lead bullet, this is about 500 fps, so kinetic energy is not in the "blows him apart range". – WhatRoughBeast Jul 25 '15 at 17:20

Most of the technical issues have been addressed in other answers, so here are a few issues that should be expanded upon:

1. A bullet made of a single material isn't going to work, since it has to deal with many different regimes during it's flight path, from space, to atmospheric entry to flying at terminal velocity to the target once it exits the plasma sheath. A dense core of something like Tungsten or Depleted Uranium will be needed to provide the high density to punch through the atmosphere without losing too much velocity. Some sort of ceramic "jacket" is needed to reduce the effects of reentry (otherwise your bullet becomes a stream of molten metal flying across the sky) and to get to the velocity needed to counter the orbital motion of the ISS and reenter the atmosphere, I would suggest a conductive layer that can interact with the electromagnetic field of a coil gun (or alternatively, a sabot with a conductive base for a railgun.

2. Since the ISS is moving at orbital velocity, you have the issue of getting your bullet to slow down enough to start reentry. An XKCD "What IF" (https://what-if.xkcd.com/58/) tells us that you could travel 1000 miles in the 3min 30 seconds it takes to sing the song "I'm Gonna Be" (which has the line "just to be the man who walked a thousand miles to fall down at your door") If you just lined up a shot, you would overshoot the target drastically, hitting someone on a ship in the Atlantic when you were aiming at the Big Bad in Rio de Janeiro. A normal rifle won't provide enough velocity for the bullet to make a successful orbital change, and even a 16" naval cannon from a battleship would be hard pressed to do so, given the limits of gunpowder as a propellant. A railgun or coilgun theoretically has the ability to do so, or you could use what already works and make the projectile a rocket...

3. What will happen if the round hits the target? Nothing good. The round is moving at hypersonic velocity when it hits, so it will have considerably more energy than a conventional rifle round (depending on how massive it is [see above], it might have the kinetic energy equal to the energy release of a stick of high explosive). As well, objects moving at that speed no longer follow the usual rules of impact and energy transfer. It is usual to think of the impacting object almost as a liquid. The best terrestrial example is a HEAT warhead from an anti tank weapon. The blast of the warhead is focused on turning a cone of copper or similar metal "inside out" and accelerating it to @ Mach 25 (coincidentally nearly orbital velocity), where a few ounces of metal forces its way through dense materials like steel armour, and virtually everything else. Protection against this involves either making the explosion far enough away the metal jet loses energy (the "slat armour" on modern vehicles) or to disrupt the jet by firing a slab of steel at it a microsecond before it strikes the main body of the tank (the brick like "explosive reactive armour" you see covering Russian tanks). These are not viable options for a person to wear on their head to protect against a similar sized projectile coming out of orbit (even a building or tank is likely not going to stop such a projectile). Even a near miss might kill the target, as a hypersonic projectile will have a massive shock wave as it passes, and the impact on the ground will spall off material like concrete or rock at considerable velocity.

More detailed answers will require experimentation.....

• The computation I did for my answer indicate that a typical high-powered rifle does have enough $\Delta v$ to put a bullet in the atmosphere. Also: here's a neat demonstration of the type of RPG countermeasure you describe. – 2012rcampion Jul 26 '15 at 21:56

Summary: Not for anything that could reasonably be called a "bullet"

Calculating the aerodynamics of a bullet moving through the atmosphere is hard. Fortunately, the energy levels are so high that we can ignore aerodynamics for the most part, and instead treat it as an impact between the bullet and the atmosphere. This means we can use Newton's approximation for impact depth:

$$D = L\frac{\rho_1}{\rho_2}$$

Depth of penetration D is the length of the penetrator L times the ratio of the penetrator's density $\rho_1$ to the target's density $\rho_2$.

The atmospheric density is varying, but we can approximate it from the surface air pressure as being equivalent to 10 meters of water. From this, we can calcluate that an osmium bullet (density: 22.6 g/cm3) fired from orbit needs to be at least 45 cm long in order to retain a reasonable velocity at impact.

Yes, you can snipe someone from orbit, but you'll be using a crowbar, not a bullet.

• Wonderful - first answer that actually takes on the "what would it take to prevent it from burning up" question! :D – Dewi Morgan Jul 25 '15 at 22:19
• retain a reasonable velocity at impact did you just completly ignored gravity? – njzk2 Jul 25 '15 at 23:10
• @njzk2, yes. The terminal velocity of a crowbar (500-600 m/s, the velocity it "wants" to fall at) is small enough in relation to orbital velocity (7000 m/s) that the dominant force for most of reentry is momentum transfer between the projectile and the atmosphere. – Mark Jul 25 '15 at 23:28
• Why not simply calculate terminal velocity? After an impact, it is assumed that the projectile is stationary. – HDE 226868 Jul 27 '15 at 15:03

Realistically, I doubt it. There's just too much air in the way, it doesn't matter how big or heavy your bullet is, without a rocket, drag is going to slow it to terminal velocity. I mean, you'll have just as much luck and the same affect as tossing a bullet out of plane. Although that might still be lethal.

Alternatively, there are advantages to using a ballistic instead of a laser, and using a smaller (15-20mm) round instead of an orbit-to-surface missile. A laser weapon would be extremely precise, but requires a lot of energy and doesn't deliver nearly as much 'punch'. A missile might be too easily tracked/intercepted. A small hypersonic round can be tracked in orbit using modern tech, but once it drops into the stratosphere it might be difficult-impossible to determine where, exactly the shot came from. A consideration for any sniper.

The most important aspect of your story is: whatever technology required to pull of something so insane, the outcome depends on some level of human skill. So we would need to rule out a completely automated process, the 'point and click' methods. To preserve the idyllic image of a sniper with a steady hand and his eye in the scope, we should consider a lazer-guided, fin-stabilized bullet delivery system.

You will need a weapon system similar to this. I like that because it boasts a 1450m/s muzzle velocity. It will likely need need to be scaled up. Notice the sabot design of the bullet. Likely, we would have a similar design to our bullet, but rather than the sabot being discarded after firing, it would be used to protect the bullet throughout re-entry. This would allow us to use nearly any material for the core/bullet, for a variety of applications. Re-entry would be faster than the times given a spaceship, designed to aerobreak. Our sabot is designed to lesson drag, and maybe even burn off some outer layers. And I'm not sure exactly at what angle this feet becomes absolutely impossible - it might be impossible to shoot straight down at a target, but we should probably assume some level of hi-tech material could allow this. I remember hearing the heat experienced by an object entering the atmosphere is inversely proportional to the drag coefficient. That is why re-entry vehicles are designed to be blunt. Our projectile is meant to cut through the atmosphere at speeds a little over the space-shuttle upon entry (~9kps), so our sabot is going to need to tolerate, and I'm just guessing here, a proportional amount of heat.

If I follow wikipedia's advice, that's OVER 9000 degrees kelvin. Yikes! Some exotic materials for sure! You might get some real academic nerd to actually go through the calc - I'm not sure exactly how well a sabot could lesson the drag and how much that would even matter.

If our sniper is deployed 100km (a 'parking' height) from the surface, assuming a 1000m/s muzzle velocity and ignoring atmospheric drag, the shortest hang time we could possibly achieve is ~1.67 minutes from firing to impact. We then factor (drag times future-tech-hand-waving). To factor drag requires such precise calculations ADD takes over . . . there are too many variables! The troposphere is thicker around the equator than the poles, and this affects the gradient density of air/elevation, and of course you must consider the size, weight, and shape of the bullet. But after all that math you might end up with the bullet finally impacting the surface at a measly terminal velocity and that's not very exciting so it is at this point we must institute some future-tech hand-waving. Perhaps the bullet has a small, in-line ramjet. But I guess that makes it an RPG. In any case estimate your sniper's bullet spends about > 2minutes en-route.

ANYWAY, the general theory is, once at a low enough altitude the sabot is discarded and the fin-stabilized unit controls the decent, guided by the laser, guided by your bad*** l337 as %!#& sniper. SOOOOOO plausible!

• Actually I'm off by a few zeros, I'm going to re-calculate – punkerplunk Jul 25 '15 at 22:37
• This, coupled with Mark's answer, and I suppose you have a feasible chance! Although it's more akin to an artillery piece than a sniper rifle. – punkerplunk Jul 25 '15 at 22:57
• I hope you mean aerobrake... usually we don't like it when our spacecraft aerobreak! – 2012rcampion Jul 25 '15 at 23:50
• Why only 1km/s muzzle velocity? – Dewi Morgan Jul 26 '15 at 8:18
• because it's an easy number to deal with. – punkerplunk Jul 26 '15 at 13:28

First off, the scope necessary to make out an individual at that distance would not fit on a standard rifle, so we are talking about a highly specialized weapon indeed.

You would need to account for:

• The speed of the ISS and it's planar motion relative to the Earth's surface.
• The rotation of the Earth itself.
• Multiple layers of atmospheric disturbance and wind speeds.
• Your angle relative to the target.
• The speed of your projectile.
• The exact altitude of the target's head.
• The target's unpredictable motion. Best to hit him while he is sitting still.
• The speed of light.

Luckily for you, the majority of these problems have already been solved! In order to get the incredible pictures of the night sky that we currently do from ground-based telescopes, we create an artificial star via laser, and use 'adaptive-optics', which are just mirrors on gimbals that react instantly to any distortion in the known location of this laser-star.

Telescopes also have systems that move them to keep stellar object within the viewfinder, compensating for the Earth's rotation.

A similar computer guiding system could line the barrel of your gun up with your target's future location with wind speeds, pressure layers and drift caused by the Earth's rotation as well as the ISS's movement all factored in. It will additionally have to take into account the difference between the laser's speed to the target and the bullet's speed.

I would use a ceramic bullet in the standard shape of a sniper bullet, made via the same process as the protective tiles on the Space Shuttles to deal with the intense heat without deforming. It is essential to be as close to vertical as possible when you fire the shot, because the Earth's atmosphere will deflect your shot otherwise. You could potentially even bounce off the atmosphere at the right angle.

Lastly and somewhat unfairly, you want to fire at a very high velocity to reduce the risk that your target avoids the shot, but too fast, and your bullet will burn up before it reaches the ground. This happens at around 74 km/second.

As to what it will do to our victim, history is of surprisingly little help. Only one person has ever been confirmed to be hit by a meteorite (one Ann Hodges in 1954), which is essentially what you are intending to do, and she was only bruised. I suspect that your bullet will travel handily through your victim to shatter on the pavement below him leaving a baseball sized crater.

• Another interesting bit of math to consider is, "How much does it perturb the orbit of the ISS to fire a projectile at 73 km/sec from it at a right angle?" – IchabodE Jul 24 '15 at 21:51
• But firing it at a right angle (straight down, IOW) probably isn't going to allow your bullet to even hit the Earth. It'll just stay in a slightly different orbit. What you'd have to do is fire it in the opposite direction to the ISS's motion. If the subsequent velocity (ISS - gun's muzzle velocity) is low enough, the resulting bullet orbit will intersect the atmosphere. It'll re-enter, and the ISS's orbit will be very slightly higher. – jamesqf Jul 24 '15 at 23:48
• Wikipedia says the mass of the ISS is "Approximately 450,000 kg (990,000 lb)", whereas a bullet is obviously less than a pound. That's 1M to 1, so I would say it is not going to perturb the orbit at all, due to it's enormous momentum. It uses about 4,000 lbs of fuel a year to maintain it's orbit from the tiny atmospheric drag it experiences. It's dang hard to move. – Dan Jul 24 '15 at 23:50
• It's not a matter of mass to mass... I'm not suggesting throwing the bullet at the ISS, I'm talking about the amount of energy it takes to ramp a bullet up to 73 km/second. That requires an equal and opposite reaction. With the average sniper bullet weighing 46.7 grams, that takes ~125 million joules of energy. That kind of force will accelerate 450,000 kg to 23.5 m/s. I guess it's not much... – IchabodE Jul 25 '15 at 4:31
• @MBurke: you should be considering momentum (which is equal and opposite in order to be conserved), not energy (a scalar quantity, not a vector). So comparing a 450tonne station to a 45g bullet, the ratio of mass is 10 million to one, which means the ratio of velocity is 10 million to one. 7.3mm/s perturbation. Since the ISS has to actively maintain orbit anyway due to drag, I believe the kickback from the gun is negligible unless you're rapid-firing these things. Bigger bullets might be another story. – Steve Jessop Jul 26 '15 at 10:53

Even if we ignored all atmospheric conditions, we had no friction at all, the space station wasn't moving, and the aim of the sniper was perfectly on target, etc., you still couldn't pull it off.

The most accurate sniper rifle ever made, the Gepárd M1 anti-material rifle, has an accuracy of 0.7 minutes of arc. The ISS when closest to the Earth, orbits at an altitude of 409 km. This results in a target with a diameter of more than 80 meters, resulting just from the properties of the gun itself, without the bullet slowing down due to the atmosphere, etc.

And this is the most accurate you can get with a rifle. The M1 version of the rifle I used as the example, trades absolutely everything for accuracy: it has almost no moving parts, is extremely heavy, holds a single bullet, and cannot be reloaded without taking it apart. A later, more practical version resembles the parameters of the famous American M82.

Although the most important reason why it's impossible is described by other answers, especially by dsollen, I included this reasoning to point out, that even if all the issues raised by the other answers were non-existent, you still had a 80 meter target area as a best case.

Perhaps he has a sci-fi-sniper-rifle. As the bullet leaves the barrel of the gun, it uses some faster-than-light/hyperspace type technology to instantaneously skip ahead, emerging just a few meters from the target. In this way, wind/gravity effects and moving targets, are no loner an issue. You only need to establish line of site with your extremely telescopic scope. Presumably the bi-pod helps stabilise, while adjusting for the earth's relative motion.

• Actually, if the bullet travelled at exactly the speed of light it could work. There would be no issues with wind or travel time. However, the bullet would need some kind of re-entry shield to prevent it from disintegrating before it reached the surface (what-if.xkcd.com/1) – Michael Jul 28 '15 at 1:04
• @Michael what you need is anti-fusion – Mathmagician Jul 15 '17 at 4:49

Using any kind of gun is redundant. All you need is something that is streamlined and has a suitable terminal velocity.

Get one of these...

...

... and, using your super aiming powers, just drop it. By the time it reaches the ground it will be going so fast that it will go straight through your target and several metres into the ground as well.

If your aim isn't so super then your best chance is to adjust the position of the target. Capture him and get someone to drive him to the estimated point of arrival of the 'bullet' as it falls. It can be tracked by a miniature embedded transmitter.

• Drop it? That's not going to work, it'll just continue to orbit right next to you. There is also no way you're going to drive that thing meters into the ground, even fired from a gun point blank. – Samuel Jul 24 '15 at 19:54
• Haha! - I forgot the bit about being in orbit! You must admit however that if someone threw that thing at you from across the living room it would hurt a heck of a lot. If it was falling at terminal velocity and landed straight on your head I don't think you'd survive. – chasly from UK Jul 24 '15 at 21:32
• Oh, I don't doubt that thing would reach a terminal velocity capable of killing a human with ease, just not driving through their body and into the ground. – Samuel Jul 24 '15 at 21:35
• @Samuel Maybe he has a magical velocity stoppifier. – PyRulez Jul 24 '15 at 21:54
• @Samuel, the video linked from Green's answer (worldbuilding.stackexchange.com/a/21230/7957) has a lovely illustration of what happens when you "just drop it," and what you have to do instead. ^_^ – Vectornaut Jul 25 '15 at 21:32

Use a really long bullet.

So long that it's mass keeps it from being blown off course. Perhaps the tail of the bullet could be engineered to predictably burn up shortly before hitting the target.

If we can have a specialized gun, I'm going to get one with a lot of barrels. It'll shoot a massive array, which will compensate for a lot of issues.

Or maybe we can just go with a specialized cluster projectile that disperses into an array before some predetermined point of impact.

These options would cause quite a bit of collateral damage.

I'm sure the projectile could be rigged with some limited guidance options to change course before impact, compensating for atmospheric changes, environmental variables, moving target, etc.