# Single-use nuke-propelled mass driver spaceship

So this is in the context where space battles are fought by ships that are little more than giant railguns (or other electromagnetic mass drivers), and therefore are fought at extreme ranges with very fast projectiles.

I have this rather fanciful idea of a single-use spaceship that is essentially a giant tube housing a projectile and a nuke, plus some aiming and targeting systems. While the nuke would cause irreparable damage to the rest of the ship, the idea is that there would be enough time for the detonation to accelerate the projectile (or what remained of it) out of the barrel. The back end of the projectile would presumably have to be made from some ablative material.

Ignoring things like expense, practicality, targeting etc, I would like to know two things:

1. Is this even a feasible design?
2. How would one estimate the velocity of such a projectile, given the projectile mass $$M_p$$, the ship mass $$M_s$$ and the yield of the nuke used (which let's say for simplicity is of the range of current nuclear weapons, from 10 kt 100 t to 25 Mt)?

I am aware of Operation Plumbbob, where they detonated a 300 t nuke underground with a steel lid on the shaft weighing 900 kg. The lid was estimated at moving above 66,000 m/s, which is six times earth escape velocity.

EDIT: comments have pointed out that some sort of transference of thermal energy to gaseous kinetic energy is needed. Below is my revised picture. Where A stands for ablator (not quite the right word I know, but I can't think of the right one off the top of my head), a mass of material that readily converts to a gas with high kinetic energy (I'm thinking of a slug of ice or solid ammonia, due to the hydrogen content) which then pushes on the projectile. I acknowledge that a megaton nuke is likely to disintegrate the entirety of the ship before any meaningful acceleration can be imparted on the projectile, but suspect that smaller ordnance will be better suited for creating appreciable acceleration on the projectile.

• What's your definition of "extreme ranges?" Most people's concept of "extreme ranges" dramatically underestimates the actual distances between objects in space. – Cort Ammon Jan 17 '19 at 22:51
• Like, tens, maybe even hundreds of feet. But seriously, I remember a quote saying that battles are fought at the furthest range possible. On that basis, the upper limit would be one of imaging the opposing spacecraft and ascertaining their position, velocity and orientation to obtain a firing solution. A softer limit would be the distance at which a ship can see an incoming projectile and reliably get out of the way. Both of those, I imagine, will be pretty extreme. – Ingolifs Jan 17 '19 at 23:31
• Those would qualify as "extreme near ranges." If you're that close, it's clear you're trying to dock with a friendly ship. Real life space distances are much further. For perspective, the ISS maintains a 4km thick 25km wide disc of space -- if any one object is predicted to enter that region, we move the station. Of course, the objects are also traveling many kilometers every second! – Cort Ammon Jan 18 '19 at 0:12
• the first sentence was ironic... – Ingolifs Jan 18 '19 at 0:15
• Did you intend to be focusing on the extreme near ranges? At that distance, I'd recommend gently lobbing the nuke over to them and getting the heck outta there! – Cort Ammon Jan 18 '19 at 0:17

There's a simple problem with this: shock and vibe.

Your projectile has to travel a very long distance before hitting. Your specification of "around the limit of detection by IR" is sufficient to be quite certain you will never hit your enemy with a dumb slug. You simply can't aim the gun that accurately. This is doubly true for the fact that you're about to set off a nuclear explosion (probably well more than doubly true).

This means you're certainly using a smart weapon, with onboard propellants, sensors, and computers, much like we see in all of our own ballistic defense programs on planet Earth.

So here's the real issue. Launching things fast (space fast) is hard. You have to accelerate rather quickly. Shock and vibe issues during launch are astonishingly difficult to deal with. The harder you accelerate, the nastier the requirements. Navy rail guns face serious challenges in hardening smart rounds so that it survives being fired (tests are typically done with dumb Tungsten slugs today, but it is recognized that they will need to be guided eventually to strike distant targets). You have invented a new level of "difficult launch conditions." Railguns can maintain a maximized shock and vibe for the length of the rail far better than a simple nuclear gun can.

Launch Conditions: Must survive being at ground zero of a 25MT thermonuclear bomb detonation, in a confined space designed to maximize the impact of the thermonuclear bomb on the projectile.

Ouch!

What you really want to do is stick to railguns. If you design a projectile that can survive the forces of a nuclear bomb, you immediately put that same projectile into a railgun, and get more performance out of it (railguns can accelerate linearly along the entire length of the rails, while this gun peters out as the pressure wave weakens)

• Good answer. There is a 'combat regime' I guess you'd call it, where you can see your enemy but do not have the ability to hit them, because the time-of-flight of your projectiles is longer than the time needed to get out of the way. If you make your projectile faster, you have a better chance of hitting, providing you do not sacrifice accuracy. Your point about projectiles needing guidance in any case is a good one. That will always limit the acceleration of the projectile to what the guidance computer system can survive. In which case, using the nukes for Orion drive would be better. – Ingolifs Jan 18 '19 at 4:14
• If I remember my rules for intercepts correctly, faster doesn't actually make it much better. Reaction times are important, and faster will make it better with respect to reactions, but once you actually get to the demanding task of moving your ship out of the way of the round, faster no lnoger helps. They have less time to move, but you have less time to respond to their movement, so it cancels out. – Cort Ammon Jan 18 '19 at 4:18

Feasibility:

Your design is almost identical to a regular gun, except, they use a small chemical reaction (a small explosion) to propel the bullet, instead of a nuclear one. So, scaling this up, there's hope for the nuclear case. Also, according to this, roughly 40~50% energy is blast energy (much better than I thought), while the rest goes to thermal energy and other kinds of energy. In a high pressure environment, with a low yield bomb, the blast energy can be further increased to 50%~70%.

Because we can engineer the environment, we can start with the projectile, staying in a tube filled with a gas/atmosphere maintained at a high pressure.

If the initial explosion is made extremely symmetric, one could even use reflections of the shock waves when hitting the wall to further increase directivity of the blast energy, and decrease damage to the ship itself.

In addition to the blast itself, the gas behind the bullet will expand, creating big pressures on it, further pushing the projectile (and the ship).

Design Calculations:

Now, this is a closed system (at least initially). We'll use the first law of thermodynamics: $$U = Q + W$$, the internal energy is work plus heat. We're interested in work. The initial internal energy of the system is negligible compared to the final one (bomb exploded), so we can consider $$U$$ to be the energy output of the bomb itself. We'll call the ratio between work and total energy as $$\eta$$. That is, $$W = \eta U$$. This work, will be effectively transformed into kinetic energy.

The remaining energy will go to heat: $$Q = (1-\eta)U = mc_v\Delta T = \rho V c_v\delta T \quad\implies\quad \Delta T = \frac{(1 - \eta)U}{\rho V}$$

Where $$c_v$$ is the specific heat of our gas at constant volume, and $$V$$ is the volume in between the projectile and the bomb. From there, we can calculate the pressure: $$\frac{P_1}{P_0} = \frac{T_1}{T_0} = 1 + \frac{\Delta T}{T_0}$$

While the initial pressure can be estimated using ideal gas law, namely $$P_0 = \rho\frac{k_B}{m_g} T_0$$, where $$m_g$$ is the mass of each molecule/particle of gas.

So, now we find final temperature and final pressure: $$\Delta T = \frac{(1 - \eta)UT_0 k_B}{P_0 m_g V},\quad\quad P_1 = 1 + \frac{(1 - \eta)U k_B}{P_0 m_g V}$$

The force applied on the projectile by the pressure is simply $$F = m_p a = A P_1$$, where $$A$$ is the cross section area of the projectile, $$m_p$$ is the mass, $$a$$ is the acceleration. Now we have a rough crude (possibly imprecise) of the acceleration, and thus, how quickly the projectile will leave the ship (as compared with, say, how quickly will be the explosion itself).

I am actually very lazy to plug some numbers and actually come up with a design that might work. I tried $$V_0=10\times 10\times 10 m^3, P_0 = 10atm, U = 25MT, \eta = 10\%$$ with a nitrogen atmosphere $$m_g = 2.341\cdot 10^{-26} Kg$$. This gives me $$T_1\approx 1.643\cdot 10^{13}K$$ and $$P_1 = 54.779 GPa$$. This gives a force of ~5470 giganewtons and acceleration of 5.47 billion meters per second squared (quite fast!) for a projectile with a ton of mass and 10x10 meter squared cross section.

With the velocity estimates below, one can calculate exactly how much time the projectile will stay in the ship. A 42.606 gigapascal pressure is already close the young's modulus of several materials, and already a bit beyond the shear modulus of, say, several titanium alloys: the ship won't survive. You might want to decrease the bomb's yield and play around with these equations. You might also want to check if the projectile itself will survive such huge accelerations (probably won't).

Conclusion: It might be feasible, but definitely not for a 25 megaton explosion. Maybe one could consider a one ton explosion and see what happens.

Velocity Estimate:

If we ignore the usual common sense explanation that one shouldn't use nuclear explosions to propel a projectile.... and focus on what speed we could get.. well.. let's start, I guess.

First, we have conservation of momentum: $$\mathbf p_s + \mathbf p_p = 0$$, assuming we are in the reference frame that the ship was initially at rest. Second, we have conservation of energy: $$U = K_S + K_P + Q$$

Where $$U$$ is the total energy of the bomb, $$K_P$$ is the kinetic energy of the projectile, and $$K_S$$ the kinetic energy of the ship (or the total sum of kinetic energy of all the wreckage left by the ship). And, lastly, $$Q$$ is the thermal energy.

Now, it is important to notice that these two equations are valid for any and all given times. Because of that, we'll take a time where the projectile is launched, but the nuclear exposion didn't have enough time to completely destroy the ship itself (by too much). Thus, the ship is still in one piece, and thus, we are talking about a two-body system. That will greatly simplify everything.

Newtonian Calculations: Let $$v_p, v_s$$ the velocity of projectile and ship. By momentum conservation, we know: $$m_s v_s = -m_p v_p \quad\implies\quad v_s = -\frac{m_p}{m_s}v_p$$

Thus, the energy: $$K = \frac{1}{2}m_s v_s^2 + \frac{1}{2}m_p v_p^2 = \frac{1}{2}m_s\left(-\frac{m_p}{m_s}v_p\right)^2 + \frac{1}{2}m_p^2$$

Thus: $$K = \frac{1}{2}\left(m_s + \frac{m_p^2}{m_s}\right)v_p^2 = \frac{1}{2}m_s\left[1 + \left(\frac{m_p}{m_s}v_p\right)^2\right]$$

There, the projectile velocity is simply: $$v_p = \frac{m_s}{m_p}\sqrt{\frac{2K}{m_s} - 1}$$

Where $$U = Q + K$$. Now it would be a matter of estimating the ratio between kinetic and total energy available, or, the efficiency $$\eta = K/U$$ of your transformation between bomb's energy to kinetic energy.

Plugging numbers: If $$U = 25MT = 1.046\cdot 10^{17}J$$, setting $$\eta = 0.1\%$$, this means $$K = 1.046\cdot 10^{14}J$$. If $$m_s = 10^9 Kg$$ and $$m_p = 10^3 Kg$$, then: $$v_p = 1.5257c$$. Faster than the speed of light!

Relativistic Calculations: Lately, our result gave something faster than $$c$$, so, we conclude newton's mechanics do not apply. We need relativistic calculations to figure out the speed.

The relativistic total energy of the system is, then: $$E = \sqrt{(m_p c^2)^2 + (p_p c)^2} + \sqrt{(m_s c^2)^2 + (p_s c)^2}$$

Where, by conservation of momentum, $$p_p + p_s = 0$$, which simplifies everything. Define $$p = p_p = -p_s$$ the relativistic momentum we wish to find. Thus, we can solve for $$p$$: $$p^2 = \frac{1}{4}\left[\frac{E}{c} + \frac{c^3}{E}\left(m_p^2 - m_s^2\right)\right]^2 - m_p^2 c^2$$

Now, using the relativistic momentum, $$p$$ is related with the velocity: $$p = \frac{m_p v_p}{\sqrt{1 - \frac{v_p^2}{c^2}}}$$

And therefore, one can easily solve for $$v_p$$: $$v_p^2 = \frac{p^2}{m_p^2 + \frac{p^2}{c^2}} = \frac{1}{\left(\frac{m_p}{p}\right)^2 + \frac{1}{c^2}}$$

Therefore, finally, the (relativistic corrected) velocity of the projectile is: $$v_p^2 = \frac1}\frac{m_p^2}{\frac{1}{4}\left[\frac{E}{c} + \frac{c^3}{E}\left(m_p^2 - m_s^2\right)\right]^2 - m_p^2 c^2} + \frac{1}{c^2}}$$

Where $$E = K + (m_s + m_p)c^2$$ and $$K = \eta U$$. Plugging the same values as before gives us the projectile speed: $$v_p = 457,386 m/s = 0.0015257c$$. This is the speed of the projectile under the given situation (ship one million tons, projectile one tons, 25 megatons explosion, 0.1% efficiency). You can use this formula to play around and plug other values.

Hopefully I made no mistakes. If I did, someone please point me out.

Be aware that all of this ignores the obvious problems of having a nuclear explosion to propel a bullet: if you have those problems solved, then this calculations gives you a rough speed estimate of the projectile. And keep in mind I completely guessed $$\eta$$ (such value will depend on the actual mechanism you have to transform the bomb's energy to kinetic energy).

• Also, these calculations are not relativistic. I might be perhaps persuaded to make a relativistic version of them if everyone wants it... – Physicist137 Jan 18 '19 at 3:21
• That's probably not necessary, Between you and JBH, a very high upper bound for the velocity of the projectile has been established. What I'm starting to suspect is even if you can fling a projectile (that isn't now a glob of molten metal or plasma spray) using a nuke, any asymmetry in the detonation will probably result in the gun being kicked off-target, resulting in the projectile missing completely. – Ingolifs Jan 18 '19 at 3:35
• @Ingolifs I was doing the relativistic calculations in a paper (just because I am too curious). I planned to only post the result. But these are huuuge. After almost half page were completely filled, there were a lot more calculations to do, so, I gave up. :). Also, notice that, a ordinary weapon is much like your design, except, they use a chemical explosion instead of a nuclear one -- and bullet weapons can be designed to be very accurate. Thus, maybe there's a way to design the nuclear one, where it could be accurate. – Physicist137 Jan 18 '19 at 3:48
• @Ingolifs I found an easy way to make the relativistic calculations. Besides, there were a mistake in my initial calculations: The speed is 1.5c, not 0.5c. The relativistic one gives a value well below speed of light, at least. – Physicist137 Jan 18 '19 at 6:32
• @Physicist137 can you include any solution as to how this will be possible, in your statement Be aware that all of this ignores the obvious problems of having a nuclear explosion to propel a bullet We obviously know that such a container might not exist, Since his question is more of the feasibility of his design and not the speed of the projectile, if we take that into account, your answer does not answer the question. – Mr.J Jan 18 '19 at 6:45

Tim's right, but let's have some fun explaining why.

Assumptions

• When you said "railgun" what you meant was "slug thrower." A railgun is something specific, and you're not describing it.

• The "gun" can't be damaged by the nuke. This is an experiment, after all.

• Neither can the shell.

• The practicality of the one-use-incredibly-expensive-hard-to-aim gun is ignored.

• The "gun" has a mass at least 1,000X of the shell. Thus, 99.9% of the final velocity belongs to the shell, while only 0.1% of the final velocity belongs to the gun (which enters a decaying orbit, devastating the southern continent of your world, but that's a story for another day). Anyway, what this does is let me reasonably assume all the energy is used to push the shell.

• Here's the whopper: we're going to assume that when they measure the explosive capability of a nuke in terms of tons of TNT, they really mean it (and the angels weep over the magnitude of this lie... but we'll get to that).

The equation of choice is F=mA. Let's be outrageous.

• The charge is equivalent to the Tsar Bomba: 50 megatons or 210 peta-joules. So, F = 210x1015.
• The shell is a useful 10,000 Kg. So, m = 1x104.
• A (acceleration) = 210x1011 m/s2.

It has no resistance (space...) and there's no more force acting upon it. So our final velocity is darn close to twice the speed of light.

Which obviously can't happen

And that's why Tim's answer is correct: puff of light and heat.

OK, what's the problem?

A whomping large amount of a nuke's energy has nothing at all to do with kinetic energy. It's radiation. Oversimplifying something awful, you only have the mass of the charge to work with in terms of expanding gas (kinetic force) and a lot (if not all) of it just got converted to pure energy.

Most of the concussive force of the nuclear explosions we're familiar with are (a) the conversion of surrounding mass (air, soil, houses, people) into expanding plasma and (b) the shock created when the vacuum caused by the explosion collapses.

Which is a fancy way of saying you don't actually have that much to work with. Since you only have whatever convert-to-plasma materials are in the nuke itself (gun and shell can't be damaged by the nuke!), your gun will glow white hot1 while the equally white hot shell kinda blurps out of the muzzle and lazily meanders away.

If I got this wrong, please correct me!

Edit:

Physicist137 pointed out the flaw in my math, and I'm grateful! I had fun, but go upvote his math! I'm still happy with my conclusions, but his ~0.5c answer is a whole lot more realistic than my 2c analysis!

1And it'll probably stay white hot for decades. Space is a terrible place to dump heat.

• You got me with Tsar Bomba. – Mr.J Jan 18 '19 at 1:19
• "So our final velocity is darn close to twice the speed of light. Which obviously can't happen" Ok so this part is fallacious for a couple of reasons, 1) because it ignores relativity, and 2) because it implies a derived speed greater than c (while ignoring relativity) counts as reducto ad absurdium. The ship is not indestructible, neither is the projectile. My question isn't about how badly the craft will blow up (it's single use, so it's already written off), but whether you can get significant velocity on the projectile that comes out of it. – Ingolifs Jan 18 '19 at 1:42
• @Ingolifs You're not having nearly enough fun. You're missing mass between the slug and the nuke that can be converted to plasma. However, what you haven't figured out is that almost anything is more useful in space than a nuke-powered slug thrower. Especially a one-use thrower. – JBH Jan 18 '19 at 1:54
• Hmm. Maybe I need to update the picture I provided with a mass of something ablative between nuke and projectile. – Ingolifs Jan 18 '19 at 2:02
• I am afraid this is not correct. You are equaling a force (from newton's second law) with an energy (the bomb), and you cannot do that. – Physicist137 Jan 18 '19 at 3:16

Your single use ship will disappear in a puff of light and heat.

The problem with what you're describing is that a nuclear bomb is exactly the wrong type of energy source for a rail gun as it generates thermal heat and radiation, not kinetic energy, in space.

We've all read about the blast waves off nuclear explosions and how powerful they are; there is even some research into this that shows that blast (kinetic) energy comprises roughly half the energy released in a nuclear explosion. The trouble is, that blast energy is essentially the atmosphere super heating from the thermal energy released, and therefore rapidly increasing in pressure in an unconfined space.

Nuclear bombs are not designed to release kinetic energy; they're designed to release thermal energy. This can still work because of a technique known as Ablative propulsion, which amounts to superheating a surface of an object to a point where the surface becomes a plasma that pushes the object rapidly in the opposite direction to that in which the heated surface is facing.

This kind of thinking about propulsion seems to be behind Nuclear Pulse Propulsion theoretical engine systems like that originally planned for Project Orion. My reading on this to date seems to indicate that the plan was to detonate nuclear bombs outside the ship and have the blast hit a pusher plate on the back of the ship, pushing it in that direction. What is unclear in all those mainstream articles is how that blast wave is generated but the best I can come up with is that the mass of the bomb itself would become a superheated plasma that would push against the rear plate of the ship. The closest I could come up with in terms of reference to how nuclear energy can be converted to kinetic energy in space was a reference to Electrical v Direct Thrust on fusion drives, but even it talks about the plasma having to be contained to sustain the reaction which to my mind seems contradictory.

If this isn't the thinking behind nuclear pulse theory and you know what is, leave it in comments and I'll integrate it into my answer with full credits. But back to the matter at hand...

In space, there's no atmosphere to heat up, so no localised pressure differential, therefore no blast wave. What WILL happen is that the heat energy will essentially vapourise your ship and the projectile through heating them all equally.

To be fair, this will result in the aforementioned ablative propulsion for your rail projectile; as the weak point in the containment, it's going to break off (or out) first but without running some serious math I have no idea just how quickly it will escape the barrel, how much of it will be left to hit the target, or even if it would escape the barrel at all before the thermal energy expands it (and contracts the barrel) before it escapes.

Before anyone says it, yes, the barrel will constrict. As the metal expands, the inner metal of the barrel also expands, meaning that the hole gets smaller through swelling.

Guns like cannon, pistols, rifles and rail guns work through the kinetic energy of expanding gases. The idea is that gunpowder, when burnt, releases a massive amount of hot gas into a confined area, forcing out the weakest point of the containment chamber, which just happens to be the bullet. This projects it suddenly in the direction of your choosing. Yes, burnt gunpowder is hot, but it's not the heat that shoots out the bullet; it's the expanding gases.

In space, your setup has no atmosphere around it to expand, and even if it did, the kinetic energy transferred to it only represents about half the energy that a nuclear device can produce, which makes it an inefficient energy source for what you're trying to do, and would require such a sturdy, heavy ship to withstand the vapourisation long enough to transfer the bulk of the energy to the projectile so as to make it uneconomic as an investment.

For my money, you'd be far better off investing in a nuclear powerplant that feeds electrical capacitors that are capable of releasing a sudden energy burst of electricity to propel your bullet. Far more efficient process of converting thermal to kinetic energy (via temporarily being electricity), reusable and less dangerous to the rest of your fleet by not creating debris fields with every shot.

• While I too question its practicality, is this not roughly analogous to the nuclear pulse rocket concept? – Cort Ammon Jan 17 '19 at 23:02
• @CortAmmon you're right; I'd forgotten about that until you mentioned it. I just did a quick check on it and most of the material I can find on this talks about specific impulses, but doesn't actually go into detail about how the blast wave in space actually works. It seems to be related to the surrounding mass being converted to plasma. I'll do some more research and edit accordingly within the next couple of hours. – Tim B II Jan 17 '19 at 23:18
• I'm not convinced by this answer. Operation Plumbbob has shown that such a thing is at least possible, though it may require a separation of gas between the nuclear explosive and the projectile, and probably a much weaker nuke than the likes of Fat Man (let alone Tsar Bomba). – Ingolifs Jan 18 '19 at 1:50
• @Ingolifs the trouble with comparing Operation Plumbbob to this scenario is that it was conducted on Earth, not space. Earth has an atmosphere that can be heated as an expanding gas and plasma, space doesn't have a medium in which to propagate the blast wave. – Tim B II Jan 18 '19 at 3:18
• I've edited my original question to include a mass of material between the projectile and the nuclear explosive, that vaporises into gas and plasma. – Ingolifs Jan 18 '19 at 3:24

While what you're describing seems to be a variation of the "Plum Bob" test, the issue is coupling the energy of the device to the projectile. The Plum Bob had a shockwave travelling the roughly 500' length of the shaft, accelerating the cap at 5X escape velocity. A much higher performance might have been achieved if the shaft had been filled with water. The X-ray radiation release from the nuclear device would have converted the water into a plasma in a tiny fraction of a second and the plasma would then have propelled the cap at a much higher acceleration. One shot devices could be made by casting a cheap 200m "barrel" out of cast regolith, filling it with water and then adding the cap and the "physics package" at opposite ends of the barrel. A few rocket thrusters to aim the device and you have a nuclear "carronade".

It will also be hopelessly unwieldy, and not very likely to be successful, unless the cap is designed to fragment and pass through the strike zone like pellets from a giant shotgun. For an impoverished space navy, or something cheap you can erect on the ground, the surface of a moon or small asteroid, that may be just the thing. However, looking at the Orion Nuclear Pulse Drive, we can make a far more compact and useful device.

What you really want is a "Third Generation" nuclear device. This utilizes various design methodologies to focus or direct the energy of the nuclear detonation to power weapons effects. While these are considered warheads, the effects can be projected a great distance, so they are effectively "guns" as well. Tough SF has most of the detailed information about the Casaba Howitzer, Nuclear Shaped Charge and EFP, and Atomic Rockets has a further section which also talks about using the nuclear device to accelerate pellets at hypervelocity (up to 100 km/sec). This is Project PROMETHEUS, and the information is buried deep in the section. Orion Nuclear Pulse Unit. The principles behind this are the basis of third generation nuclear weapons

From the various descriptions in the web sites, projecting solid pellets shotgun style at a target can be done at speeds up to 100km/sec, where even 2 gram pellets have the kinetic energy equivalent of sticks of dynamite. EFP and shaped charge weapons can project streams of liquid metal or solid slugs at speeds measurable at small fractions of c, and a Casaba Howitzer projects a stream of energetic plasma with enough speed and energy to simulate a high energy laser weapon without the heavy power generation systems, expensive optical train and large cooling systems needed for lasers.

If your basic space warship is a "Kineticstar" kicking these warheads out at 15+ km/sec (twice Earth orbital velocity) along with an array of decoys and penetration aids, once they get within 1000km of the target, then things will become very exciting...