What would a Kinetic Orbital Bombardment Strike looks like?

[RainDreamer]

Well, the question here is not concerning about the practicability of it, rather it is simply about how such a thing would look. We have seen footages of nuclear bomb explosions, but these kinds of things are simply not something we have ever seen before or going to see any time soon.

[Scotius]

And how such house-sized chunk of tungsten would be delivered to space? Every single projectile would require a separate launch of a rather hefty rocket. It would be easier, faster and cheaper to deorbit the whole station and smash it into the target.

[kStrout]

It is not going to be more powerful at all. The weapon platform simply take advantage of being an orbital platform that can attack from space with very little warning and difficult to counter, and slip through the space treaty that bans weapon of mass destruction in space (it is a conventional weapon by definition). But those are kind of negligible advantage, comparing to the cost of putting it in space, hence it is not going to appear in reality in anytime soon. And also the reason why I wonder what such an impact would look like as there is not going to be any real example.

I imagine it would look like a giant bullet hole, very deep but not much damage on either side, also that projectile in the video was moving nowhere near hypersonic speeds.

[tater]

1/2mv^2. It's all about the v. LEO to earth is 7.8 km/s, minus air resistance at impact (which is substantial). For huge amounts of damage, vs the small, long rod penetrators usually talked about (a third of a m in diameter, and a few meters long---basically a tungsten telephone pole) we're talking 10 tons TNT energy levels. You need to have ~100 metric tons at lunar return velocity to get to a 1 kt level.

[Darnok]

And how such house-sized chunk of tungsten would be delivered to space? Every single projectile would require a separate launch of a rather hefty rocket. It would be easier, faster and cheaper to deorbit the whole station and smash it into the target.

We can always use asteroids or mine metals from asteroid and built projectile in space

[Cirocco]

while I can't comment much on what it would look like at ground zero, when standing a ways away I imagine it looks something like this:

https://en.wikipedia.org/wiki/LGM-118_Peacekeeper#/media/File:Peacekeeper-missile-testing.jpg

Just with fireballs at the impact sites.

That's a long exposure shot of the re-entry of the 8 dummy warheads of a LGM-118 peacekeeper intercontinental ballistic missile. Eerily pretty. This thing actually goes up into space on a suborbital trajectory, then releases the 8 warheads and sends 'em down to go boom.

[Vaebn]

Utilizing the power of Logic I would assume the rod would release about the same amount of energy it took it to get it to orbit.

So about a pad explosion of a rocket's worth of fuel. This is a lot more boom than a rocket fired out of an airplane. This is a lot less than a nuke.

[TythosEternal]

Keeping in mind that this particular rendering is from a GI Joe film, I give it an A for effort (one of the few times Hollywood has tried to imagine just what a Rod From God would look like), an A for visuals (admit it, it looks pretty impressive), but a D for realism (I'm avoiding the 'F' solely because of the creativity required to imagine something that no one has ever seen before, even if it's beyond a worst-case scenario). All in all, not bad for a summer popcorn flick that you're not supposed to take seriously anyways. As far as doomsday weapons go, this goes right up there with the Bond villians. (Orbiting diamond-driven lasers and/or solar-powered area-effect death-rays, anyone?)

[prophet_01]

I would expect it to be somewhat like the 'grand slam' bomb regarding it's effect. No ground waves and nuke like explosions, but a pretty big shockwave that results in an earthquake like destruction.

I have no idea how much thermal energy might be created on impact and if it's enough to create a somwhat secondary effect. The high velocity impactors of kinetic anti tank weapons usually burn up while penetrating the armor which results in a significant amount of heat. I would expect that kind of thing to happen here aswell, but on a larger scale. That probably depends on the material of the impactor. As far as I know the effect is a lot more destructive with depleted uranium due to it's high density/kinetic energy

[RainDreamer]

Hmm, like a rod of light that flash down from the sky and boom, your city got a new landmark, along with large scale secondary infrastructure damage with ruptured gas pipe and stuff in an area, and whatever under it is obliterated. All happen in a flash.

Inefficiencies aside, it sure is going to be very awesome (original sense of the word).

[jwenting]

No, the video is just fancy but totally unrealistic. It assumes most of the kinetic energy would be converted to mechanical energy which is not correct. Most is converted into heat.

And that ALL energy would be released on hitting the ground, rather than most of it dissipated in the atmosphere as the projectile travels through it.

[Kilifisch]

You could shatter the rod short before impact thus crating a larger impact area wich would Dump the Energy closer to the surface 

 

 

Also there would be the way that you exceed the shatter Range and go into the vaporization Range were the rod would turn into Plasma on impact and create somewhat of an explosive impact

Edited October 25, 2016 by Kilifisch

[DrLicor]

When dropping something into the atmosphere, it get's slown down due drag, let's just ignore the heat etc. Solutions:

• make your rocket very pointy.
  Spoiler

  sorry couldn'r resist to add something from the dictator.

  

   

• Let your rocket do a burn before impact.
• Make your rocket so heavy, the drag doesn't do anything, according to the first law of newton, Force = Mass x Velocity. So more mass, moaarr Force/Impact.

 

Pointy rocket: It will still have some drag and I think it will have the same speed as a normal missile.

Burn before impact: To compensate the speedlosses due drag, let your rocket burn. In case of a 1 stage missile, the missile probably has to burn it's whole reentry. Which means it has to burn for a long time. depending on your angle of reentry. 

Make your rocket heavy: Not the cheapest way of bombing, transport etc. But yeah who cares, you have missile from space. 

 

Now since option 2 is the usefull for the biggest impact, and let's include heat now. your rocket will go faster then orbital velocity when entering the atmosphere. It will produce a massive amount of heat. So you need one hell of a heatprotector. Ofcourse you could let the rocket fire later, but during reentry you can't send a signal due the plasma, signals can be hacked too, so not done. When programming the rocket you need program it very precise. 

 

just to add something, the kinetic energy the rocket has, goes for 90% to heat due aerobreaking. 

Edited October 25, 2016 by DrLicor

[KerikBalm]

Consider the size of the explosion of a rocket that would put the rod into orbit (just the rod, not the entire weapon platform). Conservation of energy tellls you that the explosion cannot be any more powerful than that.

Now you have to consider how inefficient a rocket is... most of that energy goes into the hot exhaust gasses. Then a lot of that energy is in the form of the KE of the spent stages (which have lower velocities, but higher dry masses) or again in the propellant KE before being burned.

Then you've got losses to drag/atmospheric ionization etc. The energy release from the rod impacting is only a small fraction of the energy of the rocket on the launch pad.

So..... not nearly that bad... not even a fraction of the power of the Soviet N1 explosion... which wouldn't do that to London - sure it could destroy the parliament building, but so could a bomb in a truck or barrels of gunpowder underneath it on the 5th of Novermber... nothing like what was shown.

[p1t1o]

See Newton's approximation for impact depth. 

https://en.wikipedia.org/wiki/Impact_depth

 

It is an approximation that calculates penetration depth based only on the relative density of target and projectile and the length of the projectile.

The approximation is valid only at high velocities because at these speeds the energy required to break any chemical bond, regardless of material, is insignificant compared to the total kinetic energy involved.

This theory and approximation does in fact pan out quite accurately for a great many situations from a long-rod anti-tank kinetic penetrator impacting armor to meteorites impacting the atmosphere.

 

Note that velocity is not a component, above a certain threshold higher velocity will not result in greater penetration.

 

 

I have also read, though I cannot find a reference, that as total kinetic energy rises further above the chemical binding energy of the target, the more spherical the release of that energy becomes. what this means is basically as kinetic energy rises through an order of magnitude or two higher than the binding energy, the energy release resembles a "penetration" less and an "explosion" more.

This neatly ties into Newton's approximation in that above a certain energy, higher velocity does not increase penetration, but it does still increase the intensity of the energy release at impact.

 

Note that this applies for projectiles penetration into the atmosphere, witness the way that meteors tend to explode in the atmosphere rather than penetrating all the way.

Ergo, increasing (orbital weapon) projectile velocity only increases destruction of the target up to a point, beyond which you are just dumping more energy into the atmosphere, and further beyond which you risk premature destruction of your projectile. There will be an optimum re-entry profile for target destruction, another profile for maximum penetration and neither of which will be a "maximum attainable velocity" profile.

 

NB: Funny that Newton derived an equation almost 400 years ago that is still relevant when considering orbital weaponry!

 

Edited October 25, 2016 by p1t1o

[Darnok]

37 minutes ago, p1t1o said:

Note that velocity is not a component, above a certain threshold higher velocity will not result in greater penetration.

IMO amount of damage you can do with this weapon is not key feature. Ability to target any ground vehicle or ship (for example aircraft carrier) anywhere on planet is what makes it deadly weapon.

And I can't imagine what can you do to protect ship from being damaged from orbital bombardment.

 

[p1t1o]

19 minutes ago, Darnok said:

IMO amount of damage you can do with this weapon is not key feature. Ability to target any ground vehicle or ship (for example aircraft carrier) anywhere on planet is what makes it deadly weapon.

And I can't imagine what can you do to protect ship from being damaged from orbital bombardment.

 

Yes that is probably accurate.

The question of ships has been examined in the context of the new Chinese anti-ship ballistic missiles, whose terminal phase resembles an orbital maritime strike. Without actually trying it out, only vague conclusions can be drawn, naturally, but here are some salient points IIRC:

Yes, ships are vulnerable.

ABM systems already in service on some modern ships are quite well suited for defence against these weapons. A kinetic projectile would likely be more physically robust vs. conventional ABM, but deflection would be possible.

The "kill-chain" (Command structure, sensor network, and communication links to the weapon itself) is unusually long and is vulnerable to jamming/disruption. This is likely to be similar for orbital kinetic weapons too, complicated by the fact that ships are moving targets.

 

The best defence against ballistic or orbital weaponry is political/deterrent in nature, especially since anything streaking in from space is indistinguishable from a nuclear warhead.

Edited October 25, 2016 by p1t1o

[tater]

The crater diameter goes as the cube root of energy, though as I recall. So the penetration depth might be independent of velocity (above what is required to penetrate), but crater size scales with velocity (energy is the KE).

[p1t1o]

Yeah that fits perfectly.

[DrLicor]

14 minutes ago, tater said:

The crater diameter goes as the cube root of energy, though as I recall. So the penetration depth might be independent of velocity (above what is required to penetrate), but crater size scales with velocity (energy is the KE).

Not 100% true: excuse me I had read you comment wrong  

The diameter of the size of an impact crater is related to and can be estimated by the fundamental relationship between the mass and velocity of the impacting body. Mass and velocity can be combined to find the kinetic energy of an impactor which is defined as: KE = 0.5 x m x v2 or, alternatively: K = 1/2 m v2 [where KE or K represents kinetic energy; where m = mass of object/impacting body; and, v = velocity/speed of object/impactor].

 

Edited October 25, 2016 by DrLicor

[p1t1o]

Relevant:

https://en.wikipedia.org/wiki/CBU-107_Passive_Attack_Weapon

 

[tater]

1 hour ago, DrLicor said:

Not 100% true: excuse me I had read you comment wrong  

The diameter of the size of an impact crater is related to and can be estimated by the fundamental relationship between the mass and velocity of the impacting body. Mass and velocity can be combined to find the kinetic energy of an impactor which is defined as: KE = 0.5 x m x v2 or, alternatively: K = 1/2 m v2 [where KE or K represents kinetic energy; where m = mass of object/impacting body; and, v = velocity/speed of object/impactor].

Roughly: d³ =Kmv²

So ~ d = (Kmv²)^1/3

K is just a fudge factor, erm, proportionality constant.  

Roughly people tend to guesstimate using v^2/3 as a result.


v2

[p1t1o]

Possibly relevant:

"Cratering Capabilities of Low-Yield Nuclear Weapons"

http://www.dtic.mil/dtic/tr/fulltext/u2/a108989.pdf

Not kinetic, but it may as well be, it is still a direct energy-to-crater relationship. Tons of detail.

[tater]

21 minutes ago, p1t1o said:

Possibly relevant:

"Cratering Capabilities of Low-Yield Nuclear Weapons"

http://www.dtic.mil/dtic/tr/fulltext/u2/a108989.pdf

Not kinetic, but it may as well be, it is still a direct energy-to-crater relationship. Tons of detail.

Notice size is the cube root of energy as far as I can tell briefly skimming it.

[moronwrocket]

One way to get around the upper limit on impact velocity is to use a dual magnetic railgun to shoot two shots, one at your target, and one directly away so that your launcher doesn't get affected.  Granted, it would require lifting 2x the mass so that 1x could be thrown away, and to deorbit a rod would require a pretty massive railgun with astronomical power generation/storage/delivery requirements, but the kinetic energy delivered is only limited by your Congressional budget appropriations.