this post was submitted on 14 Nov 2023
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[–] [email protected] 43 points 1 year ago (6 children)

From a purely physical point of view, is that realistic?

If all of its energy is kinetic, it means that the energy must result from it's potential energy+any fuel it is propelled with. Ignoring air-friction and terminal velocity for free falling objects, that means that still the energy of a nuclear weapon is required to bring this thing up into space, or stored as fuel for its propulsion.

So unless the projectile is assembled in space, any rocket bringing it into space will contain at least the energy of a nuclear warhead. Gotta be a very nervous launch, knowing that any failure will result in a fire with the energy of a nuke.

[–] Brainsploosh 43 points 1 year ago (1 children)

A lot of the energy comes from orbital speeds.

The Hypervelocity Rod Bundles project proposed 6,1x0,3 m tungsten rods, weighing about 8200 kg, impacting at about 3000 m/s, meaning about 42 GJ of energy per projectile [wikipedia].

The weakest recorded nuke, the Davy Crocket Tactical Nuclear Weapon, is estimated at about twice that (84 GJ), and the largest, Tsar Bomba, at about 3 000 000x the yield (210 PJ).

[–] pennomi 35 points 1 year ago (2 children)

That’s their point, how do you get such a heavy thing to orbital speed without spending all that energy? You can’t unless you build it from materials harvested in space.

[–] Brainsploosh 29 points 1 year ago* (last edited 1 year ago) (4 children)

Oh, I apologise, I suffered some curse of knowledge there, the answer is time.

A blast is a release of energy over a short time, the whole point of building weapons is to store and handle energy in safe amounts over time.

Global electric energy consumption is about 200 PJ a day, approximately the same as the Tsar Bomba, but there's no risk for a huge explosion neither when you incinerate trash or turn off the AC.

Because time.

Although we could explode a nuke and propel things ballistically, it turns out it's a lot easier to use rockets. A rocket, although carrying frightening amounts of fuel and exploding spectacularly when it fires wrong, has several safeguards to not expend all that fuel at once. And also gives the opportunity to correct course along the way.

Now imagine that the same amount of energy has been expended many many many times over the course of the space era, and almost any mass in orbit has serious potential for damage.

For example, the MIR was 130 tons, orbiting at about 7,8 km/s, for a kinetic energy of 4 TJ, and another 235 GJ of potential energy. Totalling about a tenth of Little Boy that levelled Hiroshima.

Edit: Specifying and correcting the global energy consumption.

[–] pennomi 13 points 1 year ago (1 children)

Right, and tungsten rods are dangerous because they don’t slow down and burn up in the atmosphere like most spacecraft do (like you said, spreading out that energy over time and space). As long as you can deorbit them accurately, they are devastating since they convert the entire orbital potential energy into surface kinetic energy all at once. (Oddly, orbital potential energy and surface kinetic energy are the same thing, just from different points of reference.)

[–] Brainsploosh 6 points 1 year ago (1 children)

Agreed. On all points.

Moreover, the Tungsten rods are quite dense and thus small, and thus very hard to spot on radar or hit with countermeasures.

[–] [email protected] 3 points 1 year ago

I'm not sure countermeasures would even work. Even if you could blast it with a half dozen CWIS for the entire duration it's in the atmosphere, hitting every shot, you might change the impact zone by a few hundred meters. A high-angle trajectory would be completely unaffected.

[–] [email protected] 12 points 1 year ago (1 children)

One of the things that's stuck with me during my time on Lemmy is someone remarking that the only difference between a battery and a bomb is how controlled the release of energy is. Having seen what happens when you puncture a LiPo battery, I believe it 😰

[–] CookieOfFortune 5 points 1 year ago

There is another factor here which is the base energy level of a battery. LiPo batteries still have a relatively high base level, so even when discharged can still burn/explode. There are other battery chemistries that have a lower base and are therefore safer when fully discharged.

[–] [email protected] 8 points 1 year ago (2 children)

Wait this can't be right or I am missing something. Are you saying that the Tsar Bomba released 10 PetaJoule of energy more than our current world uses in a year?

[–] TAYRN 16 points 1 year ago* (last edited 1 year ago) (1 children)

It's.... definitely not right. Most estimates I found from a quick Google search put global energy consumption at a bit under 600,000 PJ per year, so even if they meant to say daily energy consumption or something they'd still be off by an order of magnitude.

The closest I can get to the number they gave is that global daily electricity consumption is a little over 200 PJ, so right on par with estimates for the Tsar Bomba.

[–] Brainsploosh 3 points 1 year ago

Daily electricity is right, I'll edit

[–] MacAttak8 6 points 1 year ago* (last edited 1 year ago)

Tsar Bomba released 210–240 PJ of energy according to Wikipedia. Not sure about global energy consumption.

https://en.m.wikipedia.org/wiki/Tsar_Bomba

[–] [email protected] -1 points 1 year ago (2 children)

Still you need that much energy. And it all needs to be on that rocket. So if anything goes wrong with that rocket, it will burn and release the energy of a nuclear explosion. It will be less devastating than a nuke, because it is burning fuel as opposed to a huge shockwave and temperature, but still it would insanely dangerous.

And i've yet to come across a space program that didn't include catastrophic failure rocket launches.

[–] [email protected] 4 points 1 year ago

Isn’t this system a rather normal payload? We had really large rockets with the Apollo program.

[–] [email protected] 1 points 1 year ago

We've seen this already. Starship should be capable of at least 100t to orbit, which is about 40TJ of energy on orbit. The Little Boy was 63TJ, so accounting for losses, Starship flight test 1 was exactly what that would look like.

Do note that much of the energy was lost because most of the fuel didn't burn, it just evaporated. The Beirut fertilizer explosion was 1/30th the energy, but all released at once.

[–] [email protected] 13 points 1 year ago (1 children)

The mass to orbit isn't the hard part. A reusable Falcon 9 can put 18,400 kg in low Earth orbit. That should cover two rods, plus hardware to hold and deploy them.

[–] stewsters 7 points 1 year ago (1 children)

And then you would either need to wait for your satellite to get over the target, or add a lot more weight to maneuver it to the target.

If you add wings for precision you are adding drag and heat, both sapping from your destructive power.

If your weapons satellites all start maneuvering to cross your opponents' cities then they probably would have a bit of a warning that you are planning something, and likely just shoot them down at a much cheaper cost. Anti satellite missiles have been shown to work, and it would be easy to overheat a satellite with lasers.

You also have to contend with them just nuking you in response. If Moscow were to be destroyed in a single blast they would not wait to determine if it was nukes or something else, they would fire.

[–] CookieOfFortune 2 points 1 year ago

You can definitely make some stealthy satellites.

I was thinking for release you could actually fold out wings/parachute to increase drag so it would deorbit faster and with less propellant.

Don’t think you can target a satellite with a laser, it’s too far away and you’d have to find it first.

But overall it doesn’t seem particularly efficient or useful.

[–] Red_October 11 points 1 year ago

So a little bit of looking around, and some "Close enough, fuck it" math suggests that the Saturn V over the duration of it's launch emitted about the same amount of energy (190 Gigawatts over 2.5 minutes = 2.85x10e+13 joules, close to 7000 tons of TNT at 2.93E+13 joules) as 1/3 the yield of the Fat Man dropped on Nagasaki (FM = 20,000 tons of TNT = 8.36e+13 joules).

Now I'm not math inclined, so you should take all this with more salt than your doctor recommends, but if the rocket's output is comparable to 1/3 of an actual nuke, then it's not unreasonable to think that converting all of that back into kinetic energy would get you roughly 1/3 of a nuke's output, which could be said to be "the force of a nuclear weapon." It would take a launch of something Saturn V sized or bigger to put one up there, but supposedly Starship would be up to the task if it ever stops exploding itself and/or it's launch pad.

What I'm saying is, it's plausible enough for a blurb on some article.

[–] [email protected] 9 points 1 year ago (1 children)

And friction would cost some work both way

[–] [email protected] 7 points 1 year ago

Yes but it's a 30 ft tungsten telephone pole. With that mass on size, there is minimal air resistance and free fall as it falls perpendicular

[–] [email protected] 7 points 1 year ago (1 children)

US air force put an unmanned craft into orbit for 12+ months with a large enough payload to test this.

They did it last year or year before?

[–] cloud_herder 5 points 1 year ago

TheX-37B? They definitely didn’t test that for this. The capacity weight is 227 kg.

[–] bouh 0 points 1 year ago (2 children)

Gravity is your friend. It's more when it's in orbit that you should be careful.

[–] Deme 11 points 1 year ago (1 children)

Gravity is not your friend. Getting stuff into LEO is still expensive af. A kinetic projectile dropped from space might have the same energy as a nuke, but it's still going to be a lot more expensive. Additionally, you don't have options on how that energy is released. It's going into the ground and that's that. A nuke (or any other explosive device for that matter) on the other hand can be detonated at a chosen altitude, or as a bunker buster if that's what you want.

The heavier the object, the more it's going to take to push it out of that orbit. If your weapon system is in LEO, you can realistically only drop a rod on a small envelope along the future trajectory of the weapon system. Polar orbits would have the best coverage, but fly over a target outside of polar regions only twice a day. In order to get a wider range of firing solutions, the projectile needs considerable deltaV for orbital changes. And again, gravity fucks you over here because deep within Earth's gravity well, changing the orbit of a massive tungsten rod takes a lot of fuel. Higher up these deltaV costs wouldn't be as prohibitive and you'd have more options for using the weapon, but that would increase the time from launch to impact into the regieme of hours, way too slow for anything.

The best solution would be to have a huge amount of rods in different orbits (akin to the spacejunk that is Starlink) to maximize the chances of at least one being able to fire on a target at any given place at any given time, but because those rods are still heavy af, such a plan is completely unfeasable.

Rods from gods will never happen, at least not around Earth.

[–] [email protected] 1 points 1 year ago (1 children)

High orbits are probably the way to go anyway. Not only will the payload be decently higher, the entry angle will be much steeper and more accurate, much easier hide, and much more capable of hitting anywhere in the world. This trades an hour's delay for a day's delay, but this size of weapon is a strategic weapon anyway.

Why would you need to blow up a city in an hour's notice, but not as soon as possible? If WDMs can be used in this situation (MAD doesn't apply), just use a normal ICBM. 15-30 minutes, possibly much larger booms. The advantage of orbital kinetics is stealth and immunity to countermeasures. A country-wide strike could be arranged with only seconds of warning for the targets, possibly getting ahead of most launch sites, leaving only mobile launchers to deal with.

That of course means a new intelligence war, but whoever gets there first has an unstoppable weapon, which might be important depending on how good interceptor weapons get. Even the idea of feasible hypersonic ICBMs is twisting knickers for a reason.

There are very few situations where that amount of lifting capacity is available for such a niche use. But more has been done for stupider, so something dumb very well could end up flying dangerously.

[–] Deme 1 points 1 year ago (1 children)

I have my doubts about that stealth. If it were just the tungsten rod by itself, it could well be coated in black, but the impulse needed to drop a massive projectile out of an orbit (and furthermore into a trajectory that impacts at the target) necessitates a relatively hefty maneuvering unit and that has to radiate heat to be kept operational. Cheaping out on the propulsion could be very costly because a failure in the middle of the deorbit burn would result in the rod coming down in a place you don't want it to.

The only way to hide it would be with decoys, which are already used by ICBM's. Unlike ICBM's, an orbital platform would need those decoys well before it's used because there's no terrain to hide in. A ballistic missile sub is very hard to track, satellites, not so much.

About countermeasures: Becuse the rod can't move by itself, it's stuck on a fixed trajectory after the propulsion stage is discarded. This makes it an easy target for an interceptor missile. You already know where to look so the stealthy rod isn't that hard to find, and then you just have to collide with it. And because the rod is coming down from MEO at the very least, you have ample time to do all this. An impact at such speeds would disintigrate anything and the rod isn't an exception. At the very least it would fracture into multiple less aerodynamic pieces that do much less damage than the sum of their parts.

[–] [email protected] 1 points 1 year ago

Deorbit hardware isn't big, just a small solid rocket motor would supply most of the thrust, say 100kg thruster for a 5 ton projectile. The deorbit burn is only 100m/s or so. That's some very sensitive monitoring for what amounts to ISS station keeping burns. Monitoring effectiveness could be increased by only tracking oblong objects, but such a burn on the day side might be near impossible to see anyway. This is for the LEO type the US air force is interested in.

A higher orbit projectile system would be slower but more powerful, and wouldn't need more than 3km/s to deorbit, so 4 ton-ish propulsion section (an eccentric orbit could reduce this significantly, but would narrow possible targets. The long period could allow ion engines, but the downside is big solar panels). At 30,000 km high anywhere in the sky, that's a lot of high-power telescopes tracking a lot of sky for just an exhaust plume.

Decoys would only be useful for the burns, and possibly only for false alarms. If you know a projectile is coming, you probably have a good idea about it's target, so moving to a different bunker could be good enough. In the same way, if an actual threat exists out there, a decoy burn could spur movement.

I don't think decoy satellites are useful here. If you can track these projectiles closely enough to detect plumes or small velocity changes, no amount of decoys will be enough, and orbital warfare is an entirely different ballpark.

About countermeasures, trying to intercept outside of the atmosphere requires a suborbital capable missile (probably fully orbit capable for an intercept from MEO), which will be huge and would require an incredibly precise final stage and a convenient launch location to have any chance of hitting. If you have that capability, you could just hit the projectile before it's used at all, but again, that's orbital warfare.

As for atmospheric countermeasures, a LEO type will spend maybe 20 seconds in atmosphere, mostly covered with a ball of plasma, so tracking could be a non-issue, depending on method. The issue is hitting with enough force to do anything about it. Most interceptor weapons are designed for much weaker, much much slower targets, and anything short of a direct hit will do nothing. A MEO type will be even faster, with less than 10 seconds of plasma and moving over 8 km every second. Good luck hitting that.

[–] [email protected] 4 points 1 year ago (1 children)

It is simple conservation of energy. All the energy that creates the impact, must come from somewhere. A huge metal rod that lies on the ground is not going to cause any harm, aside from stubbing toes maybe.

Leaving small height differences between the rockets launch site and the final impact site aside, the energy comes either from the rocket that brings it into space, or from propellent that the metal rod uses when launched on its target (it being a missile itself). So you end up at minimum with a rocket to transfer the whole thing into orbit, that is loaded with fuel with the same energy as the energy at the impact site. Given the rocket fuel problem, it is much more fuel, as you also need to carry the fuel for the later stages of the transport rocket up too. Then you also need additional energy for the friciton and to steer the metal rod into its designated target.

Either way you end up either having to assemble the weapon in space, or having a rocket fly into space with enough fuel to release more energy than the weapon could release on impact. So in terms of the claim of force akin to a nuclear weapon, you also need fuel with enough energy like a nuclear weapon.

Gravity does not help at all. You cannot "imbue" an object in an energy field with more energy, than you spend on changing its position gainst the field.

[–] [email protected] 2 points 1 year ago (1 children)

Conservation of Energy only applies in a closed system. If the rod is built on the Moon and then placed in Earth orbit it's quite possible to be net positive energy relative to Earth's gravity field.

[–] [email protected] 4 points 1 year ago

That is correct. But i am not aware of any Tungsten mines, let alone processing facilities on the moon within any foreseeable future.