4 ms·

The big problem with Rods of God isn't cost. We just spent $10bn on a carrier, after all. The real problem is it's destabilizing - potential adversaries have

by gozur88 10y ago

The big problem with Rods of God isn't cost. We just spent $10bn on a carrier, after all. The real problem is it's destabilizing - potential adversaries have to consider the very short time frame of that kind of attack and adjust their defensive posture accordingly, i.e. shoot-on-warning.

credit_guy 10y ago
The bigger problem with Rods of God is that they just wouldn't work. There are several reasons, let's take them one by one. 1. Kinetic energy is not that high compared to chemical energy, and is completely negligible compared to nuclear energy. For a kinetic(non-explosive) projectile to deliver the energy of its mass equivalent in TNT, it needs to travel at about 2.9km/s, or Mach 8.5. In other words, a rod weighing one ton hitting a target at 2.9km/s delivers a destructive energy equivalent to one ton of TNT. Since kinetic energy goes up with the square of the speed, such a rod traveling at the first orbital speed, 7km/s, would have a yield of about 5.8 tons TNT equivalent. For comparison, the largest non-nuclear bomb used in WW2 was the Grand Slam, with a yield of 6.5t TNT equiv, and the Hiroshima bomb had a yield of about 15000t TNT equiv. 2. Losses due to atmospheric drag. An orbital body can't magically hit the ground with its full orbital speed. It will lose a lot of its energy to air drag. The wikipedia page on the "Rods of God" [1] mentions a 2003 US Air Force study where the terminal velocity of these rods was Mach 10. That's only slightly higher than the Mach 8.5 I mentioned above, so the final conclusion was a bit underwhelming: "the practical applications of such a system are limited to those situations where its other characteristics provide a clear and decisive advantage—a conventional bomb/warhead of similar weight to the tungsten rod, delivered by conventional means, provides similar destructive capability and is far more practical and cost effective." 3. You can't simply drop something from orbit. A fun little experiment is this: you are a cosmonaut doing an EVA (extra-vehicular activity) and just for the fun of it you throw a small object towards Earth, at about 10m/s. What happens to the object? The cosmonaut's name was Leonov, who told the story to V.V. Beletskiy, who did the math, and the story is told beautifully by V.I. Arnold in [2]. The object goes initially towards Earth, but then starts turning, and does a full ellipse, and meets again the space station about 1.5 hour later, coming from above (well, technically it doesn't meet it, it has a deviation of a few dozen meters). The moral of this story is that you need some sort of launching mechanism to make the rod leave the satellite and go towards its target, maybe a rocket or some sort of a gun. If you want to deorbit something in the optimal way, you use a Hohmann transfer; I did a back of the envelope calculation and the propellant mass needed to deorbit is a bit less than 10% of the mass of the projectile. However, with a Hohmann transfer we are talking about more than an hour from the start of the strike to hitting the target. If you want to achieve minutes, you need to pay a price. I don't know how to do this calculation, maybe a KSP aficionado can help here, but it would not surprise me if the propellant ratio shot up to 30%. On top of that you need propellant to maintain the altitude of the satellite over the years. This adds mass and complexity to the satellite. Let's say you want to have a bundle of 20 Rods of Gods up there, 10t each. You get 200t, and then you add propellant, rocket engines, solar panels, etc, etc, I can't imagine you get away with less than 400t. For comparison the current mass of the ISS (the biggest satellite every built) is 370t and that cost so far $150BN. [1] https://en.wikipedia.org/wiki/Kinetic_bombardment https://en.wikipedia.org/wiki/Kinetic_bombardment [2] https://www.amazon.com/Mathematical-Understanding-Nature-Phenomena-Mathematicians/dp/1470417014 https://www.amazon.com/Mathematical-Understanding-Nature-Phe...
- gozur88 10y ago
  A reasonable impact speed for a Rod would be more than mach 20. That's what we get out of ICBMs today. And that's including atmospheric drag. As far as dropping something from orbit, the concept was each rod would have a tiny booster and be in its orbit. You'd put a few dozen (or a few hundred) in polar orbits, and that way you'd always have one lined up. The equation you're looking for in your third point is this one: https://en.wikipedia.org/wiki/Vis-viva_equation https://en.wikipedia.org/wiki/Vis-viva_equation Exactly how long it would take for the Rod to arrive on target depends on how much control authority you have at hypersonic speeds. Not really sure what's possible there.
  - credit_guy 10y ago
    A Rod at Mach 20 only has four times the energy of a Rod at Mach 10. Not really a game changer (in my first point I used 7km/s which is Mach 20.6). The vis-viva equation doesn't change what I was saying in my third point. I didn't know about it, thanks for pointing to it, but implicitly I applied it. I found a Hohmann calculator online, and I got that the delta v for going from a 400km orbit to a 100km orbit is about 175m/s. I plugged that in the rocket equation assuming a decent solid propellant (APCP used in the retrorockets of the descent stage of NASA's Curiosity and Spirit, for example, an application similar to the retrorocket needed here) and I got a propellant fraction of 6.7%. Now, at 400km you still have atmospheric drag, and you need a delta v of about 25m/s per year to maintain your orbit. If you want to go to 600km, you pay a price in the deorbiting delta v. At any rate, let's say the optimal poster that brings you down from space is between 5 and 10% of the mass of the Rod. We can agree to tall this "tiny". But this will need at least a revolution around the Earth (at 7km/s that's more than 95 min). The fact that you have plenty of Rods all over the place doesn't help you. A perfectly place Rod will still take 95 min to get from LEO to surface. If you want to hit the atmosphere in minutes instead of 1.5 hours, you need to pay a price in delta v, but you don't have a problem with the control at hypersonic speeds - we are still in vacuum. Once you hit the atmosphere you need control to hit the target, not to get to the target faster. Now let's see what happens if you put a few dozen or hundred in space. What can you do with a single Rod? Sure, you can destroy a civilian building, and that could be a skyscraper and you kill a few thousand civilians. I don't think either the US or Russia would adopt such a strategy (North Korea maybe). But in terms of military targets, what good are these? Remember, they are tiny in comparison with nuclear weapons. You can't destroy nuclear bunkers, you can't acquire mobile targets. If someone could solve the mobile target acquisition problem, these would be really good at killing tanks and armored vehicles. This could be useful, but this is hardly the application the alarmists on the internet have in mind. The typical perceived threat is some type of decapitation strike (alluded by the comment I was answering to). This simply is far from realistic.

by gozur88 10y ago

credit_guy 10y ago

The bigger problem with Rods of God is that they just wouldn't work. There are several reasons, let's take them one by one. 1. Kinetic energy is not that high compared to chemical energy, and is completely negligible compared to nuclear energy. For a kinetic(non-explosive) projectile to deliver the energy of its mass equivalent in TNT, it needs to travel at about 2.9km/s, or Mach 8.5. In other words, a rod weighing one ton hitting a target at 2.9km/s delivers a destructive energy equivalent to one ton of TNT. Since kinetic energy goes up with the square of the speed, such a rod traveling at the first orbital speed, 7km/s, would have a yield of about 5.8 tons TNT equivalent. For comparison, the largest non-nuclear bomb used in WW2 was the Grand Slam, with a yield of 6.5t TNT equiv, and the Hiroshima bomb had a yield of about 15000t TNT equiv. 2. Losses due to atmospheric drag. An orbital body can't magically hit the ground with its full orbital speed. It will lose a lot of its energy to air drag. The wikipedia page on the "Rods of God" [1] mentions a 2003 US Air Force study where the terminal velocity of these rods was Mach 10. That's only slightly higher than the Mach 8.5 I mentioned above, so the final conclusion was a bit underwhelming: "the practical applications of such a system are limited to those situations where its other characteristics provide a clear and decisive advantage—a conventional bomb/warhead of similar weight to the tungsten rod, delivered by conventional means, provides similar destructive capability and is far more practical and cost effective." 3. You can't simply drop something from orbit. A fun little experiment is this: you are a cosmonaut doing an EVA (extra-vehicular activity) and just for the fun of it you throw a small object towards Earth, at about 10m/s. What happens to the object? The cosmonaut's name was Leonov, who told the story to V.V. Beletskiy, who did the math, and the story is told beautifully by V.I. Arnold in [2]. The object goes initially towards Earth, but then starts turning, and does a full ellipse, and meets again the space station about 1.5 hour later, coming from above (well, technically it doesn't meet it, it has a deviation of a few dozen meters). The moral of this story is that you need some sort of launching mechanism to make the rod leave the satellite and go towards its target, maybe a rocket or some sort of a gun. If you want to deorbit something in the optimal way, you use a Hohmann transfer; I did a back of the envelope calculation and the propellant mass needed to deorbit is a bit less than 10% of the mass of the projectile. However, with a Hohmann transfer we are talking about more than an hour from the start of the strike to hitting the target. If you want to achieve minutes, you need to pay a price. I don't know how to do this calculation, maybe a KSP aficionado can help here, but it would not surprise me if the propellant ratio shot up to 30%. On top of that you need propellant to maintain the altitude of the satellite over the years. This adds mass and complexity to the satellite. Let's say you want to have a bundle of 20 Rods of Gods up there, 10t each. You get 200t, and then you add propellant, rocket engines, solar panels, etc, etc, I can't imagine you get away with less than 400t. For comparison the current mass of the ISS (the biggest satellite every built) is 370t and that cost so far $150BN. [1] https://en.wikipedia.org/wiki/Kinetic_bombardment https://en.wikipedia.org/wiki/Kinetic_bombardment [2] https://www.amazon.com/Mathematical-Understanding-Nature-Phenomena-Mathematicians/dp/1470417014 https://www.amazon.com/Mathematical-Understanding-Nature-Phe...

gozur88 10y ago
A reasonable impact speed for a Rod would be more than mach 20. That's what we get out of ICBMs today. And that's including atmospheric drag. As far as dropping something from orbit, the concept was each rod would have a tiny booster and be in its orbit. You'd put a few dozen (or a few hundred) in polar orbits, and that way you'd always have one lined up. The equation you're looking for in your third point is this one: https://en.wikipedia.org/wiki/Vis-viva_equation https://en.wikipedia.org/wiki/Vis-viva_equation Exactly how long it would take for the Rod to arrive on target depends on how much control authority you have at hypersonic speeds. Not really sure what's possible there.
- credit_guy 10y ago
  A Rod at Mach 20 only has four times the energy of a Rod at Mach 10. Not really a game changer (in my first point I used 7km/s which is Mach 20.6). The vis-viva equation doesn't change what I was saying in my third point. I didn't know about it, thanks for pointing to it, but implicitly I applied it. I found a Hohmann calculator online, and I got that the delta v for going from a 400km orbit to a 100km orbit is about 175m/s. I plugged that in the rocket equation assuming a decent solid propellant (APCP used in the retrorockets of the descent stage of NASA's Curiosity and Spirit, for example, an application similar to the retrorocket needed here) and I got a propellant fraction of 6.7%. Now, at 400km you still have atmospheric drag, and you need a delta v of about 25m/s per year to maintain your orbit. If you want to go to 600km, you pay a price in the deorbiting delta v. At any rate, let's say the optimal poster that brings you down from space is between 5 and 10% of the mass of the Rod. We can agree to tall this "tiny". But this will need at least a revolution around the Earth (at 7km/s that's more than 95 min). The fact that you have plenty of Rods all over the place doesn't help you. A perfectly place Rod will still take 95 min to get from LEO to surface. If you want to hit the atmosphere in minutes instead of 1.5 hours, you need to pay a price in delta v, but you don't have a problem with the control at hypersonic speeds - we are still in vacuum. Once you hit the atmosphere you need control to hit the target, not to get to the target faster. Now let's see what happens if you put a few dozen or hundred in space. What can you do with a single Rod? Sure, you can destroy a civilian building, and that could be a skyscraper and you kill a few thousand civilians. I don't think either the US or Russia would adopt such a strategy (North Korea maybe). But in terms of military targets, what good are these? Remember, they are tiny in comparison with nuclear weapons. You can't destroy nuclear bunkers, you can't acquire mobile targets. If someone could solve the mobile target acquisition problem, these would be really good at killing tanks and armored vehicles. This could be useful, but this is hardly the application the alarmists on the internet have in mind. The typical perceived threat is some type of decapitation strike (alluded by the comment I was answering to). This simply is far from realistic.