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> Just look up what happens when you drop a 10ton tungsten rod from outer-space. Thanks to gravity, when it hits the ground it has the same impact of a small nu

by elihu 5y ago

> Just look up what happens when you drop a 10ton tungsten rod from outer-space. Thanks to gravity, when it hits the ground it has the same impact of a small nuclear device. Is that actually true? I mean, you could hoist two and a half of those tungsten rods into LEO with a single Falcon 9 expendable launch, and if we're just looking at kinetic energy, you can't get more out than you put in. So, the tungsten rod's re-entry wouldn't create any more energy than it took to lift it into orbit in the first place (assuming tungsten doesn't do anything weird like combust), and there'd be a lot of energy lost in the process. I wouldn't consider a Falcon 9 launch to be "equivalent to a small nuclear device" in terms of energy. Another way to look at is is that a 10,000 kg mass at a height of 1,000 km above a planet with gravity of 9.8 m/s^2, is 10,000 * 9.8 * 1,000,000 = 9.8 * 10^10 joules. 1 kilowatt hour is 3.6 megajoules, so that comes out to 27,222 kilowatt hours of energy. A kiloton is (according to a google search) equal to 1,162,200 kilowatt hours, so that comes out to about 0.023 kiloton TNT equivalent. With some of that energy lost as atmospheric heating. That isn't to say that dropping tungsten rods from orbit isn't a serious threat, or that you couldn't impart a lot more energy by flinging them at the Earth with much greater velocity. I just don't think the comparison to nukes is quite right.

ericbarrett 5y ago
The original idea was dubbed "rods from God" and it was proposed by the U.S. Air Force to target bunkers, silos, and other hardened targets without using nuclear detonations. The rods would have orbital velocity—7 km/s at LEO—giving them, at 10,000kg, about the momentum of a jumbo jet at cruise (an unfortunate comparison, to be sure). Except the kinetic force would be targeted at an area the diameter of a telephone pole, so they would penetrate deeply (100m or more) and cause enough of a shockwave in the earth to collapse almost any buried structure. Their thin profile and solid tungsten construction ensured most of their mass would survive re-entry at close to their original speed. They also evaded the 1967 Outer Space Treaty, which forbade "objects carrying nuclear weapons or any other kinds of weapons of mass destruction"[0]. I'm not sure they actually reach the yield of a nuclear device, but "bunker-buster" nukes are relying on overkill in yield since they can't achieve that kind of ground penetration in delivery. [0] https://history.nasa.gov/1967treaty.html https://history.nasa.gov/1967treaty.html
hinkley 5y ago
You're calculating the resting potential energy, aren't you? "Dropping a tungsten rod on earth" is not picking it up and dropping it. It's firing it at the earth. Depending on what body you're firing it from, that's a lot of additional delta v. Something coming at the earth from the Moon, for instance, is moving at least 2.3 kps, plus another kps if you fire it forward. This calculator is telling me that's 55.2 gigajoules, about 15 thousand times higher. Assuming your math is right, and then my math is right, .35 megatons is still an exceedingly bad day.
- xyzzyz 5y ago
  > Something coming at the earth from the Moon, for instance, is moving at least 2.3 kps, plus another kps if you fire it forward. What do you mean "fire it forward"? If you fire it forward, it will never fall on Earth. To get from Moon to Earth, you need to fire it backward in Moon orbit, to slow it down. Essentially, you need to slow it down a lot if you want it to get to Earth surface, as opposed to orbiting it alongside Moon.
  - hinkley 5y ago
    Not if it’s in an elliptical orbit with a low enough apogee.
- elihu 5y ago
  Yeah, I was ignoring sideways velocity. I was taking "just look up what happens when you drop a 10ton tungsten rod from outer-space" perhaps a bit too literally. If we add in 1/2 * 10,000 kg * (7500 m/s)^2, to account for sideways LEO-equivalent momentum, I have it coming out as about 0.0906 kilotons: about 4 times more energy. Which isn't too surprising, since the hardest part about getting into orbit isn't the going up part, it's the going sideways fast enough enough part. My whole formula is this: (((10000 * 9.8 * 1000000) + (0.5 * 10000 * (7500 * 7500))) / 3600000) / 1162200 Basically, (mgh + 1/2 mv^2) converted from joules to kilowatt hours to get it into a larger unit I can sort of relate to, and then converting that in turn to kilotons equivalent TNT in order to compare with nuclear yields. That's assuming I didn't leave off some zeroes somewhere or apply a formula incorrectly.
- deleted 5y ago
mayama 5y ago
10^10 Joules converted to TNT directly is 2 ton TNT. All the energy should be released at the time of impact in a short period of time. Even assuming >50% energy loss due to atmospheric interference, 1ton TNT should be left.

by elihu 5y ago

ericbarrett 5y ago

The original idea was dubbed "rods from God" and it was proposed by the U.S. Air Force to target bunkers, silos, and other hardened targets without using nuclear detonations. The rods would have orbital velocity—7 km/s at LEO—giving them, at 10,000kg, about the momentum of a jumbo jet at cruise (an unfortunate comparison, to be sure). Except the kinetic force would be targeted at an area the diameter of a telephone pole, so they would penetrate deeply (100m or more) and cause enough of a shockwave in the earth to collapse almost any buried structure. Their thin profile and solid tungsten construction ensured most of their mass would survive re-entry at close to their original speed. They also evaded the 1967 Outer Space Treaty, which forbade "objects carrying nuclear weapons or any other kinds of weapons of mass destruction"[0]. I'm not sure they actually reach the yield of a nuclear device, but "bunker-buster" nukes are relying on overkill in yield since they can't achieve that kind of ground penetration in delivery. [0] https://history.nasa.gov/1967treaty.html https://history.nasa.gov/1967treaty.html

hinkley 5y ago

You're calculating the resting potential energy, aren't you? "Dropping a tungsten rod on earth" is not picking it up and dropping it. It's firing it at the earth. Depending on what body you're firing it from, that's a lot of additional delta v. Something coming at the earth from the Moon, for instance, is moving at least 2.3 kps, plus another kps if you fire it forward. This calculator is telling me that's 55.2 gigajoules, about 15 thousand times higher. Assuming your math is right, and then my math is right, .35 megatons is still an exceedingly bad day.

xyzzyz 5y ago
> Something coming at the earth from the Moon, for instance, is moving at least 2.3 kps, plus another kps if you fire it forward. What do you mean "fire it forward"? If you fire it forward, it will never fall on Earth. To get from Moon to Earth, you need to fire it backward in Moon orbit, to slow it down. Essentially, you need to slow it down a lot if you want it to get to Earth surface, as opposed to orbiting it alongside Moon.
- hinkley 5y ago
  Not if it’s in an elliptical orbit with a low enough apogee.
elihu 5y ago
Yeah, I was ignoring sideways velocity. I was taking "just look up what happens when you drop a 10ton tungsten rod from outer-space" perhaps a bit too literally. If we add in 1/2 * 10,000 kg * (7500 m/s)^2, to account for sideways LEO-equivalent momentum, I have it coming out as about 0.0906 kilotons: about 4 times more energy. Which isn't too surprising, since the hardest part about getting into orbit isn't the going up part, it's the going sideways fast enough enough part. My whole formula is this: (((10000 * 9.8 * 1000000) + (0.5 * 10000 * (7500 * 7500))) / 3600000) / 1162200 Basically, (mgh + 1/2 mv^2) converted from joules to kilowatt hours to get it into a larger unit I can sort of relate to, and then converting that in turn to kilotons equivalent TNT in order to compare with nuclear yields. That's assuming I didn't leave off some zeroes somewhere or apply a formula incorrectly.
deleted 5y ago

mayama 5y ago

10^10 Joules converted to TNT directly is 2 ton TNT. All the energy should be released at the time of impact in a short period of time. Even assuming >50% energy loss due to atmospheric interference, 1ton TNT should be left.