I’m talking about the hypothetical/fictional idea of telephone pole sized rod of Tungsten, being dropped from low earth orbit onto a ground target.
If you were standing on the ground, what would you see? Would it simply be like a meteor, a bright spot and a smoke trail? Presumably with a meteor part of the substance is being eroded as it falls to create the trail, so would a tungsten object give off a smoke trail?
Related question, would it appear to move across the sky, or would it appear to come straight down? Would this depend upon your location, say a couple of km’s from the target or standing on the roof of the target?
Anything in low orbit has a high angular velocity, so a dropped object would certainly not appear to come straight down.
If you were at the target site, it would appear near the horizon, heading toward you but initially looking like it might pass over your head. Then, as it’s slowed by the atmosphere, it winds up hitting your head.
I don’t have much to add, but there’s a description of this in Neil Stephenson’s Anathem, as a character is near ground zero when such a weapon hits. Not sure how accurate the description in the book might be.
Why wouldn’t the view be similar to the Russian dashcam shots of the Chelyabinsk meteor? Or the RV tracks from MX/Minuteman tests? Aren’t the speeds similar between an object in LEO and an ICBM in mid phase?
Someone with more math than me will have to go and calculate the temperature at the shock front for a Mach 20-whatever reentry, but isn’t it above the melting temp of tungsten?
A tungsten telegraph pole stands a pretty reasonable chance of staying mostly in one piece. It’s a good thought experiment. I bet it would make a hole in the ground. This would also make a good “What If”.
Meh, use normal steel wrapped in porcelain, but for a ground observer, I am going with something resembling a shooting star, the impact depending on the mass of the rod, should resemble a nuke. So our observer would probably not last all that long.
It would probably look like that, if the observer were in the same position as those dash cams were relative to the object. The view would be different if the projectile were headed to a location just a few feet from you.
My compressible-flow textbook has a plot of reentry velocity versus shock temperature; for 17,400 MPH, the temperature behind a normal shock wave would be about 7000K. Tungsten melts at 3695K, so yes, you would expect the tungsten to erode. Two issues though:
the projectile would slow down during its descent, so the peak temperatures would decrease.
-the air at the top of the atmosphere is very thin, so even while the normal-shock temp would be very high, there wouldn’t be much air mass available to heat the tungsten.
-A very shallow descent trajectory would allow more time to be spent at elevated temperatures, causing more erosion - but it would also decelerate the projectile more, decreasing temps for the remainder of its descent. Note that some pretty big non-tungsten chunks of the space shuttle Columbia made it to earth after its reentry breakup in 2003.
So your projectile would certainly erode on the way down, but depending on its size and flight path, it seems likely that the bulk of it would survive to deliver a killing blow.
Agree with your points, Machine Elf. I was asking about the melting point in the context of the OP wondering if it would melt tungsten at all. I thought the temp was high enough, just didn’t know for sure.
Not definitive by any means, but in Niven & Pournelle’s, “Footfall,” a character is worried about the warning time you’d have from a similar type of impact, and is told, “A meteor would flare at fifty miles up, and come in at a slant at five to six miles a second.” Resulting in about 15 to 20 seconds of warning
Aside, at a telephone pole length of 25 feet, width a touch smaller than code of 1 foot, and a density of ~.7 lb per cubic inch, I get a weight of 23-25,000 pounds for our phone pole. Seems a little light, but we’re still not going to have a lot of these things to throw around. OTOH, it’s got a KE, assuming 5 miles per second impact velocity, of around 260 billion ft./lbs. Or about 85 tons of TNT. All this assumes I did the math right, which is a pretty big assumption.
Just to clarify, one of the ideas that underpins the “Rod from God” weapon idea is that the tungsten weapon is accelerated out of its orbit into a stationary position above its target. It then drops straight down, so in fact that is what you’d see. It would be significantly different from a MIRV re-entry in that
It’s straight down and
A MIRV is, of course,a re-entry vehicle with a heat shield designed to protect the nuclear weapon inside so it’ll survive re-entry and operate correctly. The shield slows it down. A tungsten rod has no need for such a thing.
The result of impact would be a truly stunning display, nothing at all like the inert MIRVs in the video. Some estimates put the force of a large, telephone-pole-sized weapon’s impact at 10 tons of TNT, a huge bomb indeed; others guess higher still, some even a bit higher than Gray Ghost’s estimate of 85 tons, but the basic jist of it is you don’t want it to hit your house. The challenge, of course, is in building the weapon. A tungsten rod the size of a telephone pole is not easily put into space in the first place.
Maybe there’s something I’m missing, but I’m not seeing the advantage of dropping the tungsten rod from a high altitude at all, considering the ridiculous amount of energy you need to spend to get it there in the first place, and then deploy it when you’re ready to use it. “Not easily put into space in the first place” seems like a doozy of an understatement, and overlooks the cost of getting the thing down afterward.
Taking Gray Ghost’s estimate of ~11,000 kg mass for the rod, if we suspended it in LEO about 1000km above the surface, it would have gravitational potential energy of (1.1e4 kg)(9.81m/s^2)(1e6 m) ~= 1.1e11 joules, or about 26 tons TNT. That’s a substantial amount, certainly, and you wouldn’t want it to land on your house, but consider that an object orbiting at 1000km has a speed of about 7.4 km/s. That much delta-v represents at least (1/2)(1.1e4kg)(7.4e3 m/s)^2 ~= 3.0e11 joules. That alone is nearly three times as much energy as you’ll deliver to your target – and in reality, you would need much more energy depending on the mass of your fuel, the specific impulse of your rockets, and so on. Bottom line, that’s some expensive delta-v.
Add to that the cost of putting your god’s rod and all its fuel and rockets into orbit in the first place (which will be much, much higher than the cost to deorbit) and this starts to look like a lot of trouble for a firecracker. Flying a plane over and dropping a few 1000-pound bombs would be orders of magnitude cheaper.
The idea that it’s possible for an object in geostationary orbit to be made to go anything close to straight down goes against my (very limited) understanding of orbital mechanics. Can anyone explain?
The only way it can go straight down is if you’ve got a bunch of rockets strapped to it that can kill all the orbital velocity in one go. No more orbit means the only force acting on the object is earth’s gravity, which means it drops straight down.
Wouldn’t that result in a lower impact speed? As the bulk of the energy needed to get something into orbit is taken not from the getting to space bit but the going really fast to orbit bit, so it seems a bit of a waste to not use that going really fast thing that happens when you’re in orbit.
But you can see the plane coming and perhaps shoot it or its bombs/missiles down; there’s a lot more warning with a plane or even an ICBM, and a “Rod from God” would be much harder to intercept than anything but an ICBM due to its speed. And of course in the sci-fi settings where such weapons actually appear they are often made in space or a moon/asteroid, not on the planet they are orbiting so never took that much energy to put up there in the first place.
As for the expense of building such a system in the real world, the people dreaming such things up don’t usually seem to worry about that. Especially back during the Cold War era when the idea first appeared.
You’d have a fair bit of warning with the tungsten rod. They’d be easy to track, and each one’s targeting window from the time it starts a burn would be well-known. The number of conceivable targets for each weapon would be public knowledge. And you don’t drop 7 km/s of delta-v without making a spectacle of yourself, so there would be many minutes of advance notice starting the moment the rockets fired.
Granted, you’re right about countermeasures. Not many options there unless you can disable the rockets somehow before the burn completes. But a large power whose non-movable installations were threatened by such weapons would probably adopt a policy of destroying them preemptively, and would likely succeed. Basically, getting these things into space requires you to be an absolutely unopposed superpower – in which case, the people you might use them against would probably be easy to knock down using less expensive means (like ICBMs).