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Could the Orion project pusher plate model be used for asteroid deflection?

Basically, the plates (or hemispherical shells) would be deployed in the vicinity  of the asteroid , and the following nuclear blasts would push the asteroid-plates system on a different path. Depending on the composition of the asteroid , a direct nuclear blast in its vicinity (with no pusher plates) would be less efficient (and then there is the fragmentation problem).

How efficient would this deflection system be in relation to large asteroids (km range), and in relation to other deflection options?

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  • $\begingroup$ Note that you assume you can "push" on the mass of the asteroid in the first place; this might be true for a sufficiently large asteroid, but most asteroids seem to be mostly loose assemblages of dirt and such, rather than solid objects - push, and you disintegrate the asteroid, instead of changing the momentum of the bulk of its mass. This isn't a problem for deflection through a solar mirror or gravitational tractor, but a big problem for both "nuclear missile strike" and your Orion approach. $\endgroup$ – Luaan Sep 16 at 13:04
  • $\begingroup$ @Luaan Wouldn't 10 asteroids of, say, 20,000 t each be preferred over 1 asteroid weighing 200,000 tons, even if they all reach Earth (which some may not)? $\endgroup$ – Peter - Reinstate Monica Sep 16 at 14:35
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    $\begingroup$ @PeterA.Schneider There's great analysis on stardestroyer.net/Empire/Essays/Planet-Killers.html#Asteroid (and more on other pages on the site). In short, there's a few things it makes better, and a few things it makes much worse. $\endgroup$ – Luaan Sep 16 at 17:03
  • $\begingroup$ @Luaan I read the Asteroid section, but saw nothing about One Large vs. Lots of Medium asteroid hits. $\endgroup$ – RonJohn Sep 17 at 1:27
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    $\begingroup$ @RonJohn Is it really easier to deflect ten small asteroids? I don't think any of the methods discussed would work better for many small asteroids than for a single solid asteroid. In fact, that's the main reason why we don't think they would work at all, ever since we realized that most asteroids are not solid. As for the damage, compare what a single nuclear bomb does compared to an equivalent energy amount of conventional explosives does more dispersed. A single large (~100 m) asteroid can level a city. A hundred small asteroids (~10 m) can level a hundred cities. $\endgroup$ – Luaan Sep 17 at 7:31
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The challenge here is getting the pusher plate into position. The potential impactor is not generally traveling on an orbit that can be reached with low DV. So the only way with current(ish) technology to get 100-1000 tons of pusher plate alongside and stationary to an asteroid is if it uses an Orion drive. And if you have a fully working Orion then you can dock with the asteroid and do whatever you want with it.

Getting an intercept with an asteroid is a lot easier (lower DV) but that will have your pusher plate approaching the asteroid at speeds where pusher plate turns to vapour anyway, with or without one more nuclear detonation (it probably got there with an Orion drive). And still tend to disrupt the target.

The Nuclear deflection model is generally built on assumption that if you can only get 100kg onto a intercept course it might as well be nuclear, if you can launch 100-1000 ton payloads without using nuclear devices then just using them alone is simpler and probably more predictable in terms of final outcome.

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  • $\begingroup$ You don't need to deploy a large , compact Orion pusher plate in space (for large asteroids). You could consider a fleet of smaller Orion pusher plates, each one powered by Davy Crockett type nuclear warheads. The large pusher plate would be composed of many smaller elements. $\endgroup$ – Cristian Dumitrescu Sep 16 at 14:23
  • $\begingroup$ @CristianDumitrescu And the asteroid's mass would leak through the spaces between the pusher plates, and on the edges of the whole assemblage. To really get most of the mass of the asteroid, you'd need a huge pusher plate that forms a bowl with a volume about equal to the volume of the asteroid. If you can build an asteroid-sized construction in the first place, deflecting asteroids is easy :D $\endgroup$ – Luaan Sep 17 at 8:05
  • $\begingroup$ @ChristianDumitrescu for moving loosely assembled threats you probably want en.wikipedia.org/wiki/Gravity_tractor, which because you want it to be heavy might indeed be an Orion thrust plate. $\endgroup$ – GremlinWranger Sep 17 at 8:12
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This depends on the size and composition of the asteroid. Many asteroids are loose rubble piles, held together by weak gravity only. So you can't really park a small ship on the surface and start thrusting: you'd end up pushing the ship through the asteroid.

When the ship diameter is about the same as the asteroid, it could work.

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  • $\begingroup$ Do loose ruble pile asteroids need deflection? $\endgroup$ – Oxy Sep 16 at 12:26
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    $\begingroup$ @Oxy Yup. Rubble asteroids are still more than enough for large scale destruction of the Earth. The three largest known asteroids are fully differentiated (like planets) - and those start at 500km diameter. The fourth is already very oblate, though it seems solid enough. The 50km 253 Mathilde is already definitely a rubble pile - I probably don't have to tell you what the impact of such an asteroid would do to our climate :) Of course, it's possible to have a smaller asteroid that's fully solid, but for "danger that needs deflection", the upper limit is the important thing :) $\endgroup$ – Luaan Sep 16 at 13:13
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    $\begingroup$ Project Orion used the smallest bombs they could think of (which turned out to be only slightly larger than the Davy Crockett devices) to arrive at a ship weight of 800 tons at a diameter of 25 m. That's tiny compared to the kind of asteroid you'd want to deflect, and simultaneously more than we can currently launch in 1 year. $\endgroup$ – Hobbes Sep 16 at 14:49
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    $\begingroup$ I think you underestimate the mass of even a small asteroid. $\endgroup$ – Hobbes Sep 16 at 14:57
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    $\begingroup$ Asteroid Bennu is 250 m across, and weighs 14 million tons. $\endgroup$ – Hobbes Sep 16 at 15:37

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