We actually today have programs to keep an eye out for various rocks in space that may hit Earth.

It's a common sci-fi trope that when one is on course to hit us, we deflect it.

That's pretty much done "using nukes" or in some schemes, ingeniously using some sort of relatively low power rocket-like engine which is bolted on place on the Danger Object and slowly deflects it over a period of time.

All fair enough.

Can we put some numbers on this?

Say there was an incoming Danger Object of mass 1 tonne, 1000 kg.

We have a realistic "1 or a few years" to deflect it.

I'm pretty sure that although it might cost the lives of Bruce and Billy-Bob, we could probably deflect a 1-tonne object using current, real, actual technology. Rocket ships, robots, bombs and engines as we know them.

10 tonne .. I'd guess again yes. (But maybe I'm totally wrong, maybe that's hopelessly outside the scales of energy involved.)

100 tonne .. I have no clue.

1000 tonne? A million? 100 million?

A recent interesting question asked if an incoming planet-sized body could be deflected in any way at all with current actual technology; of course the answer to that is absolutely not.

But it got me to thinking.

In order of magnitude terms, in fact what is the biggest incoming Danger Object we could deflect currently?

Has this in fact already been worked out and is well-known by the various danger-object-tracking programs? Can someone here who drinks and knows stuff figure it out with a calculator?

It would seem to be a critical basic info-fact for worldbuilders!

Tag here is hard-science.

  • $\begingroup$ Is deflect the option? Because it may be easier to explode/break up an asteroid into pieces than burn up in the atmosphere. Or just move it slightly off course so it doesn't hit the earth -- deflect implies reverse direction a full 180 degrees which is much more difficult. Simply breaking it up into small pieces would achieve the prevention of it from reaching the surface, but wouldn't deflect it. $\endgroup$
    – cegfault
    Oct 20, 2019 at 15:47
  • $\begingroup$ @cegfault - sure, the question says deflect, so yeah it's about "deflecting large objects". If you "break something up" you still have to deflect the pieces around Earth. (The Armaggedon plot - two pieces which nicely go either side of Earth - was silly for this reason.) (As well as others.) $\endgroup$
    – Fattie
    Oct 20, 2019 at 17:10
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    $\begingroup$ Depends entirely on how much warning we've got. With a century or two of lead time, a single astronaut with a bucket of paint can deflect some surprisingly large asteroids. $\endgroup$
    – Mark
    Oct 21, 2019 at 0:53
  • $\begingroup$ @cegfault: Re: "deflect implies reverse direction a full 180 degrees": I'm not sure why you say that. In my experience, "deflect" typically implies a relatively small change in direction; and the definitions at merriam-webster.com/dictionary/deflect say "turn aside", not "turn back". $\endgroup$
    – ruakh
    Oct 21, 2019 at 2:07
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    $\begingroup$ For the 1-10 ton asteroids, we probably wouldn't need to do anything. They'd burn up in the atmosphere. By comparison, the Chelyabinsk meteor that hit in 2013 massed about 11,000 tons, and it caused some injuries and damage, but no fatalities. The impactor that created Meteor Crater was about 300,000 tons, and did not cause a mass extinction. The Chicxulub impactor that killed the dinosaurs had a mass of at least 10^12 tons. $\endgroup$
    – Ray
    Oct 21, 2019 at 2:19

2 Answers 2


A surprisingly large object, actually, if we decided to dust off an old cold-war era project: Project Orion. The idea was to use specially designed shaped nuclear explosives which would covert a good chunk of their blast energy into kinetic energy in the form of a wave of superheated gas, which would be used to propel the spacecraft. Given just one year to finish the project, I'm not proposing building a giant ship powered by these things. Instead, I would propose using the nukes themselves on the asteroid, to change its course. Getting a 1-ton nuclear payload anywhere in the solar system is well within our current capabilities; it's just a matter of how many we need to finish the job.

The Wikipedia article on asteroid impact avoidance, if I'm reading it correctly, suggests that you need a very tiny change in an asteroid's velocity to get it to miss earth. The equation given is 0.035/t m/s, where 't' is the number of years before impact the change is made. The article on Project Orion suggests that a 1-ton nuclear warhead can alter the course of a 1,000,000 ton object by about 10 m/s. If we assume that our nuke is only about 10% that efficient, then if we deliver that singular nuke 1 year before impact, that single 1-ton nuke should be able to divert an asteroid of size in excess of 25 million tons.

This sounds impressive, until you realize that correlates to an asteroid that is, depending on density, around 200-300 meters in diameter. City-busting, certainly, but not world-threatening. If we want to divert an asteroid the size of the one that we think killed the dinosaurs; we would need, at minimum, around 100,000 such nukes to divert such an object given only a year to prepare. Needless to say, this is beyond our current capabilities... but not by an insurmountable margin.

If we detected such an asteroid 20 years before impact, and took 10 years to build our response, we would only need about 10,000 nukes to do the job - which I estimate would cost somewhere in the ballpark of 500 billion dollars - and could launch them for (if I'm being pessimistic) a cost of about 1 trillion dollars. Double the cost of the program, and we could divert small-ish planet killing asteroid, given twenty years of warning and ten years to prepare, for about 2 trillion dollars, with the cost scaling directly to the size of the asteroid and inversely to the time we have to work.

EDIT: I underestimated the cost of getting the nukes into space. 2 trillion dollars is optimistic. 5 trillion is more realistic (which is about equivalent to the U.S. military budget for 10 years straight), and 10 trillion is probable. Almost all of that is the cost of the spacecraft. The nukes are a minor expense in comparison.

EDIT2: Man, I messed up the nuke calculations too. Not 500 billion dollars, closer to 5 trillion, for a total project cost of between 10-20 trillion dollars, spread over 10 years. Possible, but a very, very hard sell.

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    $\begingroup$ "If we detected such an asteroid 20 years before impact, and took 10 years to build our response, we would only need about 10,000 nukes to do the job" - wait, this sounds exactly like an answer. But. What size asteroid did you mean there? (They vary wildly in size.) $\endgroup$
    – Fattie
    Oct 20, 2019 at 18:30
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    $\begingroup$ @Fattie The size asteroid I was thinking of was on the lower end of the estimate for the size of the asteroid which killed the dinosaurs, around 10 kilometers in diameter, but on the upper end of the density spectrum for asteroids, around 5 grams per cubic centimeter. $\endgroup$ Oct 20, 2019 at 18:33
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    $\begingroup$ It's only 10 times the cost of the Iraq war, and it actually fits within the defense budget remit... $\endgroup$
    – pjc50
    Oct 21, 2019 at 8:19

That depends strongly on the warning time we get. Early warning allows smaller changes in the flight path to be effective, and it gives more time to launch and deploy space missions.

Keep in mind that acceptable risk is rather low these days. If one in ten, one in a hundred astronauts are likely to die, space agencies would cancel a normal mission. On the other hand, armies and air forces have ordered plenty of operations where the top brass expected 10% casualties or worse.

So, call it a year to get a Saturn-V-capacity launch system into series production, several launches per month. In parallel a modular spacecraft is designed and several units are built. Launch half a dozen, hope that most arrive in orbit, hope that most which are assembled make the flight.

Once at the target, anchor a rocket engine to the asteroid and make a relatively tiny change in the trajectory. 1 m/s delta-V applied half a year before impact would make it miss Earth.

The problem, many near-Earth asteroids are only detected after passing Earth ...

  • $\begingroup$ Wouldnt it be more efficient to just ram the asteroid instead? If the rocket and all its payload/fuel vaporizes on impact it has still done its job, rather than spending copious amounts of fuel and time to slow down, get the same speed and direction of the asteroid to land safely, anchor in place and then spend the remaining fuel trying to change its path. Otherwise good answer. $\endgroup$
    – Demigan
    Oct 20, 2019 at 17:00
  • $\begingroup$ @Demigan, shattering a big asteroid into almost-as-big parts might not help the situation much. $\endgroup$
    – o.m.
    Oct 20, 2019 at 17:03
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    $\begingroup$ The goal is not shattering the asteroid. If you slam the rocket at the left front then it'll push the asteroid to the right and slow it down a little. If it is so small that it shatters it will likely cause most debris to miss and the remainder to be small enough to burn in atmosphere. So it seems like a better option than wasting so much fuel and energy on slowing down and accelerating if it is more efficient to use that to send another rocket to impact at the same area. $\endgroup$
    – Demigan
    Oct 20, 2019 at 17:26
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    $\begingroup$ This is a common misconception. If something is moving along and it shatters (think of a bullet that fragments for more destruction) you end up with: a cloud of pieces moving in the same direction. It doesn't make a "Y". $\endgroup$
    – Fattie
    Oct 20, 2019 at 18:28
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    $\begingroup$ @o.m. When a bullet hits a wall in a shooting range, the wall is much more massive and is not really moved by the impact. When a rocket hits an asteroid in space, it's the asteroid that is much more massive. After impact, the chunks of asteroid inherit the momentum of the asteroid, plus the change in momentum from the rocket... except that the rocket's contribution is very small comparitively. It's a nudge, not a whack; the chunks left behind will be mostly on the same course as the original asteroid. $\endgroup$ Oct 21, 2019 at 8:38

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