Would Breakthrough Starshot be able to deflect an asteroid or comet? I have seen thoughts about using the lasers that would launch Breakthrough Starshot to deflect asteroids or comets: https://www.scientificamerican.com/article/how-to-stop-an-incoming-comet/ though I am not sure if the laser could deliver enough power at a great distance. My thought is to use the craft itself to deflect the asteroid / comet. Get the craft going a significant portion of the speed of light using the system, but aim at the asteroid (that part might be very hard to do). If you were able to hit the asteroid with the very high velocity projectile (the space sail space craft - even though it has low mass), would that be enough to significantly deflect it? Force = mass * velocity squared, so I think the velocity squared might be significant, though the mass is small.

Here is more on Breakthrough Starshot (its main goal is to explore other solar systems, but perhaps it could be used to save Earth in the future): https://en.wikipedia.org/wiki/Breakthrough_Starshot

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    $\begingroup$ Even very small impulses can result in large changes in trajectories. I think the answer might be yes, for cases where we have great forewarning. $\endgroup$
    – BMF
    Feb 9, 2020 at 16:47
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    $\begingroup$ For a very large asteroid, one that would be damaging to earth, you'd need a large space-craft to significantly deflect the asteroid, else you're throwing ants at an apple. Breakthrough star-shot is not intended to accelerate a payload of significant mass, so that's out. Also, it's made to accelerate a small payload over a long period of time, meaning the asteroid would need to be extremely far out. You honestly might be better off aiming star-shot at the edge of the asteroid to cause some sort of angular momentum change, and hope that the absorption properties of the asteroid are high. $\endgroup$ Feb 13, 2020 at 15:21
  • $\begingroup$ Something like the idea to paint half of an asteroid with reflective paint would be a very similar idea. Good question though, I'd love to see an answer with more credibility than my own. +1 $\endgroup$ Feb 13, 2020 at 15:22


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