

NASA Announces Asteroid Grand Challenge - bra-ket
http://www.nasa.gov/home/hqnews/2013/jun/HQ_13-188_Asteroid_Grand_Challenge.html

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calbear81
You can see a visualization of known asteroids and their orbits here:
[http://www.asterank.com/](http://www.asterank.com/)

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sneak
Once we've identified them, we should build a giant solar-powered ultraviolet
laser system for vaporizing them. We could power it from the shadow squares.

Oh, wait.

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gdubs
Bill Nye spoke at WWDC about deflecting asteroids. Interesting bit was how
little you need to move the thing for it to work (ie, not destroy earth). One
cool idea is to land a swarm of solar "mirror bees" on it -- light has
momentum, so the mirrors would push the asteroid just by collecting light.

Anyway, yea, we really should be working on this problem.

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cloudwalking
How many of these "mirror bees" would you need? How much momentum does light
have?

~~~
gjm11
(Danger: very very handwavy calculations ahead. Treat the result with caution,
or even contempt.)

One photon of light at wavelength lambda has energy hc/lambda and momentum
h/lambda; in other words momentum = energy / c. If the asteroid is near earth
then it's getting about 1kW/m^2 of light energy from the sun. So for each m^2
of area and each second of time, that's 1e3 / 3e8 kg m/s ~= 3e-7 kg m/s of
momentum; if the asteroid is, say, 99942 Apophis then its mass is 4e10 kg;
let's call it 3e10 kg, for a net acceleration of 1e-17 m/s^2 per square metre
of area. If adding the mirrors replaces absorption with perfect reflection,
then that's also the size of the change in acceleration they produce.

According to Wikipedia, which is always right (except when it's wrong),
averting a collision t years in the future requires a velocity change on the
order of 0.035/t m/s.

If we assume (this seems pretty damn dubious to me, but never mind) that the
effect of our extra momentum from sunlight accumulates "perfectly" over time,
and if we suppose we have 10 years' warning, and if we handwave away some
details, then each m^2 of mirror gets us 1e-17 m/s^2 * 3e8 s = 3e-8 m/s of
velocity change, and we need 3.5e-3 m/s. So we'd need about 10^5 m^2, or a
patch 300m on each side. That's a lot of bees.

In practice, you won't get perfect momentum, and the asteroid may not always
hold the same face to the sun, and the changes may not accumulate in the way
you want and may be in an ill-chosen direction, and all the numbers above are
only approximations, etc., etc., etc., so you probably need more than that to
have much confidence of saving the earth.

