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If there was a big asteroid with a diameter of say 50km+ in a collision course with the Earth (not orbiting), would it disintegrate into smaller chunks due to the Earth's Roche limit, or the time it will spend in the Roche radius won't be enough for the tidal forces to have an effect?

My simple calculations and assumptions of an asteroid with density as the moon will have ~9500km Roche radius with the Earth, so an asteroid with velocity of 20km/s will have about 8 minutes as soon as it enters the Roche radius until it collides with the surface of the Earth. My question here : is this time enough to disintegrate the asteroid?

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When an object comes within the Roche limit, it breaks up because of tidal stresses - the part closest to the earth feels a stronger gravitational attraction than the furthest part. Hence, the closest part will fall a little faster than the trailing parts.

As a result, "disintegration" does not mean that the body will fly apart like a bomb. Instead, it breaks up and the pieces slowly move apart. This will definitely not happen within 8 minutes, so as far as an observer on earth is concerned, the impact is the same as from a solid body. Even if the asteroid were disintegrated into dust, the effect on earth would still be the same, as all the dust particles would hit at essentially the same instant.

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This. Tidal acceleration goes by $\frac{\mathrm{d}}{\mathrm{d}R}(R^{-2})\Delta r \propto R^{-3} \Delta r$ so the loose bits of the asteroid experience micro-gee for a few minutes. Think of that astronaut's tool bag drifting off a couple of years ago. –  dmckee Feb 12 at 3:04
@dmckee "micro-gee for a few minutes" - as I said, it won't happen in 8 minutes. –  hdhondt Feb 12 at 3:20
Yes. I was agreeing with you and amplifying your assertion with some (handwavy) math. –  dmckee Feb 12 at 16:34
@dmckee, so using the relation you provided, how long should it take to make a significant change in the dimensions of such an asteroid ? –  Abanob Ebrahim Feb 12 at 17:52
@AbanobEbrahim Compute it using, say, $10^{-4}$--$10^{-3}\,\mathrm{m/s^2}$ to get a feeling for the time scale. Define "significant change" however you want, but I'd use a number on order of 1/10 the diameter at a minimum. –  dmckee Feb 12 at 20:18
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