Would a considerably big asteroid be disintegrated by the Earth's Roche limit?

Question

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?

Answer 1

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.

Answer 2

Gravitational acceleration goes as the inverse square of distance. Earth g is 9.81m/s² at the surface.

The Roche limit of the Earth is 9500km, but I believe that's measured from the centre of the earth, whose radius is 6350km. So if the asteroid holds together up to the Roche limit, it only has 3150km to impact, not 9500km. That's only 158s at 20km/s.

Supposing it starts to disintegrate at that point under tidal stress, then what you have in effect is the two sides of the body falling at different speeds due to their different distances from earth. The difference in acceleration is the difference between gravity at 9500+/-25 km. I make this difference about 0.046m/s². This will of course increase during the fall up to the difference at 6350 - 6400 km, which is 0.15m/s²

This would be responsible for a relative difference somewhere between that of falling for 158s at the lower acceleration and the higher (yeah, I'm too lazy to evaluate a simple integral ;-) That is, 570 - 1800m.

As such "yes and no". Technically the asteroid could "disintegrate", but all that means is we'd be hit by a 51km swarm of rocks containing 6% empty space, instead of a single 50km rock with no empty space. The size of the individual rocks would be whatever size can survive the tidal force, and that depends on the tensile strength of the rock, the value of which I do not know. Note that for example Jupiter's moon Metis is within Jupiter's Roche limit, is 60x40x34 km, and is probably made of ice not rock. Being inside the Roche limit means that there cannot be loose material on the surface (like there is on our Moon), since Metis's gravity would not hold it there, but the body can still be held together by its own tensile strength added to its gravity.

Presuming the asteroid did break up somewhat, its rigidity would be reduced compared with if it hadn't broken up at all. But I doubt that this would have very much effect on the dynamics of the impact, given that pretty much everything in sight would melt on impact anyway.