A-L-E-X wrote:Source of the post Can one make it into (let's say) at the same altitude as our cirrus clouds and not get burned up or hit the planet?

Maybe, if it is large (100s of meters across or more) and dense enough, but even then it might not survive the journey for other reasons (the ram pressure may fragment it or even result in an airburst.) Otherwise more typical smaller asteroids penetrating that low into the atmosphere at such a shallow angle lose their cosmic velocity.

A useful technique to apply here is

Newton's Impact Depth approximation. Basically, a high velocity projectile will be stopped after it has swept out its own mass worth of target material in its path. For high velocities the impact depth turns out to be roughly independent of the velocity, and the most important factors are the projectile's size and density relative to the target. If the target's density is uniform then this is simple to calculate: the impactor will have swept out its own mass worth of target once it has penetrated to a depth equal to its own length times the ratio of its density to the target density.

For an asteroid passing through the atmosphere, however, we must account for how the density of air decreases exponentially with altitude. The density of air at the Earth's surface is ρ

_{0} = 1.225kg/m

^{3}, and this decays with a scale height H of about 8km (decreases by a factor of e for each 8km increase in height), according to the formula

Now with a bit of calculus we can find out the depth a particular impactor should reach before losing its cosmic velocity if it comes in vertically.

Let the impactor have a mass m_{i}, approximate length L, and density ρ_{i}. The mass of a column of atmosphere it sweeps out will be its cross sectional area L^{2} times the density ρ of the air at an altitude z times a sliver of altitude dz, integrated over the range of altitudes (from the impactor's final altitude to infinity where the air density trends to zero). Setting the mass of the column of atmosphere equal to the mass of the impactor, we have:Now solving for the final altitude z

_{f},

Before we plug in any numbers, let's think about what this is saying. The scale factor is a multiplier in front, with a minus sign that might seem weird. But rest assured that for small impactors that are not stupidly dense, the answer will be positive, because the impactor's length times density will be much smaller than the scale height times the surface air density, so the natural logarithm will be negative which cancels the negative in front. That also leads to one nice piece of insight right away: the conditions needed for the impactor to be able to reach the ground before being slowed down fully. Set z

_{f}=0, and that means the argument of the natural log must be 1, and therefore Lρ

_{i} must equal Hρ

_{0}. That is, the impactor length times density must equal the atmosphere's scale height times the air density at the surface! If the impactor is even larger or more dense, then it will be able to hit the surface with some of its original cosmic velocity.

For example, this result tells us that an iron asteroid (density ~8000kg/m

^{3}) must be at least about 1.3 meters across in order to reach the surface with some of its cosmic velocity. Anything smaller or less dense than that will instead fall to the ground at a much slower terminal speed (if it reaches the ground intact at all -- it could also fully ablate or airburst.) Or consider

an icy meteoroid with a density of, say, 1000kg/m^{3}. That

would need to be at least about 10 meters across to reach the ground at a high speed.What about Earth-grazers coming in at a shallow angle, so that they might neither be destroyed by the atmosphere nor hit the surface? Once again we can use the impact depth approximation, but integrated over the whole path. The key idea is that with such a shallow angle, the path through the atmosphere will be much longer, so for a given asteroid size and composition the minimum altitude it could reach before getting seriously slowed down will be greatly increased. Unless it was something 100s to 1000s of meters across, then the only way it could survive the passage and go back out into space is if its minimum altitude was many 10s of kilometers above the surface -- much higher than cirrus clouds.A-L-E-X wrote:Source of the post Could a sufficiently large one cause tidal impacts or even influence our weather?

Weather and climate could certainly be modified if the passage deposits a lot of soot into the atmosphere. But impacts are far more effective for that, since they eject a far greater amount of vaporized target material.

As for tidal effects, no, since for the tidal effect to be significant you'd need something like a minor moon (100s of km across), but such an object is not well held together except by its own gravity, which means by being close enough to graze Earth's atmosphere it would first be torn apart near the Roche limit.