Minimum size of an asteroid to actually impact earth

As mentioned in NotAstronaut's answer, objects smaller than 25 meters will typically burn up in the atmosphere. One can very easily see why this should be the case using Newton's impact depth formula. This is based on approximating the problem by assuming that the matter in the path of the object is being pushed at the same velocity of the object, so as soon as the object has swiped out path containing the same mass as its own mass, it will have lost all of its initial momentum. All its kinetic energy will then have dissipated there, so if this happens in the atmosphere it will have burned up before reaching the ground.

This is, of course, a gross oversimplification, but it will yield correct order of magnitude estimates. We can then calculate the critical diameter as follows. The mass of the atmosphere per unit area equals the atmospheric pressure at sea level divided by the gravitational acceleration, so this is about $10^4\text{ kg/m}^2$. If an asteroid of diameter $D$ and density $\rho$ is to penetrate the atmosphere, its mass of $1/6\pi \rho D^3$ should be larger than the mass of the atmosphere it will encounter on its way to the ground, which is $5/2 \pi 10^3 D^2\text{ kg/m}^2$. Therefore:

$$ D > \frac{1.5\times 10^4}{\rho} \text{ kg/m}^2$$

If we take the density $\rho$ to be that of a typical rock of $3\times 10^3 \text{ kg}/\text{m}^3$, then we see that $D>5\text{ m}$, which is reasonably close order of magnitude estimate to the correct answer.


It needs to be more than 25m or otherwise it will burn up in the atmosphere according to this Nasa article "Space rocks smaller than about 25 meters (about 82 feet) will most likely burn up as they enter the Earth's atmosphere and cause little or no damage." https://www.nasa.gov/mission_pages/asteroids/overview/fastfacts.html


Consider a specific example. The leonids arrive at the top of the atmosphere at $72$ km/s with a maximal mass of around $0.5$ g. According to the article these particles are $0.01$ m across. They reach the ground. If such a particle strikes the $30$ km high atmosphere at $45$ degrees it must travel around $4 \times 10^4$ m before it hits the ground, mostly burning up. If it started at rest at the top of the atmosphere it would only accelerate to about $1$ km/s. As you can see it very much depends on azimuth, latitude, meteor composition, and speed.