Meteors arriving to a planet's atmosphere?

Question

Imagine a planet with little or no wind activity or precipitation, but that still has an atmosphere. When a meteoroid reaches the atmosphere and burns, it converts into dust and gas. Does this gas and dust stay in the atmosphere then? Can there be permanent solid particles (like dust) in the atmosphere that are so light that do not fall down to the ground after the meteroid burns?

Answer 1

Yes, the material (dust and gas) stays in the atmosphere, at least for a while. Larger particles of dust will tend to fall, but are strongly affected by turbulence in the lower atmosphere. Water and similar molecules will tend to remain in the atmosphere for an extended period of time.

There will be a spectrum of sizes for dust particles, at the larger end you get small meteorites. Dust in the upper atmosphere can remain for several years, and have significant climatic effects.

Answer 2

The assumption of an atmosphere with nearly no wind is difficult to find or create - but let's assume we find such body.

Gravity works on solid particles, and assuming that they are denser than the air, they eventually will fall down - but it is a question of the particle size and their (relative) density compared to the atmosphere. For simplicity let's assume spherical particles and that solid particles usually are MUCH heavier than air so that we can ignore buoyncy in a first order approximation.

Then we can calculate the sedimentation speed by a balance of the gravitational attraction with the friction onto their (downward) movement:

$$\begin{align} F_{grav} &= F_{friction}\\ m_d\cdot g &= \frac{1}{2}\varrho v^2 A_d C_d\end{align} $$ where $m_d$ is the mass, $\varrho$ the density, $A_d$ the cross-section, and $C_d$ is the friction coefficient of the considered dust-grain. For a sphere $A_d=\pi r_d^2$ and $m_d = \frac{4}{3}\pi\varrho r^3$. Thus: $$\begin{align}\frac{4}{3}\pi\varrho r^3 &= \frac{1}{2}\varrho v^2 \pi r^2 C_d\\ \frac{8}{3}r &= v^2C_d\end{align}$$ and solved for the terminal velocity $v_t$ with which the particle will fall to ground: $$v_t = \sqrt{\frac{8r}{3C_d}}$$ For a typical dust particle of micrometer size, we get for typical atmospheric conditions here on Earth a velocity of 1.6mm / s, but for a nanometer particle only 53 µm/s.

Now that we have the sedimentation speed, a typical meteorite looses most of its mass in the higher atmosphere where it is the visible meteor, at around 100km. Thus a µm-particle takes about two years to fall down to the ground and a nanometer particle about 6 years.

This calculation actually is important on Earth when considering air contamination especially inside buildings - it gives you at hand numbers at how much air ventilation you have to provide to remove dust or toxines from the air in a reasonable time or to prevent their build-up.

In nature on Earth this plays a role when looking at the time volcanic ashes play a role - residence of aerosols is of course larger than calculated here due to turbulent mixing within the atmosphere. One famous example is the eruption of the Pinatubo in 1991 which led to a noticable enrichment of the higher atmosphere with volcanic ash and droplets of sulphuric acid which in turn led to nice red sunsets, but also a measurable reduction in the insolation on the Earth's surface (thus less sun light reaching the surface) and a slight reduction in temperature.

Back to your question, assuming absolutely no mixing in the atmosphere would lead after time to a build-up of aerosols in the upper atmosphere from the inevitable flow of small particles from the planet's surrounding into the atmosphere - and would lead to a collective effect of reducing the amount of radiation reaching the ground by increasing and changing the high atmosphere's reflectivity and scattering behaviour.