# How long will it take to cover the moon with spherical miniature meteors, one meter tall, each with a diameter of 1.00xx10^-6 meters, hitting the moon at a rate of 1 meteor per second?

Mar 24, 2016

It will take $16.2$ trillion years

#### Explanation:

My first attempt to answer this was wrong, because the question was wrong. Apologies if I caused any confusion. The questioner meant to say the meteors are covering 1 square meter of the moon per second (not one meteor per second). This allows us to answer the question using just surface area.

The moon has a surface area of $510 \times {10}^{6} {\text{ km}}^{2}$ or $5.10 \times {10}^{14} {\text{ m}}^{2}$. Now imagine, instead of being the curved surface of a sphere, you are filling a plain old box, one square meter per second, with the tiny meteors. If the meteors have a diameter of

${d}_{\text{meteor"=1.00xx10^-6 " m}}$, then it will take

$\left(1.00 \text{ m")/(1.00xx10^-6 " m}\right) = 1.00 \times {10}^{6}$ meteors

That's one million meteors stacked on top of each other, to make a pillar of meteors one meter tall. So the at the rate of one square meter per second, the first layer of meteors would take

$5.10 \times {10}^{14} \text{ s}$

That is, one second for each square meter of the surface. Now you have to calculate this times one million (the number of meteors in a 1 meter tall pillar). So the answer is

$5.10 \times {10}^{14} \times {10}^{6} \text{ sec"=5.10xx10^20 " sec}$

Converting to years, this is almost $16.2$ trillion years!