# How long will it take Micrometeorites to cover the moon ?

ObviousManiac

## Homework Statement

If one micrometeorite (a sphere with a diameter of 1.50×10-6 m) strikes each square meter of the surface of the Moon each second (1 micrometeorite/m2/s), how many years will it take to cover the Moon to a depth of 1.80 m?
Assume that after impact, the micrometeorites spread evenly over the surface of the moon.

V = 4/3πr^3

## The Attempt at a Solution

I'm not sure if I'm thinking through this correctly, but here it goes.

SA of Moon (SA) = 3.8x10^13 m^2 (got this out of my textbook)
Thickness (T) = 1.8 m (Given in problem statement)
Total Volume needed to be filled by Micrometeorites = SA * T
= (3.8x10^13)(1.8) = 6.84x10^13 m^3

Volume of 1 Micrometeorite = V = 4/3πr^3 where r = (1.5x10^-6 / 2)
= 4/3π(1.5x10^-6 / 2)^3 = 1.77x10^18

Micrometeorites needed = Volume needed/Volume of Micrometeorites
= (6.84x10^13)/(1.767x10^-18) = 3.87x10^31

Since the rate at which they fall is 1 Micrometeorite/s that means a total of
3.87x10^31 seconds

which comes out to 1.22 years.

SHISHKABOB
I'm not sure how you went from 3.87 x1031 seconds to 1.22 years

should be ~1.23x1024 years

voko
You do not compute the volume of the spherical shell correctly.

ObviousManiac
True about the years - I really messed that up. How am I supposed to calculate the spherical shell?

voko
The volume of the shell is the difference of two volumes.

ObviousManiac
I keep trying but it's not working out. Could you give me a little more?

SHISHKABOB
well if the volume of the moon right now is 4/3πr3 and the volume of the moon after the micrometeorites add 1.8m to its radius is 4/3π(r + 1.8m)3

then the difference between 4/3π(r + 1.8m)3 and 4/3πr3 will be the volume added by the micrometeorites

so Vshell = 4/3π((r +1.8m)3 - r3)

and then go from there