- #1

ObviousManiac

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## 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.

## Homework Equations

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.

Apparently this is wrong (we have to enter our answers online, and we're told if we're right or not). Please help!