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How long will it take Micrometeorites to cover the moon ?

  • #1

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!
 

Answers and Replies

  • #2
537
1
I'm not sure how you went from 3.87 x1031 seconds to 1.22 years

should be ~1.23x1024 years
 
  • #3
6,054
390
You do not compute the volume of the spherical shell correctly.
 
  • #4
True about the years - I really messed that up. How am I supposed to calculate the spherical shell?
 
  • #5
6,054
390
The volume of the shell is the difference of two volumes.
 
  • #6
I keep trying but it's not working out. Could you give me a little more?
 
  • #7
537
1
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
 

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