If one micrometeorite (a sphere with a diameter of 2.3 10-6 m) struck each square meter of the Moon each second, estimate the number of years it would take to cover the Moon with micrometeorites to a depth of one meter? Hint: Consider the volume of a micrometeorite and a cubic box, 1 m on a side, on the Moon. Find how long it would take to fill the box. (Type your answer using one of the following formats, 1.2e-3 for 0.0012 and 1.20e+2 for 120.)

Question

If one micrometeorite (a sphere with a diameter of 2.3  10-6 m) struck each square meter of the Moon each second, estimate the number of years it would take to cover the Moon with micrometeorites to a depth of one meter? Hint: Consider the volume of a micrometeorite and a cubic box, 1 m on a side, on the Moon. Find how long it would take to fill the box. (Type your answer using one of the following formats, 1.2e-3 for 0.0012 and 1.20e+2 for 120.)

bobpursley · Answer

The problem is small spheres do not pack to fill a volume exactly,there is space in them. So in a cubic meter box, if it is filled, the volume of the spheres is less than that. Any Egg Packer can tell you this.

Assume the packing factor is .74, first calculated by Carl Gauss.

So, the total volume of spheres in that cubic meter box is .74m^3

Each sphere has a volume of 4/3 PI (1.15E-6m)^3, which is equal to 6.35E-18m^3

so it will take t=.74/6.35E-18 seconds to cover the moon to a depth of one meter.

If you are taking an online course, I doubt if arguing with a computer about packing factors is worthwhile, in fact,the same goes for arguing with Graduate assistants. Good luck. You do have to change the seconds to years.