# The rings around Saturn and the other jovian planets are mostly small particles of ice (solid H2O), which has a density of 1 g/cm3. Suppose an ice particle 1 mm in diameter collides with another...

The rings around Saturn and the other jovian planets are mostly small particles of ice (solid H2O), which has a density of 1 g/cm3. Suppose an ice particle 1 mm in diameter collides with another ring particle every 5 hours, and each collision knocks a billion ice molecules off the particle. How long can the ice particle exist in the ring? Show how you got your answer.

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### 1 Answer

Fun question!

First you need to figure out the volume of the 1 mm diameter ice particle in cubic centimeters. 1 mm = .1 cm, so using the formula for the volume of a sphere, 4/3 pi radius cubed, you get the volume as 0.0042 cubic centimeters. Use this and the density of ice to get the mass of this ice particle: 0.0042 cc(0.9168 g/cc) = 0.0039 g.

Now take the molecular weight of water, 18 grams/mole, and use this with the mass of the ice particle to find the number of moles in this ice particle: (0.0039 g) X (1 mole/18 g) = 2.17 X 10^-4 moles. As you know, one mole has 6.02 X 10^23 molecules, so if you multiple this number by the number of moles you get the number of molecules in the ice particle. It should be 1.306 X 10^20 molecules.

If 1 billion molecules are lost every 5 hours, your equation should look like this:

(1.306X10^20 molecules of H2O)(5 hours/ 1 X 10^9 molecules) = 6.53 X 10^11 hours. You can than convert to days, then earth years:

(6.53X10^11 hours)(1 day/24 hours)(1 year/365.25 days)=about 74 million years before the ice particle totally disappears!