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.
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!