# The rings around Saturn and the other jovian planets are mostly small particles of ice, which has a density of 1 g per cubic centimetre. Suppose an ice particle 1mm in diameter collides with...

The rings around Saturn and the other jovian planets are mostly small particles of ice, which has a density of 1 g per cubic centimetre. Suppose an ice particle 1mm in diameter collides with another ring particle every 5 hours, and each collision knocks a billion molecules off the particle. How long can the ice particle exist in the ring?` `

### 1 Answer | Add Yours

The density of the particles is 1 gram per cubic centimeter. The diameter of the particle in question is 1mm.

The equation for the volume of a sphere is (4/3)*pi*r^3.

The radius is half the diameter, so r = 0.5mm = 0.05cm. Thus, the volume of the particle is (4/3)*pi*(0.05)^3 = 0.000524 cubic centimeters.

Multiplying the volume by the density yields the mass of the particle. Since the density is 1 gram per cubic centimeter, the mass is 0.000524 g or 0.524 mg.

Now we need to calculate how many molecules of water are present in a particle weighing 0.524 mg.

The molar mass of water is 18.015 g/mol. Dividing the molar mass by Avogadro's number yields the mass per molecule. 18.015 g/mol `-:` 6.022E23 mol^-1 = 2.992E-23 g.

Dividing the mass of the particle by the mass of a single water molecule gives us the number of water molecules in the particle. In this case, 1.751E19.

Dividing the number of water molecules by 1 billion will give us the number of collisions necessary to destroy the particle. 1.751E10 collisions will be needed.

One collision occurs every five hours, so multiplying the number of collisions by 5 hours yields 8.755E10 hours, which is equivalent to 9.996E6 years.