Imagine Saturn (weight 5.68*10^26 kg) and its rings. Consider a particular ring C. Its optical depth is t = 0.1; it located between the distances of 74,658 km and 92,000 km from the center of Saturn (weight 5.68*10^26 kg). Assume that a one particle orbits in a circle outside of the ring plane (but with wery small inclination). Calculate the time in which will be in collision with a particle of the ring.
The attempt at a solution
I can said that t= 0,1 is median number of collisions in vertical passage particle through the plane of the ring in the rest of the ring system.
So, one particle passage through the plan with collision whan will be t = 1 – with the tenth passage. I know that for every other plane that crosses Saturn, the orbits are either in this plane or they cross it twice.
So, the particle cross the ring twice for five times. With the tenth passage will be the collision. The particle cross the Saturn with circular path for five times.
The radius of the particle can be from 74 658 km to 92 000 km, average 83 329 km.
I use the formula for orbital speed: v = √(G*M/r) = √(6.67*10^-11*5.68*10^26/ 83 329 000) m/s = 21 323 m/s
The period one orbit is: P=2πR/v = 2*3,14*83 329 000/21 323 s = 24 542 s
And the time of collision is 24 542 s * 5 = 122 710 s = 1, 42 days
Well, it is right?
Thank you very much for advices.