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Rings of Saturn

  1. Mar 14, 2014 #1
    The problem statement, all variables and given/known data

    Hello. I have this examples about Saturn:

    A) Calculate the radius of the inner edge of the Cassini division, if you know that this sharp transition in the density of the rings is caused by a 2:1 orbital resonance with the moon ring particles with a moon Mimas, which around Saturn (weight 5.68*10^26 kg) orbits with a period of 0.942 d.

    B) If the orbits of particle in a circular path is away from the plane of the rings (but with very small inclinations), calculate the mean time in which occurs to the collision the particle with a particle from ring (you can consider τ «1 - τ is optical depth of ring). The numerical calculation do for ring C with a typical optical depth τ ≈ 0.1. Ring C
    is between distances 74 658 km to 92 000 km from the center of Saturn.

    C) Weight of ring C is 10^18 kg and it comprises almost
    entirely water ice with a density of 900 kg*m^-3.
    Calculate based on knowledge of the measured optical depth the particle size of the ring (actually it is a estimate of the size of the largest particles in the ring rather than their typical size, but it is detail).

    The attempt at a solution

    A)
    The first example is OK, i think. This is my soulution:
    I know G and M. I don't know v. I want to know R. I can say that: v=2∏R/P. (I know that the Mimas period is twice bigger than ring period, so I can determine the P of ring)
    2∏R/P = √(G*M_J/R)
    4∏^2*R^2/P^2 = G*M_J/R
    R^3 = G*M_J*P^2/4∏^2
    R = third√(G*M_J*P^2/4∏^2) ... R = 116 737 km
    It is right?

    B)
    This is problem. No idea.

    C)
    Hmm, I can say that the all ring have this volume: V = m/ρ = 10^18/900 m^3
    Assume that the particles have a spherical shape ... than: V = 4/3 ∏r^3 * N = 10^18/900 m^3
    (assume - no gaps)
    And my teacher said that I can assume this: τ = N/V *σ*l (why?)
    where N/V is number of particles on m^3; σ is cross section and l is thickness of the ring.

    I know V = 10^18/900 m^3; ∏=3.14; τ=0,1; l=92 000 km-74 658 km (right?) and the σ ... how can I determine it?

    So, everything, it is OK? Can you help me please?

    Thank you very much.
     
  2. jcsd
  3. Mar 14, 2014 #2

    mfb

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    Staff: Mentor

    This is easy to check with Wikipedia

    B looks like it has some broken grammar. Anyway, each particle follows a circular orbit, so it has to cross the ring twice per orbit.

    C:
    Looks like the standard cross-section formula.

    If you shine parallel light on a sphere, what is the "area that hits the particle"?
     
  4. Mar 16, 2014 #3
    A: Ok, I checked it and it looks good.

    B: Broken grammar? Emh. So, I try to write it better.
    Assume that a one particle orbit in a circle outside of the ring plane.
    Calculate the time in which will be collision with a particle of ring.
    Calculate it for ring C between 74 658 km and 92 000 km from the center of Saturn, if you know optical depth t=0,1.

    C: It is a area of cross-section of sphere?
     
  5. Mar 16, 2014 #4

    mfb

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    A circle outside the ring plane is not a possible orbit. It has to cross the plane twice. What is the probability of a collision during that?

    Right.
     
  6. Mar 16, 2014 #5
    How do you know it? From which law?
    there is a assumption that a one particle orbit in a circle outside of the ring plane with really small inclination
    Twice? Why twice? Hmm. The collision have to be, so 100 %?


    so, the σ = R^2 * ∏; where R is what in this example? it is the radius of particle?
     
  7. Mar 16, 2014 #6

    mfb

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    Orbits are always in planes that have Saturn in them. For every other plane that crosses Saturn, the orbits are either in this plane or they cross it twice.

    The "collision" with the disk, but not a collision with disk particles.

    Right.
     
  8. Mar 17, 2014 #7
    Yeah, I understand the first paragraph.
    Hm. So, with disk particles... 10 %? (like the optical depth?). I don't know. However, how can I determine the time of collision by this?
     
  9. Mar 17, 2014 #8
    So, I have some solution of C. Pls, tell if it is good.

    τ = N/V *σ*l
    τ = 0,1 (by B)
    N = ? (total number of particles in ring)... N/N_A = M/M_H2O, so N = M/M_H2O * N_A (can I say this?)
    V = m/ρ
    σ = R^2 * ∏
    l = WHAT is l? it is 92 000 km-74 658 km? Or I have to say that: l = V/ (92 000^2 * 3,14 - 74 658^2 * 3,14) ?

    In both cases... the R is too small :( What is bad?
     
  10. Mar 17, 2014 #9

    mfb

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    Well the chance to hit particles with particles is higher than with light, as the particles have some size on their own.

    You can calculate the orbital period of the particles...


    What is this and how is it related to the problem? N is the number of disk particles, not the number of molecules.

    l is the length the light has to go to cross the disk. Yes, it is related to V in the way you described it, just add units to that.
     
  11. Mar 17, 2014 #10
    B)
    The orbuital period of the particles is different... I can calculate it by orbital speed: v = √(G*M_S/R)
    So, the particles 74 658 km from center of Saturn: v_1 = 22 527 m/s - period: P_1 = 2*74 658 000*3,14/22 527 s = 20 813 s = 5, 78 h
    particles 92 000 km from center of Saturn: v_2 = 20 293 m/s - period: P_2 = 2*92 000 000*3,14/20 813 s = 27 760 s = 7, 71 h
    Well. And what now?

    C)

    Hmm, so, this valueis is ok, right?
    τ = N/V *σ*l; τ = 0,1 (by B); V = m/ρ; σ = R^2 * ∏; l = V/ (92 000^2 * 3,14 - 74 658^2 * 3,14)

    The N is bad.

    However, how can I create N? By this? N=V/(4/3 * R^3 * ∏) ?
     
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