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Numeriprimi
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Homework Statement
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.
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.