Application of Granular Kinetics to Ring Processes Jim Jenkins - PDF document

Application of Granular Kinetics to Ring Processes Jim Jenkins Cornell University with Volker Simon, Brian Lawney, and Joe Burns Dilute, three-dimensional ring: repeat and extend Goldreich and Tremaine’s calculation of the relationship between optical depth and coefficient of restitution. Dense, two-dimensional ring: introduce and interpret a simple numerical simulation of the flow around a moonlet in the absence of gravitational interactions between the moonlet and the disk particles and between the disk particles.

Velocity distribution function: f ( ; ,t)d d c x c x t)   c number density: n( , f ( ; , t)d x c x c 1  c    Averages: fd c n   mean velocity: ( ,t) u c u x   velocity fluctuation: = ( ,t) C c u C x   second moment: C K C 1   ˆ    T tr K , T 1 K K 3    third moment: Q C C C 1 1           Q Q Q Q , q Q ijk ipp jk jpp ki kpp ij i ipp 5 2

Balance equations      ,    3 mass: mn d n / 6 s         0 u  t linear momentum  u R                GM u u K  3 t R second moment  K       T              u K + K u + K u    t           Q C C Explicit form   n 1        1 f (c; ,t) ex p , K det K   x C K C   1/2  2   8 K

Collisions   and c pre-collisional velocities, 1 and c c c 1 2 2 post-collisional velocities, unit vector k directed from 1 to 2, coefficient of restitution e, relative   velocity C , unit vector j in the plane of g C 1 2 g and k , perpendicular to k .       e g k g k   1 e 1 e             g k k g k k c c c c 1 1 2 2 2 2 Total change of second moment              C C C C C C C C 1 1 2 2 1 1 2 2 1 (1 e)(     ) g k 2              (1 e)( ) ( ) g k k k g j k j j k 

Collisional production of second moment 1  g k      [ ] m f f d( )d d d C C g k k c c 1 2 1 2 2   0  3/2 6 T      (1 e)  3/2 d ˆ     (1 e) 2 A B  g k    3/2 ( / T) d k A k k k Kk   0   ˆ  g k       1/2 ( / T) ( / T)d k B k i i k k Kk k Ki   0

Second moment              K K / 2T , K K / 2T 1   1 2 1 2  Cylindrical polar: r, , z cos    e e r 1         1 cos2 sin 2 0         = T sin 2 1 cos2 0 K       0 0 1 2   Nearly homogeneous          , and constant; T T(z), (z), u u u(r) 

Balance equations at lowest order 1/2   GM      u(r) (r)r, (r)  3  r           4 K Q  r rr zrr  z         K Q    r  z z        0 Q zz zzz  z  1           K 4K Q    rr r zr  2 z

Eigenvector basis  3                2 T 1 sin 2 cos Q 11 zrr  2 z             2 sin 2 Q sin Q   zr z   z z  3               2 T 1 sin 2 sin Q 22 zrr  2 z             2 sin 2 Q cos Q   zr z   z z        0 Q 33 zzz  z   1 1               T 5 3 1 cos2 sin 2 Q    zrr 2 2  z             Q cos2 Q    z zr   z  z

Integrate the last over z:  5   cos2    3 1 Integrate the isotropic part over z:   6 tr( )   A         1/2 2 2 3/2 3 sin 2 T dz 1 e T dz  3/2 d   Using this, the 33 component is     ˆ (1 e ) tr( ) A 33 With the last two, the difference between the 22 and 11 components is            ˆ 3 1 33 22 11

Approximate        4 2 2 4 2 3 9             2 2 2 3 tr( ) A  70 7 21 2 2   35 4 2 4           2 2 3 2 8 2 2              (1 e) 7 2   22 11 35 6 2 11 11       4   2 2  4 4 6 15                   ˆ 2 2 3 2 (1 e) 42 2 6 3   33 105 11 11 11   Solve the last two balance equations for α and β    in terms of 1 e 7 1013      2 2 2 396   8 3   2772 2960   9393 1/2     2    8 3 10     12 11 4851     2   2960 9393   2960 9393     1/2 Lowest order in  :      5 /14 and  5 / 2

Limitations  5   cos2    3 1 with   1/2   5 5 and     14 2 implies that   0.3688  or e 0.6312 (0.6270)

Isothermal  z         0 (1 2 ) T GM  3 z r     2       0 exp 1/2 z     2(1 2 ) T     1/2 6 tr( )T   A          2 2 3 sin 2 dz 1 e dz 3/2 d   1/2 6 T        2 3 sin 2 1 e t r( ) A  0  3/2 d 2 Optical depth 1/2   1/2   3 T        1/ 2 (z)dz 3(1 )   0  d  2  d 0 2 tr( )   A       2 3 sin 2 1 e  2 1/2 (1 )

 (z) and T(z )   2       0 T z    z (1 2 )    6 (2 ) d q       3/2 3 T sin 2 T tr( ) 2 z A  1/2 d d z       1/2 5 d (1 2 )(5 4 ) dT  1/2 q T z        2 2 4(2 ) d 2d T 4d T dz 0 1 2 ˆ 2 9 tr( ) K         d 33 49 (1 ) 0 2 28 T     4 (6 13) 4 (6 13)    d d 1 2 2 5 T 35 T

Differential Equations   2       1/2 T / T F / z / T 0 0    F       F 2       (1 2 )    C C           2 2 3 2 1 F 2 F 2 S S     C C ( ) C C ( ) 1 1 2 2     2 3 C F d  d 2   S 0   1/2 T    2 C F d 0 0 1 0 Initial Conditions         (0) 1 (0) 0 F 0 1