## Friday, March 25, 2016

### Solutions to Collisionless Shocks Problems

1) The Mach number M = 80 correlates to a lab –shocked hydrogen plasma at T =   10 5    K and ion gyroradius of 0.22 cm  with  shock value es  >  e    (which can be shown on working out es   based on the result below).

Now, the magnitude of the associated  B-field before the shock is obtained via:

e  =     qB/ m e

So:  B =    e   m e  /  q

=   (1.9 x 10 7  /s  )  (9.1 x 10 -31 kg)/   (1.6 x 10 -19 C)

=   10 -4 T

This is the equilibrium value of B before the shock is applied. But from the shock equation:

B m =  (2M – 1) Bo   =    (2(80)  – 1) 10 -4 T   =     0.015 T

Which is the magnetic field associated with the collisionless shock.

2)The pure electron plasma frequency is:

n   =   9  Ö N   =  9  Ö (10 16 /m3)  =   9 x 10 8  /s

Then:  we  =   2p n  =  2p(9 x 10 8  /s ) =   5.6 x 10 9  /s

Debye length l D,  =

[(1.38 x 10-23  )(10 6K) (8.85 x 10 -12 F/m)  / (4p)  (10 16/m3)  e2 ] ½

= 1.9 x 10 - 4   m

Plasma parameter:    L  =  n o  l3 D

=   ( 10 16 /m3) (1.9 x 10 - 4   m) 3

L  =  7.4 x 10  4

The radio emission must exceed this 900 MHz frequency for propagation in the medium, so we need:

f >  9 x 10 8  Hz = 900 MHz

3)  Let u   »  we/ k w   =  10 5 m/ sec

we =   2p (9 x 10 8  /s )  =    5.6 x 10 9 /s

k w   =   we/ u   »  (5.6 x 10 9  /s )/ 10 5 m/ s

k w =  5.6 x 10 4  m-1

4)  We need to re-arrange the equation to solve for velocity v o:

Using algebra we find:

v o    =  w/ k w   -   1/ k w  [1/(wp 2 -  (m/M)/ w2] 1/2

Substitute values:     v o    =

(10 10 s-1 )/ 5.6 x 10 4  m-1     -

1/(5.6 x 10 4 m-1) [1/(5.6 x 10 9 s-1)  2 -  (5.4 x 10 4 )/ (10 10 s-1) 2] 1/2

v o    =   7.8 x 10 4  ms-1

5)  The shock equation is:            B m =  (2M – 1) Bo

We can estimate the value of M from the ratio of the velocities: u/ v

Where u/ v   =    (10 5  ms-1)   / ( 7.8 x 10 4  ms-1»   1.3

Then:    B m =  (2M – 1) Bo   »    (2(1.3) -1 ) B   »  (2.6 - 1) B  »   1.6 Bo

6)   The form of the result yields 9 as a constant outside the root sign so we must have:

9 »  e/2p   Ö( 1 / εo m e )

Compute:     e/2p   Ö( 1 / εo m e )  =

1.6 x 10 - 19   C/ 2p  Ö( 1 / (8.5 x 10 -12 F/m)( 9.1 x 10 - 31   kg)

=  8.973 Ö N   or  n   »    9  Ö N