How to start this "Number Density of Particles" homework in Python?

Question

Part 2 - Determination of Number Density of Particles

If we say that q is the production rate of particles of a specific size, then in an interval dt the total number of particles produced is just q dt. To make things concrete in what follows, please adopt the case:

a = 0.9amax

q = 100000

Consider this number of particles at some distance r from the nucleus. The number density of particles will be number divided by volume, so to find number density we must compute the volume of a shell of radius r with a thickness that corresponds to how far the particles will travel in our time interval dt. Obviously that’s just the velocity of the particle at radius r times the time interval v(r) dt, so the volume of our shell is:

Volume = Shell Surface Area×Shell Thickness = 4πr2v(r)dt

Therefore, the number density, n, at radius r is:

n(r) = q dt /4πr2v(r)dt = q /4πr2v(r) (equation5)

You will note that our expression above will have a singularity for the number density of particles right at the surface of the nucleus, since at that position the outward velocity, v(R), is 0. Obviously this is an indication that we expect the particle density n to drop very rapidly as the dust is accelerated away from the surface. For now, let’s not worry about this point — we don’t need it later — and just graph how the number density varies with distance from the nucleus, starting with the 1st point after the surface value

• Evaluate Eqaution 5 for all calculated points using the parameters for q and a given above.

• Make a log-log graph of the number density versus radius. You should find that, after terminal velocity is achieved, the number density decreases as r−2, corresponding to a slope of -2 on a log-log plot

Current code:

% matplotlib inline
import numpy as np
import matplotlib.pyplot as pl

R = 2000 #Nucleus Radius (m)
GM_n = 667 #Nucleus Mass (m^3 s^-2)
Q = 7*10**27 #Gas Production Rate (molecules s^-1)
V_g = 1000 #Gas Velocity (m s^-1)
C_D = 4 #Drag Coefficient Dimensionless
p_d = 500 #Grain Density (kg m^-3)
M_h2o = .01801528/(6.022*10**23) #Mass of a water molecule (g/mol)
pi = np.pi
p_g_R = M_h2o*Q/(4*np.pi*R**2*V_g)
print ('Gas Density at the comets nucleus: ', p_g_R)
a_max = (3/8)*C_D*(V_g**2)*p_g_R*(1/p_d)*((R**2)/GM_n)
print ('Radius of Maximum Size Particle: ', a_max)

def drag_force(C_D,V_g,p_g_R,pi,a,v):
    drag = .5*C_D*((V_g - v)**2)*p_g_R*pi*a**2
    return drag
def grav_force(GM_n,M_d,r):
    grav = -(GM_n*M_d)/(r**2)
    return grav
def p_g_r(p_g_R,R,r):
    p_g_r = p_g_R*(R**2/r**2)
    return p_g_r
dt = 1
tfinal = 100000
v0 = 0
t = np.arange(0.,tfinal+dt,dt)
npoints = len(t)
r = np.zeros(npoints)
v = np.zeros(npoints)
r[0]= R
v[0]= v0

a = np.array([0.9,0.5,0.1,0.01,0.001])*a_max

for j in range(len(a)):
    M_d = 4/3*pi*a[j]**3*p_d
    for i in range(len(t)-1):
        rmid = r[i] + v[i]*dt/2.
        vmid = v[i] + (grav_force(GM_n,M_d,r[i])+drag_force(C_D,V_g,p_g_r(p_g_R,R,r[i]),pi,a[j],v[i]))*dt/2.           
    r[i+1] = r[i] + vmid*dt
    v[i+1] = v[i] + (grav_force(GM_n,M_d,rmid)+drag_force(C_D,V_g,p_g_r(p_g_R,R,rmid),pi,a[j],vmid))*dt
    pl.plot(r,v)

pl.show()

a_2= 0.9*a_max
q = 100000

I have never programmed anything like this before, my class is very difficult for me and I don't understand it. I have developed the above code with the help of the professor, and I am nearly out of time to finish this project. I just want help understanding the problem.

How do I find v(r) when I only have v(t), r(t)? What do I do to calculate the r values and what r values do I even use?

Answer 1

You have v as a known function of time and also r as another known function of time. You can invert these to get t vs. v and t vs. r. To get v as a function of r, eliminate t.