# Astrophysics - Accretion of Space Dust

1. Jun 18, 2009

### Granite

1. The problem statement, all variables and given/known data

Imagine a planetesimal sweeping up small particles from a dust cloud. Suppose the planetesimal, of mass M and radius R, is moving through the cloud at a velocity v. There are nd dust particles per m3, each of mass md

(a) Derive an expression for the number of collisions per second between the planetesimal and dust particles (ignore the gravity of the planetesimal).

(b) If all collisions result in the dust particles sticking to the planetesimal, derive an expression for the rate of change dM/dt of the mass of the planetesimal, and for the rate of change of the radius, dR/dt. Assume the planetesimal has constant density pp.

(c) Assume that the planetesimal does not affect the dust cloud significantly, nor change its speed relative to the dust, so that v and nd stay constant. Derive an expression for the time t required for the planetesimal to grow from a very small size to a particular radius R1.

(d) Part (d) merely encompasses calculations that rely on derivations from parts a-c.

2. Relevant equations

The formula for accretion

$$\tau = \frac{1}{\pi R^2 v n}$$

3. The attempt at a solution

Okay. So, for the first part, I figured that the number of collisions per second is just the inverse of $$\tau$$ from above, since one is the number of seconds/collision (or between them), the inverse must simply be the number of collisions per second.

So, we'll say that for part (a), I suspect the solution is:
$$\xi = \pi R^2 v n$$

For part (b), I established that the mass of the moving body, at any point, is the sum of the mass of the original planetesimal summed with the mass of the accumulated dust.

$$M_{total} = M_{planetesimal} + M_{dust}$$

Take the derivative with respect to time, and you get

$$\frac{dM}{dt} = M_{d} \cdot \xi$$

Because the total mass increases specifically with the mass from the collisions of particles.

Similarly, I used the equation for density and solved that

$$\frac{dM}{dt} = 4 \rho \pi R^2 \frac{dR}{dt}$$

However, I'm not 100% sure that this is correct, nor do I have any clue how to approach (c). I thought it had something to do with integrating, but I cannot figure out how to set up the integral, or what my integrand should even be.

Last edited: Jun 18, 2009
2. Jun 18, 2009

### ideasrule

$$\frac{dM}{dt} = 4 \rho \pi R^2 \frac{dR}{dt}$$

That seems to be right. Now you can equate it to the other dM/dt equation:
$$4 \rho \pi R^2 \frac{dR}{dt}=M_{d}\pi R^2 v n$$

So,

$$\frac{dR}{dt}=\frac{M_{d} v n}{4\rho}$$

3. Jun 18, 2009

### Granite

Okay. So, for part (c), I thought of integrating the change in radius function with respect to time, which can then be rearranged to a function that tells you time for a specific radius (treated it as integration of a constant). Can anyone tell me if this logic is flawed?

4. Jun 19, 2009

### ideasrule

It isn't flawed, but it's overly complicated. Examine the equation
$$\frac{dR}{dt}=\frac{M_{d} v n}{4\rho}$$

Notice that R isn't part of the equation, and that everything in the equation is constant. Once you realize the significance of dR/dt being a constant, part (c) is extremely trivial.