Spherical asteroid moving through a dust cloud

[Kyliticus]

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

A spherical asteroid of mass m0 and radius, R, initially moving at speed v0, encounters a stationary cloud of dust. As the asteroid moves through the cloud, it collects all the dust that it hits, and slows down as a result. Ignore the increase in radius of the asteroid, and its gravitational effect on distant dust grains. Assume a uniform average density D (mass per volume) in the dust cloud. By considering momentum conservation over an infinitesimal time interval dt, show that the velocity v of the asteroid obeys a = -k*v^3, and evaluate constant k. Also, find the velocity of the asteroid as a function of time.

2. Relevant equations

a = -k*v^3 (must evaluate for k)

3. The attempt at a solution

I am at a complete loss for this question. I recall in class we did an example with a rocket and its fuel, where the rocket loss mass as the fuel it had expelled gained mass. I know this question deals with some of the same concepts, but I really need a push to get in the right direction. Thanks!

[Quinzio]

I've got an answer to this.
This problem caught my attention as every problem linked to motion and speed.
I have troubled some times, but in the end, manipulatin the equations, I came to the formula involving the v^3. I have even done a simulation of this dust collector to convince myself that the acceleration goes with the cube of the speed.
The explanation is quite simple, of course, after you've found it :)
I'll give it to you right now, I don't like to ping-pong over.
Anyway, ..........

you've got this asteroid moving, whose momentum is mv = k = m_0v_0.
Since no force acts on the object, k is a constant.
So, differentiating the expression yelds:

dm\ v\ +\ m\ dv = 0

{dm\ \over m}\ = -{dv \over v}

And, will be useful later

{dm\ \over dt}{1 \over m}\ = -{dv \over dt}{1 \over v}

Basically what we have is an hyperbola, now we have to find how the point moves with time over this hyperbola.

In a time dt we can think m and v as constants and the object collects a mass of dust {dm } = v DS dt where DS is the density of dust per volume multiplied by the surface normal to velocity.
Then

{dm \over dt} = v DS

To find an expression of dv/dt which contains only speed as a variable (the hard parte for me), we remember that

{dm\ \over dt}{1 \over m}\ = -{dv \over dt}{1 \over v}

or

{dv \over dt} = -{dm\ \over dt}{v \over m}\

In the expressions above we find formulas to replace m and dm/dt

which give us finally

a = {dv \over dt} = -{v DS}\ v\ {v \over m_0v_0}\ = -v^3 {DS \over m_0v_0}

The expression of speed in function of time should be.

v = \sqrt { {m_0v_0^2} \over {m_0+2v_0DSt}}

Any comment or objection is welcome.

[Kyliticus]

Absolutely amazing explanation Quinzio, a lot more work shown than I would have expected too! I'm definitely going to take an hour to review this thoroughly, thanks a ton.