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Conservation of Momentum for Asteroid in Dust Cloud

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A spherical asteroid of mass [tex]m_0[/tex] is moving freely in interstellar space (no external forces) with velocity [tex]v_0[/tex] . It runs into a dust cloud whose uniform density is [tex]\rho_d[/tex] Assume that every particle of dust that hits the asteroid sticks to it and that the asteroid remains spherical at all times. (a) Obtain an expression for the velocity of the asteroid as a function of time. (b) Obtain an expression for the force exerted on the asteroid by the dust as a function of time.

    2. Relevant equations

    [tex]p_f = p_i[/tex]
    [tex]F_{ext} = 0 [/tex] which implies that [tex]m\frac{dv}{dt} + v\frac{dm}{dt}[/tex]

    3. The attempt at a solution
    What I did initially was create an expression for [tex]m(t)[/tex] which I got as:
    [tex]m(t) = \frac{4}{3} \pi r^3(t) \rho_d + m_0[/tex]

    and I took the first derivative with respect to time. Since [tex]\rho_d[/tex] is constant for the cloud, the only thing that was time dependent was [tex]r(t)[/tex]. Well, I got this

    [tex]4 \pi r^2(t) \rho_d \frac{dr}{dt}[/tex].

    I figured that, to within a proportionality constant [tex]r' = x' = v{t}[/tex]

    which also gave that

    [tex]r = r_0 + d [/tex]

    where d is the total distance traveled in the cloud and is equivalent to [tex]v(t)*t[/tex] .

    So I plugged this all into my equation and I got that

    [tex]\left[\frac{4}{3} \pi r^3(t) \rho_d + m_0 \right] \frac{dv}{dt} = V(t) \left[ 4 \pi r^2(t) \rho_d \frac{dr}{dt} \right][/tex]

    which simplified down to

    [tex]\left[\frac{4}{3} \pi r^3(t) \rho_d + m_0 \right] \frac{dv}{dt} = V(t) \left[ 4 \pi r^2(t) \rho_d v(t) \right][/tex]

    Solving this out gives us that

    [tex]v(t) = \frac{ \left( 4d^3\rho_d\pi + 3m_0 \right )t }{ 12 d^2 \rho_d \pi }. [/tex]

    This seems fair enough but the correct answer (as given by the book) is apparently

    [tex]\frac{v_0 m_0}{ \left[ m_0^{4/3} + \frac{4\rho_d\pi v_0 m_0}{3 \left( \frac{m}{r^3} \right )^{2/3} }t \right]^{3/4} } [/tex].

    I am utterly baffled as to how they solved this or got the power of [tex]\frac{3}{4}[/tex] in the denominator. Any help would be very much appreciated.
     
  2. jcsd
  3. Oct 15, 2009 #2
    I've no idea how you got [tex]
    m(t) = \frac{4}{3} \pi r^3(t) \rho_d + m_0
    [/tex]

    since how much mass the asteroid picks up depends on v and v isn't constant. You also don't
    want the volume of the asteroid, but the largest crosssectional area.

    [tex] \frac {dm}{dt} [/tex] isn't too hard to find however.

    apparently r is meant to be a constant, and the asteroid never becomes bigger no matter how
    much mass is picked up. The density of the compacted dust isn't given and r is clearly meant
    to be a constant.

    still working on it.
     
  4. Oct 15, 2009 #3
    I don't understand the book answer at all. There's still an m in there, should that be [tex] m_0 [/tex]?

    [tex] m \frac {dv} {dt} + v \frac {dm}{dt} = 0 [/tex] doesn't get you any further than [tex] m v = m_0 v_0 [/tex]


    if the asteroid travels a distance x through the cloud it will have swept up a volume [tex] \pi r^2 x [/tex] with mass [tex] \pi \rho r^2 x [/tex]

    let [tex] \pi r^2 \rho = C [/tex]

    [tex] m = m_0 + C x [/tex]

    [tex] v = \frac {dx}{dt} = \frac {v_0 m_0} {m} = \frac {v_0 m_0} {m_0 + C x} [/tex]

    this differential equation gives

    [tex] m_0 v_0 t = (1/2) C x^2 + m_0 x [/tex] after separation of variables.

    solving the quadratic for x gives [tex] x = \frac {2 m_0} {C} (\sqrt {1 + \frac {2 c v_0 t}{m_0}}-1) [/tex]

    (the solution with the negative square root has x<0)

    substituting this in the equation for v gives

    [tex] v = \frac {v_0} { \sqrt { 1 + \frac {2 v_0 C t} {m_0}}} [/tex]
     
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