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Homework Help: Planetesimal Accretion

  1. Jun 3, 2014 #1
    Kinda stumped. The question gives three equations and I don't know what to do with them, because given the equations there's always more than one unknown. Any assistance/tips would be appreciated!

    If a planetesimal has a cross-sectional area of πR2 (where R is the planetesimal's radius) and is sweeping through a cloud of smaller particles of fixed size with a velocity V, the number of collisions per second will be:

    dn/dt = (πR2N)/m

    where ρN=the space density (kg/m3) of particles in the cloud and m=the mass of each particle.

    If each collision results in the target particles sticking to the planetesimal, the planetesimal will gain mass at a rate of

    dM/dt = πR2N

    where M=the planetesimal mass.

    The time to grow to radius R is

    t = (4R/V)/(ρPN)

    where ρP=the density of the planetesimal itself. (Assume that ρN and V stay constant as particles are swept up.)

    Assuming that a reasonable value for the density of accretable material in inner part of the early solar nebula is ρN = 10-7 kg/m3, estimate the time to accrete a body of 1,000 km radius. Assume a reasonable ρP.
  2. jcsd
  3. Jun 3, 2014 #2


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    First a clarification: Are the ##P## and ##N## in ##\rho P## and ##\rho N## supposed to be subscripts? If not, what do they mean?

    Next, you have to make an attempt at solving the problem before people are allowed to help.
  4. Jun 3, 2014 #3
    Yes, the P and N are subscripts. ##\rho P## is the density of the planetesimal and ##\rho N## is the density of the nebula.

    I have tried to come up with something:

    I took the velocity of a circular orbit -- root(GM/r) -- and put in rho(nebula) x 4/3 pi R^3 for the mass.. I get a velocity of 4.838e-6 (which doesn't make any sense).

    plugging that into the first equation I get 130 billion years... which obviously isn't right.
  5. Jun 3, 2014 #4


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    Homework Helper

    The planetesimal, while travelling with speed v, sweeps a cylinder of cross sectional area equal to A=πR2. In dt time the high of the cylinder is vdt, and the planetesimal collides with every particle inside the cylinder.


    Attached Files:

  6. Jun 3, 2014 #5


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    In LaTeX you can use an underscore ('_') to indicate a subscript, thusly: ##\rho_N##, or if there is more than one character in the subscript, enclose them in curly braces: ##V_{out}##. Superscripts are done similarly, using the caret ('^').

    Yes, 10 times the life of the universe is probably not the right answer.

    The first thing to realize is that the size of the body will change, so you can't use a fixed radius. As the size increases, so will the rate of accretion. Circular orbits, or any other kind of orbit, has nothing to do with it.
  7. Jun 3, 2014 #6
    So from that I would take that the number of collisions per second per unit volume would be something like

    dn/dt = πR2vn

    would it not?
  8. Jun 3, 2014 #7

    So what I did was I took one of the givens:

    dM/dt = πR2N (EQ 1)

    and I took the equation for mass (which I differentiated):

    M = ρ (4/3) πR3
    dM/dt = 4πρR2 dR/dt (EQ 2)

    Equated EQ 1 and EQ 2 to get

    dR/dt = vρN/4ρP

    which is a constant. And that's where I got lost.
  9. Jun 4, 2014 #8
    But where do I get a velocity? That's where I'm completely stumped. There's no real body for an orbital velocity, so I don't know where to get v for m1v1 = m2v2...
  10. Jun 4, 2014 #9


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    R is the radius of the body and it travels with velocity V. I've just noticed that V is assumed constant in the OP. I deleted my previous post, you do not need to use conservation of momentum. But it is strange, ignoring the change of velocity.

    It does not matter along what orbit the planetesimal travels.

    Last edited: Jun 4, 2014
  11. Jun 4, 2014 #10


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    So what time is needed to grow the radius of the planetesimal from zero to R?

  12. Jun 4, 2014 #11
    I think I got it.

    I integrated both sides and got

    t = 4Rρ/vρN

    For a 1000km in diameter (R = 5.0e5 m) planetesimal, using the given ρN=1.0e-7 and ρP=3.5 kg/m3 (typical chondrite density), all I needed was v.

    Conservation of Energy dicates that the velocity of the impacts cannot greatly exceed the escape velocity, so I calculate Vesc, which is 22.11m/s.

    Plugging those values into the integrated formula, I get t = 1e5 years with 100% efficiency, and 1e7 years with 1% efficiency (since t and R are linearly related).

    Does that make sense?
  13. Jun 4, 2014 #12


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    Yes, it does.
  14. Jun 4, 2014 #13
    Perfect. Thanks all, much appreciated.
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