Planetesimal Accretion Time Estimation

[spinnaker]

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 = (πR²Vρ_N)/m

where ρ_N=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 = πR²Vρ_N

where M=the planetesimal mass.

The time to grow to radius R is

t = (4R/V)/(ρ_P/ρ_N)

where ρ_P=the density of the planetesimal itself. (Assume that ρ_N 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 ρ_N = 10^-7 kg/m³, estimate the time to accrete a body of 1,000 km radius. Assume a reasonable ρ_P.

 

[tms]

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.

 

[spinnaker]

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.

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.

 

[ehild]

The planetesimal, while traveling 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.


ehild

 

[tms]

Yes, the P and N are subscripts. $\rho P$ is the density of the planetesimal and $\rho N$ is the density of the nebula.

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 ('^').

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.

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.

 

[spinnaker]

The planetesimal, while traveling 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.


ehild

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?

 

[spinnaker]

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.

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

dM/dt = πR²vρ_N (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 = vρ_N/4ρ_P

which is a constant. And that's where I got lost.

 

[spinnaker]

While growing, the planetesimal will decelerate, as momentum is conserved.

ehild

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...

 

[ehild]

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.


ehild

 

[ehild]

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

dM/dt = πR²vρ_N (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 = vρ_N/4ρ_P

which is a constant. And that's where I got lost.

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


ehild

 

[spinnaker]

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


ehild

I think I got it.

I integrated both sides and got

t = 4Rρ/vρ_N

For a 1000km in diameter (R = 5.0e5 m) planetesimal, using the given ρ_N=1.0e-7 and ρ_P=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?

 

[tms]

Yes, it does.

 

[spinnaker]

Yes, it does.

Perfect. Thanks all, much appreciated.