Two Limits

There are 2 limits we identify, the first is the sparse ambient limit in which we treat the ambient as a collection of parcels of gas that are attracted to the BE sphere only, and the 2nd is the dense ambient limit where we treat the ambient as a collapsing uniform sphere. Our approximations should explain the data in this table,

​http://astrobear.pas.rochester.edu/trac/astrobear/blog/erica10152013

Sparse Cases

For a planar HSE atmosphere, with differential equation,

\frac{dP}{dz} = -\rho g,

the solution is an exponential with given scale height,

H=\frac{K_B T}{m_H g}

where

g = GM/r^2

Assuming the force is given by,

F = \frac{G M_{BE}M_{atm}(t)}{(R_{BE} + H)^2}

and Matm is the mass of the accumulated atmosphere, gravitationally attracted to the BE sphere through freefall,

Matm= \frac{4}{3} \pi R_{ff}^3 \rho_{amb}

(where rff is found by inverting the tff for r for 2 point masses)we get the expression for P = F/A at the surface (after plugging in the variables),

P(R_{BE},t) = P_{ext} + P(t) = P_{ext} + \frac{8}{3 \pi^2} \frac{G^2 M_{BE}^2 \rho_{amb}} {R_{BE}^2}\frac{1}{(R_{BE}+H)^2}t^2

Using this expression gives the following results,

Crosses indicate when a sink formed for the corresponding run. The y axis is the pressure perturbation on the surface of the sphere, above the critical pressure. The vertical line is the average ff time of the sphere (the time it takes to collapse once the sphere goes unstable). The x axis is time.

We see that as eta = rho(Rbe)/rho(amb) increases (i.e. the ambient gets sparser), it takes longer for the pressure to grow at the surface of the sphere, inducing collapse.

Dense Case

The idea is that we have an initial uniform sphere (the ambient) that is collapsing because it is jeans unstable (this is the case for the Matched, 1/3, 1/10 cases). After some time t, a shell of the uniform sphere has fallen onto the BE sphere. We can assume it piles up into a thin shell (negligible thickness) at r = Rbe. Then we can calculate the pressure from this shell as

P = F/A = \frac{GM_{be}M_{shell}(t)}{4 \pi R_{be}^2}

To get the mass of the shell we take the difference between the masses of the original uniform sphere and the sphere at some later time t'.

Here is a graphic of the situation:

Since for uniform collapse we can numerically solve for R(t) and the uniform rho(t) of the collapsing sphere, we can calculate the mass of the uniform sphere outside of the BE sphere,

M(t)= \frac {4}{3} \pi (R^3(t) - R_{BE}^3) \rho(t)

Then the total mass that has fallen "onto" the BE sphere is given by,

M_{shell} = M_{init} - M(t)

where

M_{init} = \frac{4}{3}\pi (R_0^3-R_{BE}^3) \rho_0

Since the radius of the collapsing sphere decreases more rapidly than the density is increasing, M(t) is a decreasing function. At t=0, M(t) = Minit, and Mshell = 0. As time goes on, M(t) < Minit, thus, Mshell is a (slowly) increasing function.

I used mathematica to numerically solve the equation of motion for a uniform sphere as given in Carroll and Ostlie, and checked the solution was correct against a result they published in that book. For the 1/50 case in a box of r = 90, the function R(t) looks like,

rho(t) looks like

Here you see that the R(t) drops to zero at the freefall time. You see the density diverging at this point, given the unphysical condition r=0. The collapse accelerates in time, becomming faster and faster by the end. The density increases everywhere uniformly, and is only a function of time.

For all the plots I made, the simulation time was less than the freefall time of the ambient. More importantly perhaps, R(t) never fell inside of the BE sphere.

Here are the results,

The plot was made with mathematica, hence the lack of a legend. Sorry for small font.

Compare the 2 approximations side by side