From these plots (described in blog post),

I can read off the percent error for using various equations to calculate the accretion energy. To get a sense of the error, lets consider a few different initial kinetic energies, ke(r), — some less than and some greater than the kinetic energy which would have been acquired from freefall alone (i.e. GM/r).

First, I will consider cases where the kinetic energy was greater than GM/r. This corresponds to the lower plot. I can solve the following equations for r, given various ke(r),

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.01 (\frac{GM}{R})

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.3 (\frac{GM}{R})

The LHS of these equations give the (specific) accretion energy at the surface of the star, ke®, from gas that fell from a distance r away, i.e.,

ke(R)=ke(r)-\frac{G*M}{r}+\frac{G*M}{R}

Now, solving for r given various starting ke(r) and percents of error, we have,

ke(r)	Error (%)	Distance (pc)
10*freefall	30	6.8*10-7
10*freefall	.01	.00002
100*freefall	30	7.4*10-6
100*freefall	.01	.0002
1000*freefall	30	7.5*10-5
1000*freefall	.01	.002

This shows that you get the greatest error closer into the surface of the star, and even when you are going very fast starting from a distance r = .002 you still only have a .01% error. This means that for any cells larger than this r, wouldn't expect great deviation, even in the extreme 1000*freefall case.

Now, what if a gas parcel started from rest, a distance r away from the star surface? Now we are solving,

-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})

-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})

for a .01 and 30% error. This gives the following distances,

ke(r)	Error (%)	Distance (pc)
0	.01	2.2*10-6
0	30	7.5*10-8

Again, the closer into the star we get, the worse the approximation gets — but that a cell size of r~ 2.2*10^-6 shouldn't produce much deviation.

Lastly, what if the parcel was moving slower than freefall? Now ke(r) will be a fraction of the freefall energy in the table below. In particular, what if ke(r) was ½, 1/5, 1/10 the freefall kinetic energy… at what distance, r, would we see a 0.01% error? What about a 30% error?

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})

ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})

ke(r)	Error (%)	Distance (pc)
½*freefall	.01	1.1*10-6
½*freefall	30	3.8*10-8
1/5*freefall	.01	1.8*10-6
1/5*freefall	30	6*10-8
1/10*freefall	.01	2*10-6
1/10*freefall	30	6.8*10-8

This all shows us that the majority of energy gains from a stellar source occurs very close to the source, and thus, using the equation of accretion luminosity for a particle starting from rest at infinity is a fine approximation.