Having a kinetic energy (per unit mass) at r that is higher than would have been acquired from freefall alone (which would just be GM/r), i.e. solving the equations:

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})

again where the specific kinetic energy at the surface of the star ® after having fallen from a distance r away from the center of the star is given by,

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

ke(r)	Error (%)	Distance (pc)
10*freefall	30	6.8x10-7
10*freefall	.01	.00002
100*freefall	30	7.4x10-6
100*freefall	.01	.0002
1000*freefall	30	7.5x10-5
1000*freefall	.01	.002

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})

ke(r)	Error (%)	Distance (pc)
0	.01	2.2x10-6
0	30	7.5x10-8

Lastly, what if the parcel was moving, however, it 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 I 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})