Free fall time of collapsing non-rotating star in GR according to a distant observer Suppose a star with a spherically symmetric mass distribution, mass $2\cdot M_{sun}$ and initial radius $R$, undergoes a gravitational collapse after all gas pressure has vanished. The final radius will be about 15 km, typical of a neutron star. A Newtonian derivation of the free fall time gives $$t_{ff}=\tfrac{\pi}{2}\sqrt{\frac{R^3}{2\cdot G\cdot M}}=\sqrt{\frac{3\cdot\pi}{32\cdot G\cdot\rho}}.$$
Since we end up with a compact object, general relativity might lead to a different free fall time according to a distant observer.
Question: what is the expression for the free fall time in GR in this case?
 A: An object in "freefall" spends most of its trajectory at large radii (since acceleration ensures that the velocity increases more than linearly with time), so the collapsing proto-neutron star will spend most of the collapse period close to the size of its original electron-degenerate progenitor. For a hot, degenerate iron core, this is of order $R> 1000$ km.
The dimensionless ratio $GM/Rc^2$ will then be $<0.003$ for most of the collapse, so the influence of GR on the freefall timescale will be rather negligible. Even when the neutron star is close to its final radius (and I think 10 km is too small for the collapse to be considered "free fall", because the separation of the neutrons will be $\sim 10^{-15}$m at this radius, which is comparable to the range of the strong nuclear force, and also because the neutrons will be highly degenerate and neutron degeneracy pressure would not be negligible.), the gravitational time dilation factor for freely falling object in the Schwarzschild metric, $(1 - r_s/r)^{-1}$, would only reach a factor of $\sim 2$. 
My intuition would therefore be that a Newtonian approximation for the free fall timescale is quite good here.
An analytic expression would follow an object on the surface of the (non-spinning) collapsing star. Assuming spherical symmetry, then by Birkhoff's theorem, the Schwarzschild metric can be used throughout.
The geodesic for an object freely falling from rest at $t_0, r_0$ is given by (e.g. e.g. see Chapter 25 in "Gravitation" by Misner, Thorner & Wheeler, 2017, Princeton University)
$$ \frac{c(t-t_0)}{r_s} = \ln \left| \frac{ (r_0/r_s -1)^{1/2} + \tan (\eta/2)}{(r_0/r_s -1)^{1/2} -\tan(\eta/2)}\right| + \left(\frac{r_0}{r_s}-1\right)^{1/2} \left( \eta + \frac{r_0}{2r_s}(\eta + \sin \eta)\right). \tag{1}$$
The "cycloid parameter" $\eta(r)$ is defined by
$$r = \frac{r_0}{2}(1 + \cos \eta)$$
If we assume that $r_0 \gg r$ and $r_0 \gg r_s$ and $t_0=0$, then $\eta \simeq \pi$, and the time to fall from $r_0$ to $r$ (your free fall time) is
$$ \tau \simeq \frac{\sqrt{r_s r_0}}{c} \left( \pi + \frac{\pi r_0}{2r_s}\right) \simeq \frac{\pi r_0^{3/2}}{2 r_s^{1/2} c}$$
Substituting $r_s = 2GM/c^2$, we do recover the Newtonian formula.
e.g. If $r_0 = 1000$ km, $R= 10$ km and $M = 2M_{\odot}$, the Newtonian formula gives a free fall time of 0.0680 s, but the GR formula above gives 0.0685 s - so longer by 0.7%.
