Does Einstein's geodesic equation fail close to the speed of light?

Newton's first law is simply a generalized geodesic equation in curved spacetime and/or in FLAT SPACETIME IN AN ACCELERATING/ROTATING NON-INERTIAL FRAME.

for a SINGLE ELEMENTARY TEST PARTICLE ONLY

DP(A)/ds = 0   Newton's First Law

D/ds = Levi-Civita covariant derivative with respect to particle's proper time. dtau = ds/c

one can add torsion of course

i.e.

M(A)DV(A)/ds + V(A)dM(A)/ds = 0

The  "rocket term" V(A)dM(A)/ds is I think what Jim Woodward is trying to harness? Somehow, he thinks M is controlled by the distant matter. I say it is controlled by the local Higgs field and the bound zero point energy of gluons as given in Quantum Chromodynamics in the Standard Particle Model.

DM/ds = dM/ds =/= 0 in Special Relativity even if no ejection of mass

Note however, that it's no longer a M-independent geodesic equation when dM(A)/ds =/= 0 - it's then a ROCKET EQUATION - really a form of the 2nd Law of Newton's particle mechanics where

F = V(A)dM(A)ds  is a velocity-dependent quasi DISSIPATIVE FORCE

it's more like friction than like the vxB Lorentz force.

More generally

M(A)DV(A)/ds + V(A)dM(A)/ds = F(EM, weak, strong)

or

DV(A)/ds  = F(EM, weak, strong)/M(A) - (V(A)/M(A))dM(A)/ds

(d/ds)logM = (1/M)dM/ds

Therefore, in general we have the "dissipative" form

DV(A)/ds  = F(EM, weak, strong)/M(A) - (V(A)d(logM)/ds

The question is, should M only be the rest mass m0 or the special relativistic mass m0(1 - (v/c)^2)^-12?

In SR

ds^2 = c^2dt^2 - dr^2

= c^2dt^2(1 - c^-2(dr/dt)^2)

dsxgamma = cdt

More generally

ds^2 = g00c^2dt^2 + g0idx^icdt + gijdx^idx^j

ds^2= c^2dt^2(g00 - g0ivi + gijvivj)

Therefore the SR gamma factor is replaced by the GR Gamma

Gamma = (g00 - g0ivi + gijvivj)^-1/2

Note, at a horizon g00 = 0, Gamma is still generally well defined.

also note that.

V = dx/ds

v = dx/dt