There's an important thing to consider which is handled differently by the two approaches (ie use gravity to accelerate something or a rocket to accelerate something). The link to that rocket page talks about fuel, as a rocket must be powered by something and it expands this to accelerate. To ignore the issue of fuel would be to be ignoring a physically relevant thing. But what's the gravity version? A constant gravitational field is obtained in the limit of an infinitely massive object infinitely far away and in doing this you actually throw out energy conservation. For an object a finite distance away there would be a back reaction and it would accelerate towards the rocket too. The energy lost from gravitational potential is gained in kinetic. So immediately it can be seen that if you're to consider a physically consistent system you have to be careful, as the equivalence principle applies point by point, each time readjusting if you're to be exactly correct. I used energy conservation in my last post to demonstrate that the GR parent of the SR approximation conserves energy. Bert, you've jumped to a set of equations which don't conserve energy. If the full GR model conserves energy then if your SR approximation does not then you should not put too much trust in your work because its evident you've already failed to
exactly describe the actual predictions of relativity, you've made a simplification which wasn't justified or which you're applying outside of where it is justified.
This can be seen further without too much hassle. Let's get rid of the unphysical "infinite mass infinitely far away" and consider an object of mass M at a point in space which we'll take to be our coordinate origin. As a result rather than the unphysical "The gravitational acceleration is constant" we'll let it decrease by a factor $$\lambda^{2}$$from $$r=R$$ to $$r = R' = R + D$$. In the weak field limit (which we're interested in) we'll take the gravitational acceleration to be $$g(r) = -\frac{M}{r^{2}}\hat{r}$$, ie Newtonian expression in a radial direction. Thus we have $$\frac{g(R)}{g(R')} = \frac{(R+D)^{2}}{R^{2}} = \lambda^{2}$$. Reshuffling you get $$(\lambda^{2}-1)R^{2} - 2DR - D^{2}=0$$ and so (doing some cleaning up) you get $$R = \frac{D(1 \pm \lambda)}{\lambda^{2}-1} = \frac{D}{\lambda \mp 1}$$. Since $$\lambda^{2}>1$$ and R>0 we take the positive root, $$R = \frac{D}{\lambda - 1}$$. If g(r) is to vary slowly we take $$\lambda = 1+\epsilon$$ for some very small $$\epsilon$$ and so $$R = \frac{D}{\epsilon}$$.
Therefore if we choose how slowly we wish a field to vary over an interval we also choose then for weak fields slowly varying we determine R, it is not an independent variable. We can use this to rewrite the expression for g at r=R (ie the bit of the interval which is closes to the mass), $$g(R) = -\frac{M \epsilon^{2}}{D^{2}}\hat{r}$$. This allows us to then determine either g by picking M or M by picking g, $$M = \frac{|g|D^{2}}{\epsilon^{2}}$$. Its clear that, as previously said by Bert, you can find an M which allows you to have a region of required length in a gravitational field of stipulated strength which varies at as low a rate as required.
But this is not the end of the considerations. Unlike Newtonian gravity, where in you can say all you need to say in terms of the gravitational force or potential there's more to it in GR. The case relevant is that of 'surface gravity' at the event horizon. Given a Schwarzchild black hole with event horizon area A and mass M the surface gravity on the horizon at $$r = 2M = R_{EH}$$ is $$\kappa = \frac{1}{4M} \propto \frac{r}{A}$$. With $$\frac{r}{A}$$ having units of $$m^{-1}$$ the SI expression is obtained by multiplying by $$c^{2}$$ and the usual $$M \to \frac{GM}{c^{2}}$$ so $$\kappa = \frac{c^{4}}{GM}$$ at $$r = \frac{2GM}{c^{2}} = R_{EH}$$. $$\kappa$$ is monotonic decreasing in r and so $$\kappa(r)<\kappa(R_{EH})$$ for all $$r>R_{EH}$$.
This puts a bound on the values the parameters can take. Throughout we assume a weak field, as in Newtonian physics that's enough. But in relativity you can have a weak field and yet be inside an event horizon, as can be seen by $$\kappa = \frac{1}{4M}$$, the larger the black hole the
weaker its surface gravity at the event horizon. Given any g there's an $$M_{0}$$ such that all $$M>M_{0}$$ there is a region
inside the event horizon with that weak a gravitational acceleration. This is the difference between curvature (which defines the EH) and the varying of curvature, which defines $$\kappa$$ (I can provide links to more detailed derivations of this). We require R to be above the event horizon, as its clear you cannot have your $$r \in [R,R+D]$$ interval straddling the horizon as the projectile starts inside the event horizon and cannot be projected up. More formally there are no stationary frames inside the horizon, even when considering point by point changes, ie SR is
not a good approximation.
$$R = \frac{D}{\epsilon}$$ must be above the event horizon $$R_{EH} = 2M = \frac{2|g|D^{2}}{\epsilon^{2}}$$ and so $$\frac{D}{\epsilon} > \frac{2|g|D^{2}}{\epsilon^{2}}$$ and so $$|g| < \frac{\epsilon}{2D}$$. The same result is obtained if you instead use the bound $$g(R) < \kappa(2M)$$. Putting in the factor of $$c^{2}$$ gives the dimensionally correct expression, $$|g| < \frac{2\epsilon c^{2}}{D}$$. Using your units of Earth gravity being 1.03 ly/y^2 and c=1ly/y this becomes, approximately, $$1 < \frac{2\epsilon}{D}$$, ie $$D < 2\epsilon$$. This recovers the trivial $$\epsilon \to 0$$ (ie $$\lambda \to 1$$ which implies $$g(R) = g(R+D))$$ of g=0, ie a gravitational field created by a point mass is only constant when that mass is massless!
This is a bound, in that if it is violated, ie D is too large, then the gravitational field has varied more than your stipulated level of accuracy. It is only approximate, its valid in a weak field limit (ie I could use the Newtonian expression directly), more complicated things happen if R gets too close to $$R_{EH}$$ so the schematic behaviour I've outlined won't happen in such a case but this is not a problem because if the gravitational field
isn't behaving as I just outlined then it only reinforced my point about how you can't use SR simplifications too near particular masses. This explains the graphs you have.
In your Newtonian ones a projectile moving at c is not treated any differently to that moving at 0.0001c or 9999c as Newtonian physics allows for any speed and only have the Galilean group symmetries. Hence they don't change when you increase projectile velocity other than a bit of axis scaling. This isn't the case for relativity. If the projectile moves at v<<c then it won't get very far before slowing down so D remains small. But project it at v~c then D will reach large values over enough time. Your SR picture for a fast projectile in
this involves time scales of order $$10^{0}$$ year, while your slow SR projectile is of order $$10^{-3}$$ days.
You have been working on the assumption you can dial M up as much as you like in order to make SR approximations valid. For a rocket you just accelerate slowly, there's nothing else to it. For the GR->SR approximation dialling up the mass eventually causes problems because the event horizon structure of black holes is non-trivial. For instance, if you double the radius of the Earth (keeping density $$\rho_{0}$$ the same) you get 8 times the amount of mass, ie $$M \propto Vol \propto \rho_{0}r^{3}$$. But for a black hole the mass-radius relationship is linear, $$R \propto M$$.
You've been assuming that all you need to do to get a valid SR description is to simply make the gravitational field vary slowly enough, the value of the acceleration isn't an issue. That's incorrect, as you can only drop the full GR description to SR if in a region g varies slowly
and that region is not near, on or in the event horizon. You're free to decide how slowly you wish g to vary and over what length interval it does that but you cannot then select an arbitrary g, it must be less (probably significantly so) than the surface gravity on the event horizon. By doubling g you double M and thus double the event horizon radius yet the location of your interval is unchanged, $$[\frac{D}{\epsilon},\frac{D}{\epsilon}+D]$$. A rocket doesn't suffer from this because you're not doing the accelerating via gravity, so no event horizon (though you would get s cosmic horizon).
I originally said to you that you'd blindly applied equations without considering the origin and the requirements of them. I never argued with your citation of the equivalence principle and I've demonstrated now that indeed this approximation is valid
if you're careful. But I've also demonstrated what happens if you're not careful and how obtained the implicit requirements for the approximation to be valid over any non-zero length interval (the equivalence principle is
exact at a point and an approximation over intervals) and these have squared up exactly with your results. As a result of providing this alternative explanation I have rendered your repeated "I can ignore GR with impunity and only stick with SR and it'll always be fine over any interval, speed and acceleration" excuse mute. As such if all you can do is repeat it again you'll fail to justify your case. Mind you I'd not be surprised if you make some excuse about having stopped reading after 4 lines, you seem to have a problem with your concentration (might explain how you never read a book on relativity, too busy dreaming up your magical land of make believe).
