I am not to familiar with rocket equations but I know that if you combine special relativity with classical mechanics you get "c" as a speed limit. Basically the closer you get to "c" the more you increase in relativistic momenta without increasing so much in velocity.
Now in general relativity there are two other effects that I can think of. We consider the case where the rocket is moving in a spherically symmetric gravitational field.
- The speed of light is anisotropic and varies with altitude. In the Scharzschild solution in Schwarzschild coordinates in the radial direction in coordinate time you have $v_{light}=c(1-\frac{2GM}{rc^2})$ and in the direction transverse to the radial direction you have $v_{light}=c\sqrt{1-\frac{2GM}{rc^2}}$. The fact that the speed limit varies in this way should affect the rocket equation somehow.
- The gravitational time dilation/gravitational red shift I think that in coordinate time the gravitationl acceleration for an object standing on the face of a planet such as the Earth is the same as classically, $\frac{dv}{dt}=-\frac{GM}{r^2}$. However in proper time you have $\frac{dv}{dt}=-\frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$. I think this is correct. Equivalently the energy of the fuel in the rocket will be "redshifted" so the rockets will basically not contains as much energy as classically. This might somehow affect the rocket equation. This is sometimes a confusing subject.
Question: Is there a general relativistic, relativistic rocket equation and what does it look like?
I guess effects such as those mentioned might be to miniscule to bother with in the solar system but at least you should be able to prove that you can disregard them.
Edit: While not exactly a "rocket-equation" a common way to include effects of acceleration due to relativity is to use the post-Newtonian expansion. I wrote down the 3PN-accelerations for the most simple case of one test body in a spherical field on the physics site. At the 1PN level the acceleration becomes:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$
This is an expression that JPL always uses for orbit computation. When you do not have only one massive body of spherical symmetry there are more terms. For a rocket standing still, $v=0$, on the face of the earth you get:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}\right)\hat{r} $$
and for a rocket travelling radially upwards you get:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}-3\frac{v^2}{c^2}\right)\hat{r} $$
Question: How do these two extra terms make it harder/easier for a rocket to escape the gravitational field of the Earth?
