Spacecraft acceleration and relativity I have a question regarding the acceleration of a spacecraft.
A spacecraft travels at 0.20c relative to Earth and uses exactly half of its fuel to accelerate to 0.25c.  Will it be able to accelerate from 0.25c to 0.30c using its remaining fuel?
I am thinking the answer is "no," because relative to Earth the spacecraft will appear to grow in mass due to the Lorentz factor, so acceleration will become more difficult and will take more than half of the fuel, but I am not sure.
 A: There are two effects in play: on the other hand, going from $0.25c$ to $0.30c$ requires more acceleration than going from $0.20c$ to $0.25c$, but on the other hand, the rocket's mass is lower on the second phase, so it achieves better acceleration. The answer will be: it depends.

To start, we need to know how rockets work on Newtonian mechanics. This is basically the derivation of Tsoilkovsky rocket equation. To keep things simple, I'm working in a 1D world where the rocket is accelerating in the positive direction.
The rocket works by pushing exhaust gases backwards. The average speed of these in the rocket's reference frame is $v_e$ which is (in a simple but useful model) a constant depending only on the engine and independent of the masses of the rocket and fuel remaining. When the rocket uses a mass $-dm$ (where $dm$ is negative, corresponding to change of mass of the rocket) of fuel in a short period $dt$, the momentum of the exhaust is $dm \cdot v_e$. This is also the change of the momentum of the rocket (but opposite sign). In particular, if the rocket's mass (including the fuel) is $m$, the change of its velocity is $dv = -dm \cdot v_e / m$. So we get a differential equation:
$$
 \frac{dv}{dt} = -\frac{dm}{dt} \frac{v_e}{m}
$$
Integrating this, when the rocket burns fuel so that its initial mass is $m_1$ and final mass is $m_2$, the rocket's velocity changes by
$$
 \Delta v = v_e \ln \frac{m_2}{m_1}.
$$

Now to special relativity! In the observer's reference frame, if the rocket's velocity is $v$, then there is a thing called proper velocity $w$ which is $w= \gamma v$, where $\gamma = [1-(v/c)^2]^{-1/2}$ is the Lorentz factor. This is useful because the rate of change of this, $dw/dt$, called proper acceleration, is precisely the acceleration experienced by the rocket!
When burning fuel, in the rocket's reference frame there are no relativistic effects (unless $v_e$ is ridiculously large) so the rocket experiences a proper acceleration that is the same as in the Newtonian case,
$$
\frac{dm}{d\tau} \frac{v_e}{m},
$$
where $\tau$ is time in the rocket's frame, related to $t$ by $dt = \gamma d \tau$. So we get
$$
\frac{dw}{dt} = -\frac{dm}{d\tau} \frac{v_e}{m} = -\frac{dm}{dt} \frac{\gamma v_e}{m}.
$$
On the other hand,
$$
\frac{dw}{dt} = \frac{dv}{dt} \gamma + \frac{d\gamma}{dt} v = \frac{dv}{dt} \gamma + \frac{dv}{dt} \frac{v^2}{c^2} \gamma^3.
$$
Finding $d\gamma/dt$ is straightforward but a bit long so I won't repeat it here.
Combining these, we get
$$
\frac{dv}{dt} \left( 1 + \frac{v^2}{c^2} \gamma^2 \right) = -\frac{dm}{dt} \frac{v_e}{m}.
$$
When the rocket's starting velocity is $v_1$, starting mass is $m_1$, final velocity is $v_2$ and final mass is $m_2$, integrating this yields
$$
c \left( \tanh^{-1} \frac{v_2}{c} - \tanh^{-1} \frac{v_1}{c} \right) = v_e \ln \frac{m_1}{m_2}.
$$

Now let's consider your problem. You have three parameters for the rocket: the efficiency of the rocket $v_e$, mass of the rocket with no fuel $m_\text{empty}$, and initial mass of the fuel, $m_\text{fuel}$. Actually, only $v_e$ and $m_\text{fuel}/m_\text{empty}$ matter, so you have in practice two free parameters. I'll leave you to continue from here.

Hmm. I tried some values just for fun. It seems that for the latter acceleration to be possible, $v_e$ should be above $0.03c$, which would require quite a rocket!
A: There are three factors at play here...
1) The further from the earth the spaceship gets, the less gravity works against it. The less fuel is required for acceleration.
2) The closer the spaceship gets to c, the more energy/fuel is required to accelerate the spaceship a given amount.
3) Burning off fuel equals less mass, which equals less energy required to change momentum.
The being said, you'd have to do the calculations to figure out the answer to your question.
