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.

  • $\begingroup$ This is not enough to answer the question. See the rocket equation: even in Newtonian mechanics, the total amount of velocity change (i.e. integral of acceleration, aka delta-v) is dependent not only on the amount of fuel but on the mass of the rocket without fuel. $\endgroup$ – JiK Oct 3 '16 at 21:51
  • $\begingroup$ For example a $1000 \text{ kg}$ rocket plus $100 \text{ kg}$ fuel would get 52 % of the delta-v the same rocket would have with $200 \text{ kg}$ of fuel. On the other hand, the same rocket with $1000 \text{ kg}$ of fuel it would get 63 % of the delta-v the same rocket would have with $2000 \text{ kg}$ of fuel. And a $1\text{ kg}$ rocket with $1000 \text{ kg}$ of fuel would get 90 % of the delta-v the same rocket would have with $2000\text{ kg}$ of fuel. $\endgroup$ – JiK Oct 3 '16 at 21:52
  • $\begingroup$ You probably need the relativistic rocket equation to answer this. $\endgroup$ – freecharly Oct 4 '16 at 2:47

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!

| cite | improve this answer | |
  • $\begingroup$ Small mistakes that are probably too small for a single edit so I'll leave them here for possible future edits: 1. Tsiolkovsky is Tsiolkovsky, not Tsoilkovsky. 2. After "a proper acceleration that is the same as in the Newtonian case", there should be a minus sign. 3. In the end, it is actually of course the first acceleration ($0.20c$ to $0.25c$) that requires $v_e$ to be huge because the ratio of masses can't be more than $2$. 4. In general, if using $v_e$ is too simple a model, then one can use the parameter $I_\text{sp}$ instead, which doesn't even need to be same for relativistic case. $\endgroup$ – JiK Oct 7 '16 at 19:19

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.

| cite | improve this answer | |
  • 2
    $\begingroup$ I think the OP means to neglect Earth's gravity - this is a simple relativistic version of the Tsiolkovsky problem - so probably only point 2 applies here $\endgroup$ – Selene Routley Oct 3 '16 at 20:52
  • 2
    $\begingroup$ Seriously. Compare .2c to Earth's pitiful little escape velocity. They simply aren't in the same league. $\endgroup$ – dmckee --- ex-moderator kitten Oct 3 '16 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.