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I was doing some back-of-a-postage-stamp calculations on high-isp interstellar spacecraft earlier and realised something. If I design a spaceship which according to the rocket equation has a very high Delta-V, close or even beyond the speed of light (in my calculations it was 6.7 million km/s or 22c), how do you calculate what speed that spacecraft can actually accelerate to? Obviously a spacecraft with a Delta-v of 6.7 million km/s cannot actually accelerate to that speed (with it being faster-than-light and all), but how fast could it get?

For those wondering, here’s my math on the example craft I referenced:

Assuming photon drive powered by 100% efficient antimatter reactions, 1 exawatt of power to the drive = 3 million kilonewtons of thrust or 337,000 tons thrust. This requires 1.7 million tons of antimatter (& 1.7 million tons of matter, 3.4 million total), and can fire for 10 years.

Assuming mass fraction of 90%

TWR of ~0.08

Delta V of ~6.7 million km/s (22c)

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  • $\begingroup$ If your calculations get you faster than c, then you are doing things very wrong. No matter what you do, your relativistic mass (among other things) increases without limit, so your delta-V won't matter. $\endgroup$ – Carl Witthoft Apr 2 at 14:39
  • $\begingroup$ @CarlWitthoft I think the OP has already made that point clear in the question: "Obviously a spacecraft with a Delta-v of 6.7 million km/s cannot actually accelerate to that speed..." $\endgroup$ – uhoh Apr 2 at 14:58
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    $\begingroup$ Have a look at the answer(s) to Relativistic Rocket Equation and see if that's what you need. There is also Is there a general relativistic relativistic rocket equation? but I think GR is outside the scope of your question; you're just looking for the special relativity effects. $\endgroup$ – uhoh Apr 2 at 15:00
  • $\begingroup$ But see this comment; now I am not sure that that answer is correct. $\endgroup$ – uhoh Apr 3 at 1:40
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In the case where the spacecraft can achieve relativistic speeds and no gravity is involved, $\Delta v$ is actually $c$ times the maximal change of rapidity which can be achieved. So the maximal speed the spacecraft can accelerate to, starting from zero, is $c \tanh \frac{\Delta v}{c}$. For example, if the maximal rapidity is $22$, then the maximal speed is $c \tanh 22\approx 0.9999999999999999998c$.

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  • $\begingroup$ This is interesting! Is $\Delta v$ here just the traditional $v_e \log(m_i/m_f)$? Would this help me with my confusing here and is $\gamma$ needed or not? $\endgroup$ – uhoh Apr 3 at 10:33
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    $\begingroup$ @uhoh Here, in general, $\Delta v$ is $T_s \ln(m_i/m_f)$, where $T_s$ is the total thrust per unit mass spent. In the case $v_e << c$, $T_s = v_e$, so $\Delta v = v_e \ln(m_i/m_f)$. In the case of relativistic $v_e<c$, I think that $\Delta v$ is still $v_e \ln(m_i/m_f)$, see my comments to the answer you linked to. $\endgroup$ – Litho Apr 3 at 12:21
  • $\begingroup$ Thanks for looking into these! $\endgroup$ – uhoh Apr 3 at 12:31

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