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somewhere in a galaxy a long time ago I was informed that accelerating a spaceship at one gravity (Earth value) would result in near relativistic velocity after only one day (Earth value) - I was intrigued by this and wonder how much propellant (mass) would be required to achieve this velocity please ? assume a living quarters mass of 1000 kilograms.

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  • $\begingroup$ my calculation is otherwise - one g times 24 times 60 times 60 gives 846,720 m/s - nowhere near light speed ! $\endgroup$ – Brooks Jan 7 at 22:38
  • $\begingroup$ How much propellant mass for real engines, or for hypothetical future engines? $\endgroup$ – Russell Borogove Jan 8 at 0:10
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    $\begingroup$ @Brooks that's right, by multiplication only (and not using special relativity, and therefore incorrectly) it works out to almost exactly 1 year, not 1 day. $\endgroup$ – uhoh Jan 8 at 0:58
  • $\begingroup$ Yep, 1 year, it's a neat value for discussing relativistic effects - "a 1g torch-ship to Alpha Centauri, 4ly away, would take about 6 years to get there, but in perception of the crew, it would last 4 years." $\endgroup$ – SF. Jan 8 at 13:14
  • $\begingroup$ @Russell Borogove perhaps the most efficient engines now possible? $\endgroup$ – Brooks Jan 9 at 3:17
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The thing you are describing is usually studied under the term 'relativistic rocket'. Here is a good summary of the formulae you need. All of these can be derived using Special Relativity (it is a common myth that you need General Relativity to deal with acceleration: you don't, you need it to deal with gravity or equivalently non-flat spacetimes).

One important thing to bear in mind is that the proper time, $T$, experienced by people in the rocket becomes increasingly different than the time, $t$ experienced by an inertial (unaccelerated) observer watching the rocket zoom off into the distance, which I have called inertial time below.

Velocity

That being said, the velocity of the rocket as a function of $T$ – the time experienced by people in the rocket – is given by

$$v = c \tanh\left(\frac{a T}{c}\right)$$

If we assume that $a = g$, then plotting $v/c$ for two years of proper time looks like this:

v as a function of proper time T

You can also compute $v$ as a function of $t$: the time experienced by an observer stationary with respect to the rocket when it's launched:

$$v = \frac{a t}{\sqrt{1 + (at/c)^2}}$$

And we can plot this for two years of inertial time:

v as a function of inertial time

Fuel requirements with an ideal photon drive

So, what about fuel? Well, the best rocket we can build, even in theory, is a 'photon drive': something which converts mass to photons and squirts them out of the back of the rocket. Assume we can do this with an efficiency of 1: all our reaction mass gets converted to light and squirted out of the back of the rocket. We'll carry fuel $M$ and have a payload of $m$, and we can work out $M/m$:

$$\frac{M}{m} = \gamma (1 + v/c) -1$$

Where $\gamma = 1/\sqrt{1-v^2/c^2}$ is the Lorentz factor common in Special Relativity.

Or equivalently

$$\frac{M}{m} = e^{aT/c}-1$$

So, OK, we can now just plot $M/m$ as a function of $T$, assuming $a = g$ again:

M/m as a function of proper time

So: if we assume a propulsion system which is implausibly good, things are not that bad.

Fuel requirements with a more plausible drive

[I am less sure about what follows than what's above because I did some more-or-less back of the envelope maths which may have errors.]

A photon drive, especially one with an efficiency of 1, is not terribly plausible. Well, you can generalise the Tsiolkovsky rocket equation for relativistic rockets, and you get this:

$$\Delta v = c \tanh\left(\frac{v_e}{c}\ln\left(\frac{m + M}{m}\right)\right)$$

Here

  • $\Delta v$ is the change in velocity, which is simply the final velocity $v$ is the initial velocity is 0;
  • $v_e$ is the exhaust velocity;
  • $m$ is the payload mass;
  • $M$ is the fuel mass;
  • and on the Wikipedia page, $m_0 = m + M$, the total initial mass, & $m_1 = m$, the final mass.

So, well, we know (from above, and assuming initial velocity is 0) that $\Delta v = v = c \tanh\left(a T/c\right)$, and we can thus say that

$$c \tanh \left(\frac{a T}{c}\right) = c \tanh\left(\frac{v_e}{c}\ln\left(\frac{m + M}{m}\right)\right)$$

and hence

$$\frac{a T}{c} = \frac{v_e}{c}\ln\left(\frac{m + M}{m}\right)$$

And we can rearrange this to get an expression for $M/m$:

$$\frac{M}{m} = e^{\frac{a T}{v_e}} - 1$$

I initially thought this must be wrong as there is no $c$ anywhere, but I think it is in fact OK. In fact this equation is true for any rocket of course, although since it assumes $a$ is constant the engine needs to be throttled.

So, I had a look at what $v_e$ might be for a couple of plausible technologies:

  • for chemical rockets $v_e \approx 4400\mathrm{m/s}$;
  • for ion thrusters $v_e \approx 29000\mathrm{m/s}$.

(I think these numbers come from Wikipedia somewhere). So let's assume we run the rocket for a year, with $a = g$. From above we get $\Delta v = v \approx 0.77c$ (and the distance travelled in this time is about $0.56\,\mathrm{ly}$, using a formula not given here), and

  • for a chemical rocket $M/m \approx 1.4\times 10^{30546}$;
  • for an ion drive $M/m \approx 3.9\times 10^{4634}$.

The mass of our galaxy is around $2\times 10^{42}\,\mathrm{kg}$: the amount of fuel you would need for a payload of a tonne is stupidly more than the mass of the galaxy in both these cases: more by ratios which can't be expressed with machine floating-point numbers for a 64-bit machine. In fact I am reasonably sure that the mass of fuel you need is absurdly greater than the mass of the observable universe!

Here is a plot for $M/m$ for exhaust velocities up to $0.1c$, for a rocket accelerating for a year at $a = g$. Note both the log scale and the fact that the left-hand (low $v_e$) end is missing.

M/m for 1 year at 1g as a function of exhaust velocity

For an exhaust velocity of $0.1 c$ then $M/m \approx 30000$, which is sort-of approaching a reasonable value.

This is why these kinds of rockets are not practical with any technology we have.

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A gee for a day is 9.8 m/s^2 x 86400 seconds = 846720 m/s, which is only 0.0028 of the speed of light.

The most efficient engines now possible would be hydrogen-oxygen chemical rocket engines; electric thrusters are more mass-efficient but can't produce enough thrust to achieve 1 g acceleration.

Assuming an exhaust velocity of 4400 m/s, the rocket equation says you need a fueled-mass-to-dry-mass ratio of about $3.75 \times 10^{83}$ -- all the hydrogen in the universe could maybe get a single electron up to speed if burned in a chemical rocket.

If electrical thrusters could achieve that kind of thrust-to-weight ratio, things get significantly better. Assuming an ion engine with an exhaust velocity of 50000 m/s, the needed mass ratio is only 23 million or so! Your crew cabin may be only one ton but you're going to need engines and power to run the engines and so forth, but at least this approach implies a spaceship that can be comfortably built with the resources of a single planet. It's going to take you 1535 years to get to Proxima Centauri, though, and you aren't relativistic enough for it to feel significantly shorter. And you won't have enough fuel to slow down when you get there.

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    $\begingroup$ And note that these huge numbers are to achieve $1g$ for a day: my vastly more absurd answers are to do it for a year, which is sufficient to get you to some reasonable velocity. Both answers demonstrate how absurdly out of reach getting close to $c$ is with any technically plausible rocket. $\endgroup$ – tfb Jan 9 at 10:12
  • $\begingroup$ If we need all the hydrogen in the universe there would not be enough oxygen for the burn. There is much more hydrogen than oxygen in the universe. $\endgroup$ – Uwe Jan 13 at 11:12

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