I would like to know if there is some formula / graph which would provide / show the efficiency of a certain type of propeller in space. Specifically, I'm interested in the acceleration attainable at certain speeds.

I'm writing a science fiction book and I'm trying to make it as correct as possible, fact wise. The propeller I'm talking about is the ion drive

Now, my book takes place in a world where fusion power is finally ours.

So, please, let us assume that we have unlimited energy so you could power a dozen huge ion drives non stop. OK, there's the question of argon/xenon fuel, let's assume we have 1 year of that.

So... the question is... what speed could you reach?

If a continuous acceleration of $10\frac{\mathrm{m}}{\mathrm{s}}$ is applied (I put that number because it would also constitute an advantage for my crew - living in Earth's gravity), that would mean that a ship would reach the speed of light in just 347 DAYS

But I know that's impossible because the EFFICIENCY of the ion drive would DECREASE as the ship's speed would approach the exhaust speed of the drive's "nozzle" (well, it doesn't have a nozzle per-se, as you can see in Wiki, but anyway...)

Please do not fear to elaborate on top of my question. Let's suppose for example that maybe the ion drives of the future have a much higher thrust/efficiency/nozzle exhaust speed.

This isn't only about currently POSSIBLE facts but also about THEORETICAL limitations which might be overcome in the future (such as fusion energy).

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    $\begingroup$ The efficiency of the drive cannot decrease as you increase speed - that would violate the principle of relativity. The reason that you cannot accelerate to the speed of light from someone perspective is that your ship becomes heavier and heavier as it approaches the speed of light and so harder and harder to acceelerate. $\endgroup$ – BebopButUnsteady Jul 17 '11 at 2:21
  • $\begingroup$ Ah. Right. The mass factor. Forgot about that ::- D. Sorry. So, if for example acceleration is 10 m/s at speed = 0, it would drop to 3.33 m/s at speed 100.000 Km/s, right? $\endgroup$ – Axonn Jul 17 '11 at 2:28
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    $\begingroup$ Keep in mind that the acceleration depends on the motion of the observer measuring it. Someone on the ship would measure the acceleration to be constant at $10 \frac{\mathrm{m}}{\mathrm{s}^2}$, but an external observer with respect to whom the ship is moving at $10^5\frac{\mathrm{km}}{\mathrm{s}}$ would measure $8.38\frac{\mathrm{m}}{\mathrm{s}^2}$, if my calculations are correct. The equation is $a\to a/\gamma^3$, where $\gamma = 1/\sqrt{1 - v^2/c^2}$. $\endgroup$ – David Z Jul 17 '11 at 3:50
  • $\begingroup$ Would the external observer matter? All I want is get a ship from Earth to 20 light years away in about 90 years. For this task, attaining a speed of 0.3c and maintaining it for 60 years and then slowing back to 0 would do it, right? $\endgroup$ – Axonn Jul 17 '11 at 10:58
  • $\begingroup$ 0.3c with respect to who? That's why the observer matters. $\endgroup$ – David Z Jul 17 '11 at 14:07

The principle of relativity says that we can analyze a physical situation from any reference frame, as long as it moves with some constant speed relative to a known inertial frame. Thus, the ion drive does not find it more difficult to accelerate the ship when the ship is "going fast" because the ion drive cannot physically distinguish going fast from going slow.

However, if the ion drive is going fast in the reference frame of Earth, then when the ion drive burns, say 1 kg of fuel, it picks up less speed in the Earth frame than it does in the rocket frame due to the relativistic velocity addition law.

That velocity addition law is just the angle-addition law for the hyperbolic tangent. So, suppose the ship accelerates by shooting individual ions out the back. Each time it does this, it accelerates the same amount from its own comoving frame. Then from an Earth frame, the $\textrm{arctanh}$ of the rocket's speed increases by the same amount each time.

If, as a function of the proper time $\tau$ experienced on the rocket, the acceleration of the rocket is $a(\tau)$ in a comoving frame, there is a quantity called the rapidity of the rocket which increases the way velocity does in Newtonian mechanics.

The rapidity $\theta$ will be $\theta(\tau) = \int_0^\tau a(\tau) d\tau$, and the velocity is then $v(\tau) = \tanh\theta$. Specifically, if $a = g$, the velocity is

$$v(\tau) = \tanh(g\tau)$$

When one year of time has passed on the rocket, its velocity relative to Earth will be $\tanh(1.05) = 0.78$, or 78% the speed of light. The limit of the $\tanh$ function is one as $\tau \to \infty$, so the rocket never gets to light speed.

A more important limiting factor is the fuel. If the rocket carries all its fuel, then once it burns through it all, it can't go any more. Fusion isn't a way around this because by $E=mc^2$ there is a limited energy you can get from a given mass of fuel.

If a fraction $f$ of the rocket is fuel, when the fuel is all burned, the momentum of the rocket will be $\gamma m (1-f) \beta$, with $m$ the original mass. The energy of the rocket is $\gamma m (1-f)$. Similar relations hold for the fuel. The conservation of momentum and energy give

$$m = \gamma m (1-f) + E_{fuel}$$

$$0 = \gamma m \beta (1-f) + p_{fuel}$$

$E_{fuel}$ and $p_{fuel}$ are the energy and momentum of the fuel after burning. Solving for $\beta$ gives

$$\beta = \frac{-p_{fuel}}{m - E_{fuel}}$$

The minus sign shows that the fuel and rocket go opposite directions. To maximize $\beta$, we want to make $p_{fuel}$ as large as possible subject to a fixed $E_{fuel}$. This means that we want the speed of the fuel as high as possible, so assume the fuel is massless with $\beta_{fuel} = 1$ and $p_{fuel} = -E_{fuel}$. Plugging this into the previous equations and doing some algebra, I got

$$\beta = \frac{1 - (1-f)^2}{1 + (1-f)^2}$$

Even if half the rocket's original mass were fuel, it would only get to 3/5 the speed of light.

  • $\begingroup$ Not to mention the rocket also has to slow back down to 0 speed. So that would lower your 3/5 to 0.3c. You said that when one year passed, the velocity would be 0.78c. I can't "allow" that speed because it would damage the ship (due to space particles bombardment). So for me, a 0.3c "would be great". So... bottom line is that there is no physical law which would prevent my ship from attaining 0.3c in just ~6 months if I would have a constant acceleration of 10 m/s, right? Am I missing anything? $\endgroup$ – Axonn Jul 17 '11 at 10:49
  • $\begingroup$ @Axonn No, you cannot do the math as naive arithmetic; you have to use the actual formulas. Requiring the ship to decelerate back to zero would not cut the maximum speed in half. Instead, you would go back to the formula for $\beta$ and change $f$ into $1/2(1+f)$, which comes from noting that the fraction of the ship remaining as non-fuel is squared on an acceleration/deceleration trip. Also, in 6 months at 1g you would get to about half the speed of light. $\endgroup$ – Mark Eichenlaub Jul 17 '11 at 10:59
  • $\begingroup$ Sorry ::- ). I know it's not that simple. But it's been a very long time since I've been through physics classes and I'm looking to fact-proof some things without a lot of real knowledge. I still try to do it because I'm attempting to write about a scenario which would allow humans to travel to a star 20 light years away in less than 100 years, sometimes this century, perhaps at the end of it. Even if this is a book, I'm trying to keep it 'real'. Anyway, bottom line, it could be done if you could have an acceleration of even 5 m/s for a few months, pause for about80years and then slowing to 0. $\endgroup$ – Axonn Jul 17 '11 at 11:10
  • $\begingroup$ Anything? ::- ). $\endgroup$ – Axonn Jul 22 '11 at 10:42
  • $\begingroup$ @Axonn what do you want to know? I already answered the question in detail. I believe you should be able to plug in numbers for yourself. $\endgroup$ – Mark Eichenlaub Jul 22 '11 at 11:26

Well, the structure of spacetime (i.e. special relativity) imposes an overall speed limit of the speed of light, $c = 299792458\frac{\mathrm{m}}{\mathrm{s}}$. So whatever maximum speed you do get up to, it's going to be less than that. You've done the calculation to show that, non-relativistically, the ship would pass the speed of light in less than a year; even without doing any calculations, that tells that if the ship accelerates for that long, it's going to wind up at a significant fraction of light speed. So very roughly, you could just say the maximum speed is $c$.

But suppose you want a more precise figure than that. You can calculate it, but first you have to clearly define what you're calculating. (This is what always trips people up when working with relativity) So here's the situation: suppose the ship starts out on a planet. Initially, the planet and ship are at rest with respect to each other, and let's assume that the planet does not accelerate at all (essentially, it remains at rest forever). It will define the "lab frame." That is, we're going to calculate the speed of the ship as observed by someone at rest with respect to the planet, but after 1 year as measured by people on the ship.

This is actually a well-studied problem, called the "relativistic rocket." Since I'm a little short on time, I'm just going to refer to a solution posted elsewhere, although if you like I would be happy to come back to this later and edit in an explanation of where this formula comes from. The formula we need is

$$v = c \tanh(a \tau/c)$$

where $a$ is the acceleration as measured in the ship's frame and $\tau$ is the proper time (i.e. time also measured in the ship's frame). Plugging in your numbers, $a = 10\frac{\mathrm{m}}{\mathrm{s}^2}$ and $\tau = 1\text{ year}$, we get

$$v = 0.783c = 2.35\times 10^5\frac{\mathrm{km}}{\mathrm{s}}$$

In your scenario, once your ship reaches this speed, it would run out of fuel and would be stuck at that speed.

Note that, given the way you've defined the problem, the engine parameters are completely irrelevant. In reality, you would measure the amount of fuel by its mass, not by how long it will last, and in that case you would need the engine efficiency, thrust, etc. as well as the mass of the spaceship to figure out how long a given amount of fuel would last.


The problem I've encountered is that ion thrusters produce low thrust, nothing like one gravity. This means it would take much longer time to reach even nearby star systems. This is a quote from Wikipedia:

"Electric thrusters tend to produce low thrust, which results in low acceleration. Using $g=9.81\,\frac{m}{s^2}$; $F = m\cdot a$ or $a = \frac{F}{m}$. An NSTAR thruster producing a thrust (=force) of $92\,mN$[7] will accelerate a satellite with a mass of $1000\,$kg by $0.092 / 1000 = 0.000092 \frac{m}{s^2}$ (or the low fraction of $9.38\cdot 10^{-6} g$)."

But as for possibilities, future engines might be much more powerful.

  • $\begingroup$ Well, this is true, but for long journeys the top speed is usually given by the mass-fraction of fuel available (because you generally can not carry enough fuel to burn all the way to a space-patrol style flip-turn at the middle of the trip and that is, in any case, an inefficient way to use your fuel) and the specific impulse of your drive. Ion drives have very high specific impulse by comparison to chemical rockets. $\endgroup$ – dmckee Mar 14 '13 at 21:06

If you're writing a sci-fi book, I would pick something more sci-fi than the ion drive. The ion drive is a low-thrust/high-specific impulse kind of engine. In layman terms, it goes really slow, but lasts for a long time. The kind of thrust that an ion drive can produce is around 250 mN maximum. That's micro-Newtons, or in other words 0.25 Newtons. So, by taking the simplified Force=mass*acceleration equation, to get 10 m/s^2 of acceleration from an ion drive that produces .25 N of thrust, your spacecraft would have to be 0.025 kg. Unfortunately, the ion thruster itself is about 7 kg. I hope this helps.


Alas, the conundrum of mixing a salad of fiction and physics.. Equations aside, Partial bombardment can be handled by reconfiguring your ion drive to produce an Ionic wake shield at the prow of your ship, (read the manual). If you cleverly conceive heavy mass objects in your path of travel, you can use gravitational or atmospheric/aero breaking to slow down.-or fling out a tether to change your center of gravity and combine with specific impulse for further braking effect, (also dramatic and cool for passengers). In a real pinch, sacrifice the hardware and beam down to your destination as you fly by. - NOTE - You will again need to retool your ion drive to created a pleated harmonic wave to slow the beamed partials relative to the destination point, but that is again in the manual.

Finally on fuel- It is most likely common knowledge in your book that the bright scientists who cracked fusion power, also recorded the fissionable material decay property which has effected observations of the perturbation theory of quantum field theory since the fission problem was conquered. As a stunning offshoot of the fusion technology, hitherto unknown partials (you pick the mass), are created and can be collected from the ether during fission, which in essence, resupplies the fuel mass of the ship to a sufficient fraction to negate losses from drag, and inefficiency.

I like the ION drive,..slow, but at least you'll have time to read.


As the Xenon gases used by an ion drive propulsion system exit the rear of the space craft at 1/3 the speed of light, and the forward motion generated by the thrust that is being generated out of the rear of the vehicle can never be more than 1/3 the speed at which the gases are exiting the rear of the vehicle, the absolute top speed that can ever be achieved by an ion drive propulsion system under it's own power is 1/9 the speed of light or approximately 20,698 miles per second. Alas, because the rate of acceleration of an ion drive propulsion system in a vacuum is only 2 miles per hour for every hour it will take you approximately 4,250 years to reach your top speed.

  • $\begingroup$ That is incorrect. As long as you spit fuel in a frictionless vacuum, you will continue accelerating. The top speed only depends on the total amount of fuel. $\endgroup$ – Davidmh May 13 '14 at 22:56

protected by Qmechanic May 13 '14 at 22:28

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