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).
Best Answer
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.