# How much anti-matter is needed for an anti-matter propelled rocket to reach $P$ % ($P\lt{100}$) of the speed of light (c)?

I think this is a tough one.
Imagine a rocket is produced which is propelled by a matter-anti-matter device. Its sole purpose is to reach high velocity. Which means the mass can be kept low, say $$m$$ kilograms. Let's assume it starts in an ideal, infinite, and flat spacetime. By ideal, I mean that there is no real matter, dark matter, or dark energy present.

The rocket's thrust is constant, so when the rocket approaches relativistic speed, the acceleration gets less. Let's assume further that the rocket's restmass stays constant. The mass of the matter and anti-matter (for example, electrons and positrons) is insignificant in comparison to the total restmass of the rocket. The initial velocity is zero and the initial acceleration $$a$$.

My question is simple: How much (kilogram) anti-matter is needed to reach a velocity of $$P$$% of the speed of light?
In other words, what does a general formula to calculate this for every end speed look like?

P.S I assume constant thrust (backward force) contrary to a constant acceleration. The answer of course has to be the same but it takes longer to reach the speed required.

• An antimatter rocket cannot reach a high percentage of the speed of light unless a significant percentage of it is made of matter and antimatter propellant. More generally, antimatter rockets are very speculative. Sep 11, 2020 at 4:32
• @ikrase But doesn't the ratio between the mass of the rocket and the mass of the matter-anti-matter stay the same? Sep 11, 2020 at 5:29
• For constant delta-V, yes. You will almost always have at least as much fuel as rocket, usually 3 to 10 times as much Sep 11, 2020 at 6:12
• @ikrase Antimatter-matter propulsion is 100% efficient. To reach the Moon you will need very little of this stuff. Say one gram (just a rough guess). Sep 11, 2020 at 6:16
• It's true that you need very little antimatter, but it is not 100 percent efficient (unless you mean 100 percent efficient at irradiating and heating up your engine) and you can get away with much less antimatter if you use some bulk propellant for such a short transfer. Sep 11, 2020 at 6:38

$$M/m = \gamma (1 + v/c) - 1$$
Here $$m$$ is the mass of the rocket, $$M$$ the mass of the fuel (matter/antimatter in equal quantities) $$v/c$$ is $$P/100$$ in your terms and $$\gamma = 1/\sqrt{1-v^2/c^2}$$
• Is it thät easy? I put $M=0$. Do I just have to put in a small value? Sep 10, 2020 at 12:29