3 ms·
Here's one little trick to quickly estimate the fuel you need to get to 99.9% of the speed of light. You take the rocket equation and replace the velocity with
by credit_guy 4y ago
Here's one little trick to quickly estimate the fuel you need to get to 99.9% of the speed of light.
You take the rocket equation and replace the velocity with the "rapidity", which is the inverse hyperbolic tangent of the velocity expressed as a percentage of the speed of light. That sounds like a mouthful, but you can actually estimate this in your head.
The inverse hyperbolic tangent is 1/2 log((1+x)/(1-x)). If you plug in 0.999, you get 1/2 log( 1.999/0.001), which is indistinguishable from 1/2 log(2000). If you ever did rough estimates with logarithms, you know that you only need to remember 2 logarithms, log(2) and log(10). They are 0.7 and 2.3, so log(2000) is roughly 0.7 + 6.9 = 7.6, and half of that is 3.8.
You now apply the classical (non-relativistic) rocket equation as if the final velocity is 3.8 times the speed of light (which is 300000 km/s). That's about 1 million kilometers per second.
If you use the best chemical propellants, you can reach an exhaust velocity of about 5 km/s, so the ratio is about 200000. You take exponential of that, and that's how much fuel you need for each unit of cargo you want to ship. You need to use some multiple precision library, because with normal (double) precision floating point numbers, you will get an overflow if you try to do exp(200000).
But let's say you want to use the VASIMIR engine, that reaches (theoretical) exhaust velocities of 100 km/s. Then your ratio will only be 10000. Exp(10000) is still out of reach of normal numerical libraries.
The even more theoretical fragment fission rocket engine reaches exhaust velocities of 10000 km/s. Now we get Exp(100) which is about 3 followed by 43 zeros. If you want to accelerate a space capsule weighing one ton to 99.9% the speed of light, you need that many tons of fuel (assuming a 100% efficient fragment fission rocket). That's a bit unfortunate, because it exceeds by a few orders of magnitude the mass of all the stars and planets in the galaxy.