| mathematical findings 
(extended) phonetic alphabet | 
miscellaneous | 
unicode
| |
sci-fi shenanigans
november 4th, 2023i wanted to do some research on space travel for (hard) sci-fi stories i could think of in the future. i'm not very good at physics by any means lol, but i tried to decypher some stuff i've seen and combine them into something sorta plausible.
the lorentz factor gives us the following relation : $$\frac{\text dt}{\text d\tau}=\frac1{\sqrt{1-v(\tau)^2/c^2}}=\sqrt{\frac{c^2}{c^2-(2g_0\tau)^2}}$$ with that formula, if we cruise at a speed of \((1-\varepsilon)\cdot c\), we have \({\text d\tau}/{\text dt}=\sqrt{\varepsilon}\).
we want \(|v'(\tau)|\le2g_0\) so it's somewhat bearable for the humans inside the shuttle. then, by integration, \(|v(\tau)-v(0)|\le2g_0\tau,\) though \(2g_0\tau=(1-\varepsilon)\cdot c\) iff \(\tau=(1-\varepsilon)\cdot c/(2g_0)=:\tau_{\rm end}\approx177~\text{days}.\) we can do the following table : $$\begin{array}{|l|l|l|} \hline&&\text{time it seems to take to}\\ &&\text{go from }(1-10^{-n})\cdot c\text{ to}\\ &&\left(1-10^{-(n+1)}\right)\cdot c,\text{with a}\\ 10^{-n}&1~\text{ly}/\cdots&\text{constant acceleration }2g_0\\ \hline10^{-1}&115.5~\text{days}&\approx15.9~\text{days}\\ 10^{-2}&36.52~\text{days}&\approx1.59~\text{days}\\ 10^{-3}&11.55~\text{days}&\approx3.82~\text{hours}\\ 10^{-4}&3.652~\text{days}&\approx22.9~\text{minutes}\\ 10^{-5}&1.155~\text{days}&\approx2.29~\text{minutes}\\ 10^{-6}&8.766~\text{hours}&\approx13.8~\text{seconds}\\ 10^{-7}&2.772~\text{hours}&\approx1.38~\text{seconds}\\ 10^{-8}&52.59~\text{minutes}&\approx0.14~\text{seconds}\\\hline \end{array}$$ my favorite opinion would be around the \(1~\text{ly}/\text{hour}\) mark, as it would become a bit risky to go too far below the seconds mark before we get yeeted to the end of the universe. this means we would require to strive for \((1-1.3\times10^{-8})\cdot c\), so that we have almost exactly \(1~\text{ly}/\,\text{hour}\).
\(\displaystyle|t(\tau_{\rm end})-t(0)|\le\int_0^{\tau_{\rm end}}\frac1{\sqrt{1-\frac{(2g_0\sigma)^2}{c^2}}}\,\text d\sigma\approx274~\text{days},\) instead of nearly \(177\) as how it's perceived by the astronauts themselves. we also have that \(|x(\tau_{\rm end})-x(0)|\le g_0\tau_{\rm end}^2\lt0.25~\text{ly}\), so to travel \(x~\text{ly}\) with \(x\ge0.5\), then it can take around \(\text{274 days}+(x-0.25)\cdot\text{1 year}\approx (x+\frac12)~\text{years}\) to travel there (and twice as much back), however it'll feel as though it only took \(177~\text{days}+(x-0.25)\cdot1\text{ hour}=(\frac1{8766}\cdot x+\frac12)~\text{years}\) from the astronauts' side, which is significantly less.
now for how much fuel there must be compared to the rest of the ship's and astronauts' combined mass, we can use tsiolkovsky's rocket equation, $$\frac{m_i}{m\!_f}=\exp\frac{v\!_f-v_i}{I_{\rm sp}\cdot g_0}$$ where \(m_i\) is the initial mass and \(m\!_f\) the final mass (once the fuel was depleted), \(v\!_f\) is final velocity and \(v_i\) is initial velocity, and \(I_{\rm sp}\) is specific impulse. using muon fusion catalyzing-based propulsion, we could theoretically reach \(I_{\rm sp}\) as high as \(c/g_0\approx30.56~\text{million seconds}\). let's say we get to \(20~\text{million seconds}\), and using \(v_i=0\) with \(v\!_f=(1-1.3\times10^{-8})\cdot c\), and we get that \(m_i/m\!_f\approx4.6\), meaning there needs to basically be \(360\%\) more fuel mass than the combined astronauts' and actual shuttle's mass.
|