Copyright © 2020 Henk Reints (Henk-Reints.nl)
File date: 2020-11-14 22:32 UTC

Accelerated (1st half) & decelerated (2nd half) travel through space

Symbols:
\(a \)\(=\) proper acceleration (force per mass) as observed in the starship
\(\tau \)\(=\) proper time as observed in the starship
\(x \)\(=\) coordinate distance as observed on Earth
\(t \)\(=\) coordinate time as observed on Earth
\(v_{\mathrm{eff}}\)\(=\) effective velocity \(=\) travelled coordinate distance divided by elapsed proper time
\(c \)\(=\) the speed of light
\(\quad t\left(\tau\right) = \frac{c}{a} \mathrm{sinh}\frac{a\,\tau}{c} \)
\(\quad x\left(\tau\right) = \frac{c^2}{a} \left( \mathrm{cosh}\frac{a\,\tau}{c} - 1 \right) \quad\therefore\quad \tau\left(x\right) = \frac{c}{a} \mathrm{arcosh}\left( 1 + \frac{a}{c^2} x \right) \)
\(\quad v_{\mathrm{eff}} = \frac{x}{\tau} \) which may very well exceed the speed of light!
Can you go faster than light? YES, you can! But only retrospectively from your own perspective. Nobody will ever see you exceed the speed of light. Neither can you ever see anything pass by faster than light, since the travelled path is (strongly) Lorentz contracted to you while you're speeding. But once you come to rest on arrival and look back at the then no longer contracted distance and you divide that by your own proper travel time, you may have effectively travelled WAY faster than light. From the stationary point of view, your time was largely dilated and progressed far slower than what you experienced yourself. In other words: Earth has aged far more than you have and all beings that were alive when you departed no longer are (see Polaris below). The tipping point is calculated as:
\( v_{\mathrm{eff}} = \frac{\mathrm{\Delta}x}{\mathrm{\Delta}\tau} = \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t \cdot \sqrt{1 - \frac{v^2}{c^2} } } = \frac{v}{\sqrt{1-\frac{v^2}{c^2} } } = c \)
\( \quad\therefore\quad v = c \cdot \sqrt{1-\frac{v^2}{c^2} } \quad\therefore\quad v^2 = c^2 - v^2 \quad\therefore\quad 2v^2 = c^2 \quad\therefore\quad v = \frac{c}{\sqrt{2}} \)
Hence:       \( v \gt \frac{c}{\sqrt{2}} \;\rightarrow\; v_{\mathrm{eff}} \gt c \)
With a constant acceleration equal to \(1\,g\) (Earth's gravity) you achieve that in roughly 1 year. The "only" problem is how to carry enough fuel to maintain such an acceleration. All fuel must also be accelerated, requiring even more fuel, which must be accelerated as well, requiring even more, etc.
description \(\mathrm{\Delta}x =\) distance \(a =\) acceleration \(\mathrm{\Delta}\tau =\) proper time \(\mathrm{\Delta}t =\) coordinate time \(v_{\mathrm{eff}}\) mean \(\beta\) \(1 - \beta\) \(\gamma\) max. \(\beta\) \(1 - \beta\) \(\gamma\) max. Ekin/kg
to the moon 384.4×103 km 9.80665 m/s2
= 1 \(g\)		 03:28:41.676 03:28:41.676
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (00:01.282)
+ 03:28:40.394		 30.7 km/s
= 1.024×10-4 \(c\)		 1.024×10-4 0.9999 1 4.096×10-4 0.99959 1 7.539 GJ
to the sun
= 1 astronomical unit		 149.6×106 km
= 1 au		 9.80665 m/s2
= 1 \(g\)		 2.859 dy 2.859 dy
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (08:19.004)
+ 2.853 dy		 605.6 km/s
= 2.02×10-3 \(c\)		 2.02×10-3 0.998 1 8.08×10-3 0.9919 1 2.934 TJ
outer edge of Kuiper belt 7.48×109 km
= 50 au		 9.80665 m/s2
= 1 \(g\)		 20.22 dy 20.22 dy
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (06:55:50.239)
+ 19.93 dy		 4282 km/s
= 1.428×10-2 \(c\)		 1.428×10-2 0.9857 1 5.705×10-2 0.943 1.002 146.6 TJ
estim. inner edge of Oort cloud 448.8×109 km
= 3000 au		 9.80665 m/s2
= 1 \(g\)		 5.134 mo 5.176 mo
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (17.33 dy)
+ 4.607 mo		 33.24×106 m/s
= 0.1109 \(c\)		 0.11 0.89 1.006 0.4068 0.5932 1.095 8.507 PJ
estim. outer edge of Oort cloud 4.488×1012 km
= 30×103 au		 9.80665 m/s2
= 1 \(g\)		 1.33 yr 1.436 yr
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (5.693 mo)
+ 11.54 mo		 107×106 m/s
= 0.3568 \(c\)		 0.3303 0.6697 1.059 0.8291 0.1709 1.788 70.86 PJ
= 0.3374 T.b.
to Proxima Centauri
= nearest star		 40.15×1012 km
= 4.244 ly
= 1.301 pc		 9.80665 m/s2
= 1 \(g\)		 3.541 yr 5.87 yr
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (4.244 yr)
+ 1.626 yr		 359.3×106 m/s
= 1.198 \(c\)
   superluminal		 0.723 0.277 1.448 0.9867 1.335×10-2 6.141 462.1 PJ
= 2.2 T.b.
to Sirius 81.36×1012 km
= 8.6 ly
= 2.637 pc		 9.80665 m/s2
= 1 \(g\)		 4.608 yr 10.36 yr
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (8.6 yr)
+ 1.758 yr		 559.6×106 m/s
= 1.867 \(c\)
   superluminal		 0.8303 0.1697 1.794 0.9957 4.345×10-3 10.74 875.3 PJ
= 4.168 T.b.
to Polaris 3.784×1015 km
= 400 ly
= 122.6 pc		 9.80665 m/s2
= 1 \(g\)		 11.68 yr 401.9 yr
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (400 yr)
+ 1.933 yr		 10.27×109 m/s
= 34.25 \(c\)
   superluminal		 0.9952 4.809×10-3 10.21 0.9999971 2.904×10-6 414.9 37.2 EJ
= 177.1 T.b.
= 3.72 m.i.
to centre of Milky Way 249.3×1015 km
= 26.35×103 ly
= 8079 pc		 9.80665 m/s2
= 1 \(g\)		 19.78 yr 26.35×103 yr
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (26.35×103 yr)
+ 1.937 yr		 399.3×109 m/s
= 1332 \(c\)
   superluminal		 0.999926 7.352×10-5 82.47 0.{9*9}32 6.757×10-10 27.2×103 2.445 ZJ
= 244.5 m.i.
= 3.874 AGEC
to Andromeda galaxy 24×1018 km
= 2.537 Mly
= 777.8×103 pc		 9.80665 m/s2
= 1 \(g\)		 28.63 yr 2.537 Ma
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (2.537 Ma)
+ 1.937 yr		 26.56×1012 m/s
= 88.61×103 \(c\)
   superluminal		 0.{6*9}24 7.637×10-7 809.2 0.{13*9}27 7.29×10-14 2.619×106 235.4 ZJ
= 23.54×103 m.i.
= 372.9 AGEC
Hubble distance
= "edge" of universe		 132.5×1021 km
= 14 Gly
= 4.292 Gpc		 9.80665 m/s2
= 1 \(g\)		 45.32 yr 14 Ga
= \(\mathrm{\Delta}t_{\mathrm{L}}\) (14 Ga)
+ 1.937 yr		 92.6×1015 m/s
= 308.9×106 \(c\)
   superluminal		 0.{9*9}86 1.384×10-10 60.11×103 1 2.394×10-21 14.45×109 1299 YJ
= 2.058×106 AGEC
= 3.393 L⊙ s
Energy reference levels:
T.b.=Tsar bomba = heaviest H-bomb ever detonated=210×1015J
m.i.=1 km in diameter meteorite impact at 30 km/s=10×1018J
AGEC=annual global energy consumption by man (20 terawattyears)=631.2×1018J
L⊙ s=1 second of solar luminosity=382.8×1024J
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