# Quick Turnaround to Barnard’s Star

A relativistic trip to Barnard’s Star? Those who read French will want to check out the log of such a journey as Philippe Guglielmetti sees it. Traveling at a constant 1g for acceleration and braking, the mission reaches 0.99999 c, travel time twelve years but only three as experienced by the crew. The fictionalized journey plays fast and loose with the star itself, as Adam Crowl notes in a comment below, but the trip is fun even with my rusty French. Have a look.

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• Marc January 20, 2007, 14:47

Twelve years? :o Does anyone have a translator for this report written by Guglielmetti? It sounds extremely interesting.

• Adam January 20, 2007, 16:03

Hi Paul

I can see if his destination is Barnard’s at 5.92 ly he’s not going to reach 0.99999c at 1 gee. That’s a gamma factor of 223.6 and at 1 gee gamma goes up by 1.03227 for every light-year, thus we’re looking at a journey of 215.65 ly just to get to 0.99999c – double that to brake. Round-trip time, from Earth’s point of view, would be about 865 years.

A trip of 5.92 ly hits a gamma of ~ 4 (0.969c) at turn-over. The round-trip takes 15.23 years, assuming no time to get out and look around. On board ship the trip takes 4.028 years each way, which is a bit of a saving against Earth time.

OTOH how many gees does it take to hit 0.99999c going to Barnard’s? 72.85 gee, but the ship’s trip-time decreases to just 59.3 days one-way. Quite an improvement. Earth time declines a bit to 11.89 years.

Ursula LeGuin’s fictional NAFAL star-drive can reduce the ship-time to mere hours though the stationary observer time is still roughly the light-travel time. For example, an explorer ship in “Vaster than Empires” does 256 light-years in a mere 10.65 hours ship-time – if the ship accelerated the whole way that’s over 25,000 gees.

I just ran the link through Babel-Fish (c/o AltaVista) and the log was quite interesting, though I’ve no idea what computations he did to get his figures. He fictionalises Barnard’s a bit by making it 10 ly away and Sun-like. Oh well… permis artistique

• Administrator January 20, 2007, 16:26

Thanks — I knew I could count on you guys to take a hard look at the numbers. 72.85 gee… Ouch!

• Stephen January 22, 2007, 17:18

I’d weigh 13,000 lbs. That’d mess up my diet.

How much reaction mass does it take? What’s the energy consumed, compared to the output of the Sun? Me, Myself, and I want to know. Not to meantion the rabid readers.

• Adam January 23, 2007, 7:08

Hi Stephen

Assuming a perfect photon rocket for the trip it takes a mass-ratio of 63.774 to get to Barnard’s at 5.92 ly and (+/-) 1 gee acceleration. That’s roughly 63 times the mass of the ship that must be turned into a pure photon exhaust – for a 1,000 ton ship we’re talking 63,000 tons x 9E+16 J energy = 5.67E+24 J. In megatons of TNT energy (4.2E+15 J = 1 MT) that’s 1.35 billion MT.

The jet-power from the point of view of the ship is that figure divided by the ship trip-time (128 million seconds), on average = 44,600 TW (TW being a terawatt.) Or 10.55 MT per second. Or 0.492 kg of energy per second.

A radiator, at 100,000 K, would be 100 metres across to produce that photon output, peaking in the UV part of the spectrum. Perhaps super-reflective materials might be feasible to allow that, and uranium plasma might be stable enough in a magnetic field for the radiator itself, but converting that much mass at 100% efficiency is pure voodoo. Perhaps some weird neutrino physics might allow us to make a macroscopic sphaleron field to decompose all that mass into coherent neutrino beams. Neutrinos are nearly massless and a whole lot less hot than UV photons.