In sci.physics, Dave
<dwickford@yahoo.com>
 wrote
on Tue, 18 May 2004 21:58:24 +0100
<c8dtde$72b$1@news6.svr.pol.co.uk>:

"The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net> wrote in
message news:00ion1-21a.ln1@lexi2.athghost7038suus.net...

In sci.physics, rem51204@aol.com
<rem51204@aol.com>
 wrote
on 17 May 2004 14:21:54 -0700
<b3662a91.0405171321.ebb488f@posting.google.com>:

                SEVEN MINUTE INTERSTELLAR SPACE TRAVEL

Update from The REAL Galactic Federation and The Spiritual Hierarchy
     August 19, 2003
Communicated thru Sheldan Nidle of The Planetary Activation Organization
     http://www.paoweb.com/updates.htm

     Selamat Jarin! We arrive, dear Hearts, with much to discuss with
you.

["discussion" snipped]

I'll reiterate what I have previously said about Stanislaw Lem's
Lunabus (referenced in Voyage 8) -- my previous posting, which was
quite some time back, was probably more verbose thereon.

First, the facts.

Distance to the moon: 3.84 * 10^8 m
Acceleration on Earth: 10 m/s/s (approximately)
Maximum acceleration for a human before he experiences
unconsciousness: roughly 100 m/s/s (10 g's) [*]

Now the question:

Can a lunabus take 8 minutes to get from here to the moon?

Now some formulas;

d = 1/2 a t^2 for constant acceleration a, time t,
starting from rest.

One can solve for a = 2d/t^2, or one can solve
for t = sqrt(2d/a).  Note that this is nonrelativistic,
but it doesn't really matter anyway, it turns out.

We can work either forward or backward.

Forward:  Brute-force a, knowing t and d.  Bear in mind d
is half the distance to the moon (the assumption being that
the lunabus will turn around halfway there and decelerate
at the same rate as it accelerated there in the first place)
and t is 4 minutes.  So we get

a = 3.84 * 10^8 m / (240^2) = 6667 m/s/s

Waaaaaay too much.

Backward: Work out t, knowing maximum a and d.

t = sqrt(3.84 * 10^8m / 100) = 1960 s = more than half an hour
    per trip leg; more than an hour for the whole trip.

And that's just to the moon.  Unless your friendly
interstellar space traveler/ships have a nullgrav adaptor
of some sort (a la _Perry Rhodan, Enterprise Stardust_),
don't bet on being in any shape to communicate with
the inhabitants of, say, Sirius after one's 7-minute journey
(which appears barely possible, BTW, given a highly
brutal acceleration, *lots* of energy and mass reactant,
and an interpretation of 7-minutes using subjective time
-- assuming there's a subject to experience said time,
since human meat jam doesn't have that high an IQ :-) ).

[*] there are some interesting questions regarding this but there's
    only one human I know of who volunteered for that rocket ride
    at 50-100 g; he survived but IIRC suffered some minor internal damage.

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.

Keep working at the basic physics.  I'd love to see some new laws of
physics.

So would I but I'm not all that hopeful. :-)  As it is, we already
know that wormholes would require gigantic worms (a graduate
school lecture apparently requires a computation of the size
of such a beast, from basic principles -- it turns out to be
larger than our solar system), and that near-lightspeed would
require fantastic energies.

There's also the little problem of radiation.  At .90c, one
is travelling 27,000 km/s, and being hit by 2.7 billion
atoms per square centimeter of ship surface presented
along the direction of travel.  (This in the universe reference
frame; the average matter density is apparently 1 atom / cm^3.)

Each atom has an energy of E = 1/2 mc^2 [*]
= 1.6726*10^-27 * (3*10^8)^2 = 1.51*10^-10 J.
So each square centimeter is being irradiated with about
0.4W of energy, or the mass equivalent thereof.

That's about 3x the radiation coming from old Sol.

In the ship's reference frame the problem is magnified further
because of the Lorentz shrink, which magnifies the effect
by 1/sqrt(.19) = 2.294.  As the ship gets closer to lightspeed
(or even beyond lightspeed) this effect will only get worse.

Now, I'm not up on the modern sci-fi stuff, but somehow
I don't recall an episode where Captain Kirk gets fried
to a crisp while traveling to, say, Talos IV, since the
Enterprise is traveling much faster than a mere 0.90c...

[*] there's no corrective factor here as the atoms aren't moving
    relative to the Universe.

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.