At 12:13 AM 12/20/02 +0000, Alberto Monteiro wrote:
Ronn! Blankenship wrote:
>
> So it would take about 4 weeks (2 weeks of acceleration at 1g followed by 2
> weeks of deceleration at 1g). Is that good enough for what you need?
>
BTW, 1 g = 1 light-year / year^2 [with a few percents error :-)]
3%, to be precise. ;-) (1g = 1.03 ly*y^-2)
, so any 1-g acceleration for months will require a relativistic correction
Agreed.
In this case, however, we are only talking about accelerating for 2 weeks, which would be about 4% of a year. Computed non-relativistically, the turnover speed will be v = at = (9.8 m*s^-2) * (1.22e6 s) = 1.20e7 m/s = 0.04c. The relativistic correction factor <gamma> at 0.04c is 1/sqrt(1 - 0.04^2) = 1.0008, which is probably negligible in light of the other approximations made (e.g., a straight-line flight path) . . .
FWIW, the rule of thumb physicists often use is that relativistic corrections can be ignored in a first approximation when the velocities involved are less than about 1/7*c, which is where the difference between the results of the relativistic and non-relativistic computations reaches 1%. To reach this velocity would require a bit more than 7 weeks at a constant 1g acceleration, during which time, from v^2 = vo^2 + 2as, the ship would travel 9.36e13 m = 626 AU. So unless Damon moves the jump point out to more than 1250 AU from the star, he can probably use the non-relativistic approximation without too much error . . .
-- Ronn! :)
Ronn Blankenship
Instructor of Astronomy/Planetary Science
University of Montevallo
Montevallo, AL
Disclaimer: Unless specifically stated otherwise, any opinions contained herein are the personal opinions of the author and do not represent the official position of the University of Montevallo.
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