Ah! ha!
You did not spot my deliberate mistake. <blush>

 >> accelerating at a constant 1 g.

Cannot be done. :-(
I forgot that as your speed increases your mass increases. So when you 
reach 0.9c your mass is 2.29 * your rest mass and the force needed to 
accelerate you increases as well.
0.99c > 7*m
0.999c > 22*m
0.9999c > 70 *m

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

But your approximations are good enough for me. So you passed the crew 
test.
Now I have to invent a reactionless drive using the vacuum energy.
Shouldn't take that long, I would think.


On 8/24/2015 4:32 PM, clipka wrote:
> Am 24.08.2015 um 12:15 schrieb Stephen:
>> I bought a spaceship that can approach close to the speed of light,
>> accelerating at a constant 1 g.
>> How long would it take to reach a star 10 light years away, assuming
>> both Sol and the star are at rest relative to each other? And how long
>> would it seem to take to the crew?
>
> Tell me when you plan to launch. I'd like to hitch a ride!
>
>
> Speed of light is approximately 300,000 km/s = 3*10^8 m/s, or about
> 9.5*10^15 m/year; let's use a value of 1*10^16 m/year for simplicity. A
> distance of 10 light years then equals 10*10^16 m = 1*10^17 m.
>
> You'll have to turn around and decelerate after half the distance, which
> is about 5*10^16 m.
>
> An acceleration of 1g equals approximately 9.81 m/s^2; let's use a value
> of 10 m/s^2 for simplicity.
>
> The function of acceleration vs. time is constant for the first half of
> the trip:
>
>      a(t) = a
>
> The function of speed vs. time is the integral of the acceleration
> function; the integral of a constant function is the product of the
> function's value and the parameter:
>
>      v(t) = a t
>
> The function of distance vs. time is the integral of the speed function,
> which is a linear function of the form f(x)=cx; the integral of such a
> function is F(x)=1/2 cx^2, i.e:
>
>      d(t) = 1/2 a t^2
>
> Let's compute the time at which we would reach the speed of light if
> newtonian mechanics would apply:
>
>      v(t)           = a        t
>      3*10^8 m/s     = 10 m/s^2 t
>      3*10^7         =  1  /s   t
>
> which gives us
>
>      t = 3*10^7 s = ca. 8300 hours = ca. 350 days = ca. 1 year.
>
> Next, let's compute the distance traveled in this time under newtonian
> laws of motion:
>
>      d(t)          = 1/2    a         t^2
>      d(3*10^7 s)   = 1/2 . 10 m/s^2 . (3*10^7 s)^2
>                    =        5 m/s^2 .  9*10^14 s^2
>                    =                  45*10^14 m
>
> which gives us
>
>      d = 4.5*10^15 m = ca. 1.5*10^7 light seconds = ca. 0.5 light years.
>
> Finally, let's compute the time at which we would have completed the
> half trip of 5*10^16 m, again if newtonian mechanics would apply:
>
>      d(t)          = 1/2    a       t^2
>      5*10^16 m     = 1/2 . 10 m/s^2 t^2
>      5*10^16 m     =        5 m/s^2 t^2
>      10^16         =        1  /s^2 t^2
>      10^8          =        1  /s   t
>
> which gives us
>
>      t = 10^8 s = ca. 2.8*10^4 hours = ca. 1150 days = ca. 3.2 years.
>
>
> Now what does this tell us about the situation with non-newtonian mechanics?
>
> First, it gives us a rough idea about the trip's profile under
> relativistic mechanics: As a first approximation, the first year will be
> spent accelerating to light speed, and covering the first half light
> year; the final half light year will be covered by a corresponding year
> of decelerating from light speed back to zero; while the remaining 9
> light years of the trip will be spent traveling close to the speed of
> light, which will take a time of 9 years to the outside observer. The
> total trip would thus take about 11 years.
> Note that this is just a rough estimate, giving us nothing but a lower
> bound: There's no way we can do it faster. Not at 1g acceleration.
> However, we can also derive a rough upper bound in a similar manner:
> Half the speed of light is still sufficiently relativistic within our
> margin of error; we could obviously spend only the first half year
> accelerating, bringing us up to half the speed of light and covering a
> quarter of a light year; we could then cover the next 9.75 light years
> drifting, which would take us 19.5 years; and finally a decelearation of
> just half a year would bring us back to a speed of zero while covering
> the final quarter of a light year. This profile would take 21 years in
> total, both to us and to an outside observer. Obviously a full-burn,
> turn, and full-burn profile would get us there faster.
> We could therefore estimate the duration of the trip, as witnessed by an
> outside observer, to be somewhere around 16 years; I will leave the
> corresponding exact computations as an exercise to the reader.
>
> Second, it gives us the /exact/ duration of the trip as experienced by
> the traveler (as far as our generous application of rounding allows): He
> will spend 3.2 years accelerating, then 3.2 years decelerating, for a
> total of 6.4 years. About 13 years for a round trip.
>


-- 

Regards
     Stephen

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