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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.
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