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On 8/24/2015 12:55 PM, Stephen wrote:
> 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.
>>
>
>
Nerds.
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