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?


-- 

Regards
     Stephen

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Le 24/08/2015 12:15, Stephen a écrit :
> 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?
>
>
What about the decelerating capability ? also 1 g ?

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On 8/24/2015 3:34 PM, Le_Forgeron wrote:
> Le 24/08/2015 12:15, Stephen a écrit :
>> 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?
>>
>>
> What about the decelerating capability ? also 1 g ?
>
>

Yes, I am trying to make it simple. :-)


-- 

Regards
     Stephen

Post a reply to this message

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.

Post a reply to this message

Am 24.08.2015 um 16:34 schrieb Le_Forgeron:
> Le 24/08/2015 12:15, Stephen a écrit :
>> 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?
>>
>>
> What about the decelerating capability ? also 1 g ?

With that flight profile in mind, it seems reasonable to design the
spacecraft for a turn-around halfway down the road, to accelerate in the
opposite direction for the second half of the trip.

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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

Post a reply to this message

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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Am 24.08.2015 um 18:55 schrieb Stephen:
> 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

That certainly depends on whether you look at it from the perspective of
an outside observer (who will notice your acceleration decreasing, which
to him is actually the reason why you'll never seem to reach the speed
of light), or that of the traveler. To the latter, his mass remains
constant, and acceleration at a constant 1g is perfectly possible.

When taking on the perspective of the traveler, newtonian mechanics give
perfectly good answers, no matter what aspect you're looking at. (Unless
you're looking out the window.)

> 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.

Nasa has already conducted first tests of such drives, with promising
results (to their own surprise). I kid you not.

https://en.wikipedia.org/wiki/Quantum_vacuum_thruster
https://en.wikipedia.org/wiki/RF_resonant_cavity_thruster

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Am 25.08.2015 um 08:08 schrieb clipka:

> That certainly depends on whether you look at it from the perspective of
> an outside observer (who will notice your acceleration decreasing, which
> to him is actually the reason why you'll never seem to reach the speed
> of light), or that of the traveler. To the latter, his mass remains
> constant, and acceleration at a constant 1g is perfectly possible.

This, BTW, follows directly from the most fundamental principle of
General Relativity: As long as you don't look out the window, you can't
tell whether you're accelerating, or are stationary in a gravitational
field excerting an equivalent force. In the latter case, your mass
obviously does not increase over time; therefore, neither does it in the
former case.

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Le 2015-08-25 03:09, clipka a écrit :
> Am 25.08.2015 um 08:08 schrieb clipka:
>
>  As long as you don't look out the window, you can't
> tell whether you're accelerating, or are stationary in a gravitational
> field excerting an equivalent force. In the latter case, your mass
> obviously does not increase over time; therefore, neither does it in the
> former case.
>

It depends on what you eat, and how much exercice you get.

;-)

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