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Time at 1G acceleration to travel 100000 light years
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Time at 1G acceleration to travel 100000 light years
Space travel using constant acceleration drive: Earth to EuropaIs there a simple relation between delta-v and travel time?Falcon 9 g-level/acceleration profileCan someone show me the math behind an Oberth maneuver around the sun, and acceleration to Jupiter?How can I find the time required to achieve certain velocity in orbit with changing mass?When are there launch windows to Neptune via Jupiter?If the moon's orbit decays- how would it effect the moon's phases?No need for gravity assist with Falcon Heavy?Quick and dirty way to estimate flight times for constant acceleration rocketsHow to get from the perturbative acceleration formula to the acceleration on the 3 axis in the case of the oblate Earth, considering j2?
$begingroup$
How long would it take to go 100,000 light years at a constant 1G acceleration?
orbital-mechanics
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How long would it take to go 100,000 light years at a constant 1G acceleration?
orbital-mechanics
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago
add a comment |
$begingroup$
How long would it take to go 100,000 light years at a constant 1G acceleration?
orbital-mechanics
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How long would it take to go 100,000 light years at a constant 1G acceleration?
orbital-mechanics
orbital-mechanics
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Roger P JonesRoger P Jones
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Roger P JonesRoger P Jones
261
asked 9 hours ago
Roger P JonesRoger P Jones
261
261
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Roger P Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago
add a comment |
$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago
$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nonrelativistic solution
The variables used will be
$x$ for the distance travelled
$v$ for velocity
$a$ for acceleration ($1~mathrmg$)
$t$ for the time
$c$ for the speed of light.
Non braking
Assuming the velocity you arrive at does not matter we take the equation
$$x = frac12 a t^2 .$$
Solve for $t$:
$$t = sqrtfrac2xa .$$
(Let’s discard the negative solution here)
Plugging this into Wolfram Alpha gives us
$$1.389 times 10^10~mathrms ,$$
or just over 440 years.
The velocity the object would be arriving at is be calculated by
$$v = a cdot t approx 1.362times 10^11~fracmathrmmmathrms .$$
About 454.4 times the speed of light.
So no we cannot neglect relativistic effects.
Braking
If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $t$ the same way we did above and get
$$9.822 times 10^9~mathrms ,$$
or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be
$$9.632times 10^10~mathrms ,$$
just over 321 times the speed of light.
Relativistic effects
Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.
The calculation including the relativistic effects is quite complicated.
The important thing to note here is that the traveling object and an external observer will measure the time differently.
External observer
From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:
Plugging into the equation from the linked answer:
$$t=sqrtfracx1~mathrmly^2 + 2
frac
x / 1~mathrmly
a / fracmathrmlymathrmy^2
cdot 1~mathrmy .$$
I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.
This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.
Perspective of the traveling object
From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:
Using the other answer as reference again:
$$t = fracca mathrmacoshleft(fracxcdot ac^2 + 1right)$$
We get a result of
$$3.741times10^8~mathrms ,$$
about 12 years.
Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.
Conclusion
Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago and the information would take another 100000 years to get to them.
Visualisation
The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $100~mathrmm$) you can see the traveller accelerate and decelerate.
edited 6 hours ago
Mike Harris
1034
answered 7 hours ago
HansHans
1,962522
$endgroup$
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
add a comment |
$begingroup$
Welcome to the site!
Using this tool:
Observer time: 100001 years
Traveler time: 22.4 years
Edit: Time is fixed, I blame the google calculator
answered 7 hours ago
PunintendedPunintended
2544
$endgroup$
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nonrelativistic solution
The variables used will be
$x$ for the distance travelled
$v$ for velocity
$a$ for acceleration ($1~mathrmg$)
$t$ for the time
$c$ for the speed of light.
Non braking
Assuming the velocity you arrive at does not matter we take the equation
$$x = frac12 a t^2 .$$
Solve for $t$:
$$t = sqrtfrac2xa .$$
(Let’s discard the negative solution here)
Plugging this into Wolfram Alpha gives us
$$1.389 times 10^10~mathrms ,$$
or just over 440 years.
The velocity the object would be arriving at is be calculated by
$$v = a cdot t approx 1.362times 10^11~fracmathrmmmathrms .$$
About 454.4 times the speed of light.
So no we cannot neglect relativistic effects.
Braking
If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $t$ the same way we did above and get
$$9.822 times 10^9~mathrms ,$$
or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be
$$9.632times 10^10~mathrms ,$$
just over 321 times the speed of light.
Relativistic effects
Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.
The calculation including the relativistic effects is quite complicated.
The important thing to note here is that the traveling object and an external observer will measure the time differently.
External observer
From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:
Plugging into the equation from the linked answer:
$$t=sqrtfracx1~mathrmly^2 + 2
frac
x / 1~mathrmly
a / fracmathrmlymathrmy^2
cdot 1~mathrmy .$$
I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.
This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.
Perspective of the traveling object
From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:
Using the other answer as reference again:
$$t = fracca mathrmacoshleft(fracxcdot ac^2 + 1right)$$
We get a result of
$$3.741times10^8~mathrms ,$$
about 12 years.
Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.
Conclusion
Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago and the information would take another 100000 years to get to them.
Visualisation
The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $100~mathrmm$) you can see the traveller accelerate and decelerate.
edited 6 hours ago
Mike Harris
1034
answered 7 hours ago
HansHans
1,962522
$endgroup$
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
add a comment |
$begingroup$
Nonrelativistic solution
The variables used will be
$x$ for the distance travelled
$v$ for velocity
$a$ for acceleration ($1~mathrmg$)
$t$ for the time
$c$ for the speed of light.
Non braking
Assuming the velocity you arrive at does not matter we take the equation
$$x = frac12 a t^2 .$$
Solve for $t$:
$$t = sqrtfrac2xa .$$
(Let’s discard the negative solution here)
Plugging this into Wolfram Alpha gives us
$$1.389 times 10^10~mathrms ,$$
or just over 440 years.
The velocity the object would be arriving at is be calculated by
$$v = a cdot t approx 1.362times 10^11~fracmathrmmmathrms .$$
About 454.4 times the speed of light.
So no we cannot neglect relativistic effects.
Braking
If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $t$ the same way we did above and get
$$9.822 times 10^9~mathrms ,$$
or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be
$$9.632times 10^10~mathrms ,$$
just over 321 times the speed of light.
Relativistic effects
Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.
The calculation including the relativistic effects is quite complicated.
The important thing to note here is that the traveling object and an external observer will measure the time differently.
External observer
From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:
Plugging into the equation from the linked answer:
$$t=sqrtfracx1~mathrmly^2 + 2
frac
x / 1~mathrmly
a / fracmathrmlymathrmy^2
cdot 1~mathrmy .$$
I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.
This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.
Perspective of the traveling object
From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:
Using the other answer as reference again:
$$t = fracca mathrmacoshleft(fracxcdot ac^2 + 1right)$$
We get a result of
$$3.741times10^8~mathrms ,$$
about 12 years.
Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.
Conclusion
Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago and the information would take another 100000 years to get to them.
Visualisation
The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $100~mathrmm$) you can see the traveller accelerate and decelerate.
edited 6 hours ago
Mike Harris
1034
answered 7 hours ago
HansHans
1,962522
$endgroup$
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
add a comment |
$begingroup$
Nonrelativistic solution
The variables used will be
$x$ for the distance travelled
$v$ for velocity
$a$ for acceleration ($1~mathrmg$)
$t$ for the time
$c$ for the speed of light.
Non braking
Assuming the velocity you arrive at does not matter we take the equation
$$x = frac12 a t^2 .$$
Solve for $t$:
$$t = sqrtfrac2xa .$$
(Let’s discard the negative solution here)
Plugging this into Wolfram Alpha gives us
$$1.389 times 10^10~mathrms ,$$
or just over 440 years.
The velocity the object would be arriving at is be calculated by
$$v = a cdot t approx 1.362times 10^11~fracmathrmmmathrms .$$
About 454.4 times the speed of light.
So no we cannot neglect relativistic effects.
Braking
If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $t$ the same way we did above and get
$$9.822 times 10^9~mathrms ,$$
or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be
$$9.632times 10^10~mathrms ,$$
just over 321 times the speed of light.
Relativistic effects
Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.
The calculation including the relativistic effects is quite complicated.
The important thing to note here is that the traveling object and an external observer will measure the time differently.
External observer
From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:
Plugging into the equation from the linked answer:
$$t=sqrtfracx1~mathrmly^2 + 2
frac
x / 1~mathrmly
a / fracmathrmlymathrmy^2
cdot 1~mathrmy .$$
I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.
This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.
Perspective of the traveling object
From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:
Using the other answer as reference again:
$$t = fracca mathrmacoshleft(fracxcdot ac^2 + 1right)$$
We get a result of
$$3.741times10^8~mathrms ,$$
about 12 years.
Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.
Conclusion
Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago and the information would take another 100000 years to get to them.
Visualisation
The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $100~mathrmm$) you can see the traveller accelerate and decelerate.
edited 6 hours ago
Mike Harris
1034
answered 7 hours ago
HansHans
1,962522
$endgroup$
Nonrelativistic solution
The variables used will be
$x$ for the distance travelled
$v$ for velocity
$a$ for acceleration ($1~mathrmg$)
$t$ for the time
$c$ for the speed of light.
Non braking
Assuming the velocity you arrive at does not matter we take the equation
$$x = frac12 a t^2 .$$
Solve for $t$:
$$t = sqrtfrac2xa .$$
(Let’s discard the negative solution here)
Plugging this into Wolfram Alpha gives us
$$1.389 times 10^10~mathrms ,$$
or just over 440 years.
The velocity the object would be arriving at is be calculated by
$$v = a cdot t approx 1.362times 10^11~fracmathrmmmathrms .$$
About 454.4 times the speed of light.
So no we cannot neglect relativistic effects.
Braking
If you want to arrive at that location with reasonable speeds you’d have to accelerate half the way and brake the other half. We compute $t$ the same way we did above and get
$$9.822 times 10^9~mathrms ,$$
or just over 311 years. After that time you would only have gone half the way and need to turn your spacecraft around and decelerate which takes the same time again, giving you a total of 622 and a half years. But you would stop next to your target and not shoot past it at extreme speeds. Your maximum speed (at the turning point) would now be
$$9.632times 10^10~mathrms ,$$
just over 321 times the speed of light.
Relativistic effects
Going anywhere near the speed of light (or close to a great source of gravity like a black hole for that matter) will yield a huge variety of relativistic effects, making time and space not be the same for every person.
The calculation including the relativistic effects is quite complicated.
The important thing to note here is that the traveling object and an external observer will measure the time differently.
External observer
From the perspective of an external observer (so seen from earth, the target or any other relatively static point in the universe) it would take you 100000 years to travel 100000ly at light speed (that’s kind of the definition), but the object will not be traveling faster than the speed of light:
Plugging into the equation from the linked answer:
$$t=sqrtfracx1~mathrmly^2 + 2
frac
x / 1~mathrmly
a / fracmathrmlymathrmy^2
cdot 1~mathrmy .$$
I told you it’d be complicated. Plugging it in we get 100001 years. Not surprising: as discussed above: traveling at light speed for about 100000 years plus a bit for accelerating to that speed.
This is the non-braking variant. Braking however would not take a lot longer. About 100002 years in total. So one year of accelerating to the speed of light and one year of braking – simply said.
Perspective of the traveling object
From the perspective of the object the primary effect to consider here would be length contraction. It makes the distance to the target appear smaller and smaller the faster you are going. From the perspective of that object it would thus take less than duration calculated in the nonrelativistic solution above:
Using the other answer as reference again:
$$t = fracca mathrmacoshleft(fracxcdot ac^2 + 1right)$$
We get a result of
$$3.741times10^8~mathrms ,$$
about 12 years.
Taking braking into account again we get 11.18 year for each half, so about 22.4 years in total.
Conclusion
Relativistic effects are important here. While it would be totally doable for an astronaut to complete that mission in their lifetime by the time they get there everyone they know will have died 100000 years ago and the information would take another 100000 years to get to them.
Visualisation
The other answer by user Punintended links a wonderful tool you can try out that visualizes these effects. Don't worry about the fuel parts. You can see the rocket become very short (relativistic length contraction) and time progressing at different speeds for the traveller and the observer. When choosing a less extreme example (like $100~mathrmm$) you can see the traveller accelerate and decelerate.
edited 6 hours ago
Mike Harris
1034
answered 7 hours ago
HansHans
1,962522
edited 6 hours ago
Mike Harris
1034
edited 6 hours ago
Mike Harris
1034
edited 6 hours ago
Mike Harris
1034
1034
answered 7 hours ago
HansHans
1,962522
answered 7 hours ago
HansHans
1,962522
answered 7 hours ago
HansHans
1,962522
1,962522
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
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– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
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– Hans
6 hours ago
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
add a comment |
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
Great answer! Except that it's braking, not breaking. :-)
$endgroup$
– PJNoes
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
$begingroup$
where did I misspell that, I cannot find it -_-
$endgroup$
– Hans
6 hours ago
2
2
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
oh wow, I was looking to correct in it the wrong direction. These things happen when English is not your native language (mine is German). Thanks!
$endgroup$
– Hans
6 hours ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
$begingroup$
The information that everybody they knew had died would arrive in transit, provided the traveller had the ability to receive the severely doppler shifted signal.
$endgroup$
– Omnifarious
14 mins ago
add a comment |
$begingroup$
Welcome to the site!
Using this tool:
Observer time: 100001 years
Traveler time: 22.4 years
Edit: Time is fixed, I blame the google calculator
answered 7 hours ago
PunintendedPunintended
2544
$endgroup$
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
add a comment |
$begingroup$
Welcome to the site!
Using this tool:
Observer time: 100001 years
Traveler time: 22.4 years
Edit: Time is fixed, I blame the google calculator
answered 7 hours ago
PunintendedPunintended
2544
$endgroup$
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
add a comment |
$begingroup$
Welcome to the site!
Using this tool:
Observer time: 100001 years
Traveler time: 22.4 years
Edit: Time is fixed, I blame the google calculator
answered 7 hours ago
PunintendedPunintended
2544
$endgroup$
Welcome to the site!
Using this tool:
Observer time: 100001 years
Traveler time: 22.4 years
Edit: Time is fixed, I blame the google calculator
answered 7 hours ago
PunintendedPunintended
2544
edited 6 hours ago
answered 7 hours ago
PunintendedPunintended
2544
answered 7 hours ago
PunintendedPunintended
2544
answered 7 hours ago
PunintendedPunintended
2544
2544
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
add a comment |
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
Your tool actually assumes you want to slow down: take a look at the animation. Also you somehow mistyped the observer time.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
$begingroup$
to clarify: choose a shorter distance like 100 meters and see the animation clearly speed up and slow down again. This also is the reason why the rocket expands to its full length again at the end: it is not moving at relativistic speeds anymore, but standing.
$endgroup$
– Hans
7 hours ago
add a comment |
Roger P Jones is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
With or without taking relatavistic effects into account?
$endgroup$
– Mike Brockington
8 hours ago
$begingroup$
Note, you can accelerate only on half the way. After that, you will need to decelerate, also with 1G - except if you don't want to explode on arrival like an antimatter-bomb.
$endgroup$
– peterh
8 hours ago
$begingroup$
@peterh Antimatter bomb would be at .92 c - this would be so, so much worse... In fact, if you are close enough to c, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer.
$endgroup$
– Eth
6 hours ago
$begingroup$
@Eth I think $frac1sqrt1-fracv^2c^2=2$ solves to $v=fracsqrt32c$. It is only 0.86c.
$endgroup$
– peterh
3 hours ago