If you have the energy for a constant 1G thrust, how long would it take to get to the planets in our solar system? How long for the 5 nearest solar systems?
Assuming turn over and decelerate at halfway.
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asked Jul 29, 2013 at 23:51
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4$\begingroup$ "1g of thrust" pointed straight up will balance gravity, and result in you floating. "1g" (as I read it), is the acceleration caused by the Earth's gravity; if that's how you actually define it, then your acceleration decreases as you get further (and 'feel less pull') from Earth. Of course, you don't need to point straight up, and TidalWave's assumption that what you meant is 9.8m/s/s is probably correct - but note that even so, his answer provides you with a minimum, eg assuming you could turn off gravity and the atmosphere (and the assumptions he mentions at the top). $\endgroup$– hunter2Jul 31, 2013 at 9:28
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4$\begingroup$ @hunter2, you are correct 1g of thrust will not get you off the planet. The assumption is that the starting point is in orbit, 1g of thrust during a long trip provides thrust & simulated gravity. $\endgroup$– James JenkinsJul 31, 2013 at 10:24
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$\begingroup$ See XKCD, there is a post called how to get somewhere fast. $\endgroup$– Starship - On StrikeMar 20 at 17:11
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3 Answers
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Assuming acceleration is constant, $d=(1/2) a t^2$. So plotted over time, distance traveled is a nice parabola.
If you want the time it'd take for a specific distance, it's easy to manipulate $d=(1/2) a t^2$.
$t=\sqrt{2d/a}$
If you're using meters and seconds as your units, $a=9.8 meters/sec^2$
To travel half the distance to the moon would take about 1.75 hours. The other half distance spent decelerating would take the same amount of time.
Using Days and AU (astronomical units) we can see 3 days will get about 2.5 AU (halfway to Jupiter). 4.5 days will get you 5 AU (halfway to Saturn). 9 days will get you 20 AU (more than halfway to the Kuiper belt)
It gets trickier for interstellar distances. In Newtonian mechanics v = at, so it'd take a little less than a year to reach c at 1 g acceleration. But relativity won't allow that, we can only get close to c.
Our Newtonian model is okay for nearly a year of acceleration and after that relativity wrecks this nice parabola:
After 1 year at 1 g we will have traveled .5 lightyears and our velocity will be close to maxed out. There after we're moving at close to c, so add a little more than a year for each lightyear distance.
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19$\begingroup$ Your "add a little more than a year for each lightyear distance" is correct for an outside observer, but for someone aboard the ship, the Newtonian model is correct for all distances (as measured before starting acceleration): Lorentz contraction will shrink the universe during travel to give the appearance of Newtonian physics. $\endgroup$– MarkJul 23, 2014 at 4:47
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$\begingroup$ Beautiful answer. I just want to point out that since the entire question is theoretical, why not ignore mass? if we allow ourselves to assume a=9.8m/s/s, then it's not depended on mass, so relativity isn't a big problem. $\endgroup$ Jul 24, 2014 at 14:55
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6$\begingroup$ @Mark I broke travel into 35.4 day increments, each increment accelerating .1 c. After 354 days I got about .76 c and the passengers perceiving 300 days. I'm not sure that's correct, I'm not comfortable with special relativity. I don't think either an outside observe nor the accelerating passengers would see what appears to be a Newtonian universe. $\endgroup$– HopDavidJul 25, 2014 at 2:17
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1$\begingroup$ @Mark not quite true. After all, if you observe things falling in constant gravity, you still can't see them go beyond speed of light, meaning that Newtonian physics doesn't hold even as seen by the ship. $\endgroup$ Dec 16, 2019 at 23:29
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$\begingroup$ @MaudPieTheRocktorate, if you observe things falling in constant gravity, you're the outside observer, not the person aboard the ship. $\endgroup$– MarkDec 16, 2019 at 23:52
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Not assuming any time taken for orbital maneuvering, turning halfway 180° to decelerate, assuming closest distance of planets (and Luna) to the Earth, and not accounting for fuel burn (i.e. literal constant 1g acceleration):
The Moon / Luna:
Closest to Earth (Supermoon): 356,577 km
Travel time (at 9.80665 m/s2, no deceleration): 2h 22m 12s
Travel time (at 9.80665 m/s2, decelerating halfway): 3h 20m 24sMercury:
Closest to Earth: 77.3 million km
Travel time (at 9.80665 m/s2, no deceleration): 1d 10h 52m 48s
Travel time (at 9.80665 m/s2, decelerating halfway): 2d 1h 19m 12sVenus:
Closest to Earth: 40 million km
Travel time (at 9.80665 m/s2, no deceleration): 1d 1h 5m 2s
Travel time (at 9.80665 m/s2, decelerating halfway): 1d 11h 28m 48sMars:
Closest to Earth: 65 million km
Travel time (at 9.80665 m/s2, no deceleration): 1d 7h 58m 5s
Travel time (at 9.80665 m/s2, decelerating halfway): 1d 21h 13m 1sJupiter:
Closest to Earth: 588 million km
Travel time (at 9.80665 m/s2, no deceleration): 4d 0h 11m 2s
Travel time (at 9.80665 m/s2, decelerating halfway): 5d 16h 2m 2sSaturn:
Closest to Earth: 1.2 billion km
Travel time (at 9.80665 m/s2, no deceleration): 5d 17h 25m 1s
Travel time (at 9.80665 m/s2, decelerating halfway): 8d 2h 20m 24sUranus:
Closest to Earth: 2.57 billion km
Travel time (at 9.80665 m/s2, no deceleration): 8d 9h 6m 0s
Travel time (at 9.80665 m/s2, decelerating halfway): 11d 20h 24m 0sNeptune:
Closest to Earth: 4.3 billion km
Travel time (at 9.80665 m/s2, no deceleration): 10d 20h 7m 48s
Travel time (at 9.80665 m/s2, decelerating halfway): 15d 7h 52m 48sPluto:
Closest to Earth: 4.28 billion km
Travel time (at 9.80665 m/s2, no deceleration): 10d 19h 31m 12s
Travel time (at 9.80665 m/s2, decelerating halfway): 15d 7h 1m 12s
answered Jul 30, 2013 at 2:04
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5$\begingroup$ I'll go crawl back to my corner now... ;) $\endgroup$ Jul 30, 2013 at 2:05
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5$\begingroup$ There would be some slight difference depending on the speed of each planet at the time of launch, but this should be close enough. $\endgroup$– PearsonArtPhoto ♦Jul 30, 2013 at 2:07
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$\begingroup$ @TildalWave I assume your "decelerating half way is to match earth velocity" and not the target planet velocity ? (in which case you are still moving very fast relative to the planet) $\endgroup$– AntziApr 4, 2019 at 3:25
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According to wikipedia, interstellar travel at 1G would take approximately 1 year + the distance in lightyears. Proxima Centauri (4.2 light years) for example would take 5.2 years.
But that time is from the viewpoint of stationary observers at the departure point. The trip's duration from the traveler's viewpoint would be less due to the time dilation effect predicted by Einstein's Theory of Relativity. The greater the distance, the greater the speed from the stationary observer's viewpoint. From the stationary observer's viewpoint the traveler's rate of acceleration would slow as they approached the speed of light. The traveler would see no change between their speed and the speed of light. Instead they would experience time at an increasingly slower rate which would effectively cause the distance to the destination to become shorter.
Due to the time dilation effect, 1G acceleration should be sufficient to travel anywhere in our galaxy in less than a lifetime from the viewpoint of the traveler, but not the stationary observer.
For more information on the time dilation effect read Stephen Hawking's "Brief History of Time"
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6$\begingroup$ Wikipedia article on constant acceleration interstellar travel: en.wikipedia.org/wiki/Space_travel_using_constant_acceleration $\endgroup$ Jul 20, 2014 at 17:14
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1$\begingroup$ Use an Alcubierre drive to reduce relativity (if it can..) :) $\endgroup$– MikePJun 4, 2018 at 16:28
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1$\begingroup$ @MikeP The Alcubierre drive is certainly an exciting idea, but unfortunately the most recent calculations on feasability suggest one would need "The energy output of a type III civilization" to make it work. Unfortunately thats so far advanced we barely even write SciFi about it. $\endgroup$– ShayneJun 5, 2018 at 3:04