# Acceleration to c at 1 g **URL:** \ **Category:** Factual Questions\ **Created:** [May 19, 2011, 4:14am UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412 "2011-05-19T04:14:12Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Simplicio](https://avatars.discourse-cdn.com/v4/letter/s/c37758/32.png) [@Simplicio](https://boards.straightdope.com/u/Simplicio)\ **Post date:** [May 19, 2011, 4:47pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/21 "2011-05-19T16:47:51Z")
> [@TriPolar](#): > > **Simplicio** - are you saying a ship with about 90% of it’s initial weight in fuel could reach .9c? That doesn’t sound too tough. Is this with a feasible engine that could be built now? > > ETA: Assuming the craft is at least in earth orbit initially. I think it was for a “perfect rocket”, that is something that gets the maximal energy out of its fuel (basically E=MC^2), and can do so in such a way that the acceleration is constant for the entire acceleration phase. So its more of a “best case scenario” then something anyone would ever be able to achieve. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [May 19, 2011, 4:50pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/22 "2011-05-19T16:50:09Z")
> [@Chronos](#): > > For any given target speed, any given propulsion technology, and any given desired acceleration, there is some initial value of fuel mass to payload mass that’ll do the trick. The catch is that the mass ratio grows exponentially even before you account for the relativistic effects. To clarify: Suppose that it takes a mass ratio of R (total mass to payload mass) to get to a speed V. What fuel ratio is needed to get to a speed 2V? Well, first you need to get from rest to V, and then you need to get from V to 2V. For the second leg of this, you’ll need a mass ratio of R, so if your payload has mass m, you’ll need to have a total mass of R_m at the start of the second leg. But as far as the first leg of the trip is concerned, all of the fuel you’re saving for the second leg can be considered payload. So at the start of the first leg of the trip, you’ll need to have a total mass of R_R_m. Similarly, to get to 3V, you’ll need to start with R^3_m, and so on. Better rocket technology might make this a slower-growing exponential, but it will always be exponential. Ok, am I figuring this right? Earth orbit = approx. 25000 mph = approx. 6.95 mps .9c = approx. 167400 mps fuel needed = R^(167400/6.95)_m = approx. R^24000_m ---
**Author:** ![RedSwinglineOne](https://avatars.discourse-cdn.com/v4/letter/r/a9adbd/32.png) [@RedSwinglineOne](https://boards.straightdope.com/u/RedSwinglineOne)\ **Post date:** [May 19, 2011, 5:09pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/23 "2011-05-19T17:09:33Z")
Thats interesting **Chronos**. I never thought of it that way. It reminds me of a story I heard somewhere concerning skyscrapers. The person said something like ‘if you have a blueprint for a building and you decide to make it taller, it not as easy as simply adding stories to the top. You have to think of as adding stories to the bottom.’ ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [May 19, 2011, 5:12pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/24 "2011-05-19T17:12:37Z")
> [@Simplicio](#): > > You need an exponentially increasing amount of mass depending on how close to c you want to get, as acceleration gets more difficult the closer you are to the limit. Why? If you turn off the engines for a minute in theory the ship could consider itself at rest. Turn on the engines and it would seem to be accelerating from a standstill in its own reference frame. Further, hasn’t the fuel gained mass along with the rest of the ship? Every action has an equal and opposite reaction. So while the payload may be heavier so is the reaction mass you are tossing out the back thus gaining you a harder shove forward. I would guess (read that again) these effects balance out so adding another 1000 miles/hour of speed at .99c is as easy as it is from .01c (i.e. it’d take the same amount of fuel out of the gas tank). ---
**Author:** ![Simplicio](https://avatars.discourse-cdn.com/v4/letter/s/c37758/32.png) [@Simplicio](https://boards.straightdope.com/u/Simplicio)\ **Post date:** [May 19, 2011, 5:31pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/25 "2011-05-19T17:31:55Z")
> [@Whack-a-Mole](#): > > Why? > > If you turn off the engines for a minute in theory the ship could consider itself at rest. Turn on the engines and it would seem to be accelerating from a standstill in its own reference frame. But the acceleration in the rocket frame isn’t the same as the one in the rest frame. And the transformation is such that the faster the rocket frame is moving, the less of any acceleration there will “translate” into the rest frame. This is why things appear more “massive” when the move faster, their inertia looks greater because acceleration is more difficult. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [May 19, 2011, 5:38pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/26 "2011-05-19T17:38:10Z")
Simple non-relaitivistic math, IIRC from high school physics: v^2=2ad (a=9.8m/s/s) d is distance travelled. v=at Below a significant fraction of the speed of light, relativistic effects are minimal. Accelerating at 9.8m/s^2, aw heck make it 10m/s, to get to 2/3 speed of light, or 2x10^8m/s, would be t=a/v=2x10^7 seconds. That’s 5555 hours, or about 231 days. Relativistic effects use the factor: sqrt(1-(v/c)^2) At 2/3c, this is sqrt (1-(2/3)^2) or about 3/4 of expected. So you’d need the energy that Newton say you need to get to .67c just to get to Einstein’s 0.5c (It’s more complex than that, since this would be an integral over acceleration and as others said, you have to consider the effect of consumed fuel… etc.) For 0.9 c, sqrt(1-.81) = .43 As long as we’re pulling modes of travel out of random quantum orfices, the ideal mode of travel is using direct energy to matter conversion - e=mc^2. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [May 19, 2011, 5:45pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/27 "2011-05-19T17:45:12Z")
> [@Whack-a-Mole](#): > > Why? Because of relativity. Perhaps the easiest way to understand why is in terms of the [relativistic compostion of velocities formula](http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity), though I prefer to think of it geometrically in terms of hyperbolic rotations of a body’s four-velocity (it may sound slightly esoteric, but once you start thinking that way then you stop making many of the conceptual errors that are quite easy to make in relativity). ---
**Author:** ![shiftless](https://avatars.discourse-cdn.com/v4/letter/s/e9c0ed/32.png) [@shiftless](https://boards.straightdope.com/u/shiftless)\ **Post date:** [May 19, 2011, 5:49pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/28 "2011-05-19T17:49:51Z")
What if the spaceship was on a treadmill? ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [May 19, 2011, 6:00pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/29 "2011-05-19T18:00:48Z")
> [@shiftless](#): > > What if the spaceship was on a treadmill? Don’t know exacvtly what you mean, but relativity governs the spaceship and the treadmill. ---
**Author:** ![Derleth](https://avatars.discourse-cdn.com/v4/letter/d/b9e5f3/32.png) [@Derleth](https://boards.straightdope.com/u/Derleth)\ **Post date:** [May 19, 2011, 6:49pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/30 "2011-05-19T18:49:03Z")
> [@These are my own pants](#): > > Don’t know exacvtly what you mean, but relativity governs the spaceship and the treadmill. It’s an inside joke based on a horrible, long, acrimonious thread: “If you put a plane on a treadmill that matched its velocity, would it be able to take off?” It’s fundamentally badly posed and the confusion arising lead to a massive thread going over a lot of how aircraft take off and how they work on the ground. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [May 19, 2011, 7:33pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/31 "2011-05-19T19:33:08Z")
So would we need the same amount of fuel to get the treadmill up to .9c? ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [May 19, 2011, 8:01pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/32 "2011-05-19T20:01:35Z")
> [@Simplicio](#): > > But the acceleration in the rocket frame isn’t the same as the one in the rest frame. And the transformation is such that the faster the rocket frame is moving, the less of any acceleration there will “translate” into the rest frame. > > This is why things appear more “massive” when the move faster, their inertia looks greater because acceleration is more difficult. But isn’t the rocket fuel mass also greater thus pushes harder when thrown out the back of the engine? Also, if the engine is turned off imagine it in a universe devoid of anything else. From the perspective of the spaceship it is not moving. So, to someone born on the ship (unaware that it had in the past experienced thrust from the rockets) accelerating to greater speeds would seem no different than accelerating to a given speed is to us. As far as they are concerned the ship is at a standstill. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 19, 2011, 8:07pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/33 "2011-05-19T20:07:44Z")
> [@](#): > > Why? > > If you turn off the engines for a minute in theory the ship could consider itself at rest. Turn on the engines and it would seem to be accelerating from a standstill in its own reference frame. From the perspective of the ship, this works fine. If all you’re worried about is keeping the “artificial gravity” at the right level, and don’t care about the travel time at all, you can just ignore relativity and all the calculations will come out just fine. This is not, however, because the increased mass of the fuel cancels out the increased mass of the ship. Neither ship nor fuel increases in mass. The fact that (from an external perspective) acceleration gets harder as you go faster is just an indication that you’re defining acceleration wrong, not that the mass is changing. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [May 19, 2011, 8:52pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/34 "2011-05-19T20:52:25Z")
> [@Chronos](#): > > From the perspective of the ship, this works fine. If all you’re worried about is keeping the “artificial gravity” at the right level, and don’t care about the travel time at all, you can just ignore relativity and all the calculations will come out just fine. > > This is not, however, because the increased mass of the fuel cancels out the increased mass of the ship. Neither ship nor fuel increases in mass. The fact that (from an external perspective) acceleration gets harder as you go faster is just an indication that you’re defining acceleration wrong, not that the mass is changing. So is it harder to accelerate the closer you get to light speed? Imagine an empty universe except for you and me. Neither of us ever experienced any acceleration. We were born into the universe that way. I see you approaching me at .99c. As far as I am concerned you are moving and I am standing still. Conversely, from your perspective you are the one standing still and I am the one moving towards you at .99c. If we could measure how much harder it is to accelerate when we turn on our engines wouldn’t that mean you could definitively say which one was really moving? I do not know but I suspect the universe doesn’t make it that easy for us. I thought there was no way to definitively say which one of us was moving and which was standing still. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 19, 2011, 8:56pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/35 "2011-05-19T20:56:35Z")
If we insist on using the Newtonian definition of acceleration, then from my perspective it’s harder to accelerate you, and from your perspective it’s harder to accelerate me. It’s perfectly symmetric. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [May 19, 2011, 9:07pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/36 "2011-05-19T21:07:32Z")
> [@Whack-a-Mole](#): > > So is it harder to accelerate the closer you get to light speed? > > Imagine an empty universe except for you and me. Neither of us ever experienced any acceleration. We were born into the universe that way. I see you approaching me at .99c. As far as I am concerned you are moving and I am standing still. Conversely, from your perspective you are the one standing still and I am the one moving towards you at .99c. > > If we could measure how much harder it is to accelerate when we turn on our engines wouldn’t that mean you could definitively say which one was really moving? I do not know but I suspect the universe doesn’t make it that easy for us. No. Perhaps this is easier to understand, you have intrinsic acceleration and extrinsic acceleration. The extrinsic acceleration of an object is **in a given inertial frame** is simply d[sup]2[/sup] **x** /dt[sup]2[/sup], the intrinsic or proper acceleration is the extrinsic acceleration in inertial frame at which the object is at rest at that instant (the instantaneously co-moving inertial frame). The extrinsic acceleration is not frame-invariant, so it’s value depends on the inertial frame you choose to measure it, the intrinsic acceleratrion obviously does not vary from frame to frame (as it is defined by a single frame) and is the acceleration the object ‘feels’ (i.e. inertial forces depend on the intrinsic acceleration). If we say A is the intrinsic acceleration, perhaps the best way to think of it is that for an object travelling at speed v and extrinsic accleration a **in a given inertial frame** then a-\>A as v-\>0, but a-\>0 as v-\>c. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [May 19, 2011, 9:24pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/37 "2011-05-19T21:24:04Z")
> [@Chronos](#): > > If we insist on using the Newtonian definition of acceleration, then from my perspective it’s harder to accelerate you, and from your perspective it’s harder to accelerate me. It’s perfectly symmetric. But the claims above are the faster you are moving the harder you are to accelerate. In our two person universe imagine we are in identical rocket ships. You think I am moving. I think you are moving. Now, you turn on your rockets for 1 minute and we measure the change in velocity. You then decelerate back to the original speed and I do the same (turn on my engine for 1 minute) and we measure our velocity. If one of is moving and is thus harder to move and need more fuel to accelerate a given amount then our experiment should show clearly who is the one who is really moving. The person who gained less speed for running their engine for 1 minute is the one who is moving because they gain less acceleration from a given expenditure of fuel. I did not think that would work. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [May 19, 2011, 9:31pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/38 "2011-05-19T21:31:39Z")
> [@Whack-a-Mole](#): > > But the claims above are the faster you are moving the harder you are to accelerate. > > In our two person universe imagine we are in identical rocket ships. You think I am moving. I think you are moving. > > Now, you turn on your rockets for 1 minute and we measure the change in velocity. You then decelerate back to the original speed and I do the same (turn on my engine for 1 minute) and we measure our velocity. > > If one of is moving and is thus harder to move and need more fuel to accelerate a given amount then our experiment should show clearly who is the one who is really moving. The person who gained less speed for running their engine for 1 minute is the one who is moving because they gain less acceleration from a given expenditure of fuel. > > I did not think that would work. Read my posts: it is never harder to get an object’s intrinsic acceleration to increase, but it is harder to get it’s extrinisic acceleration **in a given frame** to increase the closer your velocity is to c **in a given frame** , but both an objects velcoity and it’s extrinsic acceleration are frame-dependent so how hard depends entirely on the frame you choose to measure them in. ---
**Author:** ![iamthewalrus\_3](https://avatars.discourse-cdn.com/v4/letter/i/258eb7/32.png) [@iamthewalrus\_3](https://boards.straightdope.com/u/iamthewalrus_3)\ **Post date:** [May 19, 2011, 9:36pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/39 "2011-05-19T21:36:49Z")
> [@Whack-a-Mole](#): > > But the claims above are the faster you are moving the harder you are to accelerate. > > In our two person universe imagine we are in identical rocket ships. You think I am moving. I think you are moving. > > Now, you turn on your rockets for 1 minute and we measure the change in velocity. You then decelerate back to the original speed and I do the same (turn on my engine for 1 minute) and we measure our velocity. > > If one of is moving and is thus harder to move and need more fuel to accelerate a given amount then our experiment should show clearly who is the one who is really moving. The person who gained less speed for running their engine for 1 minute is the one who is moving because they gain less acceleration from a given expenditure of fuel. > > I did not think that would work. 1 minute according to whose clock? ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [May 19, 2011, 9:40pm UTC](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412/40 "2011-05-19T21:40:50Z")
> [@These are my own pants](#): > > Read my posts: it is never harder to get an object’s intrinsic acceleration to increase, but it is harder to get it’s extrinisic acceleration **in a given frame** to increase the closer your velocity is to c **in a given frame** , but both an objects velcoity and it’s extrinsic acceleration are frame-dependent so how hard depends entirely on the frame you choose to measure them in. I confess I will need to read up more on this to understand the distinctions you are making (it is my lack of knowledge here…not your explanation). I guess the bottom line for me here is will the rocket ship see a diminishing return on the acceleration it gets from burning a liter of fuel as it goes faster? On the one hand as it burns fuel the ship gets lighter. But assume we are getting close to light speed so relativistic effects are noticeable. Does the ship experience diminishing returns on the effectiveness of its rocket engines? Does each mile-per-hour faster cost more fuel than the last to obtain? [Previous page](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412.md?page=1) [Next page](https://boards.straightdope.com/t/acceleration-to-c-at-1-g/582412.md?page=3)