Spacetravel: t as a function of a and d

1. Aug 13, 2007

DaveC426913

I'd like a formula that returns the duration of a journey between parts of the solar system assuming a constant acceleration (and deceleration).

Seems to me, simplistically, this is t as a function of a and d. All I need to do is determine the distance between the two points of interest (and then halve that to account for accel/decel), pick a constant acceleration value, and resolve for t.

Assumptions:
- above a certain acceleration (say, 1g), the f of the sun will have a negligible effect on trajectory and thus the duration of the trip.(i.e. trajectories can be treated as mostly straight lines)
- planetary gravity wells will be ignored for now, so start v and stop v are zero. Considering the velocities involved in these trips, planetary orbital velocity is close enough to zero to have small effect on the duration.

Are these safe assumptions?

In H.S. Physics, we learned these rules of motion generally as SAVTU. (s=distance,a=acc,v=init. vel.. etc.) but it seems to have changed its name.

Where will I find these formulae: eg. t as a function of a and d?

2. Aug 13, 2007

Sean Powell

t as a function of a and d

Sounds like a simple question and I'm bored at work so I'll take a shot at it.

Assuming (falsely) that both start velocity and end velocity are zero (or zero relative to each other). A ship in transit will spend half of it’s time accelerating, a relatively small portion flipping end for end (either with or without thrust engaged) and then the 2nd half of the trip decelerating. Distance for an object under constant acceleration and no starting velocity is:

D=1/2*A*t^2

So the formula for the first half of the distance and first half of the time is:

1/2*D=1/2*A*(1/2*t)^2

A little rearranging:

D=4*A*t^2

Or:

t = (D/(4*A))^(1/2)

So one quarter the distance takes half as long and four times as far takes only twice as long. Four times the acceleration is half the duration and a quarter the acceleration is twice the duration.

Remember to check your units and also to check that your peak velocity is well below the speed of light or you will start having relativistic errors in your time calculation.

Happy flying,
Sean

3. Aug 13, 2007

arivero

Is realistic to ignore gravity wells? Some trajectories can be shorter by using them.

4. Aug 13, 2007

Timo

That's wrong. You have an equality A=B. From this, it does not follow that A/2 = B/4 (where the 4 stems from the 0.5²). You screwed up the follow-up step, too. The violation of the equality is due to your implicit assumption that acoording to this formula, half the distance would take half the time. That's not correct.

Sidenote: It's very uncommon to use a captial A for acceleration; the standard letter is a small a.

EDIT: To expand on arivero: You (=topic starter) should at some (possibly later) stage also reconsider if the idea of starting and ending with v=0 (in the same frame of reference) is a realistic boundary condition. After all, I'd expect the destination planet to move relative to the planet that you started at in most cases.

Last edited: Aug 13, 2007
5. Aug 13, 2007

Staff: Mentor

Given the assumption of uniform acceleration/deceleration, I don't see anything wrong with Sean's analysis. Not sure what you mean by "half the distance would take half the time"--I don't see where he says that. Are you comparing two trips of different total distance? Or are you referring to the fact that the first half (acceleration) of any journey takes the same time as the second half (deceleration)?

6. Aug 13, 2007

Timo

I'm refering to $$D = \frac{1}{2} a t^2 \Rightarrow \frac{1}{2} D = \frac{1}{2} a \left( \frac{1}{2} t \right)^2$$, which is wrong and the two equations I quoted seemed to suggest. Might have misunderstood it, though (which could explain why later the factor 0.25 suddenly becomes a factor 4).

EDIT: Anyways, the final result is wrong exactly by the 0.25->4 error. I begin to understand the original statement, but the notation is highly misleading (would have been better to use a capital T in the 2nd equation and from there on to avoid confusion).

Last edited: Aug 13, 2007
7. Aug 13, 2007

Integral

Staff Emeritus
It needs to be said:

Constant acceleration trips are not the way you navigate the solar system. This is a very bad assumption and will lead to results that are meaningless for a trip inside the solar system.

8. Aug 13, 2007

DaveC426913

Thanks. OK, let's back up and talk about some assumptions and you'll see why I'm exploring this.

I'm positing an Earth's future where
- fuel is dirt cheap - i.e. a non-issue by whatever technological means we come up with (fusion, antimatter, whatever).
- super-light velocity drives have not been created
- artifical gravity (or other acceleration-nullifying technology) has not been created
- the ships are human-piloted

With the possible exception of the last one, I see these assumptions as very-well-founded in practical engineering of the next few centuries.

So, from my assumptions to my extrapolation. In the economy of a bustling solar system, time is money, so you'd still want to minimize time at the expense of fuel. (Otherwise all current commercial jetliners would have glide capability, wouldn't they?)

With these assumptions, the fastest way to traverse the solar system is limited by the occupants' ability to tolerate the g forces. A comfortable trip will be 1 g all the way, an emergency trip could be multiple g's, up to a limit imposed by the technology helping the occupants survive it (say, 10+g's). Yet, except for a few circumstances, all journeys will be long enough that multiple g trips will drive the occupants stir-crazy, since they'll likely be heavily restrained in some way. So really, I can't see sustained high g.

But, I suppose there is a middle ground, you could hit the 10+g mark in the first and last few hours of the journey. But still, I don't see that, with cheap fuel, you wouldn't do 1g all the way anyway. The trips are long enough that it would make a significant difference.

So, how long are we talking, ballpark? At 1g, how long might it take to make an interplanetary journey of, say, 100 milliion miles? (The planets aren't always ideally aligned.)

9. Aug 13, 2007

Staff: Mentor

50 million miles = d = 1/2 g t^2

Then double the time, since that is just for the first (accelerating) part of the trip.

10. Aug 13, 2007

rcgldr

Note that at 1 g of acceleration for even just a few months will get into relativistic effects.

11. Aug 13, 2007

Staff: Mentor

Good point Jeff. I didn't work any numbers in Dave's question. So he can only use the simple kinematic equations of motion up to something like 0.9c, right? Then he needs to start using an integral instead, and include the Lorentz factor for the mass increase? Is that how he should approach this?

(Sorry, I don't work with this much, but want to try to help out)

http://en.wikipedia.org/wiki/Special_relativity#Mass.2C_momentum.2C_and_energy

.

12. Aug 13, 2007

pervect

Staff Emeritus
Relativistic effects would be computed by using the relativistic rocket equation

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

If we are interested in distance as a function of stay-at-home time, the relevant relativistic equation is:

$$\frac{c^2}{a} \left( \sqrt{1 + \left( \frac{at}{c} \right)^2} -1 \right)$$

This can be series expanded (using Maple) as:

$$d \approx \frac{a}{2} t^2 - \frac{a^3}{8 c^2} t^4 = \frac{a t^2}{2} \left(1 - \frac{a^2 t^2}{4 c^2} \right)$$

So if you accelerate at 9.8 m/s^2 for 2 months, you'll have an error of less than a percent by using the Newtonian distance formula.

At this point, you'll be at 1.3*10^11 km, or about 900 au away from the sun, well into the Oort cloud, going at .17c. Double the distance if you want to stop.

Relativistic effects will be detectable here, but not really significant, with a peak gamma of about 1.015.

Note that google calculator helps a lot in doing these routine calculations, for instance:

and

show how the distance formula can be calculated.

I should also add that at least one well-known science fiction(!) author (Heinlein) has gone through most of this in some detail, using the same assumption of human-limited accelerations. (He didn't address the relativistic aspects). Unfortunately, I don't recall exactly where, and he is a rather prolific author. It might have been "Space cadet". It would probably be better to work out the details for oneself rather than to use a work of fiction such as "Space Cadet" as a reference on PF, however.

However, even with fusion power, the initial assumptions that it is possible to accelerate at 1g for extended periods of time is questionable.

Last edited: Aug 13, 2007
13. Aug 13, 2007

DaveC426913

Huh. So, if I wanted to travel from Earth to Mars when they were 200Mkm apart, I could do that in 6 1/2 days at one g!

.5 * (9.8 (m / (s^2))) * ((40 hour)^2) = 101 606 400 kilometers x 2

Did I do that right?

That's extremely practical!

Last edited: Aug 13, 2007
14. Aug 13, 2007

DaveC426913

The way I see it, the only thing tha'ts questionable is whether we will discover this conventional drive with dirt-cheap and/or compact fuel before we discover gravity-control or ftl drive or a space warping drive.

If you don't put much stock in those last three being discovered anytime soon, then the conventional drive is a virtual certainty - it's just a matter of time.

15. Aug 13, 2007

Integral

Staff Emeritus
Not really, you assume a zero initial and final velocity, neither is correct. When you leave earth you have it's orbital velocity, when you land on Mars you had BETTER have its orbital velocity.

You really HAVE to consider orbital mechanics to navigate the solar system.
This over simplified model cannot be used for real world solutions.

16. Aug 14, 2007

rcgldr

Say you want to get to the nearest star, which is 4.2 light years away. If a ship can maintain 9.8 m/s^2 acceleration (relative to it's own frame of reference, basically so the human inside experiences forces corresponding to 1 g of non-relativistic acceleration) by the time the ship reaches the halfway point, it's going over .5c, if I remember correctly.

17. Aug 14, 2007

K.J.Healey

The next step you should do is to figure out the energy needed to make these trips, and how economical they would be.

18. Aug 14, 2007

DaveC426913

I have a feeling some people are not following the thread from its beginning.

1] This is about travel within the solar system.

2] This assumes a level of technology where fuel is cheap like jet fuel - you don't sacrifice the duration of the journey for the sake of fuel savings*. We don't have 747s with glide capability for the same reason (well, you get my point) - it's all about "time is money".

(*Let's not get into a debate about fuel economy in commercial travel. Yes, they economize all over the place, but the one thing they never compromise is duration of the trip.)

Last edited: Aug 14, 2007
19. Aug 14, 2007

DaveC426913

With no limit on fuel, and accelerations in multiple gs, matching orbital v would be a matter of minutes.

And besides, with this much power on-hand you could just power right up and down from the ground.

Last edited: Aug 14, 2007
20. Aug 14, 2007

pervect

Staff Emeritus

As far as initial and final velocity goes, Mar's orbital velocity is 24 km/sec.

http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

Worst case, we'd have to make up double that difference, let's call it 50 km/sec. At 1g, it would take us only 1.5 hours to accelerate to 50 km/sec.

While the gravitational acceleration of the sun would also affect our trajectory, at 1 au that is only (correction) .005 m/s^2.

Thus the initial / final velocity difference would not significantly affect our trip time. We'd have to adjust the flip over time by a few hours at worst. The sun's gravity is also negligible in comparison to our ship's gravity - it also does not significantly affect the trip time.

Thus I generally agree with the OP that it's possible to ignore planetary motions, or rather that they don't have a major impact on trip time, given the assumption of 1g accelerations. One would have to be aware of one's distance and velocity relative to Mars to make the "flip over" at the proper time, but the total trip time required would not be greatly impacted by either the initial/final velocity distance or by the acceleration caused by the sun's gravity.

Small errors in the flip-over time could be compensated for by adjusting the deacceleration profile - i.e. slightly over or slightly under 1g.

However, I am not so convinced that it is possible to accelerate at 1g for extended periods of time. Doing so requires a high specific impulse, i.e. a high exhaust velocity. High exhaust velocity with a rocket requires very high chamber temperatures. High chamber temperatures implies a lot of thermal radiation. This is a problem I have not been able to work out in any detail. For some background, see the thread: