Spacetravel: t as a function of a and d

  • Thread starter DaveC426913
  • Start date
  • Tags
    Function
In summary: I'm not sure where you're getting the idea of constant acceleration. With the exception of a small bit of the trip, obviously you'd want to accelerate for the first half and decelerate for the second half. Unless you're using a continuous rocket motor (which isn't practical) it wouldn't be truly constant anyway. The point I'm (apparently not) making is that the 2nd half of the trip (where you're decelerating) only takes as long as the first half of the trip (where you're accelerating).For a given acceleration, it takes less than half the time to go a given distance at a given higher speed. That's the only point I'm trying to make.
  • #1
DaveC426913
Gold Member
22,432
6,106
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?
 
Astronomy news on Phys.org
  • #2
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
DaveC426913 said:
- planetary gravity wells will be ignored for now, so start v and stop v are zero.

Is realistic to ignore gravity wells? Some trajectories can be shorter by using them.
 
  • #4
Sean Powell said:
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
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:
  • #5
Timo said:
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.
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
I'm referring to [tex] D = \frac{1}{2} a t^2 \Rightarrow \frac{1}{2} D = \frac{1}{2} a \left( \frac{1}{2} t \right)^2 [/tex], 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:
  • #7
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
Integral said:
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.

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
DaveC426913 said:
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.)

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
Note that at 1 g of acceleration for even just a few months will get into relativistic effects.
 
  • #11
Jeff Reid said:
Note that at 1 g of acceleration for even just a few months will get into relativistic effects.

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

Relativistic effects would be computed by using the relativistic rocket equation

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

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

[tex]
\frac{c^2}{a} \left( \sqrt{1 + \left( \frac{at}{c} \right)^2} -1 \right)
[/tex]

This can be series expanded (using Maple) as:

[tex]
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)
[/tex]

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:

http://www.google.com/search?hl=en&q=.5*(9.8+m/s^2)*(2+month)^2=&btnG=Search

and

http://www.google.com/search?hl=en&q=.5*(9.8+m/s^2)*(2+month)^2+in+au&btnG=Search

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 by a moderator:
  • #13
pervect said:
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:

http://www.google.com/search?hl=en&q=.5*(9.8+m/s^2)*(2+month)^2=&btnG=Search

and

http://www.google.com/search?hl=en&q=.5*(9.8+m/s^2)*(2+month)^2+in+au&btnG=Search

show how the distance formula can be calculated.
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:
  • #14
pervect said:
=However, even with fusion power, the initial assumptions that it is possible to accelerate at 1g for extended periods of time is questionable.
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
DaveC426913 said:
Huh. So, if I wanted to travel from Earth to Mars when they were 100Mkm apart, I could do that in 56 hours (4 days 8 hours) at one g!

.5 * (9.8 (m / (s^2))) * ((28 hour)^2) = 49 787 136 kilometers

Did I do that right?


That's extremely practical!

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
Note that at 1 g of acceleration for even just a few months will get into relativistic effects.
Relativistic effects will be detectable here, but not really significant, with a peak gamma of about 1.015.
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
The next step you should do is to figure out the energy needed to make these trips, and how economical they would be.
 
  • #18
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:
  • #19
Integral said:
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.
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:
  • #20
Integral said:
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.

A couple of comments:

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:

https://www.physicsforums.com/showthread.php?t=120096

If we could use the stefan-boltzman law for the total radiation from plasma, it would be easy to do the calculations and we would find that total radiation went up with the 8th power of the ISP. (Exhaust velocity, equivalent to ISP, goes up very roughly with the square root of the chamber temperature, and radiated power goes up with the 4th power of the temperature when the Stefan-Boltzman law applies).

A consequnce of this is that the cooling problems would be unmanagable for a high-ISP 1g rocket - and a high ISP is required to be able to accelerate at 1g for long periods of time, otherwise one will run out of reaction mass.

To put this in perspective - to just double the ISP of the shuttle, would, if this calculation were correct, require 256 times as much cooling capacity. We would have to do a LOT better than double the ISP of the space shuttle to accelerate for days.

However, radiation from plasma is not such a simple thing, the plasma is almost transparent rather than a black body, and it takes some rather elaborate codes to compute it, and I was never able to get the codes to even run on my PC.

Fortunately, the transparancy of the plasma means that the radiated power is a lot lower than that from a black body (which would have the maximum radiated power), meaning that the problem is not quite so bad. But it's still unclear if it is possible to build the 1g rockets from science fiction. Note that current designs do NOT generate such thrust levels for nuclear rockets, not even close.
 
Last edited:
  • #21
pervect said:
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.
Do you think it is possible that this problem will be solved in, say, the next 500 years? 1000 years?

More to the point, do you think this problem will be solved sooner than we develop a hyperdrive, a warp drive or a gravity-compensator?

Because if your answer is yes, then, presuming the continued existence of our race in this solar system, the result is, as I said before, a virtual certainty - it's just a matter of when.
 
  • #22
I think it is a rather big assumption at this point to assume that it is even possible to create a rocketship that can accelerate at 1g continuously for weeks.

I'd actually be interested in seeing more on what's involved and if it is possible in theory or not- as I mentioned, I've attempted to think about this somewhat, but the problem is complex.

Another thing I should mention - one should probably also look at designs that are simpler (but less efficient) than a rocket, such as orion-type designs with pusher plates.

However, what I've seen on Orion talks about ISP's of 10,000 seconds or so. or so. To run for a few weeks at 1g, you'd want an ISP of around 2 weeks or so, say a million seconds.

To digress a little bit to explain where this figure came from:

The delta-v capability of a rocket will be given by the rocket equation:

http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

v = v_e ln(M) = g * ISP * ln(M)

where M is the mass ratio. With a 5:1 mass ratio of fuel+payload:payload, you would have a total maximum delta-v of 1.6 * g * ISP. If you assume that your rocket can be throttled over a wide enough range of accelerations to have a constant 1g acceleration fully fuelled and empty, you can see that you need an ISP that is equal to your acceleration time at 1g divided by 1.6 for a 5:1 mass ratio. In general:

ISP = (time at 1g) / ln (mass ratio)

By setting ISP = time at 1g, I've basically assumed that the ln(mass ratio) is near unity, somewhat of an oversimplifiation, but this is a very rough calculation.

You might want to do some more research to see what sort of ISP's have been proposed for theoretical designs, and what sort of acceleration capabilities these designs have. You will find that in general high ISP designs have very low accelerations - it is only the low-ISP designs that can accelerate at 1G.
 
  • #23
But don't you think that we'll get past this with new technology? I don't mean deux ex machina type technology - but there's got to be a lot of grey area between "chemical rockets" and "warp drive".

How about a nice matter/antimatter drive?

Now, before we go too far down this road - this thread is not about <I>what</I> that technology is, it's about the assumption that this technology will merely arrive <I>before</I> gravity-compensators or another technology comes along and possibly renders it moot.
 
Last edited:
  • #24
Let's be a little careful with the warp drive and anti-gravity stuff. We've been letting this thread keep going because of some of the interesting 1g and ISP thoughts. Let's try to steer clear of the science fiction angles if possible. Thanks.
 
  • #25
berkeman said:
Let's be a little careful with the warp drive and anti-gravity stuff. We've been letting this thread keep going because of some of the interesting 1g and ISP thoughts. Let's try to steer clear of the science fiction angles if possible. Thanks.
Absolutely, that's my point. I'm presuming we won't get any sci-fi stuff and am looking for a more realistic - if distant - future.


Still, it seems to me, just off the cuff, that a 200Mkm journey should take longer than 6 days! That's almost too good to be true. I'd better check that.
 

What is "Spacetravel: t as a function of a and d"?

"Spacetravel: t as a function of a and d" refers to a mathematical equation that calculates the amount of time it takes for a spacecraft to travel from one point to another in space, based on the acceleration (a) and distance (d) of the journey.

Why is "Spacetravel: t as a function of a and d" important?

This equation is important because it allows scientists and engineers to accurately predict and plan space missions, taking into account the variables of acceleration and distance.

How is "Spacetravel: t as a function of a and d" derived?

The equation is derived from the basic principles of physics, particularly Newton's laws of motion, and can be represented as t = √(2d/a), where t is time, d is distance, and a is acceleration.

What are the units of measurement for "Spacetravel: t as a function of a and d"?

The units of measurement for this equation will depend on the units used for acceleration and distance. For example, if acceleration is measured in meters per second squared and distance is measured in meters, then time will be measured in seconds.

How accurate is "Spacetravel: t as a function of a and d" in predicting travel time?

The accuracy of this equation will depend on the accuracy of the input values for acceleration and distance. However, it is a widely accepted and reliable formula used in space mission planning and has been verified through numerous experiments and observations.

Similar threads

Replies
15
Views
694
Replies
5
Views
2K
Replies
86
Views
4K
Replies
13
Views
2K
  • Differential Equations
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • STEM Educators and Teaching
7
Replies
233
Views
18K
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top