# I Duration of a round-trip to Alpha Centauri

#### Warp

(No, this is not homework assignment. This is something I'm curious about.)

Assume an astronaut were to travel from Earth to Alpha Centauri (about 4.37 light years distance) in a rocket. During the first half of the distance the rocket constantly accelerates at a comfortable 9.8 m/s2. At the half-way point the rocket turns the other way around and starts decelerating at that same 9.8 m/s2 until it stops at Alpha Centauri. The rocket immediately starts the journey back to Earth, and does the same thing on the way back.

When the rocket finally arrives back to Earth, how long did the journey take from the point of view of the astronaut? How much time will have passed on Earth?

For simplicity we can restrict ourselves to special relativity.

• Vick
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#### Ibix

Science Advisor
Equations 8 and 9 here give you the coordinate time $T$ (as experienced by Earth) and the distance travelled $X$ (as measured by the Earth) (edit: on closer inspection, I see they use $T_p$ and $X_p$ - p for parameter, I guess?) in terms of the proper acceleration $\alpha$ and proper time $\tau$ (experienced by the ship) during a period of constant acceleration. Some factors of $c$ have been ignored, but if you measure distance in light years and time in years then $c=1$ anyway so it doesn't matter, and conveniently one gravity is also 1ly/year2 to within a few percent, so you can set $\alpha=1$.

You can insert the distance travelled on the outbound accelerating leg to get the proper time and then use that to get the coordinate time - then just multiply by four, since the deceleration phase is the same as the acceleration phase backwards, and the return journey is the same as the outbound journey. Your phone calculator app or computer desktop calculator can almost certainly do hyperbolic sines and cosines and their inverses if it has a scientific mode.

Post if you get stuck.

#### pervect

Staff Emeritus
Science Advisor
You can find the needed equations at http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html, the "relativistic rocket FAQ".

It's convient for me to modify the problem so that we measure time in years, distance in light years, and acceleration in light years/years^2. Making a = 9.5 m/s^2 instead of 9.8 makes the acceleration 1 light year/ year^2, and using years for time and light years for distance makes c=1.

Then we can quadruple the time it takes the rocket to get to the halfway point (2.18 light years) to get the round trip time.

The time it takes to travel 2.18 light years at 1 light-year / year^2 is just

$$T = \frac{\cosh^{-1}\, \left(ad + 1 \right)}{a} = cosh^{-1} 3.18 = 1.82 \,years$$

So , the round trip takes 4 times that, our about 7.3 years.

It'll take a little less if you accelerate at 9.8 m/s^2 rather than 9.5, but I'll leave the details to you. Hopefully, I didn't make any math errors, but - I can't really guarantee that.

$\cosh^{-1}$ is the hyperbolic arc cosine, see for example the wiki << link>>

• Dale

#### phyzguy

Science Advisor
It might be interesting to try and calculate how much fuel would be required to accelerate at 1g to Alpha Centauri and back. You won't like the answer.

#### Vick

Your answer based on the equations posted by pervect:
1) Ship time for round trip = 7.165 years
2) Earth time " = 12.005 years

For the question of fuel it would take approx. 80 kg of fuel for each kg of payload.

#### phyzguy

Science Advisor
Your answer based on the equations posted by pervect:
1) Ship time for round trip = 7.165 years
2) Earth time " = 12.005 years

For the question of fuel it would take approx. 80 kg of fuel for each kg of payload.
It will take vastly more fuel than this. How did you arrive at this number?

#### Vick

It will take vastly more fuel than this. How did you arrive at this number?
I only said 80 kg for each kg of payload.....that would depend on the actual mass of the spacecraft!

The equation is EXP(a*T/1)-1, T being ship time. This is quoted in your reference to Baez article.

#### Warp

For the question of fuel it would take approx. 80 kg of fuel for each kg of payload.
Is that taking into account the weight of the fuel itself?

#### Vick

Is that taking into account the weight of the fuel itself?
No, this would defeat the purpose of the calculation. For each kg of mass of your spacecraft, you need 80 kg of fuel. This includes the theoretically most efficient fuel generation engine, even though we don't have the technical know-how for now (matter/antimatter reactor). So 80 kg of fuel is needed to take each kg of spacecraft to Alpha Centaurus and back.

#### phyzguy

Science Advisor
So if you have a 100% efficient antimatter engine, and there are a string of "gas stations" along the way where you can re-fuel with more anti-matter, then the 80kg fuel per kg of payload is a reasonable number. Why don't you try doing a more realistic calculation?

#### Nugatory

Mentor
So if you have a 100% efficient antimatter engine, and there are a string of "gas stations" along the way where you can re-fuel with more anti-matter, then the 80kg fuel per kg of payload is a reasonable number. Why don't you try doing a more realistic calculation?
The calculation on the Baez page (http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html) is assuming a total conversion photon drive but no refueling en route. 80 kg per delivered kg looks about right.

• Vick

Science Advisor

#### Vick

So if you have a 100% efficient antimatter engine, and there are a string of "gas stations" along the way where you can re-fuel with more anti-matter, then the 80kg fuel per kg of payload is a reasonable number. Why don't you try doing a more realistic calculation?
If you are thinking of stopping at each station, then you won't even have the number of years elapsed for Earth and in the ship as you won't be able to maintain a steady 1 g acceleration! It is as Nugatory said, these calculations do not assume re-fueling en route.

Actually I think I was mistaken in identity! It was not you who posted the Baez article but Pervect.

#### Warp

No, this would defeat the purpose of the calculation. For each kg of mass of your spacecraft, you need 80 kg of fuel. This includes the theoretically most efficient fuel generation engine, even though we don't have the technical know-how for now (matter/antimatter reactor). So 80 kg of fuel is needed to take each kg of spacecraft to Alpha Centaurus and back.
But if that calculation does not take into account the weight of the fuel itself, doesn't it become meaningless? To move 1 kg of mass to Alpha Centauri you need 80 kg of fuel. But to move the entire thing, which now weighs 81 kg, you need 81*80 kg of fuel, which in turn means that the entire thing now weighs 81+81*80 kg... and so on and so forth. Am I missing something?

#### Vanadium 50

Staff Emeritus
Science Advisor
Education Advisor
You are correct that you need fuel to carry the fuel. But you don't need an infinite amount since you don't have to carry all of it the entire way, because you burn it as you go along.

#### jbriggs444

Science Advisor
Homework Helper
But if that calculation does not take into account the weight of the fuel itself, doesn't it become meaningless? To move 1 kg of mass to Alpha Centauri you need 80 kg of fuel. But to move the entire thing, which now weighs 81 kg, you need 81*80 kg of fuel, which in turn means that the entire thing now weighs 81+81*80 kg... and so on and so forth. Am I missing something?
You do not have to carry all the fuel all the way. You expend the first little bit of fuel to accelerate the whole craft. Then the next bit of fuel has to accelerate a bit less than the whole craft and so on.

With plain old classical mechanics, the result is the Tsiolkovsky rocket equation.

[Drat. Pause to get the Wiki reference and that pesky @Vanadium 50 sneaks in]

#### Vick

But if that calculation does not take into account the weight of the fuel itself, doesn't it become meaningless? To move 1 kg of mass to Alpha Centauri you need 80 kg of fuel. But to move the entire thing, which now weighs 81 kg, you need 81*80 kg of fuel, which in turn means that the entire thing now weighs 81+81*80 kg... and so on and so forth. Am I missing something?
That calculation is to take the 1 kg of spacecraft to AC and back to Earth with zero fuel at the end! The fuel is for the acceleration that will provide us with "distance travelled". The more distance you seek, the more acceleration is needed and therefore more fuel. It is simple really! The fuel is not needed to carry itself. We are not sending fuel that distance but the spacecraft.

#### DaveC426913

Gold Member
If you are thinking of stopping at each station, then you won't even have the number of years elapsed for Earth and in the ship as you won't be able to maintain a steady 1 g acceleration!
So we assume an infrastructure of gas stations with long skyhook-like extensions that can accelerate to match velocities with the passing ship.  #### Vick

So we assume an infrastructure of gas stations with long skyhook-like extensions that can accelerate to match velocities with the passing ship. View attachment 247766
That would be a nightmare in space, especially with relativistic velocities! A slight impact, would be lethal. On another score, any relativistic spacecraft with a hypothetical matter/antimatter engine reactor, which is able to achieve velocities in excess of 75% the speed of light, would need very strong deflection and shielding abilities due to kinetic energies of space particles. A magnetic field in excess of 100 Tesla would be needed. In comparison, the Earth generates between 25 to 65 microteslas.

#### Warp

That calculation is to take the 1 kg of spacecraft to AC and back to Earth with zero fuel at the end! The fuel is for the acceleration that will provide us with "distance travelled". The more distance you seek, the more acceleration is needed and therefore more fuel. It is simple really! The fuel is not needed to carry itself. We are not sending fuel that distance but the spacecraft.
Yes, we are not sending fuel that distance, but we are sending fuel part of that distance. I don't know (haven't checked the math) how much fuel is needed for the latter half of that trip (ie. from AC to Earth), but surely it must be something like at least 10 kg or so, maybe more? This means that 10kg or whatever it is, is dead weight that needs to be carried in the first half of the trip, from Earth to AC.

But maybe the math does work like that. Maybe sending 11 kg (or whatever the amount may be) from Earth to AC does need 80-11 kg (or whatever it may be) of fuel.

#### Vick

Yes, we are not sending fuel that distance, but we are sending fuel part of that distance. I don't know (haven't checked the math) how much fuel is needed for the latter half of that trip (ie. from AC to Earth), but surely it must be something like at least 10 kg or so, maybe more? This means that 10kg or whatever it is, is dead weight that needs to be carried in the first half of the trip, from Earth to AC.

But maybe the math does work like that. Maybe sending 11 kg (or whatever the amount may be) from Earth to AC does need 80-11 kg (or whatever it may be) of fuel.
Visualize it another way, let's say spacecraft (SC) is 10,000 kg in mass and the fuel needed to journey from Earth to AC and back is 80 kg per kg of mass, which equals to 10000x80= 800,000 kg of fuel. So the total mass of both fuel and SC is 810,000 kg.

So in this case 800,000 kg of fuel is needed to make SC which is 10000 kg go to AC and back.

#### jbriggs444

Science Advisor
Homework Helper
Yes, we are not sending fuel that distance, but we are sending fuel part of that distance. I don't know (haven't checked the math) how much fuel is needed for the latter half of that trip (ie. from AC to Earth), but surely it must be something like at least 10 kg or so, maybe more? This means that 10kg or whatever it is, is dead weight that needs to be carried in the first half of the trip, from Earth to AC.
The important thing is the mass ratio (total mass to payload). Whatever fraction of the mass must be fuel to get a payload to AC, that same fraction must be fuel to get the [smaller] payload back. The mass ratio for the round trip will be the square of the mass ratio for the two one-way segments.

One can take that reasoning one step farther. The mass ratio for the first half of the first leg (accelerating outbound from Earth) will be the same as the mass ratio for the second half (decelerating inbound to AC). The mass ratio for the complete trip will be the fourth power of the mass ratio for each of the four half-segments.

The only thing that remains is to calculate the mass ratio for one of the half-segments.

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• Ibix

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