I am working on a sci-fy short story and I need to use a somewhat realistic value for the time in transit for going to a star. Assume that energy on the space ship is not a problem. The space ship accelerates at some amount of G-force until reaching some fraction of the speed-of-light. It then coasts until reaching the midpoint and then reverses the procedure (including the coasting) until reaching the destination star.

So what would be the formula to compute the time to the coast segment and distance covered?

Assume G = .1, D = 50 light-years, Max-speed .5 of light-speed. How do I compute D(s) - distance covered until coasting T(s) - time taken to reach D(s)

Added bonus; would there be any time dilatation at this speed and if so how to I compute it?


I'm going to piggyback on anorton's answer, especially the key observation that the acceleration time is irrelevant, and then turn on (special) relativity.

I will also follow the assumptions that he made, including $300$ m/s$^2$ acceleration and no significant gravity problems. If I understand his work correctly, he also did away with the flight procedure you suggest for the pilot. This is because if he coasts until reaching the midpoint, and then accelerates in reverse, he's going to "coast" at $0$ m/s which is not going to get us anywhere— or $\frac{c}{2}$ in the opposite direction, which would be even worse!

Okay, so the primary ambiguity that arises in the question when considering relativity is whose time frame we are measuring. However, if we take it from the outside observer's time, then anorton is exactly correct. So the only interesting addition is "How much do the passengers age?"

Relativity (for our purposes) shrinks distances in space by the reciprocal of a number called the Lorentz factor, which for a given velocity $v$ happens to be $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$. Basically any text on relativity, including many online sources, will explain where this comes from, but basically it's what you get when the dust settles after examining the behavior of a light clock, which uses light to tell time (which is a natural clock to consider because light has a universal speed).

If you look carefully at the expression for $\gamma$, you will see that it must always be greater than $1$. So if you accept that space should shrink distances at high velocity, then we must divide by it, i.e. multiply by the denominator.

If $L = 4.73\times 10^{17}$ m is the distance as measured from the outside, then the distance measured from the ship is $$L' = \frac{L}{\gamma}= L\sqrt{1-(v/c)^2} = (4.73\times 10^{17}\text{ m})(0.866) = 4.096\times 10^{17}\text{ m}.$$

Furthermore, the ship sees the planet rushing toward it at $V=1.5\times 10^8$ m/s, and thus it calculates that it will crash into the ship at time $t'=L'/V=2.731\times 10^{9}$ s. Doing the unit conversions gets you $86.6$ years.

So relativity is not entirely irrelevant here, although there are certainly more efficient ways to look 13 years younger.

To answer the question directly: time dilation also happens under the Lorentz factor, which means that every second that passes in the ship is seen as $0.866$ seconds to the outsiders.

[An interesting side note, which may have implications for your purposes: If you are storing not-massless fuel for this ship, then more than half the fuel will be spent in the acceleration than the deceleration. That's because $F=ma$ means that a more massive object takes more force to maintain a constant acceleration. But since you have to accelerate all of the fuel and decelerate less than half of it, the mass difference could be quite noticeable.]

  • $\begingroup$ Eric,thanks for your help. I hadn't thought about the effect of fuel mass reduction. I was not every clear in my initial question, but my assumption would be to accelerate for time T1, then coast for time T2 and finally decelerate for time T3. Where T1 = T3. I see from you answer that I should plan to use more fuel during T1 then in time T3. My ships engine would adjust the fuel flow to maintain the 0.1 G force on the ship. $\endgroup$ – Kent Byerley Jan 17 '13 at 5:27

Simplifications for Modeling

Let's assume that gravity is not an issue here. (Once you include the gravitational attraction from one planet, you're dealing with an improper integral, typically dealt with in second semester calculus. With 3 or more planets, you're dealing with a graduate level problem, that, IIRC, is still open.)

Now, you're dealing with a period of acceleration, followed by a a period of constant velocity, followed by a period of negative acceleration.

One problem with your setup is that you can't accelerate by some amount of force without knowing the mass of the spacecraft. It's better to say that you accelerate at, say, a rate of $10^{-6}$ of the speed of light every second (that is, $\sim 300 m/s^2$).

Working the Problem

So, we accelerate at a constant rate of $300$ meters per second per second (meaning, we advance around 1 Mach number every second) until we reach $1.499x10^8$ meters per second. We use this formula, which calculates velocity as a function of time, assuming constant acceleration: $$v = at + v_0$$ We start at rest, so $v_0 = 0$. The acceleration is 300, and the final velocity (v) is $1.499\times10^6$ Thus, we have: $$1.499\times 10^6 = 300t$$ Solving for $t$ yields 500,000 seconds.

So, it takes 500,000 seconds to accelerate to $0.5c$, where $c$ is the speed of light.

It will also take 500,000 seconds to decelerate at the end, as our acceleration constant is the same.

So, we have $1,000,000$ seconds accounted for in the flight. We want to travel a distance of 50 light-years, denoted $D$.

How far will those $1,000,000$ seconds take us? Well, we have a position formula under constant acceleration: $$x = \frac{a}{2}t^2 + v_0t$$ So, we have: $$x = 150t^2$$ $$x = 150(500,000)^2$$ $$x = 3.75\times 10^{13}$$

So, on the first leg of the journey (we haven't coasted yet), we'll travel $3.75 \times 10^{13}$ meters.

While we're coming to rest: $$x = 150t^2 + 1.499\times 10^6t$$ Plugging in as above yields a distance traveled of approx $3.82\times10^{13}$ meters.

So, we have to travel $50 \text{ light years} - 3.82\times10^{13}\text{meters} - 3.75 \times 10^{13} \text{meters} \approx 49.99 \text{light years}$. (calculation here)

This is essentially 50 light years, so the acceleration time could have been ignored.

Traveling 50 light years at one half the speed of light will take 100 years.


Note that the mathematical model that we used was overly complex. Because of the distances and speeds involved, it doesn't really matter if you accelerate or instantly assume the target speed.

One could account for relativity because of the speed involved, but (per my physics teacher), it doesn't have much effect until you're hitting around 70%+ the speed of light.

Hope this helps, let me know if you have questions.

  • $\begingroup$ ANorton, thanks for trying to make me understand the issues. $\endgroup$ – Kent Byerley Jan 17 '13 at 4:49
  • $\begingroup$ ANorton, thanks for trying to make me understand the issues. But why not 0.1 G? Woundn't that be about 1m/s/s? My space ship (although having a good supply of energy) does not have the engines to push the craft more than that. Can I take the rest of you formulas and solve for T? I think that it would take 1.5E-08 seconds or 4.763 years. I followed your answer on distance up to the 1,000,000 seconds for the accel/decel period, but lost you when you started using 10,000 seconds. Can you try again explaining this. $\endgroup$ – Kent Byerley Jan 17 '13 at 5:11
  • $\begingroup$ The 10,000 looks like a typo; he doesn't use that number in the remaining calculation. However, as he said in the intro, $0.1$ G is a force, not an acceleration, so he substituted an acceleration for you. It was pretty big compared to what you seem to be suggesting, but that's what happened. $\endgroup$ – Eric Stucky Jan 17 '13 at 5:20
  • $\begingroup$ Yes, the 10K was a typo... (corrected now). It should be 1,000,000 Eric basically said exactly what I would in response to your comment: 0.1G isn't an acceleration, but rather a force. I picked something fairly big, but it doesn't really matter--due to the times and distances involved, it would be an suitable model to consider that you move at a constant velocity of $\frac{c}{2}$. Unless you're actually planning to build the ship and travel there, that is... :P $\endgroup$ – apnorton Jan 17 '13 at 12:08

When you say that $G = 0.1$, I will assume that you mean that you are saying that the ship will be accelerating at one tenth of the apparent acceleration felt on the Earth at sea level or so, i.e. about $1 \; \text{m}/\text{s}^2$. Light travels at about $300\,000\,000 \;\text{m}/\text{s}$, so it would take your ship about $150\,000\,000$ seconds to get up to half the speed of light (and were you to slow down, the same amount of time to slow down).

This works out to be around 47 years to speed up to that speed. How far have you gone in that time? You've traveled $\frac{1}{2} a t^2$ total distance, or about $\frac{1}{2} (47\cdot 365 \cdot 24\cdot 60\cdot 60)^2$ meters. (I'm keeping everything in meters and seconds, and regretting it every time I have to write down an equation). You'll go the same distance when you slow down. Unfortunately, having done this, you overshoot your goal, so I suspect you'll increase the rate of acceleration.

Does this process make sense (i.e. how to calculate the times)?

Finally, for this journey, the time dilation would be by somewhere around the order of magnitude of $10\%$, so that this 100 year journey would 'take' 110 years or so. (I didn't calculate it out fully since we overshot the goal).

  • $\begingroup$ I can expand on this if you need to - I'm writing from an ipad and so much slower. $\endgroup$ – davidlowryduda Jan 17 '13 at 3:18

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