When accelerating at a human-friendly 1G or less, how long would it take to get to four percent of the speed of light? I’ve been trying to work it out for myself but I kind of suck at maths.
v = at
t = v/a
if a = 1G (9.8ms^2) and v = .04c or 1.2 * 10^9 m/s, then t =122,450,000 seconds, or 3.88years.
Ignoring relativistic effects.
ETA, 4%, not 10%
At only 4% c, we can probably ignore relativistic effects and just use the simple Newtonian formula v=at. 1 G acceleration is 9.8 m/s[sup]2[/sup], and you want
v = 0.04c = 12,000,000 m/s. So
t = v/a = (12,000,000 m/s) / (9.8 m/s[sup]2[/sup]) = 1,224,000 s = 14.16 days
Yeah, I missed a decimal place:
v = at
t = v/a
if a = 1G (9.8ms^2) and v = .04c or 1.2 * 10^7 m/s, then t =1,224,500 seconds, or 340 hours
I thought c was 300,000km/sec or 300000000m/s
Yes that’s correct. 4% of 300,000,000 m/s is 12,000,000 m/s.
- 14.1589 days ship time
- 14.1626 days earth time
You would travel 13.6 light hours (about 9.12 billion miles). About twice as far as Pluto.
I used this calculator to find the answer: Greg's Space Calculations: Relativistic Rocket and how far is 13.6 light hours - Wolfram|Alpha
We can ignore relativistic effects, unless you want a precise answer:
From the point of view of the observer on the rocket, the time taken will be tanh[sup]-1/sup ÷ 0.04 ≈ 1.0005 times longer than the approximately 14 days estimated from ignoring relativity.
From the point of view of the observer who is observing the rocket accelerate from zero to 0.04c, the time taken will be (1 - 0.04[sup]2[/sup])[sup]-0.5[/sup] ≈ 1.0008 times longer than the approximately 14 days estimated from ignoring relativity.
Conveniently, 1 g is just about equal to c divided by one year (this is a pure coincidence, but a useful one). This does not, of course, mean that you could reach c by accelerating at g for one year, because relativistic effects interfere. For only 4% of c, though, relativistic effects are small, and so it takes 4% of a year.
As an aside, four percent sounds like an oddly specific number. Is there any particular reason for the question?
1 fricking g? Get out of the way old lady I’m passing you on the right. ;).
Nice easy formula…but howcome? I thought the guy on the rocket wouldn’t observe any relativistic effects, since he’s his own frame of reference? Where does this dilation come from? (How does one observe one’s own time dilation anyway? What are you gonna measure it with, given your stopwatch is delayed the same degree your thoughts are.)
I’m confuzled.
(And, of course, so far, no one has wanted to take fuel consumption into consideration, but we can cheat and imagine a ramscoop spaceship. But then there’s friction to compensate for… Scientists have it much easier than engineers do!)
The thing about relativity is…it is relative to each observer (go figure).
So, each person lives in their own reference frame. To each person everything seems normal. All their clocks tick at one second per second. The issue arises when you compare those clocks.
So, the person in the spaceship accelerating will have their clock slow down. That person will not notice it…they see the clock ticking same as always. When they come back to earth though they will see less time elapsed on their clock than the clock on earth.
And yeah, the reason we are not zipping off to far flung solar systems at 4% the speed of light is because we cannot carry that much fuel. Doesn’t mean we can’t ask the question though.
I don’t think you understood Trinopus’s question. He’s asking why Asympotically fat gave dilation numbers for both the earth bound observer and the observer on the space ship. The values he gave were not dilation values, in the sense that it is the difference between two observers, it’s the ERROR in the answers which assume Newtonian physics and ignore relativity.
The I’d refer him to post #7 which gives ship time and earth time answers. If their clocks are synchronized at the start it shows how much they will differ once 4% light speed is achieved (a bit more than five minutes over the 14 days).
At 4% light speed though time dilation effects are pretty small so most have ignored it since it barely changes the answer.
Let’s say you start in frame A and accelerate to 0.02c relative to frame A so that you are now in frame B and then you accelerate again in the same direction to 0.02c relative to frame B, so that you are now in frame C. As you are probably aware. velocities in special relativity aren’t additive and you can see from the relativistic composition of velocities formula that in frame C you are travelling slightly less than 0.04c relative to frame A and you would have to accelerate for a little bit more time to reach 0.04c relative to frame A. To get the answers I got in my last post you need to consider the composition of infinitesimal changes in velocity, but the same idea applies.
The formulae I used are not time dilation formulae, they are the relativistic corrections to the non-relativistic formula (the time dilation factors can be got by dividing the two corrections by each other).
A hand-waving way to see why t = v/a can’t be right for the observer on the rocket is that otherwise v can go to c with only a finite acceleration and a finite amount of time. The correction factor must diverge as as v goes to c.
I absolutely get the second paragraph. . . .
ETA: And…ah! Now I get the first paragraph. Even a guy in a spaceship realizes that he hasn’t yet got to 4% of tsol yet, and can wonder why. Forgive my obtusity!
Can you do 0 to 60 in 2.7 seconds? That’s how fast 1G acceleration is.
I think he’s pointing out that many space-faring launching devices accelerate at more than 1G. Of course, the body might be a bit put out if you tried accelerating for the requisite time at 2G. :dubious: