How long to reach 0.9 c, accelerating at 1 gravity?

Question

Assume that at time t=0, I am in a space ship starting at rest relative to the earth, and then I accelerate so that I continuously experience an acceleration equivalent to 1 earth gravity (about 9.81 Newtons).

At what time would I observe that I am travelling at 90 percent of the speed of light, relative to the earth? That is, how long after I started accelerating would I reach the speed of 0.9 c ?

Please note:

A few people have been answering 45 weeks. This is wrong. I don't know the right answer, but I know 45 weeks is wrong. The reason is, those who calculated 45 weeks did not use relativity, they only used Newtonian equation (velocity=acceleration x time). The answer 45 weeks is absurd, because if the Newtonian method worked, then I would be able to accelerate 90 weeks and go twice as fast, 1.8 times the speed of light. The theory of relativity is based on the fact that it is impossible to accelerate to faster than light.

Please notice that I specified the acceleration as 1 gravity. That is, at any time, I could swing a pendulum on my spaceship and check that my acceleration is 9.81 Newtons. But this does not tell me how fast my speed is changing. When I look out my space ship window with my telescope, and watch the earth, as I get close to the speed of light, time begins to dilate and space begins to contract.

I have no idea how to do the math for this.

Answer 1

Boy, you asked a doozy of a question. Well, as you noted, you have to account for relativistic mass. If invariant mass is m, then the relativistic mass M is km, where k is the Lorentz factor, k = c / (c^2 - v^2), where c is of course the speed of light.

Now, we usually say F = ma, but we can also say F = dp/dt, where p is momentum, Mv. v is still the velocity in a stationary frame of reference. F = dp/dt = d/dt(Mv) = M(dv/dt) + v(dM/dt) = Ma + v(dM/dt) = ma(c/(c^2-v^2)^(-1/2)) + v*d/dt(mc/(c^2-v^2)^(-1/2) = ma(c/(c^2-v^2)^(-1/2)) + v(-mc/(2(c^2-v^2)^(-3/2)))(-2v)*dv/dt = ma(c/(c^2-v^2)^(-1/2)) + mav^2(c/(c^2-v^2)^(-3/2)). Complicated enough yet? Now, we've defined the invariant acceleration as gravity, i.e., F = mg. So dividing by the invariant mass, g = a[(c/(c^2-v^2)^(-1/2)) + v^2(c/(c^2-v^2)^(-3/2))]. Rearranging, a = g/[(c/(c^2-v^2)^(-1/2)) + v^2(c/(c^2-v^2)^(-3/2))]. Rewriting a as dv/dt and rearranging again, g*dt = [(c/(c^2-v^2)^(-1/2)) + v^2(c/(c^2-v^2)^(-3/2))]dv If we integrate, we get gt equal to the integral of the longer, extremely difficult expression. I, for one, can't compute it, and I'm not even sure if it's possible.

I'll offer you an approximation, instead, by calculating it incrementally. I can take the space ship as it reaches speeds at increments of 0.1c, and calculate how long it takes to reach the next increment, by using the relativistic mass at the previous increment. For a given velocity of 0.1cn, the relativistic mass M is mc/(c^2-(0.1cn)^2), giving an acceleration equal to (g/c)(c^2-(0.1cn)^2)^(1/2). Using t = v/a, the time required to reach a velocity of 0.1c(n+1) is (0.1c)c/(g*(c^2-(0.1cn)^2)^(1/2)). Setting n equal to 0, 0.1, 0.2, and so on up to 0.8 (used to calculate the final acceleration up to 0.9), and creating an Excel spreadsheet to do the heavy lifting, including converting the resulting t from seconds into days, I found that the time required for each segment, in days, was as follows:

0 to 0.1c = 35.4
0.1c to 0.2c = 35.6
0.2c to 0.3c = 36.2
0.3c to 0.4c = 37.1
0.4c to 0.5c = 38.7
0.5c to 0.6c = 40.9
0.6c to 0.7c = 44.3
0.7c to 0.8c = 49.6
0.8c to 0.9c = 59.1

Summing these, the approximate time required is 377 days, slightly longer than a year. You could get a more accurate result by making the increments smaller. I hope that helps.

Answer 2

0.9 x speed of light is approx. 269820000 m/s. For a body starting from rest, velocity = acceleration x time. Acceleration = 9.81 m/s, therefore time = velocity/acceleration = 269820000/9.81 = 27502548 seconds = 7640 hours = 318 days. It's meaningless to try to be more accurate because you're only using the acceleration due to gravity correct to 3 sig. figures.

Answer 3

About 45 weeks After that your rate of acceleration will slow down more and more and your mass will increase while time slows down.

Answer 4

.9 the speed of light is
.9 * 299 792 458 = 269813212. we will call it 270000000
at 10 m/sec^2 we will get there in 270 000 000 / 10 = 27 000 000 seconds or about 45 weeks

(((27 000 000 / 60) / 60) / 24) / 7 = 44.6428571

Answer 5

Hmm
Do you ask 'cuz you read about this in a sci fi book?
I seem to remember a stroy about this
anyways
45.5 weeks to reach 90% of c
you want the excell file?